Sylow applications

CONSEQUENCES OF THE SYLOW THEOREMS(MATH 315, FALL 2005)

KEITH CONRAD

1. Statement of the Sylow theorems

We recall here the statement of the Sylow theorems.

Theorem 1.1 (Sylow I). A finite group G has a p-Sylow subgroup for everyprime p.

Theorem 1.2 (Sylow II). For each prime p, any p-subgroup of G lies in ap-Sylow subgroup of G and the p-Sylow subgroups of G are conjugate.

Theorem 1.3 (Sylow III). Let np be the number of p-Sylow subgroups ofG. Write #G = pkm, where p doesn’t divide m. Then

np|m and np ≡ 1 mod p.

Theorem 1.4 (Sylow III*). Let np be the number of p-Sylow subgroups ofG. Then np = [G : N(P )], where P is any p-Sylow subgroup and N(P ) is itsnormalizer.

The information in the Sylow theorems leads to basic structure theoremsabout finite groups, some of which we treat below. Because of the Sylowtheorems, groups of prime power size take on a heightened significance, andthese theorems suggest a careful study of such groups [4].

We will not have too much use for Sylow III* in this handout. It is usedin Theorems 2.3, 5.2, and A.4.

Corollary 1.5. The condition np = 1 means a p-Sylow subgroup is a normalsubgroup.

Proof. All p-Sylows are conjugate by Sylow II, so np = 1 precisely when ap-Sylow of G is self-conjugate, i.e., is a normal subgroup of G. �

Be sure you understand that reasoning. We will often shift back and forthbetween the condition np = 1 (if it holds) and the condition that G has anormal p-Sylow subgroup. In particular, the Sylow theorems are a tool forproving a group has a non-trivial normal subgroup because we can try toshow np = 1 for some p. However, there are groups which have non-trivialnormal subgroups but no non-trivial normal Sylow subgroups, such as S4.(See Example 6.6.)

1

2 KEITH CONRAD

Corollary 1.6. If p and q are different prime factors of #G and np = 1and nq = 1 then the elements of the p-Sylow subgroup commute with theelements of the q-Sylow subgroup.

Proof. Let P be the p-Sylow subgroup and Q be the q-Sylow subgroup.Since P and Q have relatively primes sizes, P ∩Q = {e} by Lagrange. Thesubgroups P and Q are normal in G since np = 1 and nq = 1 by hypothesis.For a ∈ P and b ∈ Q,

aba−1b−1 = (aba−1)b−1 = a(ba−1b−1) ∈ P ∩Q = {e},

so ab = ba. �

Note Corollary 1.6 is not saying the p-Sylow and q-Sylow subgroups of Gare abelian, but rather that any element of either subgroup commutes withany element of the other subgroup if the two Sylow subgroups are the onlysubgroups of their size.

2. Applications to specific groups

Theorem 2.1. The groups A5 and S5 each have 6 subgroups of size 5.

Proof. Since #A5 = 60 = 22 · 3 · 5, the 5-Sylow subgroups have size 5. BySylow III, n5|12 and n5 ≡ 1 mod 5. The only choices are n5 = 1 or 6. SinceA5 has at least two subgroups of size 5 (the subgroups generated by (12345)and by (21345) are different), n5 > 1 and therefore n5 = 6.

The argument for S5 is analogous, since the only divisors of 24 which are≡ 1 mod 5 are 1 and 6. �

Theorem 2.2. In the group Aff(Z/5Z), there are five 2-Sylow subgroupsand the 5-Sylow subgroup is normal.

Proof. This group has size 20, so the 2-Sylows have size 4 and the 5-Sylowshave size 5.

By Sylow III, n2|5, so n2 = 1 or 5. The matrices ( 2 00 1 ) and ( 2 1

0 1 ) generatedifferent 2-Sylow subgroups, so n2 = 5.

Now we turn to the 5-Sylow subgroups. By Sylow III, n5|4 and n5 ≡1 mod 5. The only choice is n5 = 1, which means there is one 5-Sylowsubgroup, so it must be normal. �

Let’s explore Aff(Z/5Z) a little further. Since we know the number of 2-Sylow and 5-Sylow subgroups, we can search for all the Sylow subgroups andknow when to stop. There are five 2-Sylow subgroups and the five matrices( 2 j

0 1), where j ∈ Z/5Z, generate different subgroups of size 4, so these are

all of the 2-Sylow subgroups (and they are cyclic). The matrix ( 1 10 1 ) has

order 5 and therefore generates the unique 5-Sylow subgroup.As an illustration of Sylow II in Aff(Z/5Z), any element of 2-power order

is conjugate to an element of the subgroup 〈( 2 00 1 )〉. For instance, ( 2 1

0 1 ) has

CONSEQUENCES OF THE SYLOW THEOREMS (MATH 315, FALL 2005) 3

order 4 and an explicit search reveals(2 10 1

)=

(3 40 1

) (2 00 1

) (3 40 1

)−1

.

The matrix ( 4 40 1 ) has order 2 and(

4 40 1

)=

(3 20 1

) (2 00 1

)2 (3 20 1

)−1

.

Theorem 2.3. For a prime p, any element of GL2(Z/pZ) with order p isconjugate to a strictly upper-triangular matrix ( 1 a

0 1 ). The number of p-Sylowsubgroups is p + 1.

Proof. The size of GL2(Z/pZ) is (p2−1)(p2−p) = p(p−1)(p2−1). Thereforea p-Sylow subgroup has size p. The matrix ( 1 1

0 1 ) has order p, so it generatesa p-Sylow subgroup P = 〈( 1 1

0 1 )〉 = {( 1 ∗0 1 )} Since all p-Sylow subgroups are

conjugate, any matrix with order p is conjugate to some power of ( 1 10 1 ).

The number of p-Sylow subgroups is [GL2(Z/pZ) : N(P )] by Sylow III*.We’ll compute N(P ) and then find its index. For ( a b

c d ) to lie in N(P ) meansit conjugates ( 1 1

0 1 ) to some power ( 1 ∗0 1 ). Since(

a bc d

) (1 10 1

) (a bc d

)−1

=(

1− ac/∆ a2/∆−c2/∆ 1 + ac/∆

),

where ∆ = ad − bc 6= 0, the condition to lie in N(P ) is precisely c = 0.Therefore N(P ) = {( ∗ ∗

0 ∗ )} in GL2(Z/pZ). The size of N(P ) is (p − 1)2p.Since GL2(Z/pZ) has size p(p−1)(p2−1), the index of N(P ) is np = p+1. �

Corollary 2.4. The number of elements of order p in GL2(Z/pZ) is p2−1.

Proof. Each p-Sylow subgroup has p − 1 elements of order p. Different p-Sylow subgroups intersect trivially, so the number of elements of order p is(p− 1)np = p2 − 1. �

Theorem 2.5. There is a unique p-Sylow subgroup of Aff(Z/p2Z).

