CONTAINMENT OF SUBGROUPS IN A DIRECT PRODUCT OF GROUPS
BY
DANDRIELLE C. LEWIS
BS, Winston-Salem State University, 2001MS, University of Iowa, 2006
DISSERTATION
Submitted in partial fulfillment of the requirements forthe degree of Doctor of Philosophy in Mathematics
in the Graduate School ofBinghamton University
State University of New York2011
c© Copyright by Dandrielle Lewis 2011
All Rights Reserved
Accepted in partial fulfillment of the requirements forthe degree of Doctor of Philosophy in Mathematics
in the Graduate School ofBinghamton University
State University of New York2011
April 14, 2011
Ben Brewster, Chair and Faculty AdvisorDepartment of Mathematical Sciences, Binghamton University
Fernando Guzman, MemberDepartment of Mathematical Sciences, Binghamton University
Marcin Mazur, MemberDepartment of Mathematical Sciences, Binghamton University
Joseph Petrillo, Outside ExaminerDivision of Mathematics, Alfred University
iii
Abstract
Given a direct product of groups G and H and information on the subgroups of G and
H, what can one say about the subgroup structure of G ×H? In 1889, Edouard Goursat
proved a theorem that allows us to obtain the subgroup structure of a direct product by
examining the sections of the direct factors.
This dissertation provides a containment relation property between subgroups of a direct
product. Namely, if U1, U2 ≤ G × H, we provide necessary and sufficient conditions for
U2 ≤ U1. We also show when two subgroups of a direct product are isomorphic. Using
these results, we construct the subgroup lattices of two groups of order 64, Q × Q and
Q × D8, where Q and D8 are the quaternion and dihedral groups of order 8 respectively.
Lastly, we use these subgroup lattices to obtain and provide the subgroup lattices of the
two extraspecial groups of order 32.
iv
Acknowledgements
It is said that ”the journey is more important than the destination.” This has been an
amazing journey full of many lessons learned and full of many people who have supported
me, motivated me and who have helped me reach this destination.
To my advisor and friend, Ben Brewster, thank you for taking me on as your student, for
taking a chance with me and for guiding me on this project. You have taught me so much in
the past few years. I have learned how to do Group Theory, how to teach more effectively,
how to write Mathematics better, how to give better presentations, and how to have fun
doing and learning it all. You helped me see the bigger picture, and helped shape me into
the Mathematician that I am today. Thank you for keeping me focused and for listening to
all my ideas no matter how crazy some of them were!
To my committee members, Professors Fernando Guzman and Marcin Mazur and Professor
Joseph Petrillo of Alfred University, thank you for taking the time to serve on my committee,
for taking interest in this project and for giving me suggestions. I would also like to thank
Professor L.C. Kappe for taking the time to help prepare me for the job search and for
carefully reading all my material.
To my family, thank you for always being there for me, for supporting me in my many
adventures, for listening to my ideas, for motivating me when I felt like giving up and for
believing in me when no one else did. To my mentors, thank you for exposing me and
introducing me to the wonderful world of Mathematics and for giving me great advice. To
my friends in the math department, thank you so much for helping me with latex!! To Q,
thank you for being a great friend along this journey, and thank you for introducing me
to Geometer’s Sketchpad. Without that, I do not know how these subgroup lattices would
have made it in my dissertation. To all my other friends, thank you for all your help and
for all the motivating conversations.
v
All of you have challenged, inspired and motivated me along this great journey, and now
we are finally here. This is to new beginnings.
vi
Contents
Notation ix
Introduction 1
0.1 Automorphisms of Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
0.2 Automorphisms of V4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1 Characterization of Containment of Subgroups of a Direct Product A×B 6
1.1 Goursat’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 Conditions for Projections, Intersections, and Isomorphisms Between Sec-
tions of Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.1 Background on Isomorphism Theorems . . . . . . . . . . . . . . . . 7
1.2.2 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Main Results: Containment of Subgroups of a Direct Product . . . . . . . . 10
2 Construction of the Subgroup Lattice for Q×Q 12
2.1 Using Goursat to Calculate the Number of Subgroups of Q×Q . . . . . . 13
2.2 The Subgroups of Orders 2 and 4 . . . . . . . . . . . . . . . . . . . . . . . 15
2.3 The Subgroups of Order 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.4 The Subgroups of Order 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.5 The Subgroups of Order 32 . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.6 Subgroup Lattice of Q×Q . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3 Further Results about Constructions and Subgroup Lattices 49
3.1 Q×D8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.2 Extraspecial Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.3 Subgroup Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
vii
Conclusions 60
Bibliography 62
viii
Notation
Notation that is not cited here is consistent with that used in [2]. Functions are usually
written on the left. However, in this dissertation there are times where we write them on
the right as exponents. We trust that the reader will be able to decipher this.
V4 the Klein 4-group
Q the quaternion group of order 8
D8 the dihedral group of order 8
Sn the symmetric group on n letters
Cn the cyclic group of order n
Z(G) the center of G
Φ(G) the Frattini subgroup of G
G′ the commutator subgroup of G
Aut(G) the automorphism group of G
GF (p) the integers modulo p
Zn the groups of order 2 in lattice of Q×Q
πG the natural projection of G×H onto G
πH the natural projection of G×H onto H
|x| the order of a group element x
G H the central product of G with H [2]
G H a subdirect product of G and H with amalgamated factor group [7] P. 50
ix
Introduction
The main focus of this work is on a characterization of containment of subgroups in a direct
product of finite groups. In particular, our primary goal is to establish this characterization
and to give applications of this result. Suppose G and H are finite groups. Although
subgroups of G × H, their embeddings and their interactions with other subgroups have
been investigated before (see [4], [6], [11]), finding properties which can further characterize
the containment relation between subgroups remain open. This dissertation provides such
a property.
Please note that throughout this dissertation, all of the groups discussed will be finite. Also
note that our convention is that automorphisms will act on the right.
Our general approach to giving a characterization of containment of subgroups of a direct
product and to providing applications of it is given in the outline to follow. Chapter 1
initiates this by providing background on the subgroup structure of a direct product and
notation that will be used. In Section 1.1, we begin this discussion by presenting a theorem
by Edouard Goursat, which will serve as a starting point for investigating containment of
subgroups. Goursat’s Theorem provides a nice way of understanding the structure of a
subgroup of a direct product. The reader should recall that the natural projections πA and
πB are homomorphisms of A×B onto A and B respectively.
In Section 1.2, we expand on the notation introduced in Goursat’s Theorem, by presenting
results that will provide a better understanding of mappings that will be used in our con-
tainment characterization. This section may seem very technical, and it is indeed. To follow
the information presented here, the reader should recall the Isomorphism Theorems [3] and
the projections and intersections that will be introduced in Section 1.1 after Goursat’s
Theorem. To unify notation, we are explicit.
Section 1.3 concludes the chapter. This is the theoretical backbone of the dissertation.
1
There is a characterization of containment of subgroups of a direct product and a theorem
that will allow one to see when two subgroups are isomorphic and have the same subgroup
lattice structure. Suppose U1 and U2 are subgroups of A×B. The characterization provides
necessary and sufficient conditions for U2 ≤ U1. The theorem that follows the character-
ization will then say when U1∼= U2 under an automorphism of a direct product. We use
Aut(A) to denote the automorphism group of A.
More technically, the two results given in Section 1.3 are very applicable for those who
study subgroup lattices because they provide us with another way of constructing subgroup
lattices which have a large number of subgroups. In particular, using these results will allow
us to construct two subgroup lattices of groups of order 64; namely, Q × Q and Q × D8,
where Q and D8 denote the quaternion and dihedral groups of order 8.
In order to demonstrate this, we construct the subgroup lattice of Q×Q. The details of this
construction lie within Chapter 2. In Section 2.1, we use Goursat’s Theorem to calculate
the number of subgroups in Q×Q, and we provide a formula to calculate the order of the
subgroups. We do this by first giving a presentation of Q and the notation we will use for
its subgroups. Then we use these subgroups of Q and its subgroup lattice to examine the
sections in each direct factor of Q ×Q. The reader should recall here that a section [2] of
a group A is just a quotient of a subgroup of the group A. After examining the sections of
each factor we then calculate the orders of the groups.
In Sections 2.2 − 2.5, the details of the subgroups of Q × Q of orders 2 and 4, 8, 16 and
32 respectively will be given. The details of each section include providing notation for
each of these subgroups, their group structures, and arguments which will give the maximal
subgroups of each of these subgroups. Being that the factors of Q × Q are the same, one
should expect symmetry among the groups and also expect that many subgroups will be
isomorphic. Because we are in a 2-group, it is enough to determine the maximal subgroups
from one level of the subgroup lattice to the next by order consideration.
In constructing the subgroup lattice, we will determine, by transitivity, only the maximal
subgroups of the subgroups of Q × Q. We lessen the amount of details involved by deter-
mining the maximal subgroups of one subgroup, say U , and then finding automorphisms
which move the subgroup U to another subgroup, say W . We then use this automorphism
to determine the maximal subgroups of W .
2
Since we are in a p-group, p = 2, we can use Burnside’s Basis Theorem to determine the
number of generators of each subgroup of Q×Q. For a group G, we define Φ(G) =⋂M |
M is a maximal subgroup of G . In our case, namely a 2-group, we know that Φ(G) =⟨x2|x ∈ G
⟩([7] Satz 3.14 P.272). A version of Burnside’s Basis Theorem is given below.
Theorem 0.1. [7]
Let G be a p-group with
∣∣∣∣ G
Φ(G)
∣∣∣∣ = pd. Then G is generated by exactly d elements, and
no set with less than d elements generates G. Every element of G \ Φ(G) belongs to some
minimal generating set.
Consequence 1:G
Φ(G)is an elementary abelian p-group ([7] Satz 3.14 P. 272). Hence, it
is isomorphic to a vector space over GF (p).
Consequence 2: From linear algebra, we then know that the number of maximal subgroups
of G ispd − 1
p− 1.
These consequences will be heavily used throughout the construction of the lattice of Q×Q
in Chapter 2.
0.1 Automorphisms of Q
In Section 2.3, the reader should know that Aut(Q) ∼= S4, where S4 is the symmetric group
on 4 numbers. Because these 24 automorphisms will be essential to knowing what group
we are considering, we will list them in the following section.
In this section, we list the 24 automorphisms of Q. They will be denoted by τn, where
1 ≤ n ≤ 24, and x and y are the generators for Q. The list is given below.
The inner automorphisms are:
τ1 : x 7→ x , τ2 : x 7→ x , τ3 : x 7→ x−1 , τ4 : x 7→ x−1
y 7→ y y 7→ y−1 y 7→ y y 7→ y−1
Note that the Out(Q) ∼= S3. So, we should have 6 cosets of 4. Hence the outer automor-
phisms are as follows:
3
τ5 : x 7→ x , τ6 : x 7→ x , τ7 : x 7→ x−1 , τ8 : x 7→ x−1
y 7→ xy y 7→ xy−1 y 7→ x−1y y 7→ x−1y−1
τ9 : x 7→ y , τ10 : x 7→ y−1 , τ11 : x 7→ y , τ12 : x 7→ y−1
y 7→ x y 7→ x y 7→ x−1 y 7→ x−1
τ13 : x 7→ y , τ14 : x 7→ y−1 , τ15 : x 7→ y , τ16 : x 7→ y−1
y 7→ xy y 7→ xy−1 y 7→ x−1y y 7→ xy
τ17 : x 7→ xy , τ18 : x 7→ xy−1 , τ19 : x 7→ x−1y, τ20 : x 7→ xy
y 7→ y y 7→ y−1 y 7→ y y 7→ y−1
τ21 : x 7→ xy , τ22 : x 7→ xy−1 , τ23 : x 7→ x−1y, τ24 : x 7→ xy
y 7→ x y 7→ x y 7→ x−1 y 7→ x−1
0.2 Automorphisms of V4
In Section 2.4, the reader should know that Aut(V4) ∼= S3, where V4 is the Klein 4-group and
S3 is the symmetric group on 3 numbers. Because these 6 automorphisms will be essential
to knowing what group we are considering, we will list them in the following Section.
In this section, a list of the 6 automorphisms of V4, denoted by βp, where 1 ≤ p ≤ 6, is
given below. If one would like, think of S3 as GL(2, 2). The automorphishms of GL(2, 2) are:
β1 =
1 0
0 1
β2 =
1 0
1 1
β3 =
0 1
1 0
β4 =
0 1
1 1
β5 =
1 1
0 1
β6 =
1 1
1 0
4
For V4 = 〈x, y〉, when writing automorphisms on the right, we then have:
β1 : x 7→ x β2 : x 7→ x β3 : x 7→ y
y 7→ y y 7→ xy y 7→ x
β4 : x 7→ y β5 : x 7→ xy β6 : x 7→ xy
y 7→ xy y 7→ y y 7→ x
Section 2.6 concludes the chapter with the subgroup lattice of Q×Q. This subgroup lattice
uses the details given in Sections 2.1− 2.5.
In Section 3.1, we calculate the number of subgroups and their orders in Q×D8. We do not
include the details of how we constructed the subgroup lattice of Q×D8 for the sake of the
reader and because the details would be similar, in nature, to the details in construction of
the subgroup lattice of Q×Q. The subgroup lattice of Q×D8 will be given in Section 3.3.
So, why construct the subgroup lattices of Q×Q and Q×D8, two groups of order 64? Our
reason for constructing these two subgroup lattices is that there are exactly two extraspecial
groups of order 32. They can be recognized as the quotient of the group by the diagonal
of the centers; that is, Q Q and Q D8. In Section 3.2, we discuss extraspecial groups
and give a well known classification of them. This classification will reiterate why we chose
these two subgroup lattices. A p-group P is extraspecial [2] if Z(P ) = P ′ = Φ(P ).
Chapter 3 ends with the subgroup lattices of Q × Q, Q × D8, and the two extraspecial
groups of order 32 along with information about them. If one is interested in further
information about groups of order 2n, where n ≤ 6, then [8] is a great reference. Opening
that double oversized book was very motivating and inspirational for many reasons. The
authors commitment to and passion for understanding groups of order 2n was revealed
by the adversity they endured to complete the work they had started before World War
II. Mostly, this book allowed us to see how mathematicians studied groups and subgroup
lattices that we are currently interested in. Many of the techniques used to construct the
tables and lattices which appear in that book are attributed to Philip Hall.
5
Chapter 1
Characterization of Containment of Subgroups of a DirectProduct A×B
Given finite groups G and H and information on the subgroups of G and H, what can
one say about the subgroups of G × H? In this chapter, we give a characterization of
containment of subgroups in a direct product.
In Section 1.1, we state a well known result by Edouard Goursat which gives the subgroup
structure of a direct product. In Section 1.2, we present results that will give rise to the
mappings and notation introduced in Section 1.3. In Section 1.3, we give the main results,
which include a characterization of containment of subgroups. The theorems presented in
Section 1.3 are essential in the construction of the subgroup lattices of Q×Q and Q×D8,
done in Chapters 2 and 3, where Q is the quaternions and D8 is the dihedral group of order
8.
1.1 Goursat’s Theorem
Much is known about subgroups of a direct product and a great deal of the properties that
we know about such subgroups depend heavily on applications of Goursat’s theorem. In
1889 [5], Edouard Goursat proved a theorem which describes the subgroup structure of a
direct product in terms of the sections of the factor groups.
Theorem 1.1. (Goursat) Let A and B be groups. Then there exists a bijection between
the set S of all subgroups of A × B and the set T of all triples
(I
J,L
K, σ
), where
I
Jis a
section of A,L
Kis a section of B and σ :
I
J→ L
Kis an isomorphism.
To prove this theorem, one can make the following observations. The projections πA :
6
A × B → A given by πA(a, b) = a and πB : A × B → B given by πB(a, b) = b, for a ∈ A
and b ∈ B, are homomorphisms.
Let U ≤ A × B. Since A × B is a direct product and each factor is normal in A × B we
can say that U ∩ A C U and U ∩ B C U . Furthermore since πA and πB are projections
we can in fact say that given U ≤ A × B, there is a unique triple which corresponds to
it. Namely, since U ≤ A × B, U ∩ A C πA(U) and U ∩ B C πB(U). Let (a, b) ∈ U . Then
σ :πA(U)
U ∩A→ πB(U)
U ∩Bgiven by (a(U ∩ A))σ = b(U ∩ B) is an isomorphism. Then one can
show that U = (a, b)|a ∈ πA(U), b ∈ πB(U), (a(U ∩A))σ = b(U ∩B). Conversely, given a
triple, one obtains U as defined above.
