Finite Groups & Subgroups

Order of a group

• Definition:The number of elements of a group (finite or infinite) is called its order.

• Notation:We will use |G| to denote the order of group G.

Examples

|D4| =

|Dn| =

|<R90>| =

|Zn| =

|U(8)| = |U(11)| =

|Z| =

82n4n410∞

Order of an element

• Definition:The order of an element g in a group G is the smallest positive integer n such that gn = e (In additive notation, ng = 0).If no such integer exists, we say g has infinite order.

• Notation:The order of g is denoted |g|.

Examples

In D4, |R90| =

In D4, |H| =

In Z10, |4| =

In Z11, |4| =

In U(8), |5| = In U(9), |5| =

In Z, |1| =

4 ( R490 = R0

)

2 ( H2 = R0)

5 (5•4 mod 10 = 0)11 (11•4 mod 11 =

0)2 (52 mod 8 = 1)6 {5, 7, 8, 4, 2, 1}∞ (n•1 ≠ 0 for n>0)

Group G (•mod 35)

|G| =e =

|5| =|10| =|15| =|20| =|30| =

61566123

• 5 10 15 20 25 30

5 25 15 5 30 20 10

10 15 30 10 25 5 20

15 5 10 15 20 25 30

20 30 25 20 15 10 5

25 20 5 25 10 30 15

30 10 20 30 5 15 25

Subgroups

• Definition:If a subset H of a group G is itself a group under the operation of G, then we say that H is a subgroup of G.

Notation:

• We write H ≤ G to mean H is a subgroup of G.

• If H is not equal to G, we write H < G.We say H is a proper subgroup of G.

• {e} is called the trivial subgroup.All other subgroups are nontrivial.

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

€

R90 < D4

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

€

R180 < R90 < D4

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

€

R0 < R180 < R90 < D4

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

€

{R0,R180,H,V} ≤ D4

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

€

{R0,R180,D,D'} ≤ D4

• R0 R90 R180 R270 H V D D'

R0 R0 R90 R180 R270 H V D D'

R90 R90 R180 R270 R0 D' D H V

R180 R180 R270 R0 R90 V H D' D

R270 R270 R0 R90 R180 D D' V H

H H D V D' R0 R180 R90 R270

V V D' H D R180 R0 R270 R90

D D V D' H R270 R90 R0 R180

D' D' H D V R90 R270 R180 R0

€

{R0,H,V} not a subgroup of D4

Subgroup tests

• Three important tests tell us if a nonempty subset of a group G is a subgroup of G.

• One-Step Subgroup Test• Two-Step Subgroup Test• Finite Subgroup Test

One-Step Test

Let H be a nonempty subset of a group G.

If ab-1 belongs to H whenever a and b belong to H, then H is a subgroup of G.

(In additive groups: If a–b belongs to H whenever a and b

belong to H, then H ≤ G.)

Proof of One Step Test.

• Let G be a group, and H a nonempty subset of G. Suppose ab-1 is in H whenever a and b are in H. (*)We must show:1. In H, multiplication is associative

2. The group identity e is in H3. H has inverses4. H is closed under the group multiplication.Then, H must be a subgroup of G.

(1) Multiplication is Associative:

• Choose any elements a, b, c in H. • Since H is a subset of G, these

elements are also in the group G, so(ab)c = a(bc) as required.

(2) H contains e

• Choose any x in H. (Since H is nonempty there has to be some element x in H)

• Let a = x and b = x. Then a and b are in H, so by (*)ab-1 = xx-1 = e is in H, as required.

(3) H has inverses

• Choose any x in H.• Let a = e and b = x.

Since a and b are in H,ab-1 = ex-1 = x-1 must be in H as well.

(4) H is closed

• Choose any x and y in H.• Let a = x and b = y-1.• Since a and b are in H,

ab-1 = x(y-1)-1 = xy is also in H.We have shown that H is closed under the

multiplication in G, and that H is associative, contains the identity, and has inverses.

Therefore, H is a subgroup of G.

To use the One-Step Test

1. Identify the defining property P that distinguishes elements of H.

2. Prove the identity has property P. 3. Assume that two elements have

property P4. Show that ab-1 has property P.Then by the one-step test, H ≤ G.

H≠

a,b in H

ab-1 in H

Example: One Step• Prove: Let G be an Abelian group with

identity e. Let H = {x |x2 = e}. Then H ≤ G.

• Proof: e2 = e, so that H is nonempty.Assume a, b in H. Then(ab-1)2 = a(b-1a)b-1 = aab-1b-1 (G is Abelian)

= a2b-2 = a2(b2)-1 = ee-1 (since a and b in H) = e.

By the one-step test, H ≤ G.

Example One-Step

• Prove: The set 3Z = {3n | n in Z} (i.e. the integer multiples of 3) under the usual addition is a subgroup of Z.

• Proof: 0 = 3•0, so 3Z is not empty.• Assume 3a and 3b are in 3Z. • Then 3a – 3b = 3(a–b) is in 3Z. • By the One-Step test, 3Z ≤ Z.

Terminology• Let H be a nonempty subset of a group G with

operation *. • We say

"H is closed under *" or "H is closed"when we mean

"ab is in H whenever a and b are in H"• We say

"H is closed under inverses"when we mean

"a-1 is in H whenever a is in H"

Two Step Test

• Let H be a nonempty subset of group G with operation *. If (1) H is closed under * and (2) H is closed under inverses, then H ≤ G

• Proof: Assume a and b are in H. By (2), b-1 is in H. By (1) ab-1 is in H.By the one-step test, H ≤ G.

Finite Subgroup Test

• Let H be a nonempty finite subset of a group G. If H is closed under the operation of G, then H ≤ G.

• Proof. Choose any a in H. By the two step test, it only remains to show that a-1 is in H.

To show a-1 is in HIf a = e, then a-1 (= e) is in H.If a ≠ e, consider the sequence a,a2,a3…Since H is closed, all are in H. Since H is finite, not all are unique.Say ai = aj where i < j. Cancel ai to get e = aj-i. Since a ≠ e, j-i > 1.Let b = aj-i-1. Then ab = a1 aj-i-1 = aj-i = eSo b = a-1 and b belongs to H.

Definition

• Let a be an element of a group G. The cyclic group generated by a, denoted <a> is the set of all powers of a.That is, <a> = {an | n is an integer}

• In additive groups, <a> = {na | n is a integer}

<a> is a subgroup

• Let G be group, and let a be any element of G.Then <a> is a subgroup of G.

• Proof: a is in <a>, so <a> is not empty.Choose any x = am and y = an in <a>.xy–1= am(an)-1 = am-n which belongs to <a> since m–n is an integer.By the one-step test, <a> is a subgroup of G.

Example

• <25> =• 5 10 15 20 25 30

5 25 15 5 30 20 10

10 15 30 10 25 5 20

15 5 10 15 20 25 30

20 30 25 20 15 10 5

25 20 5 25 10 30 15

30 10 20 30 5 15 25

Example

• <25> = {25, 30, 15}

• 5 10 15 20 25 30

5 25 15 5 30 20 10

10 15 30 10 25 5 20

15 5 10 15 20 25 30

20 30 25 20 15 10 5

25 20 5 25 10 30 15

30 10 20 30 5 15 25