Proof. The group has size p2ϕ(p2) = p3(p − 1), so a p-Sylow subgroup hasorder p3.

Letting np be the number of p-Sylow subgroups, Sylow III says np|(p− 1)and np ≡ 1 mod p. Therefore np = 1.

As an alternate proof, we can locate a p-Sylow subgroup of Aff(Z/p2Z)explicitly, namely the matrices (

a b0 1

)where ap = 1 in (Z/p2Z)×. (There are p choices for a and p2 choices for b.)This subgroup is the kernel of the homomorphism Aff(Z/p2Z) → (Z/p2Z)×

given by ( a b0 1 ) 7→ ap, so it is a normal subgroup, and therefore is the unique

p-Sylow subgroup by Sylow II. �

4 KEITH CONRAD

Note the unique p-Sylow subgroup of Aff(Z/p2Z) is a nonabelian groupof size p3. It has an element of order p2, namely ( 1 1

0 1 ), and therefore is notisomorphic to Heis(Z/pZ) when p 6= 2, since every non-identity element ofHeis(Z/pZ) has order p. (It can be shown that every nonabelian group ofsize p3 for an odd prime p is isomorphic to Heis(Z/pZ) or to this p-Sylowsubgroup of Aff(Z/p2Z).)

Can we characterize Heis(Z/pZ) as the unique p-Sylow subgroup of somelarger group? Yes.

Theorem 2.6. For any prime p, Heis(Z/pZ) is the unique p-Sylow subgroupof the group of invertible upper-triangular matrices

(2.1)

d1 a b0 d2 c0 0 d3

in GL3(Z/pZ).

Proof. This matrix group, call it U , has size (p−1)3p3, so Heis(Z/pZ) is a p-Sylow subgroup of U . To show it is the only p-Sylow subgroup, the relationsin Sylow III are not adequate. They tell us np|(p − 1)3 and np ≡ 1 mod p,but it does not follow from this that np must be 1. For instance, (p − 1)2

satisfies these two conditions in place of np.To show np = 1, we will prove Heis(Z/pZ) C U . Projecting a matrix

in U onto its 3 diagonal entries is a function from U to the 3-fold directproduct (Z/pZ)× × (Z/pZ)× × (Z/pZ)×. This is a homomorphism withkernel Heis(Z/pZ), so Heis(Z/pZ) C U . �

3. Characterizing Cyclic Groups

Here is a nice application of Sylow I.

Theorem 3.1. If a finite group has at most one subgroup of any size, thenit is a cyclic group.

The converse is of course true and we saw it earlier in the course.

Proof. Our argument has two steps: use Sylow I to reduce the theorem tothe prime power case, and then settle the prime power case.

Step 1: Let G be a group with a unique subgroup of each size. In particu-lar, for each prime p we obtain by Sylow I that G has one p-Sylow subgroup.Each Sylow subgroup is normal since np = 1. Then, for different primes pand q dividing #G, the elements of the p-Sylow and q-Sylow subgroupscommute with each other by Corollary 1.6.

Any subgroup of G has at most one subgroup of any size (otherwiseG itself would have two subgroups of the same size). Suppose we knewthe theorem for all groups with prime-power size. Then, for each prime pdividing #G, the p-Sylow subgroup of G has to be cyclic. Choose a generatorap of the p-Sylow subgroup of G. The order of ap is the size of the p-Sylowsubgroup of G. These ap’s commute as p varies, by the previous paragraph,


and their orders are relatively prime, so the product of the ap’s has orderequal to the product of the sizes of the Sylow subgroups of G. This productof sizes is #G, so G is cyclic.

Step 2: We are now reduced to verifying our theorem for groups withprime-power size. The Sylow theorems will not be used further.

Let #G = pk where p is prime, k ≥ 1, and assume G has at most onesubgroup of each size. To show G is cyclic, we argue by induction on k. Ifk = 1 then G has prime size so it is cyclic. We now suppose that k ≥ 2 andthe theorem is proved for all groups with p-power size less than pk whichhave at most one subgroup of each size.

Since G is a non-trivial group with p-power size, it has a non-trivial center.Pick (by Cauchy) an element a of order p in the center of G. Then 〈a〉 is asubgroup of G with order p, so it is the unique such subgroup. Since everynon-trivial subgroup of G contains a subgroup of size p by Cauchy, everynon-trivial subgroup of G contains 〈a〉.

Since a lies in the center of G, 〈a〉 is a normal subgroup of G. We thereforecan consider the group G/〈a〉, whose size is pk−1. Let’s show G/〈a〉 has atmost one subgroup of any size. For any subgroup H of G/〈a〉, let H ′ be theinverse image of H in G (all the elements of G which reduce to H). Then H ′

contains 〈a〉, has p times as many elements as H, and H = H ′/〈a〉. If K isa subgroup of G/〈a〉 with the same size as H, then #K ′ = #H ′ (we defineK ′ from K in the same way as H ′ is defined from H), so H ′ = K ′ since G isassumed to have at most one subgroup of any size. Reducing back modulo〈a〉, we get H = H ′/〈a〉 = K ′/〈a〉 = K in G/〈a〉.

By induction, G/〈a〉 is a cyclic group:

G/〈a〉 = 〈b〉.Then every element of G has the form biaj for some i and j, and b 6= esince G/〈a〉 is non-trivial. Since 〈b〉 is a non-trivial subgroup of G, it mustcontain 〈a〉. Therefore a ∈ 〈b〉, so biaj is a power of b. This implies G iscyclic, which settles the theorem for groups of prime-power size. �

Remark 3.2. If every Sylow subgroup is cyclic, the group need not becyclic. Try S3. However, this is true if there is just one Sylow subgroupfor each p dividing #G, since that is the basic reasoning in the proof ofTheorem 3.1.

Here is another application of Sylow I to prove a similar theorem.

Theorem 3.3. Let G be a finite group such that, for each n dividing #G,the equation xn = 1 in G has at most n solutions. Then G is cyclic.

Proof. We again argue in two steps: reduction to the prime power case usingSylow I and then the prime power case.

Step 1: Let p be a prime dividing #G and pk be the largest power of p

in #G. Every g ∈ G of p-power order in G has order dividing pk (all ordersdivide #G), so g is a solution to xpk

= 1. Let P be a p-Sylow subgroup of

6 KEITH CONRAD

G. It provides us with pk solutions to this equation, so by assumption theseare all the solutions. Therefore all elements of p-power order are in P , so Pis the only p-Sylow subgroup.

The hypothesis on G obviously passes to any of its subgroups, such as itsSylow subgroups. If we knew the theorem for groups of prime-power sizethen we get cyclicity of the Sylow subgroups, so G is cyclic by the sameargument as in Step 1 of the proof of Theorem 3.1.