When U ≤ A× B, we use the following notation: I = πA(U), J = U ∩ A, L = πB(U) and
K = U ∩B.
In proving Goursat’s Theorem, one sees that the subgroups of A×B are really established
by their projections onto A and B and their intersections with A and B. So, it is then
reasonable to believe that these projections and intersections will come up often when we
discuss containment of subgroups in a direct product.
One can use this theorem in counting the number of subgroups of a direct product; see [10]
on how to do this. Provided that one is able to determine all the sections of A and B and
the isomorphisms between the isomorphic sections, one can count the number of subgroups
by counting the number of triples.
1.2 Conditions for Projections, Intersections, and Isomor-phisms Between Sections of Direct Products
First, we will provide isomorphism theorems that will motivate the isomorphism introduced
in Lemma 1.6. Then we will continue the use of the notation introduced in Section 1.1 to
provide conditions for containment of subgroups of a direct product.
1.2.1 Background on Isomorphism Theorems
In this section, we provide isomorphism theorems that will motivate the notation introduced
in Lemma 1.6. If ϕ : G → H is a homomorphism, we will denote the inverse image of a
7
subgroup N as Nϕ−1
.
Theorem 1.2. [12]
If ϕ : G → H is a homomorphism, Gϕ = H, K C H and N = Kϕ−1
, then ϕ :G
N→ H
Kgiven by gN = gϕK, where g ∈ G, is an isomorphism.
We now recall the following result from Dummit and Foote.
Theorem 1.3. [3]
Let G be a group and let N and M be normal subgroups of G with M ≤ N . ThenN
MCG
M
andG/M
N/M∼=G
N.
In conjunction with Goursat’s theorem, we use these results to establish notation for the
rest of the dissertation.
FACT: If ϕ : G → H is an isomorphism and K CH with N = Kϕ−1
, then Nϕ = K and
there is an isomorphism ϕ :G
N→ H
Nϕ, as specified in Theorem 1.2. In addition, using
Theorem 1.3, if M CG, M ≤ N , then there is another isomorphism ϕ :G/M
N/M→ H/Mϕ
(N/M)ϕ
given by
(gM
(N
M
))ϕ= gϕMϕ
(N
M
)ϕ.
1.2.2 Lemmas
The goal of this section is to present results that will be used to prove the main results
of this chapter. Using the notation introduced in Goursat’s theorem, which describes the
projections, intersections, isomorphisms and subgroups of A×B, we will provide background
information that will be an integral part of the proof of the main results to follow in section
1.3.
Recall the following notation introduced in Goursat’s theorem. For U1, U2 ≤ A × B,
In = πA(Un), Jn = Un ∩A, Ln = πB(Un) and Kn = Un ∩B, where
Un = (a, b)|a ∈ In, b ∈ Ln, (aJn)σn = bKn
for n = 1, 2. To further understand the subgroup structure of a direct product we will
examine the isomorphisms σn :InJn→ Ln
Kn.
8
Lemma 1.4. For n = 1, 2 let Un ≤ A × B, In = πA(Un), Jn = Un ∩ A, Ln = πB(Un),
Kn = Un ∩ B and σ1 :I1
J1→ L1
K1be the isomorphism from the triple associated to U1.
Suppose further that the following conditions hold:
(i) I2 ≤ I1
(ii) L2 ≤ L1 and
(iii)
(I2J1
J1
)σ1
=L2K1
K1.
Define σ1 :I2J1
J1→ L2K1
K1to be the restriction of σ1. Then σ1 is an isomorphism.
Lemma 1.5. For n = 1, 2 suppose Un ≤ A × B with U2 ≤ U1. Suppose further that
In = πA(Un), Jn = Un ∩A, Ln = πB(Un) and Kn = Un ∩B. Then
(i) J2 ≤ I2 ∩ J1
(ii) K2 ≤ L2 ∩K1.
The proof of this lemma is routine. Parts (i) and (ii) are easy consequences of the following:
U2 ≤ U1 implies that I2 ≤ I1, L2 ≤ L1, J2 ≤ J1 and K2 ≤ K1.
Observation: To understand the following lemma it suffices to make a couple observations.
First, observe that for n = 1, 2, Jn C In, I2 ≤ I1 and so I2 ∩ J1 C I2 with J2 ≤ I2 ∩
J1. Then by Theorem 1.3,I2 ∩ J1
J2CI2
J2and the mapping ηA :
(I2
J2
)/(I2 ∩ J1
J2
)−→
I2
I2 ∩ J1is an isomorphism. Second, making a similar observation in B, we will have ηB :(
L2
K2
)/(L2 ∩K1
K2
)−→ L2
L2 ∩K1is an isomorphism.
Lemma 1.6. For n = 1, 2 let Un ≤ A×B with U2 ≤ U1. Suppose further that In = πA(Un),
Jn = Un ∩ A, Ln = πB(Un), Kn = Un ∩B and σ2 :I2
J2→ L2
K2is an isomorphism such that(
I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2. Then, from the FACT following Theorems 1.2 and 1.3,
(i) There is an isomorphism σ2 :
(I2
J2
)/(I2 ∩ J1
J2
)−→
(L2
K2
)/(L2 ∩K1
K2
)as de-
fined in those consequences.
(ii) We define σ2 :I2
I2 ∩ J1→ L2
L2 ∩K1by σ2 = ηA
−1σ2ηB, where composition is read
left to right, and it is an isomorphism. (This is a slight deviation from notation in
FACT.)
Note that for (a, b) ∈ U2, (a(I2 ∩ J1))σ2 = b(L2 ∩K1).
9
1.3 Main Results: Containment of Subgroups of a DirectProduct
The goal of this section is to give a characterization of containment of subgroups in a product
of groups. This is accomplished in Theorem 1.7. In Theorem 1.8, we then state when two
subgroups of a direct product are isomorphic under an automorphism of that direct product.
Recall the following. for n = 1, 2, let Un ≤ A × B. Suppose further that I2 ≤ I1, J2 ≤ J1,
L2 ≤ L1 and K2 ≤ K1. Notice that J1 C I1 and K1 C L1. Then observe that the mapping
θ1 :I2J1
J1−→ I2
I2 ∩ J1given by (aJ1)θ1 = a(I2 ∩ J1) is an isomorphism. Similarly, observe
that θ2 :L2K1
K1−→ L2
L2 ∩K1given by (bK1)θ2 = b(L2 ∩K1) is an isomorphism.
Theorem 1.7. Suppose U1, U2 ≤ A×B. Then U2 ≤ U1 ⇐⇒
(i) I2 ≤ I1, J2 ≤ J1, L2 ≤ L1 and K2 ≤ K1
(ii)
(I2J1
J1
)σ1
=L2K1
K1
(iii)
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2and
(iv) σ2 θ1 = θ2 σ1, where θn and σn are defined as before and n = 1, 2.
An equivalent statement for (iv) is that the following diagram is a commutative diagram.
I2J1
J1
θ1
σ1
// L2K1
K1
θ2
I2
I2 ∩ J1
σ2 // L2
L2 ∩K1
Proof:(=⇒) Suppose U2 ≤ U1. It is obvious that I2 ≤ I1, J2 ≤ J1, L2 ≤ L1, K2 ≤ K1,(I2J1
J1
)σ1
=L2K1
K1, and
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2. Hence, it suffices to show that the
diagram commutes. More specifically, that θ2 σ1 = σ2 θ1. Let c ∈ I2 = πA(U2).
Then there exists a d ∈ L2 = πB(U2) such that (c, d) ∈ U2. Since U2 ≤ U1, we know
(cJ1)σ1 = dK1. By Lemma 1.4, σ1 is a restriction of σ1. Hence, we can consider (cJ1)σ1 .
Then ((cJ1)σ1)θ2 = (dK1)θ2 = d(L2 ∩ K1). On the other hand, using Lemma 1.6, we get
((cJ1)θ1)σ2 = (c(I2 ∩ J1))σ2 = d(L2 ∩ K1). Therefore, θ2 σ1 = σ2 θ1, and the diagram
commutes.
(⇐=) Conversely, suppose the containments hold and the diagram commutes. Our aim is
10
to show U2 ≤ U1. Let (c, d) ∈ U2. Then (cJ2)σ2 = dK2, where c ∈ I2 and d ∈ L2. Consider
cJ1. We can consider this coset because the containments hold. Because of Lemma 1.6 we
can consider (cJ1)σ2 . Then (cJ1)σ1 = (((cJ1)θ1)σ2)θ−12 = ((c(I2 ∩ J1))σ2)θ
−12 . Using Lemma
1.6, we then see that ((c(I2 ∩ J1))σ2)θ−12 = (d(L2 ∩ K1))θ
−12 = dK1. So, (cJ1)σ1 = dK1.
Therefore, by Lemma 1.4 we know (cJ1)σ1 = dK1 and U2 ≤ U1.
Theorem 1.8. Suppose U1, U2 ≤ A×B with Jn C In, Kn C Ln and σn : In/Jn → Ln/Kn
such that Un = (a, b)|a ∈ In, b ∈ Ln, (aJn)σn = bKn, n = 1, 2. Suppose further that
α ∈ Aut(A), γ ∈ Aut(B) such that
α : I1 → I2 and J1 → J2
γ : L1 → L2 and K1 → K2
α : I1/J1 → I2/J2
γ : L1/K1 → L2/K2
such that σ2α = γσ1. Then (α, γ) ∈ Aut(A×B) and U(α,γ)1 = U2.
The proof of this theorem is routine to check.
Remark: Later we will point out that conditions (i)−(iv) of Theorem 1.7 are independent.
This will be done at the end of Section 2.3.
11
Chapter 2
Construction of the Subgroup Lattice for Q×Q
In this chapter, we apply the theorems from Section 1.3 to construct the subgroup lattice of
Q×Q. Throughout each section, there will be a detailed explanation of the group structures
of specific orders included with the lattice construction. If one is interested in the lattice
only, see the tables with groups and maximal subgroups in each section. As one will see,
every group in the construction is determined by certain projections and intersections in
each factor of Q×Q. To determine the maximal subgroups of these groups, we will verify
the hypothesis of Theorem 1.7. If the sections of each factor of Q × Q are trivial then
the showing that the diagram commutes is trivial, and we only need to verify the other
hypotheses of Theorem 1.7. If the sections of each factor are nontrivial then we must verify
all of the hypotheses of Theorem 1.7
In Section 2.1, we will introduce Q × Q and describe how we used Goursat’s theorem to
calculate the number of subgroups. Section 2.2 will introduce the groups of order 2 and
the groups of order 4. Section 2.3 will introduce the groups of order 8 and determine the
maximal subgroups of each group. Section 2.4 will introduce the groups of order 16 and
determine the maximal subgroups of each group. Section 2.5 will introduce the groups of
order 32 and determine the maximal subgroups of each group. Section 2.6 will show the
subgroup lattice of Q × Q. Let H be a subgroup in Q × Q. In the subgroup lattice the
subgroups that were called←−H will be written as ∗H.
12
2.1 Using Goursat to Calculate the Number of Subgroupsof Q×Q
In the following example we will use the quaternion group of order 8, written Q, to construct
the subgroup lattice of Q×Q. Although Q is a well known group, it suffices to introduce our
notation for it and its subgroups because it will be used heavily and consistently throughout
the remainder of the chapter.
The presentation of the quaternions is Q =< x, y|x4 = y4 = 1, x2 = y2 >. There are 3 cyclic
groups of order 4 contained in Q. Namely, F1 = < x >, F2 = < y > and F3 = < xy >.
There is one cyclic group of order 2 contained in Q, and we will call it Z =< z >, where
z = x2. The subgroup lattice of Q can be found below.
Q
F1 F3 F2
Z
1
Example. Using the notation for Q introduced above, let E = Q,Fi, Z, 1|1 ≤ i ≤ 3. Our
aim is to construct the subgroup lattice for G = Q×Q. Using Goursat’s Theorem, it first
suffices to find the number of subgroups in G by examining the sectionsI
Jand
L
K, where
I, J,K,L ∈ E. In order to calculate the orders of the subgroups we will use the following:
If U ≤ A×B, then |U | =|I||L||I/J |
.
Case 1. Let I = Q, and let J ∈ E. Then
13
I, J I/J Aut(I/J) L, K No. of subgroups
Q, Q 1 1
Q, Q 1 of order 64
Fj , Fj 3 of order 32
Z, Z 1 of order 16
1, 1 1 of order 8
Q, Fi C2 1
Q, Fj 9 of order 32
Fj , Z 9 of order 16
Z, 1 3 of order 8
Q, Z V4 S3 Q, Z 6 of order 16
Q, 1 Q S4 Q, 1 24 of order 8
Case 2. Let I = Fi, and let J ∈ E \ Q. Then
I, J I/J Aut(I/J) L, K No. of subgroups
Fi, Fi 1 1
Q, Q 3 of order 32
Fj , Fj 9 of order 16
Z, Z 3 of order 8
1, 1 3 of order 4
Fi, Z C2 1
Q, Fj 9 of order 16
Fj , Z 9 of order 8
Z, 1 3 of order 4
Fi, 1 C4 C2 Fj , 1 18 of order 4
Case 3. Let I = Z, and let J ∈ E \ Q,Fi. Then
I, J I/J Aut(I/J) L, K No. of subgroups
Z, Z 1 1
Q, Q 1 of order 16
Fj , Fj 3 of order 8
Z, Z 1 of order 4
1, 1 1 of order 2
Z, 1 C2 1
Q, Fj 3 of order 8
Fj , Z 3 of order 4
Z, 1 1 of order 2
Case 4. Let I = 1 = J . Then
14
I, J I/J Aut(I/J) L, K No. of subgroups
1, 1 1 1
Q, Q 1 of order 8
Fj , Fj 3 of order 4
Z, Z 1 of order 2
1, 1 1 of order 1
Hence, there are 133 subgroups of Q×Q, which include the trivial subgroup, 3 subgroups
of order 2, 31 subgroups of order 4, 47 subgroups of order 8, 35 subgroups of order 16, 15
subgroups of order 32 and the group itself.
2.2 The Subgroups of Orders 2 and 4
In order to construct this subgroup lattice, it suffices to introduce a bit of notation and
to determine the maximal subgroups on each level. Clearly, the groups of order 2 are all
copies of C2, and the only subgroup it contains is the trivial subgroup. For the sake of
construction and reference, the groups are listed in the table below.
Subgroups of Order 2
Notation I, J I/J Aut(I/J) L, K No. of subgroups group structure
Z1 Z, Z 1 1 1, 1 1 C2
Z2 Z, 1 C2 1 Z, 1 1 C2
Z3 1, 1 1 1 Z, Z 1 C2
Subgroups of Order 4
The groups of order 4 are either copies of C4, the cyclic group of order 4, or V4 the Klein-4
group. The groups of order 4 are given in the table below, and for these groups 1 ≤ i, j ≤ 3.
Notation I, J I/J Aut(I/J) L, K No. of subgroups group structure←−Fi Fi, Fi 1 1 1, 1 3 C4
−→Fj 1, 1 1 1 Fj , Fj 3 C4
←−Xi Fi, Z C2 1 Z, 1 3 C4
−→Xj Z, 1 C2 1 Fj , Z 3 C4
←−Fij ,−→Fij Fi, 1 C4 C2 Fj , 1 18 C4
V Z, Z 1 1 Z, Z 1 V4
15
Observe that all of the groups of order 4 are copies of C4, with the exception of V . To
verify these group structures, one can examine the generators of the group in question.
Then determining its maximal subgroup(s) is a routine check, consisting of verifying the
hypotheses of Theorem 1.7.
For the groups←−Fij and
−→Fij , Aut(I/J) = C2.
←−Fij is the group that corresponds to the identity
automorphism, and−→Fij is the group that corresponds to the automorphism which sends x
to x−1. As an example of how one would verify the group structure of one of these groups
of order 4 and determine its maximal subgroup(s), read below. Otherwise, one can examine
the table at the end of this section to see a list of the maximal subgroups of the groups of
order 4.