Step 2: We now verify the theorem for p-groups. The Sylow theoremsare not going to be used. Let #G = pk, where k ≥ 2. (The case k = 1is trivial.) Assume xn = 1 has at most n solutions in G whenever n|pk.We want to show G is cyclic. If N C G and #(G/N) > p then G/N has anon-trivial normal subgroup of order p (such as a subgroup of order p in itscenter). Lifting this subgroup of G/N back to G gives a normal subgroupH with N ⊂ H ⊂ G and [H : N ] = p. Therefore a maximal proper normalsubgroup of G has index p. Let M be such a subgroup, so #M = pk−1.The equation xpk−1

= 1 has pk−1 solutions in M , so by hypothesis these arethe only solutions to this equation in G. Therefore any element of G −Mdoes not have order dividing pk−1, so its order must be pk, which shows Gis cyclic. �

The only sizes that entered into the proofs of Theorems 3.1 and 3.3 areprime-powers. It turns out that for a group with prime power size, there isa necessary and sufficient condition for cyclicity which appears much weakerthan the hypotheses of either Theorem 3.1 or Theorem 3.3.

Theorem 3.4. Let p be prime. A group with size pk is cyclic if and only ifit has one subgroup of order p (or, equivalently, only p solutions to xp = 1),provided pk is not a power of 2 greater than 4. If p = 2 and k ≥ 3 then agroup with size pk is cyclic if and only if it has one subgroup of order 4 (or,equivalently, only 4 solutions to x4 = 1).

We mention Theorem 3.4 since it is nice to know, but we omit a proof.The fact that a unique subgroup of order 2 is insufficient to characterizecyclic groups of 2-power size above 4 is connected to the quaternion group:Q8 is nonabelian of size 8 but has just one subgroup of order 2: {±1}. Moregenerally, for any 2-power 2k ≥ 8 there is a generalized quaternion group oforder 2k, denoted Q2k , which is nonabelian with one subgroup of order 2.

Using Theorem 3.4, we get a stronger form of Theorems 3.1 and 3.3, usingsubgroups of order p (or order 4 if p = 2) and solutions to xp = 1 (or x4 = 1if p = 2) instead of subgroups of all sizes or equations of all degrees.

4. Commutativity properties based on #G

The class equation shows all groups of order p2 are abelian. (See thehandout on conjugation in a group.) Cauchy’s theorem can be used to showall groups of order pq with primes p < q and q 6≡ 1 mod p (e.g., pq = 15)


are abelian (and in fact cyclic). The Sylow theorems provide further toolsto show all groups of a given size are abelian.

Theorem 4.1. Any group of size 45 is abelian.

Proof. Let G have size 45. In G, a 3-Sylow subgroup has size 9 and a 5-Sylowsubgroup has size 5. Using Sylow III,

n3|5, n3 ≡ 1 mod 3 =⇒ n3 = 1

andn5|9, n5 ≡ 1 mod 5 =⇒ n5 = 1.

Therefore G has normal 3-Sylow and 5-Sylow subgroups. Denote them byP and Q respectively, so #P = 9 and #Q = 5. Then P is abelian and Q iscyclic (thus abelian).

The set PQ = {ab : a ∈ P, b ∈ Q} is a subgroup of G since P and Qare normal subgroups (we really only need one of them to be normal forPQ to be a subgroup). Since PQ contains P and Q as subgroups, Lagrangetells us #PQ is divisible by both 9 and 5. Therefore 45|#PQ, so PQ = G.Since P and Q are both abelian, we will know G is abelian once we showeach element of P commutes with each element of Q. This commutativityis Corollary 1.6. �

Remark 4.2. The reader can check that the same argument shows anygroup of size p2q with primes p < q and q 6≡ 1 mod p is abelian. Examplesinclude 99 = 22 · 11 and 175 = 52 · 7.

Theorem 4.3. Let p and q be primes where p < q and q 6≡ 1 mod p. Thenany group of size pq is cyclic.

This was already proved in the handout on consequences of Cauchy’stheorem. The proof we give here is in the same spirit, but it uses the Sylowtheorems to handle more efficiently certain parts of the proof. Compare thetwo proofs.

Proof. Let #G = pq, where p < q and q 6≡ 1 mod p. By Cauchy’s theorem,G has an element a of order p and an element b of order q. Let P = 〈a〉 andQ = 〈b〉.

These subgroups have size p and q and are, respectively, p-Sylow and q-Sylow subgroups of G. Using the Sylow theorems, we will show P and Qare both normal subgroups of G. It then will follow from Corollary 1.6 thatelements of P commute with elements of Q. Then, since a and b commuteand they have relatively prime order, their product ab has order pq. As#G = pq, G is cyclic.

By Sylow III, np|q and np ≡ 1 mod p. The only choices are np = 1 or q.Since q 6≡ 1 mod p (by hypothesis) we must have np = 1, so P is the onlyp-Sylow subgroup and is thus normal in G.

By Sylow III, nq|p and nq ≡ 1 mod q. The only choices are nq = 1 or p.Since 1 < p < q, the congruence condition on nq implies nq = 1. Therefore

8 KEITH CONRAD

Q is the only q-Sylow subgroup and is thus normal in G. (Compare this tothe proof in the handout on consequences of Cauchy’s theorem that G hasonly one subgroup of size q.) �

Remark 4.4. The preceding theorem gives a condition on a positive integerm such that all groups of size m are cyclic. One can classify all m for whichall groups of size m are cyclic or all groups of size m are abelian. Allgroups of size m are cyclic if and only if m is squarefree and, for all pairs ofdistinct primes p and q dividing m, we have q 6≡ 1 mod p. (This conditionis equivalent to (m,ϕ(m)) = 1.) All groups of size m are abelian if and onlyif m is cubefree and, for all pairs of distinct primes p and q dividing m, wehave q 6≡ 1 mod p and, if q2|m too, then q 6≡ −1 mod p.

5. Normalizers of Sylow subgroups

The normalizers of Sylow subgroups naturally occur in Sylow III: thenumber of p-Sylow subgroups is the index of the normalizer of a p-Sylowsubgroup. We record here some additional properties of Sylow normalizers.

Recall we write Sylp(G) for the set of p-Sylow subgroups of G.

Theorem 5.1. Let P ∈ Sylp(G). Then P is the unique p-Sylow subgroupof N(P ) and N(P ) is the largest subgroup of G with this property.

Proof. Since P ∈ Sylp(N(P )), Sylow II for N(P ) says any p-Sylow subgroupof N(P ) is gPg−1 for some g ∈ N(P ). By the definition of N(P ), gPg−1 = P .So P is the unique p-Sylow subgroup of N(P ). (This kind of argument wasused in the proof of Sylow III to show np ≡ 1 mod p.)

Now suppose P ⊂ H ⊂ G and the only p-Sylow subgroup of H is P . Forh ∈ H, hPh−1 is a p-Sylow subgroup of H, so hPh−1 = P . Thus h ∈ N(P ),so H ⊂ N(P ). �

Theorem 5.2. Let P ∈ Sylp(G). If N(P ) ⊂ H ⊂ G, then [G : H] ≡ 1 mod pand N(H) = H. In particular, N(N(P )) = N(P ).