Consider←−F11. In this case, σ1 is the isomorphism that maps F1 → F1. So,
←−F11 =
(x, x)|x ∈ F1 is a cyclic group of order 4, written C4. To determine its maximal sub-
group we could verify the hypotheses of Theorem 1.7. However, since this group is cyclic
of order 4, notice that its unique subgroup of order 2 is generated by the square of (x, x),
and this group is Z2 given in the section discussing groups of order 2.
Let us now explain why V ∼= V4, where V4∼= C2 × C2. Notice that V =
(a, b)|a ∈ Z, b ∈ Z ∼= C2 ×C2. This group has 3 maximal subgroups of order 2, and they
are Z1, Z2 and Z3. The maximal subgroups of all the groups of order 4 are listed in the
table below.
Groups Maximal Subgroups Groups Maximal Subgroups←−Fi Z1
−→Fj Z3
←−Xi Z1
−→Xi Z3
←−Fij Z2
−→Fij Z2
V Z1, Z2, Z3
2.3 The Subgroups of Order 8
The groups of order 8 that appear in this lattice will be copies of Q, C4 × C2 or C2 × C4.
We are familiar with the subgroup lattice of Q. In order to identify C4 × C2 in Q × Q, it
suffices to give its subgroup lattice. It is given below.
16
C4 × C2
C4 C4 C2 × C2
C2 C2 C2
1
Although not used within the explanations to follow, knowing the subgroup lattice above
was very helpful in determining and checking that the groups with the above group structure
indeed had the correct maximal subgroups. The table that follows lists the groups of order
8 and its corresponding group structure. For these groups, 1 ≤ i, j ≤ 3 and 1 ≤ n ≤ 24.
Notation I, J I/J Aut(I/J) L, K No. of subgroups group structure←−Q1 Q, Q 1 1 1, 1 1 Q−→Q2 1, 1 1 1 Q, Q 1 Q←−Wi Q, Fi C2 1 Z, 1 3 Q−→Wj Z, 1 C2 1 Q, Fj 3 Q←−Yi Fi, Fi 1 1 Z, Z 3 C4 × C2
−→Yj Z, Z 1 1 Fj , Fj 3 C2 × C4
Fij Fi, Z C2 1 Fj , Z 9 C4 × C2
∆n Q, 1 Q S4 Q, 1 24 Q
Now, let us begin to determine the maximal subgroups of the groups of order 8. Using the
table above, we will begin by determining the maximal subgroups of the groups that have
trivial sections, and then we will determine the maximal subgroups of the groups that have
nontrivial sections.
Consider←−Q1 = Q × 1. Its maximal subgroups are
←−Fi = Fi × 1, 1 ≤ i ≤ 3. Similarly,
−→Q2 = 1 × Q, and its maximal subgroups are
−→Fj = 1 × Fj , 1 ≤ j ≤ 3. A table with the
maximal subgroups of←−Q1 and
−→Q2 is given below.
Groups Maximal Subgroups←−Q1
←−F1,←−F2,←−F3
−→Q2
−→F1,−→F2,−→F3
Consider−→Y1, which is clearly isomorphic to C2 × C4. As shown at the beginning of this
17
section, C2×C4 has 3 maximal subgroups, and we claim that the maximal subgroups of−→Y1
are−→X1,−→F1 and V . Since these groups have order 4, they are indeed maximal. To determine
that these are subgroups, we must verify the hypotheses of Theorem 1.7. Examine the tables
below to see that the containment of the projections and intersections of these proposed
maximal subgroups are satisfied.
−→X1
−→Y1
I2 = Z I1 = Z
J2 = 1 J1 = Z
L2 = F1 L1 = F1
K2 = Z K1 = F1
−→F1
−→Y1
I2 = 1 I1 = Z
J2 = 1 J1 = Z
L2 = F1 L1 = F1
K2 = F1 K1 = F1
V−→Y1
I2 = Z I1 = Z
J2 = Z J1 = Z
L2 = Z L1 = F1
K2 = Z K1 = F1
Now, we must show that parts (ii) and (iii) of Theorem 1.7, namely
(I2J1
J1
)σ1
=L2K1
K1and(
I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2respectively, are satisfied for each proposed maximal subgroup. In
order to do this, we must know σ1 and σ2. In our case, σ1 :Z
Z→ F1
F1, which is the
identity, and verifying part (ii) of Theorem 1.7 is trivial. Observe that σ2 changes for
each subgroup that we are considering. Hence, it suffices to examine each of the proposed
maximal subgroups, determine σ2 and verify part (iii) of Theorem 1.7.
For−→X1, σ2 :
Z
1→ F1
Zand (z)σ2 = xZ. Then
(I2 ∩ J1
J2
)σ2
=
(Z ∩ Z
1
)σ2
= (Z)σ2 = 〈xZ〉
andL2 ∩K1
K2=F1 ∩ F1
Z= F1Z = 〈xZ〉. Hence, part (iii) is satisfied.
For−→F1 and V , the associated σ2 is the identity map. Hence, part (iii) is obviously satisfied.
18
Showing that the diagram commutes is trivial because the sections which define−→Y1 are
trivial. Therefore, the hypotheses of Theorem 1.7 are satisfied, and the maximal subgroups
of−→Y1 are
−→X1,−→F1 and V .
One can show that−→Yj are isomorphic, where 1 ≤ j ≤ 3, by applying Theorem 1.8. Namely,
one can use an automorphism of Q × Q to show that the subgroups are isomorphic. Let
us now show that−→Y1∼=−→Y2. The associated σ1 for
−→Y1 and
−→Y2 is the identity. So, to show
−→Y1∼=−→Y2, we need to recall τ9 ∈ Aut(Q) given by (x)τ9 = y. Then (1, τ9) ∈ Aut(Q × Q)
and(−→Y1
)(1,τ9)=−→Y2. Similarly, to show that
−→Y1∼=−→Y3, recall that τ17 ∈ Aut(Q) given by
(x)τ17 = xy. Then (1, τ17) ∈ Aut(Q×Q) and(−→Y1
)(1,τ17)=−→Y3. Hence
−→Yj ’s are isomorphic.
To determine the maximal subgroups of−→Y2 and
−→Y3, we will use the same automorphisms
of Q×Q from above that we used to show−→Yj were isomorphic. Then we will examine the
maximal subgroups of−→Y1 under these automorphisms to attain the maximal subgroups of
the isomorphic subgroup we are considering. Recall that the maximal subgroups of−→Y1 are
−→X1,−→F1 and V .
Let us now determine the maximal subgroups of−→Y2. Consider
−→X1. Then
(−→X1
)(1,τ9)=
〈(z, xZ)〉(1,τ9) = 〈(z, yZ)〉 =−→X2. Consider
−→F1. Then
(−→F1
)(1,τ9)= (1× F1)(1,τ9) = 1 × F2 =
−→F2. Consider V . Then V (1,τ9) = 〈(z, z)〉(1,τ9) = V . Therefore, the maximal subgroups of
−→Y2
are−→X2,−→F2 and V . Similarly, using (1, τ17), one can show the maximal subgroups of
−→Y3 are
−→X3,−→F3 and V .
Analogously, we can produce a similar argument for←−Yi , 1 ≤ i ≤ 3, by noticing that this
group is defined by reversing the roles of I, J and L,K in−→Yj . Also, we can make this
observation for←−Yi by noticing that ρ : (a, b) 7→ (b, a) is an automorphism of Q × Q. The
maximal subgroups of←−Yi are
←−Xi,←−Fi and V . A table with the maximal subgroups of
←−Yi and
−→Yj , where 1 ≤ i, j ≤ 3, is given below.
Groups Maximal Subgroups Groups Maximal Subgroups←−Y1
←−X1,←−F1, V
−→Y1
−→X1,−→F1, V
←−Y2
←−X2,←−F2, V
−→Y2
−→X2,−→F2, V
←−Y3
←−X3,←−F3, V
−→Y3
−→X3,−→F3, V
Consider←−W1 = (a, b)|a ∈ Q, b ∈ Z, (aF1)σ1 = b, where σ1 :
Q
F1→ Z is given by (yF1)σ1 =
19
z. So, (y, z), (x, 1) ∈←−W1,←−W1 is nonabelian, (y, z)4 = 1 = (x, 1)4 and (y, z) 6= (x, 1). Hence,
←−W1 = 〈(y, z), (x, 1)〉 ∼= Q. We propose that its maximal subgroups are
←−F1,←−X2 and
←−X3.
To determine that these are subgroups, we must verify the hypotheses of Theorem 1.7.
Examine the tables below to see that the containment of the projections and intersections
of these proposed maximal subgroups are satisfied.
←−F1
←−W1
I2 = F1 I1 = Q
J2 = F1 J1 = F1
L2 = 1 L1 = Z
K2 = 1 K1 = 1
←−X2
←−W1
I2 = F2 I1 = Q
J2 = Z J1 = F1
L2 = Z L1 = Z
K2 = 1 K1 = 1
←−X3
←−W1
I2 = F3 I1 = Q
J2 = Z J1 = F1
L2 = Z L1 = Z
K2 = 1 K1 = 1
Now, we must verify that parts (ii), (iii) and (iv), of Theorem 1.7, are satisfied for each
proposed maximal subgroup. Namely that
(I2J1
J1
)σ1
=L2K1
K1,
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2,
and that the diagram commutes respectively. To verify parts (ii) and (iii), we must know
σ1 and σ2. In our case, σ1 :Q
F1→ Z and (yF1)σ1 = z. Observe that σ1 will remain the
same for each proposed maximal subgroup, and σ2 changes. To verify part (iv), that is,
σ2 θ1 = θ2 σ1, for each proposed subgroup we will give the associated maps θ1, σ2, σ1
and θ2. Then we will verify that the diagram commutes. In verifying part (iv) it is enough
to check that the diagram commutes using the generators because our maps are linear. Let
us now examine each of the proposed maximal subgroups, determine σ2 and verify parts
(ii), (iii) and (iv) of Theorem 1.7.
20
Consider←−F1. Then
(I2J1
J1
)σ1
=
(F1F1
F1
)σ1
= 1 andL2K1
K1= 1. So, part (ii) is satisfied.
The associated σ2 for←−F1 is the identity, and hence part (iii) is obviously satisfied. To
show the diagram commutes for←−F1, observe that
L2
L2 ∩K1= 1. This means that all of the
associated maps must be the identity. One can verify this is indeed true. Therefore, the
diagram commutes and the hypotheses of Theorem 1.7 are satisfied. Hence,←−F1 is a maximal
subgroup.
Consider←−X2. Then
(I2J1
J1
)σ1
=
(F2F1
F1
)σ1
= 〈yF1〉σ1 = 〈z〉 = Z andL2K1
K1= Z.
So, part (ii) is satisfied. Observe that σ2 :F2
Z→ Z is given by (yZ)σ2 = z. Then(
I2 ∩ J1
J2
)σ2
=
(F2 ∩ F1
Z
)σ2
=
(Z
Z
)σ2
= 1 andL2 ∩K1
K2=Z ∩ 1
1= 1. Hence, part (iii)
is satisfied.
For←−X2, we have the following associated maps:
θ1 :F2F1
F1=
Q
F1−→ F2
F2 ∩ F1=F2
Zis given by (yF1)θ1 = yZ;
σ2 :F2
F2 ∩ F1=F2
Z−→ Z
Z ∩ 1= Z is given by (yZ)σ2 = z;
σ1 :F2F1
F1=
Q
F1−→ Z is given by (yF1)σ1 = z;
θ2 : Z −→ Z
Z ∩ 1= Z is given by (z)θ2 = z.
Then σ2 (θ1 (yF1)) = σ2 (yZ) = z and θ2 (σ1 (yF1)) = θ2 (z) = z. Hence the diagram
commutes, and←−X2 is a maximal subgroup.
Consider←−X3. Then
(I2J1
J1
)σ1
=
(F3F1
F1
)σ1
= 〈yF1〉σ1 = 〈z〉 = Z andL2K1
K1= Z.
So, part (ii) is satisfied. Observe that σ2 :F3
Z→ Z is given by (xyZ)σ2 = z. Then(
I2 ∩ J1
J2
)σ2
=
(F3 ∩ F1
Z
)σ2
=
(Z
Z
)σ2
= 1 andL2 ∩K1
K2=Z ∩ 1
1= 1. Hence, part (iii)
is satisfied.
For←−X3, we have the following associated maps:
θ1 :F3F1
F1=
Q
F1−→ F3
F3 ∩ F1=F3
Zis given by (yF1)θ1 = xyZ;
σ2 :F3
F3 ∩ F1=F3
Z−→ Z
Z ∩ 1= Z is given by (xyZ)σ2 = z;
σ1 :F3F1
F1=
Q
F1−→ Z is given by (yF1)σ1 = z;
θ2 : Z −→ Z
Z ∩ 1= Z is given by (z)θ2 = z.
Then σ2 (θ1 (yF1)) = σ2 (xyZ) = z and θ2 (σ1 (yF1)) = θ2 (z) = z. Therefore, the diagram
commutes, and the hypotheses of Theorem 1.7 are satisfied for all of our proposed maximal
21
subgroups. Hence the maximal subgroups of←−W1 are
←−F1,←−X2 and
←−X3.
One can show that←−Wi are isomorphic, where 1 ≤ i ≤ 3, by applying Theorem 1.8. Namely,
one can use an automorphism of Q × Q to show that the subgroups are isomorphic. Let
us now show that←−W1∼=←−W2. For
←−W1, σ1 :
Q
F1→ Z is given by (yF1)σ1 = z. For
←−W2,
σ1 :Q
F2→ Z is given by (xF2)σ1 = z. So, for
←−W1 to be isomorphic to
←−W2, we need
two automorphisms of Q. The first automorphism we need should send y to x and x to
y. The second automorphism we need is just the identity. Recall τ9 ∈ Aut(Q) given by
(y)τ9 = x and (x)τ9 = y. Then (τ9, 1) ∈ Aut(Q × Q) and(←−W1
)(τ9,1)=←−W2. To show that
←−W1∼=←−W3, observe that, for
←−W3, σ1 :
Q
F3→ Z and is given by (yF3)σ1 = z. Then recall
that τ17 ∈ Aut(Q) given by (y)τ17 = y and (x)τ17 = xy. So, (τ17, 1) ∈ Aut(Q × Q) and(←−W1
)(τ17,1)=←−W3. Hence,
←−Wi’s are isomorphic.
To determine the maximal subgroups of←−W2 and
←−W3, we will use the same automorphisms
of Q×Q from above that we used to show←−Wi were isomorphic. Then we will examine the
maximal subgroups of←−W1 under these automorphisms to attain the maximal subgroups of
the isomorphic subgroup we are considering. Recall that the maximal subgroups of←−W1 are
←−F1,←−X2 and
←−X3.
Let us now determine the maximal subgroups of←−W2. Consider
←−F1. Then
(←−F1
)(τ9,1)=
(F1 × 1)(τ9,1) = F2 × 1 =←−F2. Consider
←−X2. Then
(←−X2
)(τ9,1)= 〈(yZ, z)〉(τ9,1) = 〈(xZ, z)〉 =
←−X1. Consider
←−X3. Observe that (xy)τ9 = x−1y, and that xZ = x−1Z in Q. Then by our
observation,(←−X3
)(τ9,1)= 〈(xyZ, z)〉(τ9,1) =
⟨(x−1yZ, z)
⟩= 〈(xyZ, z)〉 =
←−X3. Therefore, the
maximal subgroups of←−W2 are
←−F2,←−X1 and
←−X3. Similarly, using (τ17, 1), one can show the
maximal subgroups of←−W3 are
←−F3,←−X1 and
←−X2.
Analogously, we can produce a similar argument for−→Wj , 1 ≤ j ≤ 3, by noticing that this
group is defined by reversing the roles of I, J and L,K in←−Wi. The maximal subgroups of
−→Wj are
−→Fj and
−→Xk, where 1 ≤ j, k ≤ 3 and k 6= j. A table with the maximal subgroups of
←−Wi and
−→Wj is given below.
Groups Maximal Subgroups Groups Maximal Subgroups←−W1
←−F1,←−X2,←−X3
−→W1
−→F1,−→X2,−→X3
←−W2
←−F2,←−X1,←−X3
−→W2
−→F2,−→X1,−→X3
←−W3
←−F3,←−X1,←−X2
−→W3
−→F3,−→X1,−→X2
22
Consider F11 = (a, b)|a ∈ F1, b ∈ F1, (aZ)σ1 = bZ, where σ1 :F1
Z→ F1
Zand (xZ)σ1 =
xZ. So, (x, x), (z, z) ∈ F11, (x, x)4 = 1 = (z, z)2 and (x, x) 6= (z, z). Hence, F11 =
〈(x, x), (z, z)〉 ∼= C4 × C2. As discussed at the beginning of this section, this group has 3
maximal subgroups, and we propose that they are←−F11,
−→F11 and V . Recall that
←−F11 is the
group which sends x to x, and−→F11 is the group which sends x to x−1. To determine that
these are subgroups, we must verify the hypotheses of Theorem 1.7. Examine the tables
below to see that the containment of the projections and intersections of these proposed
maximal subgroups are satisfied.
←−F11 F11
I2 = F1 I1 = F1
J2 = 1 J1 = Z
L2 = F1 L1 = F1
K2 = 1 K1 = Z
−→F11 F11
I2 = F1 I1 = F1
J2 = 1 J1 = Z
L2 = F1 L1 = F1
K2 = 1 K1 = Z
V F11
I2 = Z I1 = F1
J2 = Z J1 = Z
L2 = Z L1 = F1
K2 = Z K1 = Z
Now, we must verify that parts (ii), (iii) and (iv), of Theorem 1.7, are satisfied for each
proposed maximal subgroup. Namely that
(I2J1
J1
)σ1
=L2K1
K1,
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2,
and that the diagram commutes respectively. To verify parts (ii) and (iii), we must know
σ1 and σ2. In our case, σ1 :F1
Z→ F1
Zis given (xZ)σ1 = xZ. Observe that σ1 will remain
the same for each proposed maximal subgroup, and σ2 changes. To verify part (iv), that
is, σ2 θ1 = θ2 σ1, for each proposed subgroup we will give the associated maps θ1, σ2, σ1
and θ2. Then we will verify that the diagram commutes. Recall that in verifying part (iv),
it is enough to check that the diagram commutes using the generators because our maps
23
are linear. Let us now examine each of the proposed maximal subgroups, determine σ2 and
verify parts (ii), (iii) and (iv) of Theorem 1.7.
Consider←−F11. Then
(I2J1
J1
)σ1
=
(F1Z
Z
)σ1
=
(F1
Z
)σ1
= 〈xZ〉σ1 = 〈xZ〉 andL2K1
K1=
F1Z
Z=
F1
Z= 〈xZ〉. So, part (ii) is satisfied. Observe that σ2 : F1 → F1 is given by
(x)σ2 = x. Then
(I2 ∩ J1
J2
)σ2
=
(F1 ∩ Z
1
)σ2
= (Z)σ2 = Z andL2 ∩K1
K2=F1 ∩ Z
1= Z.
Hence, part (iii) is satisfied.
For←−F11, we have the following associated maps:
θ1 :F1Z
Z=F1
Z−→ F1
F1 ∩ Z=F1
Zis given by (xZ)θ1 = xZ;
σ2 :F1
F1 ∩ Z=F1
Z−→ F1
F1 ∩ Z=F1
Zis given by (xZ)σ2 = xZ;
σ1 :F1Z
Z=F1
Z−→ F1Z
Z=F1
Zis given by (xZ)σ1 = xZ;
θ2 :F1Z
Z=F1
Z−→ F1
F1 ∩ Z=F1
Zis given by (xZ)θ2 = xZ.
Since all of these maps are essentially the identity and defined in the same way, it is easy
to see that the diagram commutes. Hence←−F11 is a maximal subgroup.
For−→F11, parts (ii), (iii) and (iv) are the same as
←−F11, done above. The only difference is
that x gets sent to x−1, but in Q, xZ = x−1Z. Hence, parts (ii), (iii) and (iv) are satisfied.
Consider V . Then
(I2J1
J1
)σ1
=
(ZZ
Z
)σ1
=
(Z
Z
)σ1
= 1σ1 = 1 andL2K1
K1=ZZ
Z=Z
Z= 1.
So, part (ii) is satisfied. Observe that σ2 in this case is the identity map. Hence, part
(iii) is obviously satisfied. Part (iv) is also satisfied because, for V , all of the maps are the
identity.
Therefore the hypotheses of Theorem 1.7 are satisfied for all of our proposed maximal
subgroups. Hence the maximal subgroups of F11 are←−F11,
−→F11 and V .
One can show that Fij are isomorphic, where 1 ≤ i, j ≤ 3, and one can determine the
maximal subgroups of the isomorphic groups by applying Theorem 1.8. Namely, one can
use an automorphism of Q×Q to show that the subgroups are isomorphic. Then one uses
this same automorphism to determine its maximal subgroups.
Let us now show that F11∼= F23. First it suffices to make the following observation: for
F11, σ1 :F1
Z→ F1
Zis given by (xZ)σ1 = xZ, and for F23, σ1 :
F2
Z→ F3
Zis given by
(yZ)σ1 = xyZ. So, to show these two groups are isomorphic we need two automorphisms
of Q. We need one to send x to y and the other to send x to xy. Referring back to the 24
24
automorphisms of Q listed in the introduction, we can use τ9 and τ17 respectively. Then
(τ9, τ17) ∈ Aut(Q×Q), and (F11)(τ9,τ17) = F23. Therefore these two groups are isomorphic.
To determine the maximal subgroups of F23, we can examine the maximal subgroups of F11
under (τ9, τ17). Let us now determine the maximal subgroups of F23. Consider←−F11. Then(←−
F11
)(τ9,τ17)= 〈(x, x)〉(τ9,τ17) = 〈(y, xy)〉 =
←−F23. Consider
−→F11. Then since
−→F11 indicates
we are using the automorphism that sends x to x−1, we know that(−→F11
)(τ9,τ17)=−→F23.
Consider V . Since V has trivial sections and there are no other isomorphic copies of V4,
we know V is definitely a maximal subgroup. Therefore, the maximal subgroups of F23 are←−F23,
−→F23 and V .
Similarly, one can show the remaining Fij ’s are isomorphic and determine their maximal
subgroups. A list of the Fij and their maximal subgroups can be found in the table below.
Groups Maximal Subgroups Groups Maximal Subgroups
F11←−F11,
−→F11, V F23
←−F23,
−→F23, V
F12←−F12,
−→F12, V F31
←−F31,
−→F31, V
F13←−F13,
−→F13, V F32
←−F32,
−→F32, V
F21←−F21,
−→F21, V F33
←−F33,
−→F33, V
F22←−F22,
−→F22, V
Consider ∆n, 1 ≤ n ≤ 24. These groups are determined by τn, where τn ∈ Aut(Q). Recall
that τn are defined and given in Section 0.1. Observe that ∆n ≤ Q × Q, and for any ∆n,
its corresponding projection map (introduced in Goursat’s Theorem), is π : Q×Q→ Q.
Consider ∆n = (a, aτn)|a ∈ Q. Then π|∆n: ∆n → Q is an isomorphism because it is a
restriction of π and because Ker(π|∆n) = 1. Hence ∆n
∼= Q.
To understand the following table, it suffices to recall that each ∆n is defined by its corre-
sponding τn, where 1 ≤ n ≤ 24 and τn are listed in the Introduction. Considering inverse
images and using the appropriate τn, one determines the 3 maximal subgroups of ∆n are
〈(x, xτn)〉, 〈(y, yτn)〉 and 〈(xy, xyτn)〉. A table which lists the maximal subgroups of the ∆n,
1 ≤ n ≤ 24 is given below.
25
Groups Maximal Subgroups Groups Maximal Subgroups
∆1←−F11,
←−F22,
←−F33 ∆13
←−F12,
←−F23,
←−F31
∆2←−F11,
−→F22,
−→F33 ∆14
←−F31,
−→F12,
−→F23
∆3←−F22,
−→F11,
−→F33 ∆15
←−F12,
−→F23,
−→F31
∆4←−F33,
−→F11,
−→F22 ∆16
←−F23,
−→F12,
−→F31
∆5←−F11,
←−F23,
−→F32 ∆17
←−F13,
←−F22,
−→F31
∆6←−F11,
←−F32,
−→F23 ∆18
−→F13,
−→F22,
−→F31
∆7−→F11,
−→F23,
−→F32 ∆19
←−F22,
←−F31,
−→F13
∆8←−F23,
←−F32,
−→F11 ∆20
←−F13,
←−F31,
−→F22
∆9←−F12,
←−F21,
−→F33 ∆21
←−F13,
←−F21,
←−F32
∆10←−F21,
←−F33,
−→F12 ∆22
←−F21,
−→F13,
−→F32
∆11←−F12,
←−F33,
−→F21 ∆23
←−F32,
−→F13,
−→F21
∆12−→F12,
−→F21,
−→F33 ∆24
←−F13,
−→F21,
−→F32
Now, we will give an example to show that conditions (i) − (iv) of Theorem 1.7 are inde-
pendent. That is, we will give an example that shows (i) − (iii) can be satisfied without
(iv) being satisfied. Consider the subgroup of order 8, ∆1, given by the triple (Q,Q, τ1),
where τ1 is the identity automorphism in Aut(Q). Also consider the subgroup of order 4,−→F11, given by the triple (F1, F1, σ2), where (x)σ2 = x−1. We will now show that
−→F11 ∆1.
It is easy to see that the containment of the projections and intersections hold. Hence part
(i) is satisfied. To verify part (ii), observe that σ1 = τ1. Because σ1 is the identity, part
(ii) is obviously satisfied. To verify part (iii), observe that
(I2 ∩ J1
J2
)σ2
=
(F1 ∩ 1
1
)σ2
= 1
andL2 ∩K1
K2=F1 ∩ 1
1= 1. Since both are 1, part (iii) is satisfied. To see that part (iv)
is not satisfied, observe that (x)σ1 = x, (x)θ2 = x, (x)θ1 = x and (x)σ2 = x−1. Therefore,
σ2(θ1(x)) = σ2(x) = x−1 and θ2(σ1(x)) = θ2(x) = x. Since x 6= x−1, the diagram does
not commute and part (iv) is not satisfied. Hence, parts (i) − (iv) of Theorem 1.7 are
independent.
2.4 The Subgroups of Order 16
The groups of order 16 which appear in this lattice will be copies of Q×C2, C2×Q, C4×C4
or C4 o C4. The table that follows lists the groups of order 16 and their corresponding
26
group structure. For these groups 1 ≤ i, j ≤ 3 and 1 ≤ p ≤ 6.
Notation I, J I/J Aut(I/J) L, K No. of subgroups group structure
Q1 Q, Q 1 1 Z, Z 1 Q× C2
Q2 Z, Z 1 1 Q, Q 1 C2 ×Q
Fij Fi, Fi 1 1 Fj , Fj 9 C4 × C4
←−−Mij Q, Fi C2 1 Fj , Z 9 C4 o C4
−−→Mij Fi, Z C2 1 Q, Fj 9 C4 o C4
Bp Q, Z V4 S3 Q, Z 6 Q× C2
Now, let us begin to determine the maximal subgroups of the groups of order 16. Using the
table above, we will begin by determining the maximal subgroups of the groups that have
trivial sections, and then we will determine the maximal subgroups of the groups that have
nontrivial sections.
Consider Q1, which is clearly isomorphic toQ×C2. Observe that this group has 3 generators,
and because of this, Consequence 2 implies this group has 7 maximal subgroups. We claim
the 7 maximal subgroups are←−Q1,←−Yi and
←−Wi, where 1 ≤ i ≤ 3. To determine that these
are subgroups, we must verify the hypotheses of Theorem 1.7. Examine the tables below
to see that the containment of the projections and intersections of these proposed maximal
subgroups are satisfied.
←−Q1 Q1
I2 = Q I1 = Q
J2 = Q J1 = Q
L2 = 1 L1 = Z
K2 = 1 K1 = Z
←−Yi Q1
I2 = Fi I1 = Q
J2 = Fi J1 = Q
L2 = Z L1 = Z
K2 = Z K1 = Z
27
←−Wi Q1
I2 = Q I1 = Q
J2 = Fi J1 = Q
L2 = Z L1 = Z
K2 = 1 K1 = Z
Now, we must show that parts (ii) and (iii) of Theorem 1.7, namely
(I2J1
J1
)σ1
=L2K1
K1and(
I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2respectively, are satisfied for each proposed maximal subgroup. In
order to do this, we must know σ1 and σ2. In our case, σ1 :Q
Q→ Z
Z, which is the
identity, and verifying part (ii) of Theorem 1.7 is trivial. Observe that σ2 changes for
each subgroup that we are considering. Hence, it suffices to examine each of the proposed
maximal subgroups, determine σ2 and verify part (iii) of Theorem 1.7.
For←−Q1 and
←−Yi , 1 ≤ i ≤ 3, the associated σ2 is the identity map. Hence, part (iii) is
obviously satisfied.
For←−Wi, 1 ≤ i ≤ 3, σ2 will change for each i. So, we will consider each
←−Wi individually
and verify part (iii) of Theorem 1.7 in each case. For←−W1, σ2 :
Q
F1→ Z
1is given by
(yF1)σ2 = z. Then
(I2 ∩ J1
J2
)σ2
=
(Q ∩QF1
)σ2
=
(Q
F1
)σ2
= 〈yF1〉σ2 = 〈z〉 = Z and
L2 ∩K1
K2=Z ∩ Z
1= Z. Hence, part (iii) is satisfied.
For←−W2, σ2 :
Q
F2→ Z
1is given by (xF2)σ2 = z. Then
(I2 ∩ J1
J2
)σ2
=
(Q ∩QF2
)σ2
=(Q
F2
)σ2
= 〈xF2〉σ2 = 〈z〉 = Z andL2 ∩K1
K2=Z ∩ Z
1= Z. Hence, part (iii) is satisfied.
For←−W3, σ2 :
Q
F3→ Z
1is given by (yF3)σ2 = z. Then
(I2 ∩ J1
J2
)σ2
=
(Q ∩QF3
)σ2
=(Q
F3
)σ2
= 〈yF3〉σ2 = 〈z〉 = Z andL2 ∩K1
K2=Z ∩ Z
1= Z. Hence, part (iii) is satisfied.
Showing that the diagram commutes is trivial because the sections which define Q1 are
trivial. Therefore, the hypotheses of Theorem 1.7 are satisfied, and the maximal subgroups
of Q1 are←−Q1,←−Yi and
←−Wi, where 1 ≤ i ≤ 3.
Similarly, one can show the maximal subgroups of Q2, an isomorphic copy of C2 × Q, are−→Q1,−→Yj and
−→Wj , where 1 ≤ j ≤ 3. A table with the maximal subgroups of Q1 and Q2 is
given below.
28
Groups Maximal Subgroups
Q1←−Q1,←−W1,←−W2,←−W3,←−Y1,←−Y2,←−Y3
Q2−→Q2,−→W1,−→W2,−→W3,−→Y1,−→Y2,−→Y3
Consider F12, which is clearly isomorphic to C4 ×C4. Because this group has 2 generators,
Consequence 2 implies it has 3 maximal subgroups. We claim these maximal subgroups are←−Y1,−→Y2 and F12. To determine that these are subgroups, we must verify the hypotheses of
Theorem 1.7. Examine the tables below to see that the containment of the projections and
intersections of these proposed maximal subgroups are satisfied.
←−Y1 F12
I2 = F1 I1 = F1
J2 = F1 J1 = F1
L2 = Z L1 = F2
K2 = Z K1 = F2
−→Y2 F12
I2 = Z I1 = F1
J2 = Z J1 = F1
L2 = F2 L1 = F2
K2 = F2 K1 = F2
F12 F12
I2 = F1 I1 = F1
J2 = Z J1 = F1
L2 = F2 L1 = F2
K2 = Z K1 = F2
Now, we must show that parts (ii) and (iii) of Theorem 1.7, namely
(I2J1
J1
)σ1
=L2K1
K1and(
I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2respectively, are satisfied for each proposed maximal subgroup. In
order to do this, we must know σ1 and σ2. In our case, σ1 :F1
F1→ F2
F2, which is the
identity, and verifying part (ii) of Theorem 1.7 is trivial. Observe that σ2 changes for
each subgroup that we are considering. Hence, it suffices to examine each of the proposed
maximal subgroups, determine σ2 and verify part (iii) of Theorem 1.7.
29
For←−Y1 and
−→Y2, the associated σ2 is the identity map. Hence, part (iii) is obviously satisfied.