Proof. Since N(P ) ⊂ H, the normalizer of P in H is also N(P ) (symbolically,NH(P ) = NG(P )). By Sylow III and III* applied to H and to G, we have

np(G) = [G : N(P )] ≡ 1 mod p, np(H) = [H : N(P )] ≡ 1 mod p.

Thus their ratio, which is [G : H], is ≡ 1 mod p.To show N(H) = H, we just show N(H) ⊂ H; the other inclusion is

trivial. Pick g ∈ N(H). Since P ⊂ H, gPg−1 ⊂ gHg−1 = H. Both P andgPg−1 are p-Sylow subgroups of H, so by Sylow II there is an h ∈ H suchthat gPg−1 = hPh−1. Thus h−1gPg−1h = P , so h−1g ∈ N(P ) ⊂ H, sog ∈ H. �

Theorem 5.3. If N CG and P ∈ Sylp(N) then G = N ·N(P ). In particular,if P C N then P C G.

Here P is Sylow in N , not necessarily in G, while N(P ) is the normalizerof P in G, not N .


Proof. Pick g ∈ G. Since P ⊂ N and N C G, gPg−1 ⊂ N . Then by SylowII for N there is an n ∈ N such that gPg−1 = nPn−1, so n−1gPg−1n = P .That means n−1g ∈ N(P ), so g ∈ n N(P ). Thus G = N ·N(P ).

If P C N , then N ⊂ N(P ), so N · N(P ) = N(P ). Thus G = N(P ), soP C G. �

Normality is not usually transitive: if N1 C N2 and N2 C G, it need notfollow that N1 C G. (This illustrated by 〈s〉 C 〈r2, s〉 C D4.) Theorem 5.3gives a setting where something like this is true: a normal Sylow subgroupof a normal subgroup is a normal subgroup.

Example 5.4. Let G = GL2(Z/3Z), N = SL2(Z/3Z), and P be the unique2-Sylow subgroup of N . Since P is normal in N , it is normal in G too. ThusP lies in every 2-Sylow subgroup of G (but P is not a 2-Sylow subgroup ofG).

We will apply Theorem 5.3 several times in the next section.

6. Non-trivial normal subgroups

The consequences of the Sylow theorems in this section are cases wherethe size of G forces G to have a non-trivial normal subgroup (usually, butnot always, a normal Sylow subgroup). This topic is a popular source ofexercises in algebra textbooks, in part because it can be used to determineall groups of various sizes up to isomorphism. For instance, at the end ofthis section we will see there are just two groups of size 105.

Theorem 6.1. If #G = 20 or 100 then G has a normal 5-Sylow subgroup.

Proof. By Sylow III, n5|4 and n5 ≡ 1 mod 5. Thus n5 = 1. �

This proof is identical to part of the proof of Theorem 2.2, which wasconcerned with a specific group of size 20.

Theorem 6.2. If #G = pq, where p < q are distinct primes, then G has anormal q-Sylow subgroup.

Proof. Read the proof of Theorem 4.3, where it is shown that nq = 1. Thispart of the proof did not use the congruence condition q 6≡ 1 mod p fromthat theorem, so nq = 1 whether or not that congruence condition holds. �

The following lemma does not involve the Sylow theorems, but will beused in conjunction with the Sylow theorems to prove more theorems aboutthe existence of normal subgroups.

Lemma 6.3. If G has k subgroups of size p, it has k(p − 1) elements oforder p.

Proof. In a subgroup of size p, all nonidentity elements have order p. Con-versely, any element of order p generates a subgroup of size p. By Lagrange,distinct subgroups of size p must intersect trivially, so their nonidentity el-ements are disjoint from each other. Therefore each subgroup of size p has

10 KEITH CONRAD

its own p− 1 elements of order p, not shared by any other subgroup of sizep. The number of elements of order p is therefore k(p− 1). �

Theorem 6.4. If #G = 12, then G has a normal 2-Sylow or 3-Sylow sub-group.

Proof. By Sylow III, n2|3, so n2 = 1 or 3. Also n3|4 and n3 ≡ 1 mod 3, son3 = 1 or 4. We want to show n2 = 1 or n3 = 1.

Assume n3 6= 1, so n3 = 4. Since the 3-Sylows have size 3, Lemma 6.3says G has n3 ·2 = 8 elements of order 3. The number of remaining elementsis 12−8 = 4. A 2-Sylow subgroup has size 4, and thus fills up the remainingelements. Therefore n2 = 1.

Since n3 6= 1 implies n2 = 1, either n2 or n3 = 1. �

In Theorem 6.4, it need not happen that both n2 and n3 equal 1 when#G = 12. For example, A4 has n2 = 1 and n3 = 4, while D6 has n2 = 3and n3 = 1.

Theorem 6.5. If #G = 24 then G has a normal subgroup of size 4 or 8.

Proof. Let P be a 2-Sylow subgroup, so #P = 8. Consider the left mul-tiplication action ` : G → Sym(G/P ) ∼= S3. Set K to be the kernel of `.Then

• K ⊂ P , so #K|8,• G/K embeds into S3, so [G : K]|6. That is, 4|#K.

This tells us #K = 4 or 8. Since K is the kernel of `, K C G. �

Example 6.6. Let G = S4. The number of 2-Sylow subgroups is 3, so S4

does not have a normal subgroup of size 8. Theorem 6.5 then says S4 musthave a normal subgroup of size 4. Indeed, one is

{(1), (12)(34), (13)(24), (14)(23)}.

Example 6.7. Let G = SL2(Z/3Z). This group has size 24 and a normal2-Sylow subgroup. (Since we can write down more than one subgroup of size4, such as 〈( 1 1

1 −1 )〉 and 〈( 0 −11 0 )〉, there must be just one 2-Sylow subgroup.

Theorem 6.8. If #G = 30 then G has normal 3-Sylow and 5-Sylow sub-groups.

Proof. Pick g ∈ G of order 2. Left multiplication `g : G → G is a productof 15 transpositions, so its sign is −1. Therefore the composite sgn ◦` : G →{±1} is onto, so the kernel is a (normal) subgroup of G with size 15. Call itN . Then N is cyclic (Theorem 4.3). Its 3-Sylow and 5-Sylow subgroups arenormal in N (since N is abelian), so they are also normal in G by Theorem5.3. �

Remark 6.9. Up to isomorphism there are four groups of order 30: Z/30Z,D15, Z/3Z×D5, and Z/5Z×D6.


Theorem 6.10. Every group of size 105 has normal 5-Sylow and 7-Sylowsubgroups. In other words, every group of size 105 has unique subgroups ofsize 5 and 7.

Proof. We will first prove n5 = 1 or n7 = 1. Then we will refine this ton5 = 1 and n7 = 1.

By Sylow III,

n3|35, n3 ≡ 1 mod 3 =⇒ n3 = 1 or 7,

n5|21, n5 ≡ 1 mod 5 =⇒ n5 = 1 or 21,

n7|15, n7 ≡ 1 mod 7 =⇒ n7 = 1 or 15.

Could n5 > 1 and n7 > 1? If so, then n5 = 21 and n7 = 15. Using Lemma6.3, the number of elements with order 5 is 21 · 4 = 84 and the numberof elements with order 7 is 15 · 6 = 90. Since 84 + 90 > #G, we have acontradiction, so n5 = 1 or n7 = 1. In either case we will show G has asubgroup with order 35.

Suppose n5 = 1 and let N5 be the (normal) 5-Sylow subgroup of G. ThenN5 C G, so G/N5 is a group of size 21. The pullback (i.e., inverse image)of the 7-Sylow of G/N5 under the natural homomorphism G → G/N5 is asubgroup of G with size 7 · 5 = 35.

Now suppose n7 = 1 and let N7 be the (normal) 7-Sylow subgroup ofG. The group G/N7 has size 15. Under the natural map G → G/N7, thepullback of a 5-Sylow of G/N7 is a subgroup of G with order 5 · 7 = 35.

We have proved, whether n5 = 1 or n7 = 1, that G has a subgroup of order35. Such a subgroup has index 105/35 = 3 in G. This is the smallest primefactor of G, so the subgroup is normal in G. (See the handout on groupactions.) Denote it as N . Any group of size 35 is cyclic, by Theorem 4.3, soN is cyclic. In particular, any Sylow subgroup of N is a normal subgroup ofN , so Theorem 5.3 tells us any Sylow subgroup of N is also normal subgroupof G. A 5-Sylow or 7-Sylow of N is also a 5-Sylow or 7-Sylow of G, andtherefore their normality in G implies n5 = 1 and n7 = 1. �

Theorem 6.11. Any group of size 105 is isomorphic to Z/5Z ×H, where#H = 21.

Proof. Let #G = 105. By Theorem 6.10 we have n5 = n7 = 1. Let N5 andN7 denote the 5-Sylow and 7-Sylow subgroups of G. Let P be a 3-Sylowsubgroup of G. Then the product set H = PN7 is a subgroup of G with size21. Since N5 C G, HN5 is a subgroup of G with size 105, so G = HN5. Itremains to show G ∼= H×N5, that is, elements of H commute with elementsof N5.

Consider the conjugation action of H on N5 (this makes sense since N5 CG). It gives a homomorphism H → Aut(N5) ∼= (Z/5Z)×. Since H hassize 21, this homomorphism is trivial, elements of H fix elements of N5 byconjugation. Thus elements of H commute with elements of N5. �

Corollary 6.12. Up to isomorphism there are two groups of size 105.

12 KEITH CONRAD

Proof. From the handout on applications of Cauchy’s theorem, there aretwo groups of size 21 up to isomorphism. One is abelian (namely the cyclicgroup) and one is non-abelian. Using these for H gives two groups of size105, one being abelian (in fact cyclic) and the other being non-abelian. �

Theorem 6.13. If #G = p2q2, where p < q are distinct primes, then Ghas a normal q-Sylow subgroup unless #G = 36, in which case G has eithera normal 2-Sylow or 3-Sylow subgroup.

That groups of size 36 are different can be seen with Z/3Z × A4, whichhas nq = n3 = 4.

Proof. Without loss of generality, p < q. By the Sylow theorems, nq|p2 andnq ≡ 1 mod q, so nq = 1 or p2. If nq = 1 then the q-Sylow subgroup isnormal. Now suppose nq = p2. Then p2 ≡ 1 mod q, so p ≡ ±1 mod q. Sincep 6≡ 1 mod q we have p ≡ −1 mod q. Thus q|(p + 1). Since q > p we musthave q = p + 1, so p and q are consecutive primes. Therefore p = 2 andq = 3, which shows for #G 6= 36 that nq = 1. (The reader who doesn’t caretoo much about this theorem can skip the rest of the proof, which analyzesthe remaining case #G = 36.)

For the rest of the proof, let #G = 36. We will show that if n3 = 4 thenn2 = 1. Assume n3 = 4 and n2 > 1. Since n2 > 1, G has no subgroupof size 18 (it would automatically be normal so n2 = 1). Since n2 > 1, Gis non-abelian. Our goal is to get a contradiction. We will try to countelements of different orders in G and find the total comes out to more than36 elements. That will be our contradiction.

Let Q be a 3-Sylow in G, so [G : Q] = 4. The left multiplication actionof G on G/Q is a homomorphism G → Sym(G/Q) ∼= S4. Since #G > #S4,there is a nontrivial kernel. Call it K. Since K ⊂ Q, either #K = 3 orK = Q. Since Q is not normal in G it does not equal K, so #K = 3.

Since K C G, we can make G act on K by conjugations. This is a ho-momorphism G → Aut(K) ∼= Z/2Z. If this homomorphism is onto (that is,some element of G conjugates on K in a nontrivial way then the kernel is asubgroup of G with size 18, which G does not have. So the homomorphismis trivial, which means every element of G commutes with the elements ofK, so K ⊂ Z(G). Then 3|#Z(G), so size of Z(G) is one of the numbers in{3, 6, 9, 12, 18, 36}. Since G is non-abelian and a group is abelian when thequotient by its center is cyclic, #Z(G) can’t be 12, 18, or 36. Since n3 > 1there is no normal subgroup of size 9, so #Z(G) 6= 9. If #Z(G) = 6 thenthe product set Z(G)Q is a subgroup of size 18, a contradiction. So we musthave #Z(G) = 3, which means Z(G) = K.

Now we start counting elements with various orders. The center is a3-subgroup of G, so by the conjugacy of 3-Sylow subgroups every 3-Sylowsubgroup contains K. Any two different 3-Sylow subgroups have K as theirintersection, so we can count the total number of elements of 3-power order:#K + n3(9− 3) = 27.


Let g ∈ G have order 2. Then the product set K〈g〉 is abelian of size 6, socyclic. It has a unique element of order 2, which must be g. Therefore wheng and g′ are different elements of order 2 in G, the groups K〈g〉 and K〈g′〉have K has their intersection. So each element of order 2 provides us with 2new elements of order 6. Let n be the number of elements of order 2 in G, sothere are at least 2n elements of order 6, giving at least 3n elements in totalwith order 2 or 6. Since we already found 27 elements with 3-power order(including the identity), 3n ≤ 36 − 27, or n ≤ 3. We can get an inequalityon n in the other direction: n ≥ 2. Indeed, no element of order 2 lies in thecenter K, so a suitable conjugate of an element of order 2 provides us witha second element of order 2. So n = 2 or 3.