For F12, σ2 :F1
Z→ F2
Zis given by (xZ)σ2 = yZ. Then
(I2 ∩ J1
J2
)σ2
=
(F1 ∩ F1
Z
)σ2
=(F1
Z
)σ2
= 〈xZ〉σ2 = 〈yZ〉 = Z andL2 ∩K1
K2=F2 ∩ F2
Z=F2
Z= 〈yZ〉. Hence, part (iii) is
satisfied.
Showing that the diagram commutes is trivial because the sections which define F12 are
trivial. Therefore, the hypotheses of Theorem 1.7 are satisfied, and the maximal subgroups
of F12 are←−Y1,−→Y2 and F12.
One can show that Fij are isomorphic, where 1 ≤ i, j ≤ 3, and one can determine the
maximal subgroups of the isomorphic groups by applying Theorem 1.8. Namely, one can
use an automorphism of Q×Q to show that the subgroups are isomorphic. Then one uses
this same automorphism to determine its maximal subgroups.
Let us now show that F12∼= F23. Notice that for these two groups, the associated σ1
is the identity map. In order to show these groups are isomorphic, it suffices to recall
τ13 ∈ Aut(Q), which is given by xτ13 = y and yτ13 = xy. Then (τ13, τ13) ∈ Aut(Q×Q) and(F12
)(τ13,τ13)= F23. Therefore these two groups are isomorphic.
To determine the maximal subgroups of F23, we can examine the maximal subgroups of
F12 under (τ13, τ13). Let us now determine the maximal subgroups of F23. Consider←−Y1. Then
(←−Y1
)(τ13,τ13)= 〈(x, x), (z, z)〉(τ13,τ13) = 〈(y, y), (z, z)〉 =
←−Y2. Consider
−→Y2. Then(−→
Y2
)(τ13,τ13)= 〈(z, z), (y, y)〉(τ13,τ13) = 〈(z, z), (xy, xy)〉 =
−→Y3. Consider (F12)(τ13,τ13) =
〈(xZ, yZ)〉(τ13,τ13) = 〈(yZ, xyZ)〉 = F23. Therefore, the maximal subgroups of F23 are←−Y2,
−→Y3 and F23.
Similarly, one can show the remaining Fij ’s are isomorphic and determine that their maximal
subgroups are←−Yi ,−→Yj and Fij . A list of the Fij and their maximal subgroups can be found
in the table below.
30
Groups Maximal Subgroups Groups Maximal Subgroups
F11←−Y1, F11,
−→Y1 F23
←−Y2, F23,
−→Y3
F12←−Y1, F12,
−→Y2 F31
←−Y3, F31,
−→Y1
F13←−Y1, F13,
−→Y3 F32
←−Y3, F32,
−→Y2
F21←−Y2, F21,
−→Y1 F33
←−Y3, F33,
−→Y3
F22←−Y2, F22,
−→Y2
Before considering the remaining groups of order 16, it suffices to make an observation.
Observe that C4 o C4 =⟨x, y|x4 = y4 = 1, xy = x−1
⟩. Because C4 o C4 is noncyclic and
has a minimal generating set of two elements, Consequence 2 implies it contains 3 maximal
subgroups. Although the group C4 o C4 is not crucial to this dissertation, the number of
generators it has is crucial.
Consider←−−M22 = (a, b)|a ∈ Q, b ∈ F2, (aF2)σ1 = bZ, where σ1 :
Q
F2→ F2
Zis given by
(xF2)σ1 = yZ. So, (y, 1), (x, y) ∈←−−M22, (y, 1)4 = (x, y)4 = 1, 〈(y, 1)〉 C 〈(y, 1), (x, y)〉 and
(y, 1)(x,y) = (yx, 1) =(y−1, 1
). So,
←−−M22 = 〈(y, 1), (x, y)〉, and one can believe that the
group structure of←−−M22 is C4 o C4. Therefore,
←−−M22 is not cyclic and can be generated
by 2 elements. Hence, it has 3 maximal subgroups, and we propose that these maximal
subgroups are←−Y2, F12 and F32. To determine that these are subgroups, we must verify the
hypotheses of Theorem 1.7. Examine the tables below to see that the containment of the
projections and intersections of these proposed maximal subgroups are satisfied.
←−Y2
←−−M22
I2 = F2 I1 = Q
J2 = F2 J1 = F2
L2 = Z L1 = F2
K2 = Z K1 = Z
F12←−−M22
I2 = F1 I1 = Q
J2 = Z J1 = F2
L2 = F2 L1 = F2
K2 = Z K1 = Z
31
F32←−−M22
I2 = F3 I1 = Q
J2 = Z J1 = F2
L2 = F2 L1 = F2
K2 = Z K1 = Z
Now, we must verify that parts (ii), (iii) and (iv), of Theorem 1.7, are satisfied for each
proposed maximal subgroup. Namely that
(I2J1
J1
)σ1
=L2K1
K1,
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2,
and that the diagram commutes respectively. To verify parts (ii) and (iii), we must know
σ1 and σ2. In our case, σ1 :Q
F2→ F2
Zis given by (xF2)σ1 = yZ. Observe that σ1 will
remain the same for each proposed maximal subgroup, and σ2 changes. To verify part (iv),
that is, σ2 θ1 = θ2 σ1, for each proposed subgroup we will give the associated maps θ1, σ2,
σ1 and θ2. Then we will verify that the diagram commutes. Recall that in verifying part
(iv), it is enough to check that the diagram commutes using the generators because our
maps are linear. Let us now examine each of the proposed maximal subgroups, determine
σ2 and verify parts (ii), (iii) and (iv) of Theorem 1.7.
Consider←−Y2. Then
(I2J1
J1
)σ1
=
(F2F2
F2
)σ1
=
(F2
F2
)σ1
= 1 andL2K1
K1=ZZ
Z=Z
Z= 1. So,
part (ii) is satisfied. The associated σ2 for←−Y2 is the identity. Hence part (iii) is obviously
satisfied.
For←−Y2, the associated maps are θ1 :
F2F2
F2−→ F2
F2 ∩ F2, σ2 :
F2
F2 ∩ F2−→ Z
Z ∩ Z, σ1 :
F2F2
F2−→ ZZ
Z, and θ2 :
ZZ
Z−→ Z
Z ∩ Z. Notice that all of these maps are the identity.
Hence, the diagram commutes, and←−Y2 is a maximal subgroup.
Consider F12. Then
(I2J1
J1
)σ1
=
(F1F2
F2
)σ1
=
(Q
F2
)σ1
= 〈xF2〉σ1 = 〈yZ〉 andL2K1
K1=
F2Z
Z=
F2
Z= 〈yZ〉. So, part (ii) is satisfied. Observe that σ2 :
F1
Z→ F2
Zis given by
(xZ)σ2 = yZ. Then
(I2 ∩ J1
J2
)σ2
=
(F1 ∩ F2
Z
)σ2
=
(Z
Z
)σ2
= 1 andL2 ∩K1
K2=F2 ∩ ZZ
=
Z
Z= 1. Hence, part (iii) is satisfied.
For F12, the following associated maps are:
θ1 :F1F2
F2=
Q
F2−→ F1
F1 ∩ F2=F1
Zis given by (xF2)θ1 = xZ;
σ2 :F1
F1 ∩ F2=F1
Z−→ F2
F2 ∩ Z=F2
Zis given by (xZ)σ2 = yZ;
σ1 :F1F2
F2=
Q
F2−→ F2Z
Z=F2
Zis given by (xF2)σ1 = yZ;
32
θ2 :F2Z
Z=F2
Z−→ F2
F2 ∩ Z=F2
Zis given by (yZ)θ2 = yZ.
Then σ2 (θ1 (xF2)) = σ2 (xZ) = yZ and θ2 (σ1 (xF2)) = θ2 (yZ) = yZ. Hence, the diagram
commutes, and F12 is a maximal subgroup.
Consider F32. Then
(I2J1
J1
)σ1
=
(F3F2
F2
)σ1
=
(Q
F2
)σ1
= 〈xF2〉σ1 = 〈yZ〉 andL2K1
K1=
F2Z
Z=
F2
Z= 〈yZ〉. So, part (ii) is satisfied. Observe that σ2 :
F3
Z→ F2
Zis given
by (xyZ)σ2 = yZ. Then
(I2 ∩ J1
J2
)σ2
=
(F3 ∩ F2
Z
)σ2
=
(Z
Z
)σ2
= 1 andL2 ∩K1
K2=
F2 ∩ ZZ
=Z
Z= 1. Hence, part (iii) is satisfied.
For F32, the following associated maps are:
θ1 :F3F2
F2=
Q
F2−→ F3
F3 ∩ F2=F3
Zis given by (xF2)θ1 = xyZ;
σ2 :F3
F3 ∩ F2=F3
Z−→ F2
F2 ∩ Z=F2
Zis given by (xyZ)σ2 = yZ;
σ1 :F3F2
F2=
Q
F2−→ F2Z
Z=F2
Zis given by (xF2)σ1 = yZ;
θ2 :F2Z
Z=F2
Z−→ F2
F2 ∩ Z=F2
Zgiven by (yZ)θ2 = yZ.
Then σ2 (θ1 (xF2)) = σ2 (xyZ) = yz and θ2 (σ1 (xF2)) = θ2 (yZ) = yZ. Hence, the diagram
commutes, and F32 is a maximal subgroup.
Therefore, the hypotheses of Theorem 1.7 are satisfied for all of our proposed maximal
subgroups. Hence the maximal subgroups of←−−M22 are
←−Y2, F12 and F32.
One can show that←−−Mij are isomorphic, where 1 ≤ i, j ≤ 3, by applying Theorem 1.8.
Namely, one can use an automorphism of Q×Q to show that the subgroups are isomorphic.
Let us now show that←−−M22
∼=←−−M12. First it suffices to make the following observation: for
←−−M22, σ1 :
Q
F2→ F2
Zis given by (xF2)σ1 = yZ, and for
←−−M12, σ1 :
Q
F1→ F2
Zis given by
(yF1)σ1 = yZ. So, to show these two groups are isomorphic, we need two automorphisms
of Q. One should send x to y and y to x, and the other should send y to y. Referring back
to the 24 automorphisms of Q given in the introduction, we can use τ9 and τ17 respectively.
Then (τ9, τ17) ∈ Aut(Q × Q) and(←−−M22
)(τ9,τ17)=←−−M12. Therefore, these two groups are
isomorphic.
To determine the maximal subgroups of←−−M12 we examine the maximal subgroups of
←−−M22
under the automorphism (τ9, τ17). Let us now determine the maximal subgroups of←−−M12.
Consider←−Y2. Then
(←−Y2
)(τ9,τ17)= 〈(y, y), (z, z)〉(τ9,τ17) = 〈(x, x), (z, z)〉 =
←−Y1. Consider
F12. Then (F12)(τ9,τ17) = 〈(xZ, yZ)〉(τ9,τ17) = 〈(yZ, yZ)〉 = F22. Consider F32 and recall
33
that xZ = x−1Z in Q. Then (F32)(τ9,τ17) =⟨(x−1yZ, yZ)
⟩(τ9,τ17)= 〈(xyZ, yZ)〉 = F32.
Therefore, the maximal subgroups of←−−M12 are
←−Y2, F22 and F32.
Similarly, one can show the remaining←−−Mij ’s are isomorphic and determine that their max-
imal subgroups are←−Yi and Fij , where j is fixed, i 6= j and 1 ≤ i, j ≤ 3.
Analogously, we can produce a similar argument for−−→Mij by noticing that this group is
defined by reversing the roles of I, J and L,K in←−−Mij . Namely, in reversing the roles, we are
using the fact that ρ : (a, b) 7→ (b, a) is an automorphism of Q×Q. The maximal subgroups
of−−→Mij are
−→Yi and Fij , where i is fixed, i 6= j and 1 ≤ i, j ≤ 3. A table with the maximal
subgroups of←−−Mij and
−−→Mij is given below.
Groups Maximal Subgroups Groups Maximal Subgroups←−−M11
←−Y1, F21, F31
−−→M11
−→Y1, F21, F31
←−−M12
←−Y1, F22, F32
−−→M12
−→Y1, F22, F32
←−−M13
←−Y1, F23, F33
−−→M13
−→Y1, F22, F23
←−−M21
←−Y2, F11, F31
−−→M21
−→Y2, F11, F31
←−−M22
←−Y2, F12, F32
−−→M22
−→Y2, F12, F32
←−−M23
←−Y2, F13, F33
−−→M23
−→Y2, F13, F33
←−−M31
←−Y3, F11, F21
−−→M31
−→Y3, F11, F21
←−−M32
←−Y3, F12, F22
−−→M32
−→Y3, F12, F22
←−−M33
←−Y3, F13, F23
−−→M33
−→Y3, F13, F23
Consider Bp, 1 ≤ p ≤ 6. These groups are defined using the automorphisms, βp, of V4
that were given in Section 0.2. Since the Frattini subgroup of a 2-group is generated by
the squares, we know that Φ(Bp) = 〈(z, z)〉, where |〈(z, z)〉| = 2. Notice that for Bp,
σ1 :Q
Z→ Q
Z, where σ1 is defined using the definition of the corresponding βp. Hence,∣∣∣∣ Bp
Φ(Bp)
∣∣∣∣ =|Bp||Φ(Bp)|
=16
2= 8. Therefore, Consequence 2 implies this group has 3 generators
and, as a consequence, 7 maximal subgroups.
Consider B1, which is given by the triple
(Q
Z,Q
Z, β1
), where xβ1 = x and yβ1 = y. We
propose that its maximal subgroups are ∆k, where 1 ≤ k ≤ 4 and Fii, where 1 ≤ i ≤ 3.
To determine that these are subgroups, we must verify the hypotheses of Theorem 1.7. We
will show that ∆k, 1 ≤ k ≤ 4, satisfies the hypotheses first. Then we will show that Fii,
1 ≤ i ≤ 3, satisfies the hypotheses. Examine the table below to see that the containment
34
of the projections and intersections of ∆k is satisfied.
∆k B1
I2 = Q I1 = Q
J2 = 1 J1 = Z
L2 = Q L1 = Q
K2 = 1 K1 = Z
Before verifying the remaining hypotheses of Theorem 1.7, it suffices to make a few observa-
tions. First, because the corresponding isomorphism for B1 is given by xβ1 = x and yβ1 = y,
in order for ∆k to be a maximal subgroup of B1, the τk which define ∆k must fix x, y and
z or must send these elements to their inverses. Second, it is alright to include the elements
being sent to their inverses because, in Q, xZ = x−1Z, yZ = y−1Z and xyZ = (xy)−1Z.
Now, we must show that parts (ii) and (iii) of Theorem 1.7, namely
(I2J1
J1
)σ1
=L2K1
K1
and
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2respectively, are satisfied for each proposed ∆k. In order to
do this, we must know σ1 and σ2. In our case, σ1 :Q
Z→ Q
Zis given by (xZ)σ1 = xZ,
(yZ)σ1 = yZ and (xyZ)σ1 = xyZ. Observe that σ2 changes for each k. Hence, it suffices to
examine each of the proposed maximal subgroups, determine σ2 and verify parts (ii) and
(iii) of Theorem 1.7.
Consider ∆1, and let us verify parts (ii) and (iii). For part (ii),
(I2J1
J1
)σ1
=
(QZ
Z
)σ1
=(Q
Z
)σ1
, which is given by 〈xZ〉σ1 = 〈xZ〉, 〈yZ〉σ1 = 〈yZ〉 and 〈xyZ〉σ1 = 〈xyZ〉. Also,
L2K1
K1=
QZ
Z=
Q
Z= 〈xZ〉, 〈yZ〉 or 〈xZ〉, respectively. So, part (ii) is satisfied. The
associated σ2 for ∆1 is the identity. Hence part (iii) is obviously satisfied.
We can show that ∆2, ∆3 and ∆4 also satisfy parts (ii) and (iii) by using a similar process
as done above for ∆1 and using the second observation given above.
To show the diagram commutes for ∆1, observe that its associated maps are as follows:
θ1 :QZ
Z=Q
Z−→ Q
Q ∩ Z=Q
Z
σ2 :Q
Q ∩ Z=Q
Z−→ Q
Q ∩ Z=Q
Z
σ1 :QZ
Z=Q
Z−→ QZ
Z=Q
Z
θ2 :QZ
Z=Q
Z−→ Q
Q ∩ Z=Q
Z.