Since #{g ∈ G : g2 = e} is even (by McKay’s proof of Cauchy’s theorem)and this number is 1 + n, we see n is odd, so n = 3 and there are atleast 3n = 9 elements of order 2 or 6. Adding this to 27 from before give9 + 27 = 36 so we have accounted for the whole group: each element of Ghas 3-power order or order 2 or 6. In particular, the 2-Sylow subgroup of Gmust be isomorphic to Z/2Z×Z/2Z instead of Z/4Z (no elements of order4). But then two different 2-Sylow subgroups meet at most in a group oforder 2, which gives us 5 elements of order 2 from both groups. This is ourfinal contradiction. �

7. Sylow numbers of subgroups and quotient groups

How do the numbers np(G) behave when we pass to subgroups and quo-tient groups? Intuitively one might expect the numbers drop (or stay thesame). We will show this is true.

Theorem 7.1. Let H be a subgroup of G. For any P ∈ Sylp(G), there is aconjugate gPg−1 such that gPg−1 ∩H ∈ Sylp(H).

Proof. We give two proofs. The first will use the Sylow theorems for H (andG). The second, somewhat longer, will not use the Sylow theorems for H.

For our first proof, note P ∩ H is a p-subgroup of H, so by the Sylowtheorems for H there is a p-Sylow of H, say Q, containing P ∩H. As Q liesin a p-Sylow of G, Q ⊂ gPg−1 for some g ∈ G. Therefore Q ⊂ gPg−1 ∩H.This intersection is a p-subgroup of H, so it has to equal Q.

For our second proof, we start by noting that for any g ∈ G the inter-section gPg−1 ∩ H is a p-subgroup of H. For it to be a p-Sylow subgroupof H (our goal) we need the index [H : gPg−1 ∩ H] not to be divisible byp. Consider the left multiplication action of H (not G!) on G/P . The setG/P has size [G : P ], which is not divisible by p. Therefore some left cosetgP has an H-orbit with size not divisible by p. We are going to show theH-orbit of gP has size [H : gPg−1∩H], so gPg−1∩H is a p-Sylow subgroupof H.

14 KEITH CONRAD

To count the size of the H-orbit of gP we compute the stabilizer subgroupof gP :

Stab{gP} = {h ∈ H : hgP = gP}= {h ∈ H : g−1hg ∈ P}= {h : h ∈ gPg−1}= gPg−1 ∩H.

By the orbit-stabilizer formula, the H-orbit of gP has size [H : Stab{gP}] =[H : gPg−1 ∩H]. �

Example 7.2. Let G = D6 = 〈r, s〉, of size 12. Let H = 〈r2, s〉, of size6. A 2-Sylow subgroup of H has size 2. One 2-Sylow subgroup of G isP = 〈r3, rs〉. While P ∩H is trivial, rPr−1 ∩H = {1, s} is a 2-Sylow in H.

Example 7.3. For a prime p, let G = GL2(Z/pZ) and H = {( ∗ 0∗ ∗ )}. One

p-Sylow subgroup of G is P = {( 1 ∗0 1 )}, which meets H trivially. However,

since ( 0 1−1 0 )( 1 x

0 1 )( 0 1−1 0 )−1 = ( 1 0

−x 1 ), ( 0 1−1 0 )P ( 0 1

−1 0 )−1 is a p-Sylow subgroupof H.

Remark 7.4. Since the second proof of Theorem 7.1 did not use the Sylowtheorems for H, it means that the existence of Sylow subgroups of a groupimplies the existence of Sylow subgroups of any subgroup. In particular, wecan show a finite group has a p-Sylow subgroup by embedding it in a largergroup where it might be easier to write down a p-Sylow subgroup.

For example, every finite group can be embedded in a symmetric group(Cayley’s theorem). To be precise, the left multiplication action of G on Ggives an embedding of G into Sym(G) ∼= Sn, where n = #G. Therefore ifone can construct, for any prime p, a p-Sylow subgroup of each symmetricgroup we obtain the existence of a p-Sylow subgroup of every finite group.Exercises 15 and 16 in [3, p. 84]. give a construction of Sylow subgroupsof symmetric groups using wreath products. A construction of a p-Sylowsubgroup of the symmetric groups Spk actually suffices, since Sn naturallyembeds into Spk for pk ≥ n; the construction in this special case can befound in the discussion of the Sylow theorems in [2, pp. 95–97].

Corollary 7.5. Let N CG. For any p-Sylow P of G, P ∩N is a p-Sylow ofN and all p-Sylows of N arise in this way. In particular, np(N) ≤ np(G).

Proof. By Theorem 7.1, there is a g such that gPg−1∩N is a p-Sylow in N .Since N is a normal subgroup of G,

gPg−1 ∩N = gPg−1 ∩ gNg−1 = g(P ∩N)g−1.

Therefore P ∩N is a p-Sylow subgroup of g−1Ng = N .Let Q be a p-Sylow subgroup of N . Pick a p-Sylow of G, say P , which

contains Q. Then Q ⊂ P∩N , and P∩N is a p-Sylow of N , so Q = P∩N . �


There is an extension of Corollary 7.5 to subgroups which are not nec-essarily normal. Suppose H C K C G (perhaps H is not normal in G). IfP is a Sylow subgroup of G, then P ∩ K is a Sylow subgroup of K, so(P ∩K) ∩H = P ∩H is a Sylow subgroup of H. This can be extended toany “subnormal” subgroup (a subgroup which is at the bottom of a chainof subgroups increasing up to G with each one normal in the next): theintersection of a Sylow subgroup of G with a subnormal subgroup of G is aSylow subgroup of the subnormal subgroup.

We now show the inequality in Corollary 7.5 is true for non-normal sub-groups too.

Theorem 7.6. Let G be a finite group and H a subgroup. Choose a primep. Distinct p-Sylow subgroups of H do not lie in a common p-Sylow subgroupof G. In particular, np(H) ≤ np(G).

Proof. Let Q and Q′ be distinct p-Sylow subgroups of H. If they lie in acommon p-Sylow subgroup of G, then the group 〈Q,Q′〉 is a p-subgroup ofH. However its size is too large, since it properly contains Q.

If we associate to each p-Sylow subgroup of H a p-Sylow subgroup ofG it lies inside of (there is no canonical way to do this if we have choicesavailable), then this correspondence from Sylp(H) to Sylp(G) is one-to-one,so np(H) ≤ np(G). �

Theorem 7.7. Let N CG. For any p-Sylow P of G, PN/N is a p-Sylow ofG/N and all p-Sylows of G/N arise in this way. In particular, np(G/N) ≤np(G).

Proof. First, we will show for every p-Sylow P of G that PN/N is a p-Sylowof G/N . The group PN/N is a p-group (either because every element hasp-power order or because PN/N ∼= P/(P ∩N)). Using the inclusions

G ⊃ PN ⊃ N, G ⊃ PN ⊃ P,

the first one shows [G/N : PN/N ] = [G : PN ] and the second one shows[G : PN ] 6≡ 0 mod p. Therefore PN/N is a p-Sylow of G/N .

Now we show every p-Sylow of G/N has the form PN/N for some p-SylowP of G. Let Q ∈ Sylp(G/N) and write Q = H/N for some subgroup H ⊂ Gcontaining N . Then [G : H] = [G/N : Q] 6≡ 0 mod p. Choose P ∈ Sylp(H),so P ∈ Sylp(G) too by the previous congruence. Then PN/N is a subgroupof Q. It is also a p-Sylow subgroup of G/N by the previous paragraph, soQ = PN/N . �

Corollary 7.5 and Theorem 7.7 tell us the maps Sylp(G) → Sylp(N) andSylp(G) → Sylp(G/N) given by P 7→ P ∩ N and P 7→ PN/N are surjec-tive. By comparison, although # Sylp(G) ≥ # Sylp(H) for a subgroup H,there are no natural maps between Sylp(G) and Sylp(H). (The functionSylp(H) → Sylp(G) in the proof of Theorem 7.6 is not natural in any way.)

Theorem 7.7 gives another proof of Corollary 7.5: since PN/N is a p-Sylow subgroup of G/N , its size [PN : N ] = [P : P ∩ N ] is the highest

16 KEITH CONRAD

power of p in [G : N ]. Since #P is the highest power of p in #G, weconclude that #(P ∩ N) is the highest power of p in #N , so P ∩ N is ap-Sylow subgroup of N .

The proof of the next theorem is a nice application of the preservationof the Sylow property when intersecting with a normal subgroup (Corollary7.5).

Theorem 7.8. Write #G = 2nm, where m is not divisible by 2. If G hasa cyclic 2-Sylow subgroup then G has a normal subgroup of size m.

Proof. Every 2-Sylow subgroup of G is cyclic (they are isomorphic to eachother), so every 2-subgroup of G is cyclic since they all lie in a 2-Sylowsubgroup.

We prove our theorem by induction on n.First we show G has a subgroup of index 2 (necessarily normal). Consider

the action of G on G by left multiplication. This is a homomorphism ` : G →Sym(G). There is an element of order 2n in G. Left multiplication bythis on G is a permutation consisting of m disjoint 2n-cycles. Since m isodd, this permutation is odd. Thus the composite map sgn ◦` → {±1} isonto. Its kernel is a normal subgroup of index 2. Let N be this kernel, so#N = 2n−1m.

If n = 1 we are done; N is a normal subgroup of G with size m. (This partrepeats the case when 2m = 30 from the proof of Theorem 6.8.) Supposen ≥ 2. The intersection H ∩N is a 2-Sylow subgroup of N (Corollary 7.5)so it must be cyclic. Therefore, by induction, N has a normal subgroupM C N of size m. We will show M C G, which will end the proof. NoteM is the unique subgroup of N with order m: if M ′ is another one, then(since M C N) the set MM ′ ⊂ N is a subgroup with odd order (its sizedivides #M#M ′) but its order exceeds m. That’s too big. For any g ∈ G,gMg−1 ⊂ gNg−1 = N , so gMg−1 = M by the uniqueness of M as asubgroup of N with its size. �

Remark 7.9. Theorem 7.8 is also true if #G is odd and 2 is replaced bythe smallest prime factor of #G. The proof in this case is different. See [5,p. 138].

Corollary 7.10. If #G = 2m where m is odd then G contains a normalsubgroup of size m and all elements of order 2 in G are conjugate to eachother.

Proof. A 2-Sylow subgroup of G has size 2, which must be cyclic. Therefore,by Theorem 7.8 (the base case n = 1), there is a normal subgroup of size m.

Since the 2-Sylow subgroups of G have size 2, any two elements of order2 generate conjugate subgroups by Sylow II, and therefore the elementsthemselves are conjugate. �

Example 7.11. A group of size 70 has a normal subgroup of size 35.


Example 7.12. For odd m, all reflections in Dm are conjugate and thereis a normal subgroup of size m. Of course this is something we alreadyknow by explicit calculation in dihedral groups, but Corollary 7.10 puts thissituation into a larger context.

Appendix A. Simple groups of order 60

We call a group simple when it is nontrivial and its only normal subgroupsare the trivial subgroup and the whole group. For example, a group of primesize is simple for the crude reason that it has no subgroups at all besidesthe whole group and the trivial subgroup. An abelian group of non-primesize is not simple, since it always has a proper non-trivial subgroup, whichis necessarily normal. Thus any simple group other than a group of primesize is nonabelian.

Simple groups can be characterized in terms of group homomorphisms,as follows.

Theorem A.1. A nontrivial group G is simple if and only if any non-trivialgroup homomorphism out of G is an embedding.

Proof. Suppose G is simple. Let f : G → H be a homomorphism, withf(g) 6= e for some g. Then the kernel of f is a proper normal subgroup of G.Since G is simple, its only proper normal subgroup is trivial, so the kernelof f is trivial, which means f is an embedding. Conversely, suppose all non-trivial homomorphisms out of G are embeddings. If N CG and N 6= G, thenthe reduction map G → G/N is a homomorphism with kernel N . The imageis not just the identity, so by hypothesis this is an embedding. Thereforethe kernel N is trivial, so G is simple. �

Theorem A.2. The group A5 is simple.

Proof. We want to show the only normal subgroups of A5 are (1) and A5.There are 5 conjugacy classes in A5, with representatives and sizes as

indicated in the following table.

Rep. (1) (12345) (21345) (12)(34) (123)Size 1 12 12 15 20

If A5 has a normal subgroup N , then N is a union of conjugacy classes –including (1) – whose total size divides 60. However, no sum of the abovenumbers which includes 1 is a factor of 60 except for 1 and 60. ThereforeN is trivial or A5. �

The proof of Theorem A.2 required knowledge of the conjugacy classes inA5. We now prove A5 is simple using much less information: its size andthat it has more than one 5-Sylow subgroup. (cf. Theorem 2.1). The resultwill apply to any group with the same two properties. Our discussion isbased on [1, pp. 145–146].

Theorem A.3. If #G = 60 and n5 > 1, then G is a simple group.

18 KEITH CONRAD

Proof. Assume G is not simple, so there is N C G with 1 < #N < 60. Thatmeans

#N ∈ {2, 3, 4, 5, 6, 10, 12, 15, 20, 30}.

We will get a contradiction. Our argument will use many of the previousconsequences we drew from the Sylow theorems (to groups of size 12, 15,20, and 30).