35
Then since all of these maps are essentially the identity map and since ∆1 fixes x, y and z,
this diagram commutes.
Similarly, using the second observation given above we can conclude that the diagram
commutes for ∆2, ∆3 and ∆4. Therefore, the hypotheses of Theorem 1.7 are satisfied, and
∆1, ∆2, ∆3 and ∆4 are maximal subgroups of B1.
Since B1 has 7 maximal subgroups, we must show it has 3 more. Examine the table below
to see that the containment of the projections and intersections for Fii are satisfied, where
1 ≤ i ≤ 3.
Fii B1
I2 = Fi I1 = Q
J2 = Z J1 = Z
L2 = Fi L1 = Q
K2 = Z K1 = Z
Consider F11, and let us now verify parts (ii) and (iii) of Theorem 1.7. For part (ii),(I2J1
J1
)σ1
=
(F1Z
Z
)σ1
=
(F1
Z
)σ1
= 〈xZ〉 andL2K1
K1=F1Z
Z=F1
Z= 〈xZ〉. So, part (ii)
is satisfied. Observe that σ2 :F1
Z→ F1
Zis given by (xZ)σ2 = xZ. Then
(I2 ∩ J1
J2
)σ2
=(F1 ∩ ZZ
)σ2
=
(Z
Z
)σ2
= 1 andL2 ∩K1
K2=F1 ∩ ZZ
=Z
Z= 1. Hence, part (iii) is satisfied.
To show the diagram commutes for F11, observe that its associated maps are as follows:
θ1 :F1Z
Z=F1
Z−→ F1
F1 ∩ Z=F1
Z
σ2 :F1
F1 ∩ Z=F1
Z−→ F1
F1 ∩ Z=F1
Z
σ1 :F1Z
Z=F1
Z−→ F1Z
Z=F1
Z
θ2 :F1Z
Z=F1
Z−→ F1
F1 ∩ Z=F1
Z.
Then since all of these maps are essentially the identity and are defined by sending xZ to
xZ, the diagram commutes. Therefore, F11 is a maximal subgroup of B1.
Similarly, we can show that F22 and F33 are maximal subgroups. Therefore the maximal
subgroups of B1 are ∆1, ∆2, ∆3, ∆4, F11, F22 and F33.
One can show that Bp are isomorphic, where 1 ≤ p ≤ 6, by applying Theorem 1.8. Namely,
one can use an automorphism of Q×Q to show that the subgroups are isomorphic. Let us
now show that B1∼= B4. For B1, σ1 :
Q
Z→ Q
Zis given by (xZ)σ1 = xZ and (yZ)σ1 = yZ.
36
For B4, σ1 :Q
Z→ Q
Zis given by (xZ)σ1 = yZ and (yZ)σ1 = xyZ. So, to show these two
groups are isomorphic, we need two automorphisms of Q. One should send x to x and y to
y. The other should send x to y and y to xy. Referring back to the 24 automorphisms of Q
given in the introduction, we can use τ1 and τ13 respectively. Then (τ1, τ13) ∈ Aut(Q×Q)
and (B1)(τ1,τ13) = B4. Therefore, these two groups are isomorphic.
To determine the maximal subgroups of B4 we examine the maximal subgroups of B1
under the automorphism (τ1, τ13). Let us now determine the maximal subgroups of B4.
Consider ∆1. Then (∆1)(τ1,τ13) = 〈(x, x), (y, y)〉(τ1,τ13) = 〈(x, y), (y, xy)〉 = ∆13. Con-
sider ∆2. Then (∆2)(τ1,τ13) =⟨(x, x), (y, y−1)
⟩(τ1,τ13)=⟨(x, y), (y, x−1y)
⟩= ∆15. Con-
sider ∆3. Then (∆3)(τ1,τ13) =⟨(x, x−1), (y, y)
⟩(τ1,τ13)=⟨(x, y−1), (y, xy)
⟩= ∆16. Consider
∆4. Then (∆4)(τ1,τ13) =⟨(x, x−1), (y, y−1)
⟩(τ1,τ13)=⟨(x, y−1), (y, xy−1)
⟩= ∆14. Con-
sider F11. Then (F11)(τ1,τ13) = 〈(x, x), (z, z)〉(τ1,τ13) = 〈(x, y), (z, z)〉 = F12. Consider F22.
Then (F22)(τ1,τ13) = 〈(y, y), (z, z)〉(τ1,τ13) = 〈(y, xy), (z, z)〉 = F23. Consider F33. Then
(F33)(τ1,τ13) = 〈(xy, xy), (z, z)〉(τ1,τ13) = 〈(xy, x), (z, z)〉 = F31. Therefore, the maximal sub-
groups of B4 are ∆13, ∆14, ∆15, ∆16, F12, F23 and F31.
Similarly, one can show the remaining Bp’s are isomorphic and determine their maximal
subgroups, where 1 ≤ p ≤ 6. A table with the maximal subgroups of Bp is given below.
Groups Maximal Subgroups
B1 ∆1, ∆2, ∆3, ∆4, F11, F22, F33
B2 ∆5, ∆6, ∆7, ∆8, F11, F23, F32
B3 ∆9, ∆10, ∆11, ∆12, F12, F21, F33
B4 ∆13, ∆14, ∆15, ∆16, F12, F23, F31
B5 ∆17, ∆18, ∆19, ∆20, F13, F22, F31
B6 ∆21, ∆22, ∆23, ∆24, F13, F21, F32
2.5 The Subgroups of Order 32
The groups of order 32 which appear in this lattice will be copies of Q × C4, C4 × Q or
Q Q. Q Q is called a subdirect product of Q and Q with amalgamated factor group H,
where H is isomorphic to a quotient of a set by a direct product of the kernels of Q. In
our case, the kernels are isomorphic to C4 (ala Goursat’s Theorem). For more information
37
on this notation, see [2] P. 73. The table that follows lists the groups of order 32 and their
corresponding group structure. For these groups, 1 ≤ i, j ≤ 3.
Notation I, J I/J Aut(I/J) L, K No. of subgroups group structure←−Aj Q, Q 1 1 Fj , Fj 3 Q× C4
−→Ai Fi, Fi 1 1 Q, Q 3 C4 ×Q
Sij Q, Fi C2 1 Q, Fj 9 Q Q
Consider←−A1, which is clearly isomorphic toQ×C4. Observe that this group has 3 generators,
and because of this, Theorem 0.1 implies it has 7 maximal subgroups. We propose these
7 maximal subgroups are Q1, Fi1,←−−Mi1, where 1 ≤ i ≤ 3. To determine that these are
subgroups, we must verify the hypotheses of Theorem 1.7. Examine the tables below to
see that the containments of the projections and intersections of the proposed maximal
subgroups are satisfied.
Q1←−A1
I2 = Q I1 = Q
J2 = Q J1 = Q
L2 = Z L1 = F1
K2 = Z K1 = F1
Fi1←−A1
I2 = Fi I1 = Q
J2 = Fi J1 = Q
L2 = F1 L1 = F1
K2 = F1 K1 = F1
←−−Mi1
←−A1
I2 = Q I1 = Q
J2 = Fi J1 = Q
L2 = F1 L1 = F1
K2 = Z K1 = F1
Now, we must show that parts (ii) and (iii) of Theorem 1.7, namely
(I2J1
J1
)σ1
=L2K1
K1and(
I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2respectively, are satisfied for each proposed maximal subgroup. In
38
order to do this, we must know σ1 and σ2. In our case, σ1 :Q
Q→ F1
F1, which is the identity,
and so verifying part (ii) is trivial. Observe that σ2 changes for each subgroup we consider.
Hence, it suffices to examine each of the proposed maximal subgroups, determine σ2 and
verify part (iii) of Theorem 1.7.
For Q1 and Fi1, 1 ≤ i ≤ 3, the associated σ2 is the identity. Hence, part (iii) is obviously
satisfied.
For←−−Mi1, σ2 will change for each i. However,
L2 ∩K1
K2=F1 ∩ F1
Z=F1
Z= 〈xZ〉 will not
change. So, to complete verifying part (iii), we will consider each←−−Mi1 individually and
examine
(I2 ∩ J1
J2
)σ2
. For←−−M11, σ2 :
Q
F1→ F1
Zand is given by (yF1)σ2 = xZ. Then(
I2 ∩ J1
J2
)σ2
=
(Q ∩QF1
)σ2
=
(Q
F1
)σ2
= 〈yF1〉σ2 = 〈xZ〉. For←−−M21, σ2 :
Q
F2→ F1
Zand is
given by (xF2)σ2 = xZ. Then
(I2 ∩ J1
J2
)σ2
=
(Q ∩QF2
)σ2
=
(Q
F2
)σ2
= 〈xF2〉σ2 = 〈xZ〉.
For←−−M31, σ2 :
Q
F3→ F1
Zand is given by (yF3)σ2 = xZ. Then
(I2 ∩ J1
J2
)σ2
=
(Q ∩QF3
)σ2
=(Q
F3
)σ2
= 〈yF3〉σ2 = 〈xZ〉. Hence, in all 3 cases, part (iii) is satisfied.
Showing that the diagram commutes is trivial because the sections which define←−A1 are
trivial. Therefore, the hypotheses of Theorem 1.7 are satisfied, and the maximal subgroups
of←−A1 are Q1, F11, F21, F31,
←−−M11,
←−−M21 and
←−−M31.
One can show that←−Aj are isomorphic, where 1 ≤ j ≤ 3, by applying Theorem 1.8. Namely,
one can use an automorphism of Q × Q to show that the subgroups are isomorphic. Let
us now show that←−A1∼=←−A2. The associated σ1 for
←−A1 and
←−A2 is the identity. So, to show
←−A1∼=←−A2, we need to recall τ9 ∈ Aut(Q) given by (x)τ9 = y. Then (1, τ9) ∈ Aut(Q × Q)
and(←−A1
)(1,τ9)=←−A2. Similarly, to show that
←−A1∼=←−A3, recall that τ17 ∈ Aut(Q) given by
(x)τ9 = xy. Then (1, τ17) ∈ Aut(Q×Q) and(←−A1
)(1,τ17)=←−A3. Hence,
←−Aj are isomorphic.
To determine the maximal subgroups of←−A2 and
←−A3, we will use the same automorphisms
of Q×Q from above that we used to show←−Aj were isomorphic. Then we will examine the
maximal subgroups of←−A1 under these automorphisms to attain the maximal subgroups of
the isomorphic subgroup we are considering. Recall that the maximal subgroups of←−A1 are
Q1, F11, F21, F31,←−−M11,
←−−M21 and
←−−M31.
39
Let us now determine the maximal subgroups of←−A2. Consider Q1. Then
(Q1
)(1,τ9)=
〈(x, 1), (y, 1), (1, z)〉(1,τ9) = 〈(x, 1), (y, 1), (1, z)〉 = Q1. Consider F11. Then(F11
)(1,τ9)=
〈(x, 1), (1, x)〉(1,τ9) = 〈(x, 1), (1, y)〉 = F12. Consider F21. Then(F21
)(1,τ9)=
〈(y, 1), (1, x)〉(1,τ9) = 〈(y, 1), (1, y)〉 = F22. Consider F31. Then(F31
)(1,τ9)=
〈(xy, 1), (1, x)〉(1,τ9) = 〈(xy, 1), (1, y)〉 = F32. Consider←−−M11. It is easy to see that (y, x) and
(x, z) generate←−−M11. Then
(←−−M11
)(1,τ9)= 〈(y, x), (x, z)〉(1,τ9) = 〈(y, y), (x, z)〉 =
←−−M12 since
(y, y) and (x, z) ∈←−−M12 and in a finite group under an automorphism, generating sets go to
generating sets. Consider←−−M21. Then
(←−−M21
)(1,τ9)= 〈(x, x), (y, z)〉(1,τ9) = 〈(x, y), (y, z)〉 =
←−−M22. Consider
←−−M31. Then
(←−−M31
)(1,τ9)= 〈(y, x), (xy, z)〉(1,τ9) = 〈(y, y), (xy, z)〉 =
←−−M32.
Therefore, the maximal subgroups of←−A2 are Q1, F12, F22, F32,
←−−M12,
←−−M22 and
←−−M32. Similarly,
using (1, τ17), one can show the maximal subgroups of←−A3 are Q1, F13, F23, F33,
←−−M13,
←−−M23
and←−−M33.
Analogously, we can produce a similar argument for−→Ai, 1 ≤ i ≤ 3, by noticing that this
group is defined by reversing the roles of I, J and L,K in←−Aj . Namely, in reversing the roles,
we are using the fact that γ : (a, b) 7→ (b, a) is an automorphism of Q × Q. The maximal
subgroups of−→Ai are Fij ,
−−→Mij and Q2, where i is fixed and 1 ≤ j ≤ 3. A table with the
maximal subgroups of←−Aj and
−→Ai, where 1 ≤ i, j ≤ 3, is given below.
Groups Maximal Subgroups←−A1 F11, F21, F31,
←−−M11,
←−−M21,
←−−M31, Q1
←−A2 F12, F22, F32,
←−−M12,
←−−M22,
←−−M32, Q1
←−A3 F13, F23, F33,
←−−M13,
←−−M23,
←−−M33, Q1
−→A1 F11, F12, F13,
−−→M11,
−−→M12,
−−→M13, Q2
−→A2 F21, F22, F23,
−−→M21,
−−→M22,
←−−M23, Q2
−→A3 F31, F32, F33,
−−→M31,
−−→M32,
−−→M33, Q2
Consider S11. Before determining the maximal subgroups of S11, we will show that S11 has
3 generators.
Claim 1. Φ(S11) = V , where V is the subgroup of order 4 that is isomorphic to V4. As
a conseqence, S11 has three generators. Then Consequence 2 of Burnside’s Basis Theorem
will imply this group has 7 maximal subgroups.
40
Proof: Since the Frattini subgroup of a 2-group is generated by the squares, we know that
Φ(U) ≤ V , for all U ≤ Q × Q since a, b ∈ Q and (a, b)2 = (a2, b2) ≤ V . Observe that
(x, 1) ∈ S11. So, (x, 1)2 = (z, 1) ∈ Φ(S11). Observe that (1, x) ∈ S11. So, (1, x)2 = (1, z) ∈
Φ(S11). Therefore, Φ(S11) = 〈(z, 1), (1, z)〉 = V .
As a consequence
∣∣∣∣ S11
Φ(S11)
∣∣∣∣ =
∣∣∣∣S11
V
∣∣∣∣ =32
4= 8. Therefore, Burnside’s Basis Theorem
implies this group has 3 generators and 7 maximal subgroups. This concludes the proof of
this claim.
Now, we can determine the maximal subgroups contained in S11. We propose that its
maximal subgroups are F11,←−−M12,
←−−M13,
−−→M21,
−−→M31, B1 and B2. To determine that these are
the maximal subgroups, we must verify the hypotheses of Theorem 1.7 for each proposed
maximal subgroup.
Consider F11. Examine the table below to see that the containment of the projections and
intersections is satisfied.
F11 S11
I2 = F1 I1 = Q
J2 = F1 J1 = F1
L2 = F1 L1 = Q
K2 = F1 K1 = F1
Now, we must verify that parts (ii), (iii) and (iv), of Theorem 1.7, are satisfied. Namely that(I2J1
J1
)σ1
=L2K1
K1,
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2, and that the diagram commutes respectively.
To verify parts (ii) and (iii), we must know σ1 and σ2. In our case, σ1 :Q
F1→ Q
F1is given
by (yF1)σ1 = yF1, and σ2 is the identity. So,
(I2J1
J1
)σ1
=
(F1F1
F1
)σ1
=
(F1
F1
)σ1
= 1 and
L2K1
K1=F1F1
F1=F1
F1= 1. Hence part (ii) is satisfied. Since σ2 is the identity, part (iii) is
obviously satisfied.
To verify part (iv), that is, σ2 θ1 = θ2 σ1, let us make an observation. Observe thatL2
L2 ∩K1=
F1
F1 ∩ F1= 1. So, for this diagram to commute, all of the associated maps must
be the identity. One can easily verify that this is indeed true. Hence the hypotheses of
Theorem 1.7 are satisfied, and F11 is a maximal subgroup of S11. Since there are 7 maximal
subgroups, we need 6 more.
41
Consider←−−M12. Examine the table below to see that the containment of the projections and
intersections is satisfied.
←−−M12 S11
I2 = Q I1 = Q
J2 = F1 J1 = F1
L2 = F2 L1 = Q
K2 = Z K1 = F1
Now, we must verify that parts (ii), (iii) and (iv), of Theorem 1.7, are satisfied. Namely that(I2J1
J1
)σ1
=L2K1
K1,
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2, and that the diagram commutes respectively.
To verify parts (ii) and (iii), we must know σ1 and σ2. In our case, σ1 :Q
F1→ Q
F1is
given by (yF1)σ1 = yF1, and σ2 :Q
F1→ F2
Zis given by (yF1)σ2 = yZ. So,
(I2J1
J1
)σ1
=(QF1
F1
)σ1
=
(Q
F1
)σ1
= 〈yF1〉σ1 = 〈yF1〉 andL2K1
K1=
F2F1
F1=
Q
F1= 〈yF1〉. Hence
part (ii) is satisfied. For part (iii),
(I2 ∩ J1
J2
)σ2
=
(Q ∩ F1
F1
)σ2
=
(F1
F1
)σ2
= 1 and
L2 ∩K1
K2=F2 ∩ F1
Z=Z
Z= 1. Hence part (iii) is satisfied.
To verify part (iv), that is, σ2 θ1 = θ2 σ1, observe that the associated maps for←−−M12 are:
θ1 :QF1
F1=
Q
F1−→ Q
Q ∩ F1=
Q
F1is given by (yF1)θ1 = yF1;
σ2 :Q
Q ∩ F1=
Q
F1−→ F2
F2 ∩ F1=F2
Zis given by (yF1)σ2 = yZ;
σ1 :QF1
F1=
Q
F1−→ F2F1
F1=
Q
F1is given by (yF1)σ1 = yF1;
θ2 :F2F1
F1=
Q
F1−→ F2
F2 ∩ F1=F2
Zis given by (yF1)θ2 = yZ .
Then σ2 (θ1 (yF1)) = σ2 (yF1) = yZ and θ2 (σ1 (yF1)) = θ2 (yF1) = yZ. Hence the diagram
commutes, and the hypotheses of Theorem 1.7 are satisfied. Therefore,←−−M12 is a maximal
subgroup of S11.
Similarly, one can show that←−−M13 is a maximal subgroup of S11. One can do this by replacing
F2 with F3 in the argument for←−−M12 given above. Now we need 4 more maximal subgroups.
Consider−−→M21. Examine the table below to see that the containment of the projections and
intersections is satisfied.
42
−−→M21 S11
I2 = F2 I1 = Q
J2 = Z J1 = F1
L2 = Q L1 = Q
K2 = F1 K1 = F1
Now, we must verify that parts (ii), (iii) and (iv), of Theorem 1.7, are satisfied. Namely that(I2J1
J1
)σ1
=L2K1
K1,
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2, and that the diagram commutes respectively.
To verify parts (ii) and (iii), we must know σ1 and σ2. In our case, σ1 :Q
F1→ Q
F1is
given by (yF1)σ1 = yF1, and σ2 :F2
Z→ Q
F1is given by (yZ)σ2 = yF1. So,
(I2J1
J1
)σ1
=(F2F1
F1
)σ1
=
(Q
F1
)σ1
= 〈yF1〉σ1 = 〈yF1〉 andL2K1
K1=
QF1
F1=
Q
F1= 〈yF1〉. Hence
part (ii) is satisfied. For part (iii),
(I2 ∩ J1
J2
)σ2
=
(F2 ∩ F1
Z
)σ2
=
(Z
Z
)σ2
= 1 and
L2 ∩K1
K2=Q ∩ F1
F1=F1
F1= 1. Hence part (iii) is satisfied.
To verify part (iv), that is, σ2 θ1 = θ2 σ1, observe that the associated maps for−−→M21 are:
θ1 :F2F1
F1=
Q
F1−→ F2
F2 ∩ F1=F2
Zis given by (yF1)θ1 = yZ;
σ2 :F2
F2 ∩ F1=F2
Z−→ Q
Q ∩ F1=
Q
F1is given by (yZ)σ2 = yF1;
σ1 :F2F1
F1=
Q
F1−→ QF1
F1=
Q
F1is given by (yF1)σ1 = yF1;
θ2 :QF1
F1=
Q
F1−→ Q
Q ∩ F1=
Q
F1is given by (yF1)θ2 = yF1 .
Then σ2 (θ1 (yF1)) = σ2 (yZ) = yF1 and θ2 (σ1 (yF1)) = θ2 (yF1) = yF1. Hence the diagram
commutes, and the hypotheses of Theorem 1.7 are satisfied. Therefore,−−→M21 is a maximal
subgroup of S11.
Similarly, one can show that−−→M31 is a maximal subgroup of S11. One can do this by replacing
F2 with F3 in the argument for−−→M21 given above. Now we need 2 more maximal subgroups.
Consider B1, which is given by the triple
(Q
Z,Q
Z, β1
), where xβ1 = x and yβ1 = y. Let
us now verfiy that this group satisfies the hypotheses of Theorem 1.7. Examine the table
below to see that the containment of the projections and intersections is satisfied.
43
B1 S11
I2 = Q I1 = Q
J2 = Z J1 = F1
L2 = Q L1 = Q
K2 = Z K1 = F1
Now, we must verify that parts (ii), (iii) and (iv), of Theorem 1.7, are satisfied. Namely that(I2J1
J1
)σ1
=L2K1
K1,
(I2 ∩ J1
J2
)σ2
=L2 ∩K1
K2, and that the diagram commutes respectively.
To verify parts (ii) and (iii), we must know σ1 and σ2. In our case, σ1 :Q
F1→ Q
F1is given
by (yF1)σ1 = yF1, and σ2 :Q
Z→ Q
Zis given by (xZ)σ2 = xZ and (yZ)σ2 = yZ. Then(
I2J1
J1
)σ1
=
(QF1
F1
)σ1
=
(Q
F1
)σ1
= 〈yF1〉σ1 = 〈yF1〉 andL2K1
K1=QF1
F1=
Q
F1= 〈yF1〉.
Hence part (ii) is satisfied. For part (iii),
(I2 ∩ J1
J2
)σ2
=
(Q ∩ F1
Z
)σ2
=
(F1
Z
)σ2
=
〈xZ〉σ2 = 〈xZ〉 andL2 ∩K1
K2=Q ∩ F1
Z=F1
Z= 〈xZ〉. Hence part (iii) is satisfied.
To verify part (iv), that is, σ2 θ1 = θ2 σ1, observe that the associated maps for B1 are:
θ1 :QF1
F1=
Q
F1−→ Q
Q ∩ F1=
Q
F1is given by (yF1)θ1 = yF1;
σ2 :Q
Q ∩ F1=
Q
F1−→ Q
Q ∩ F1=
Q
F1is given by (yF1)σ2 = yF1;
σ1 :QF1
F1=
Q
F1−→ QF1
F1=
Q
F1is given by (yF1)σ1 = yF1;
θ2 :QF1
F1=
Q
F1−→ Q
Q ∩ F1=
Q
F1is given by (yF1)θ2 = yF1 .
Since all of these maps are essentially the identity, the diagram commutes. Hence, the
hypotheses of Theorem 1.7 are satisfied, and B1 is a maximal subgroup of S11.
Similarly, one can show that B2 is a maximal subgroup of S11, where B2 is given by the
triple
(Q
Z,Q
Z, β2
)and xβ2 = x and yβ2 = xy. Therefore, the maximal subgroups of S11 are
F11,←−−M12,
←−−M13,
−−→M21,
−−→M31, B1 and B2.
One can show that Sij are isomorphic, where 1 ≤ i, j ≤ 3, by applying Theorem 1.8.
Namely, one can use an automorphism of Q×Q to show that the subgroups are isomorphic.
Let us now show that S11∼= S23. For S11, σ1 :
Q
F1→ Q
F1is given by (yF1)σ1 = yF1. For S23,
σ1 :Q
F2→ Q
F3is given by (xF2)σ1 = yF3. So, to show these two groups are isomorphic, we
need two automorphisms of Q. One should send x to y and y to x. The other should send
y to y and x to xy. Referring back to the 24 automorphisms of Q given in the introduction,
44
we can use τ9 and τ17 respectively. Then (τ9, τ17) ∈ Aut(Q × Q) and (S11)(τ9,τ17) = S23.
Therefore, these two groups are isomorphic.
To determine the maximal subgroups of S23 we examine the maximal subgroups of S11 under
the automorphism (τ9, τ17). Let us now determine the maximal subgroups of S23. Consider
F11. Then(F11
)(τ9,τ17)= 〈(x, 1), (1, x)〉(τ9,τ17) = 〈(y, 1), (1, xy)〉 = F23. Consider
←−−M12. Then(←−−
M12
)(τ9,τ17)= 〈(y, y), (x, z)〉(τ9,τ17) = 〈(x, y), (y, z)〉 =
←−−M22. Consider
←−−M13, and recall that
xZ = x−1Z in Q. Then(←−−M13
)(τ9,τ17)= 〈(y, xy), (x, z)〉(τ9,τ17) =
⟨(x, x−1), (y, z)
⟩=←−−M21.
Consider−−→M21. Then
(−−→M21
)(τ9,τ17)= 〈(y, y), (z, x)〉(τ9,τ17) = 〈(x, y), (z, xy)〉 =
−−→M13. Con-
sider−−→M31, and recall that xyZ = x−1yZ in Q. Then
(−−→M31
)(τ9,τ17)= 〈(xy, y), (z, x)〉(τ9,τ17) =⟨
(x−1y, y), (z, xy)⟩
=−−→M33. Consider B1, the group whose corresponding isomorphism is
given by xβ1 = x and yβ1 = y. Then (x, x)(τ9,τ17) = (y, xy) and (y, y)(τ9,τ17) = (x, y). Since
we are in a finite group and (τ9, τ17) is an automorphism, it takes a generating set to a
generating set. So, the group given by the triple whose corresponding isomorphism is the
automorphism of V4 which sends y to xy and x to y is the image of B1 under (τ9, τ17).
This automorphism is β4. Hence (B1)(τ9,τ17) = B4. Consider B2, the group whose corre-
sponding isomorphism is given by xβ2 = x and yβ2 = xy. Then (x, x)(τ9,τ17) = (y, xy) and
(y, xy)(τ9,τ17) = (x, x−1). Again, since we are in a finite group and (τ9, τ17) is an automor-
phism, it takes a generating set to a generating set. So, the group given by the triple whose
corresponding isomorphism is the automorphism of V4 which sends y to xy and x to x is
the image of B2 under (τ9, τ17). Observe that we can use x instead of x−1 because in Q,
xZ = x−1Z. This automorphism is β2. Hence (B2)(τ9,τ17) = B2. Therefore, the maximal
subgroups of S23 are F23,←−−M22
←−−M21,
−−→M13,
−−→M33, B4 and B2.
Similarly, one can show the remaining Sij ’s are isomorphic and determine that their maximal
subgroups are Fij ,←−−Mik where i is fixed and k 6= i, and
−−→Mkj where j is fixed and k 6= j,
1 ≤ k ≤ 3. A table with the maximal subgroups of Sij is given below.
45
Groups Maximal Subgroups
S11 F11, B1, B2,←−−M12,
←−−M13,
−−→M21,
−−→M31
S12 F12, B3, B4,←−−M11,
←−−M13,
−−→M22,
−−→M32
S13 F13, B5, B6,←−−M11,
←−−M12,
−−→M23,
−−→M33
S21 F21, B3, B6,←−−M22,
←−−M23,
−−→M11,
−−→M31
S22 F22, B1, B5,←−−M21,
←−−M23,
−−→M12,
−−→M32
S23 F23, B2, B4,←−−M21,
←−−M22,
−−→M13,
−−→M33
S31 F31, B4, B5,←−−M32,
←−−M33,
−−→M11,
−−→M21
S32 F32, B2, B6,←−−M31,
←−−M33,
−−→M12,
−−→M22
S33 F33, B1, B3,←−−M31,
←−−M32,
−−→M13,
−−→M23
46
2.6 Subgroup Lattice of Q×Q
In concluding this chapter, we use the construction given in Sections 2.1 - 2.5 to give the
subgroup lattice of Q × Q. We were able to construct this subgroup lattice and all the
subgroup lattices in this dissertation using The Geometer’s Sketchpad Version 5 [1].
47
QxQ
*A1
*A2
*A3
S11
S12
S13
S21
A*3
A*2
A*1
S33
S32
S31
S23
S22
*F1
*F2
*F3
*X1
*X2
*X3
*F11
*F12
*F13
*F21
*F22
*F23
*F31
*F32
*F33
F*3
F*2
F*1
X*3
X*2
X*1
F*33
F*32
F*31
F*23
F*22
F*21
F*13
F*12
F*11
Z3
Z1
Z2
1V4
Q^2
M*33
M*32
M*31
M*23
M*22
M*21
M*13
M*12
M*11
B6
B4
B1
F^33
F^32
F^31
F^23
F^22
F^21
F^13
F^12
F^11
B5
B3
B2
*M33
*M32
*M31
*M23
*M22
*M21
*M13
*M12
*M11
Q^1
Q*2
W*3
W*2
W*1
Y*3
Y*2
Y*1
∆ ∆∆∆1
81
81
81
8∆ ∆∆∆
12
12
12
12
∆ ∆∆∆7 777
∆ ∆∆∆2
42
42
42
4∆ ∆∆∆
23
23
23
23
∆ ∆∆∆2
22
22
22
2∆ ∆∆∆
16
16
16
16
∆ ∆∆∆1
51
51
51
5∆ ∆∆∆
14
14
14
14
∆ ∆∆∆4 444
∆ ∆∆∆3 333
∆ ∆∆∆2 222
F33
F32
F31
F23
F22
F21
F13
F12
F11
∆ ∆∆∆2
02
02
02
0∆ ∆∆∆
19
19
19
19
∆ ∆∆∆1
71
71
71
7∆ ∆∆∆
11
11
11
11
∆ ∆∆∆1
01
01
01
0∆ ∆∆∆
9 999∆ ∆∆∆
8 888∆ ∆∆∆
6 666∆ ∆∆∆
5 555∆ ∆∆∆
21
21
21
21
∆ ∆∆∆1
31
31
31
3∆ ∆∆∆
1 111*Y3
*Y2
*Y1
*W3
*W2
*W1
*Q1
48
Chapter 3
Further Results about Constructions and Subgroup Lattices
In this chapter, we would like to discuss Q × D8 and give its subgroup lattice, where D8
denotes the dihedral group of order 8. This group is of interest particularly because it
contains an extraspecial group of order 32, and we would like to identify it. Section 3.1 will
introduce Q×D8 and describe how we used Goursat’s theorem to calculate the number of
subgroups. The details of the subgroup lattice construction of Q×D8 will not be included
in this dissertation. As one can imagine, the details would be similar to the construction of
Q×Q. In Section 3.2, we will discuss extraspecial groups and give a classification of them.
In Section 3.3, we will give the subgroup lattices of Q×Q and Q×D8. Then we will give
the subgroup lattices of the two extraspecial groups of order 32 that were found using the
subgroup lattices of Q×Q and Q×D8.
3.1 Q×D8
Before giving the subgroup lattice of Q×D8, let us introduce notation and show how one
would count the number of subgroups it has. The quaternions will have the presentation
Q =< x, y|x4 = y4 = 1, x2 = y2 >. Let F1 =< x >, F2 =< y >, and F3 =< xy >
be the subgroups of order 4 in Q. Let the subgroup of order 2 be Z1 = < x2 >. Let
E = Q,Fi, Z1, 1|1 ≤ i ≤ 3. The dihedral group of order 8 will have the presentation
D8 =< r, s|r4 = s2 = 1, rs = r−1 >. Let G1 =< r2, rs >, G2 =< r2, s >, and G3 =< r > be
the subgroups of order 4 in D8. Let H1 =< rs >, H2 =< sr >, H3 =< sr2 >, H4 =< s >,
and Z2 =< r2 > be the subgroups of order 2 in D8. Let F = Gi, Hl, Z2|1 ≤ i ≤ 3 ,
1 ≤ l ≤ 4. The subgroup lattice of D8 can be found below.