First we show #N is not divisible by 5. Assume 5|#N , so N contains a5-Sylow subgroup, which is also a 5-Sylow subgroup of G since 60 = 5 · 12.Because N CG, Sylow II shows all the 5-Sylow subgroups of G lie in N . Letn5 be the number of 5-Sylows in G (which we know are all subgroups of N).Since n5|12 and n5 ≡ 1 mod 5, n5 = 1 or 6. Because n5 > 1 by hypothesis,n5 = 6. Therefore N contains six different subgroups of size 5. Countingelements of N with orders 1 or 5, Lemma 6.3 says

#N ≥ n5 · 4 + 1 = 25.

Since #N is a proper factor of 60, #N = 30. But then, by Theorem 6.8, Nhas only one 5-Sylow subgroup. This is a contradiction of n5 = 6, so #N isnot divisible by 5. This means

#N ∈ {2, 3, 4, 6, 12}.

If #N equals 6, then Theorem 6.2 says N contains a normal 3-Sylowsubgroup. If #N equals 12, then Theorem 6.4 says N contains a normal2-Sylow or 3-Sylow subgroup. A normal Sylow subgroup of N is a normalsubgroup of G by Theorem 5.3. Because such a normal subgroup of G hassize 3 or 4, which is one of the possibilities already under consideration for#N , we are reduced to eliminating the possibility that #N = 2, 3, or 4.

If #N equals 2, 3, or 4, let G = G/N , so G is a group with size 30, 20,or 15. By Theorem 4.3, a group of size 15 is cyclic and thus has a normal5-Sylow subgroup. By Theorem 6.1, a group of size 20 has a normal 5-Sylow subgroup. By Theorem 6.8, a group of size 30 has a normal 5-Sylowsubgroup. Therefore in all cases G contains a normal 5-Sylow subgroup, sayP , with #P = 5.

Consider the projection π : G → G. Set H = π−1(P ). Since P CG, HCG.Since H 6= G, H is a proper normal subgroup of G. Since π sends H ontoP , #H is divisible by 5. But we showed earlier that G contains no propernormal subgroups of size divisible by 5.

Since all choices for #N have been eliminated, there is no such N . ThusG is simple. �

The next result shows that A5 is the only simple group of size 60, up toisomorphism. (In total, there are 13 groups of size 60 up to isomorphism.)The proof will use Sylow III*.

Theorem A.4. Every simple group of size 60 is isomorphic to A5.


Proof. Let G be a simple group of size 60. To prove G is isomorphic to A5,we will make G act on a set of 5 objects and then show this action is givenby the even permutations of the 5 objects.

We seek an action on cosets. Suppose G has a subgroup H with [G : H] =5 (i.e., #H = 12), so the left multiplication action of G on the coset spaceG/H gives a homomorphism

ϕ : G → Sym(G/H) ∼= S5.

The kernel of ϕ is a normal subgroup of G, and therefore is trivial or is Gsince G is simple. If g ∈ G is in the kernel of ϕ, then gH = H, so g ∈ H. Inparticular, the kernel of ϕ is a subgroup of H and therefore the kernel can’tbe G. Thus the kernel of ϕ is trivial, so ϕ is an embedding of G into S5;G is isomorphic to its image ϕ(G). In particular, ϕ(G) is a simple group ofsize 60. Let’s prove this image is A5.

If ϕ(G) 6⊂ A5, then ϕ(G) contains an odd permutation. That means thesign homomorphism

sgn: ϕ(G) → {±1}is surjective, so its kernel is a normal subgroup of ϕ(G) with index 2. How-ever, ϕ(G) doesn’t have such normal subgroups since it is simple. (Remem-ber, ϕ gives an isomorphism of G with ϕ(G).) We conclude that all elementsof ϕ(G) have sign 1, so ϕ(G) ⊂ A5. Both groups have size 60, so ϕ(G) = A5.

We have shown that if G has a subgroup H with index 5 then the leftmultiplication action of G on the coset space G/H gives an isomorphism ofG with A5. The rest of the proof is devoted to showing G has a subgroupwith index 5.

Step 1: For any proper subgroup H ⊂ G, [G : H] ≥ 5. Thus #H ≤ 12.

Let t = [G : H]. The left multiplication action of G on G/H gives ahomomorphism G → Sym(G/H) ∼= St. Since H is a proper subgroup andG is simple, this homomorphism has trivial kernel. (The reason follows asbefore, when we were only concerned with index 5 subgroups: the kernelis a subgroup of H and therefore is a proper normal subgroup of G, whichmust be trivial since G is simple.) Therefore we have an embedding of Ginto St, so 60|t!. This can happen only when t ≥ 5.

Step 2: G has a subgroup with index 5.

We use Sylow III for the primes 2, 3, and 5. They tell us that

n2 ∈ {1, 3, 5, 15}, n3 ∈ {1, 4, 10}, n5 ∈ {1, 6}.Since G is simple, the non-trivial Sylow subgroups are not normal, so n2, n3,and n5 all exceed 1. Moreover, because Sylow III* says each np is the indexof a subgroup of G, Step 1 tells us n2, n3, n5 ≥ 5. Therefore

n2 ∈ {5, 15}, n3 = 10, n5 = 6.

If n2 = 5 then Sylow III* says there is a subgroup of G with index 5 andwe’re done. What should we do now: show the only other possibility, that

20 KEITH CONRAD

n2 = 15, leads to a contradiction? Instead we will show that if n2 = 15 thenthere is a second way to show G has a subgroup with index 5.

Assume n2 = 15. By Lemma 6.3, G has n3 ·2 = 20 elements of order 3 andn5 · 4 = 24 elements of order 5. This is a total of 44 elements, which leavesat most 60 − 44 = 16 elements that can lie in the 2-Sylow subgroups of G.Each 2-Sylow subgroup of G has size 4 (and thus is abelian), so if n2 = 15then we have 15 different subgroups of size 4 squeezed into a 16-elementsubset of G. These 2-Sylow subgroups can’t all pairwise intersect trivially(otherwise there would be 3 · 15 = 45 non-identity elements among them).Pick two different 2-Sylows, say P and Q, which intersect nontrivially. LetI = P ∩ Q. Both P and Q are abelian (they have size 4), so I is normalin each. Therefore the normalizer of I in G contains both P and Q, so ithas size properly divisible by 4. The normalizer of I is not all of G sinceG has no proper nontrivial normal subgroups. Since proper subgroups ofG have size 1, 2, 3, 4, 6, or 12, the normalizer of I has size 12 and thus[G : I] = 5. �

Since n2(A5) = 5, we know after the proof that the assumption n2 = 15in the last paragraph does not actually occur.

References

[1] D. Dummit and R. Foote, “Abstract Algebra,” 3rd ed., Wiley, 2004.[2] I. Herstein, “Topics in Algebra,” 2nd ed., Wiley, 1975.[3] N. Jacobson, “Basic Algebra I,” 2nd ed., W. H. Freeman & Co., New York, 1985[4] C. Leedham-Green, “The Structure of Groups of Prime Power Order,” Oxford Univ.

Press, Oxford, 2002.[5] W. R. Scott, “Group Theory,” Dover, New York, 1987.

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Sylow applications