49
D8
G1 G3 G2
H1 H2 Z2 H3 H4
1
Observe that our aim is to determine the number of subgroups in Q ×D8 and the orders
of the respective subgroups. To do this we use Goursat’s theorem to examine the quotientsI
Jand
L
K, where I, J ∈ E and L,K ∈ F. For a reference on counting subgroups, see [10].
In order to calculate the orders of the subgroups we will use the following: If U ≤ A × B,
then |U | =|I||L||I/J |
.
In the following 4 cases, the number of subgroups and their respective orders will be deter-
mined. For the notation used, 1 ≤ i, k ≤ 3, 1 ≤ l ≤ 4, and 1 ≤ j ≤ 2. If j = 1, then t = 1
or 2. If j = 2, then t = 3 or 4. (i.e., t depends on j) Also note that the choice of I and J
will come from E and the choice of L and K will come from F.
Case 1. Let I = Q, and let J ∈ E. Then
I, J I/J Aut(I/J) L, K No. of subgroups
Q, Q 1 1
D8, D8 1 of order 64
Gk, Gk 3 of order 32
Hl, Hl 4 of order 16
Z2, Z2 1 of order 16
1, 1 1 of order 8
Q, Fi C2 1
D8, Gk 9 of order 32
Gj , Ht 12 of order 16
Gk, Z2 9 of order 16
Hl, 1 12 of order 8
Z2, 1 3 of order 8
Q, Z1 V4 S3
D8, Z2 6 of order 16
G1, 1 6 of order 8
G2, 1 6 of order 8
Case 2. Let I = Fi, and let J ∈ E \ Q. Then
50
I, J I/J Aut(I/J) L, K No. of subgroups
Fi, Fi 1 1
D8, D8 3 of order 32
Gk, Gk 9 of order 16
Hl, Hl 12 of order 8
Z2, Z2 3 of order 8
1, 1 3 of order 4
Fi, Z1 C2 1
D8, Gk 9 of order 16
Gj , Ht 12 of order 8
Gk, Z2 9 of order 8
Hl, 1 12 of order 4
Z2, 1 3 of order 4
Fi, 1 C4 C2 G3, 1 6 of order 4
Case 3. Let I = Z1, and let J ∈ E \ Q,Fi. Then
I, J I/J Aut(I/J) L, K No. of subgroups
Z1, Z1 1 1
D8, D8 1 of order 16
Gk, Gk 3 of order 8
Hl, Hl 4 of order 4
Z2, Z2 1 of order 4
1, 1 1 of order 2
Z1, 1 C2 1
D8, Gk 3 of order 8
Gj , Ht 4 of order 4
Gk, Z2 3 of order 4
Hl, 1 4 of order 2
Z2, 1 1 of order 2
Case 4. Let I = 1 = J . Then
I, J I/J Aut(I/J) L, K No. of subgroups
1, 1 1 1
D8, D8 1 of order 8
Gk, Gk 3 of order 4
Hl, Hl 4 of order 2
Z2, Z2 1 of order 2
1, 1 1 of order 1
51
Hence, there are 189 subgroups of Q×D8, which include the trivial subgroup, 11 subgroups
of order 2, 39 subgroups of order 4, 71 subgroups of order 8, 51 subgroups of order 16, 15
subgroups of order 32 and the group itself.
As in the construction of Q × Q, one then uses this information to examine the groups of
orders 2, 4, 8, 16 and 32 to construct its subgroup lattice. The details of this construction
will not be included in this dissertation. However, the subgroup lattice of Q ×D8 will be
given in Section 3.3 for comparison with Q×Q.
3.2 Extraspecial Groups
In this section, we define two types of groups, and we discuss how these groups are relevant
to the subgroup lattices we have constructed. Two definitions from [2] are given here.
Definition. A group G is called a central product of its subgroups A and B if
(i) G = A×B
(ii) [A,B] = 1
This definition implies that A,B C G and that A ∩ B ≤ Z(A) ∩ Z(B). When A and B
have isomorphic centers, we may also refer to a central product as a quotient of G with
amalgamated centers, and write it as A B.
Definition. Let P be a p-group. We call P extraspecial if Z(P ) = P ′ = Φ(P ).
There is a well known classification of extraspecial groups. In order to understand the
classification of extraspecial groups which will follow in the next Theorem, the reader should
be aware of the following notation and observations, given in [2].
For an odd prime p, let
E = 〈x, y|xp = yp = 1, [x, y] ∈ Z(E)〉, and
F =⟨x, y|xp2 = yp = 1, [x, y] = xp
⟩.
These are the two extraspecial groups of order p3.
When p = 2, the extraspecial groups of order 23 are D8 and Q. In the following Theorem,
which classifies extraspecial groups, the notation and observations given above will be used.
52
Theorem 3.1. [2]
Let p be a prime, and let P be an extraspecial group of order p2t+1. Then exactly one of the
following four cases arises:
(i) p 6= 2, Exp(P ) = p, and P is a central product of t copies of E;
(ii) p 6= 2, Exp(P ) = p2, and P is a central product of t−1 copies of E with a copy of F ;
(iii) p = 2, and P is a central product of t copies of D8;
(iv) p = 2, and P is a central product of t− 1 copies of D8 with a copy of Q.
Using the above theorem, one is able to see that there are exactly two extraspecial groups
of order 32. They are given by Q Q and Q D8. The subgroup lattices of these two
extraspecial groups of order 32 lie in the subgroup lattices of Q×Q and Q×D8. The reader
can recognize them as the quotient by the diagonal of the centers. In Q × Q, the group
that identifies the centers is Z2, a cyclic group of order 2. Being able to apply Theorems
1.7 and 1.8 to construct these subgroup lattices of Q × Q and Q ×D8 and to identify the
extraspecial groups that lie within them was our goal. In the next section these subgroup
lattices will be given.
3.3 Subgroup Lattices
In this section, the subgroup lattices of Q × Q and Q ×D8 are given so one can compare
them. Also, the subgroup lattices of the extraspecial groups of order 32 are given.
53
The subgroup lattice of Q×Q is:
54
QxQ
*A1
*A2
*A3
S11
S12
S13
S21
A*3
A*2
A*1
S33
S32
S31
S23
S22
*F1
*F2
*F3
*X1
*X2
*X3
*F11
*F12
*F13
*F21
*F22
*F23
*F31
*F32
*F33
F*3
F*2
F*1
X*3
X*2
X*1
F*33
F*32
F*31
F*23
F*22
F*21
F*13
F*12
F*11
Z3
Z1
Z2
1V4
Q^2
M*33
M*32
M*31
M*23
M*22
M*21
M*13
M*12
M*11
B6
B4
B1
F^33
F^32
F^31
F^23
F^22
F^21
F^13
F^12
F^11
B5
B3
B2
*M33
*M32
*M31
*M23
*M22
*M21
*M13
*M12
*M11
Q^1
Q*2
W*3
W*2
W*1
Y*3
Y*2
Y*1
∆ ∆∆∆1
81
81
81
8∆ ∆∆∆
12
12
12
12
∆ ∆∆∆7 777
∆ ∆∆∆2
42
42
42
4∆ ∆∆∆
23
23
23
23
∆ ∆∆∆2
22
22
22
2∆ ∆∆∆
16
16
16
16
∆ ∆∆∆1
51
51
51
5∆ ∆∆∆
14
14
14
14
∆ ∆∆∆4 444
∆ ∆∆∆3 333
∆ ∆∆∆2 222
F33
F32
F31
F23
F22
F21
F13
F12
F11
∆ ∆∆∆2
02
02
02
0∆ ∆∆∆
19
19
19
19
∆ ∆∆∆1
71
71
71
7∆ ∆∆∆
11
11
11
11
∆ ∆∆∆1
01
01
01
0∆ ∆∆∆
9 999∆ ∆∆∆
8 888∆ ∆∆∆
6 666∆ ∆∆∆
5 555∆ ∆∆∆
21
21
21
21
∆ ∆∆∆1
31
31
31
3∆ ∆∆∆
1 111*Y3
*Y2
*Y1
*W3
*W2
*W1
*Q1
55
The subgroup lattice of Q×D8 is:
56
QxD8
O6
C7
B7
A7
Z6
Y6
X6
W6
V6
U6
T6
S6
R6
Q6
P6
N6
M6
L6
K6
J6
I 6H6
G6
F6
E6
D6
C6
B6
A6
Z5
Y5
X5
W5
V5
U5
T5
S5
R5
Q5
P5
O5
N5
M5
L5
K5
J5
I 5H5
G5
F5
E5
D5
C5
B5
A5
Z4
P4
Q4
R4
S4
T4
U4
V4
W4
X4
Y4
K1
C4
P1
O1
N1
M1
G1'
L1
B4
A4
Z3
Y3
X3
W3
V3
U3
T3
S3
R3
Q3
P3
O3
N3
M3
L3
K3
J3
I 3H3
G3
F3
E3
D3
C3
B3
A3
Z2
Y2
X2
W2
V2
U2
T2
S2
R2
Q2
P2
O2
N2
M2
L2
K2
J2
I 2H2
G2
F2
E2
D2
C2
B2
A2
Z1
Y1
X1
W1
V1
U1
T1
S1
R1
Q1
Q'
R'
S'
T'
U'
V'
W'
X'
Y'
Z'
A1'
B1'
C1'
D1'
E1'
F1'
H1'
I 1'
I 1H1
G1
F1
E1
D1
C1
B1
A1
ZY
XW
VU
TS
RQ
C
NM
LK
JI
HG
FE
D
A
57
The extraspecial group of order 32 from Q×Q is:
Q Y Q
*A1 *A2 *A3 S11S12 S13 S21 A*3A*2A*1S33S32S31
S23S22
*F11 *F12 *F13 *F21 *F22 *F23 *F31 *F32 *F33 F*33F*32F*31F*23F*22F*21F*13F*12F*11
Z2
V
Q^2M*33M*32M*31M*23M*22M*21M*13M*12M*11B6B4B1F^33F^32F^31F^23F^22F^21F^13F^12F^11B5B3B2*M33*M32*M31*M23*M22*M21*M13*M12*M11Q^1
Y*3Y*2Y*1∆∆∆∆18181818∆∆∆∆12121212∆∆∆∆7777∆∆∆∆24242424∆∆∆∆23232323∆∆∆∆22222222∆∆∆∆16161616∆∆∆∆15151515∆∆∆∆14141414∆∆∆∆4444∆∆∆∆3333∆∆∆∆2222F33F32F31F23F22F21F13F12F11∆∆∆∆20202020∆∆∆∆19191919∆∆∆∆17171717∆∆∆∆11111111∆∆∆∆10101010∆∆∆∆9999∆∆∆∆8888∆∆∆∆6666∆∆∆∆5555∆∆∆∆21212121∆∆∆∆13131313∆∆∆∆1111
*Y3*Y2*Y1
To determine the number of subgroups in QQ, we use the table from Section 2.1 to count
the subgroups that satisfy the following property: Z2 is contained in a subgroup if and only
if Z ≤ I, Z ≤ L and Z ≤ J if and only if Z ≤ K.
Q Q has 110 subgroups, consisting of the following:
15 subgroups of order 16
35 subgroups of order 8
39 subgroups of order 4
19 subgroups of order 2, and
the trivial subgroup and the group itself.
58
The extraspecial group of order 32 from Q×D8 is:
ΑΑΑΑ
ΒΒΒΒ ΧΧΧΧ ∆∆∆∆ ΕΕΕΕ ΦΦΦΦ ΓΓΓΓ ΗΗΗΗ ΙΙΙΙ ϑϑϑϑ ΚΚΚΚ ΛΛΛΛ
ΜΜΜΜ ΝΝΝΝ ΟΟΟΟ ΠΠΠΠ ΘΘΘΘ ΡΡΡΡ ΣΣΣΣ ΤΤΤΤ ΥΥΥΥ ςςςς ΩΩΩΩ ΞΞΞΞ ΨΨΨΨ ΖΖΖΖ ΑΑΑΑ1111
ΒΒΒΒ1111
ΧΧΧΧ1111 ∆∆∆∆1111 ΕΕΕΕ1111 ΦΦΦΦ1111 ΓΓΓΓ1111 ΗΗΗΗ1111 ΙΙΙΙ1111 ϑϑϑϑ1111 ΚΚΚΚ1111 ΛΛΛΛ1111 ΜΜΜΜ1111 ΝΝΝΝ1111 ΟΟΟΟ1111 ΠΠΠΠ1111 ΘΘΘΘ1111 ΡΡΡΡ1111 ΣΣΣΣ1111 ΤΤΤΤ1111 ΥΥΥΥ1111 ςςςς1111 ΩΩΩΩ1111 ΞΞΞΞ1111 ΨΨΨΨ1111 ΖΖΖΖ1111 ΑΑΑΑ2222 ΒΒΒΒ2222 ΧΧΧΧ2222 ∆∆∆∆2222 ΕΕΕΕ2222 ΦΦΦΦ2222 ΓΓΓΓ2222 ΗΗΗΗ2222 ΙΙΙΙ2222 ϑϑϑϑ2222
ΚΚΚΚ2222 ΛΛΛΛ2222 ΜΜΜΜ2222 ΝΝΝΝ2222 ΟΟΟΟ2222 ΠΠΠΠ2222 ΘΘΘΘ2222 ΡΡΡΡ2222 ΣΣΣΣ2222 ΤΤΤΤ2222 ΥΥΥΥ2222 ςςςς2222 ΩΩΩΩ2222 ΞΞΞΞ2222 ΨΨΨΨ2222
Q Y D8
To determine the number of subgroups in Q D8, we use the table from Section 3.1 to
count the subgroups that satisfy the following property: The group whose projections and
intersections are given by Z1, 1 and Z2, 1 is contained in a subgroup if and only if Z1 ≤ I,
Z2 ≤ L and Z1 ≤ J if and only if Z2 ≤ K.
Q D8 has 78 subgroups, consisting of the following:
15 subgroups of order 16
35 subgroups of order 8
15 subgroups of order 4
11 subgroups of order 2, and
the trivial subgroup and the group itself.
59
Conclusions
We finish by discussing future projects and questions that our current work gave rise to. In
the future, we would like to continue our work in characterizing subgroups of products of
groups. The methods used to construct these lattices involved explicit details. Particularly,
the details included repeated application of the results given in Section 1.3 to determine
containment of subgroups from one level to the next in the subgroup lattices and knowledge
of the subgroup structures of groups of orders 16 and 32. It would be interesting to know if
this information can be generalized to provide more insight about subgroup lattices without
doing such explicit work and also if the work already done might have deeper implications.
After investigating lattice properties of extraspecial groups in the case when p = 2, it would
be interesting to explore lattice properties of extraspecial groups in the case when p > 2.
When we started on this project of further characterizing subgroups in products of groups,
we first studied and characterized subgroups of a central product. Our motivation for this
dissertation was determined by our original question and goal, which was to determine
when two central products were isomorphic. There are more generalized versions of a
central product in [9] that make this question a plausible one. We are still working to
answer this question. Studying the subgroup lattices of Q×Q, Q×D8 and the extraspecial
groups of order 32 provided more insight and led to other questions that are worth pursuing.
For example, what can one say about two nonisomorphic groups with the same subgroup
lattice structure? Are the lattices of normal subgroups of the extraspecial groups of order
32 the same as their dual? I intend to revisit my original question and continue these
characterizations. For comparison, with the information we have found, it would also be an
interesting project to construct the subgroup lattice of D8×D8. Although we already know
what the extraspecial group is going to be, it would be interesting to see if one could use
the details from our constructions to learn more information about constructing subgroup
lattices and to see if more information can be obtained from within that subgroup lattice.
60
In working with extraspecial groups, it was important to understand p-groups. Because
p-groups play such a significant role in the study of group theory it would be interesting to
study their structure.
Another interesting future project would be to see if a computer program could be written to
construct these subgroup lattices because the processes used in our constructions were very
methodical. Constructing the subgroup lattices was indeed tedious, as one can imagine,
because of the number of subgroups involved. Although this work was tedious, seeing
how the results could be applied to construct subgroup lattices that contain a significant
amount of information and making this contribution to the area of finite group theory was
gratifying. It was very rewarding to provide a property that further characterizes subgroups
of a direct product and to find the subgroup lattices of the extraspecial groups of order 32.
Providing them served as a bridge from our original question involving central products to
the classification of extraspecial groups of order p2t+1.
61
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