Math 594. Some consequences of the Sylow theorems

The Sylow theorems concern subgroups with prime power size, especially with maximal prime power size.

When #G = p^{k}m and p^{k} is as large as possible (that is, p doesn’t divide m), we call any subgroup of G with size p^{k} a p-Sylow subgroup. (When p doesn’t divide #G, p^{k} = 1 and the trivial subgroup is the p-Sylow subgroup.) A subgroup that is a p-Sylow subgroup for some p is called a Sylow subgroup. For example, the Sylow subgroups of S_{5} (whose size is 120 = 2^{3} · 3 · 5) are the subgroups with size 1, 8, 3, or 5.

Theorem 0.1 (Sylow I). If p^{d}|#G, then G has a subgroup of size p^{d}. In particular, G has a p-Sylow subgroup for every prime p.

Theorem 0.2 (Sylow II). For each prime p, any p-subgroup of G lies in a p-Sylow subgroup, and the p-Sylow subgroups of G are conjugate.

Theorem 0.3 (Sylow III). Let n_{p} be the number of p-Sylow subgroups of G. Write #G = p^{k}m, where p doesn’t divide m. Then

n_{p}|m, n_{p} ≡ 1 mod p.

Theorem 0.4 (Sylow III*). Let n_{p} be the number of p-Sylow subgroups of G. Then n_{p} = [G : H] for some subgroup H.

The information in the Sylow theorems leads to many structure theorems about finite groups, some of which we treat below. Because of the Sylow theorems, groups of prime power size take on a heightened significance, and these theorems might motivate a mathematician to undertake a special study of groups with prime power size.

For each prime p, the p-Sylow subgroups in G are conjugate by Sylow II, and therefore n_{p} = 1 precisely when a p-Sylow subgroup is a normal subgroup. Be sure you understand that reasoning. We will constantly shift back and forth between the condition n_{p} = 1 (if it holds) and the condition that G has a normal p-Sylow subgroup. In particular, the Sylow theorems are a tool for proving a group has a nontrivial normal subgroup because we can try to show n_{p} = 1 for some prime p dividing #G.

The last theorem above, which we call Sylow III*, is not traditionally stated as part of the Sylow theorems, although it does follow from the proof of Sylow III. Sylow III* gives a condition on the number n_{p} which is somewhat less useful in basic applications than the conditions on n_{p} in Sylow III. We will not have a use for Sylow III* until the final application of this handout where we use it to give a proof of simplicity of A_{5} which avoids mucking around with 3-cycles and instead is of more “structural” type.

We first apply the Sylow theorems to some concrete groups.

Corollary 0.5. The group A_{5} has 6 subgroups of size 5.

Proof. Since #A_{5} = 60 = 2^{2} ·3·5, the 5-Sylow subgroups have size 5. By Sylow III, n_{5}|12 and n_{5} ≡ 1 mod 5. The only choices are n_{5} = 1 or 6. Since A_{5} has at least two subgroups of size 5 (the subgroups generated by (12345) and by (21345) are different), n_{5} > 1 and therefore n_{5} = 6. ■

The next few corollaries show that knowledge of the size of G can sometimes force G to have a unique p-Sylow subgroup for some p, which is thus a normal subgroup.

We start with a lemma which, while not a consequence of the Sylow theorems, will be used in conjunction with the Sylow theorems.

Lemma 0.6. If G has k subgroups of size p, it has k(p − 1) elements of order p.

Proof. In a subgroup of size p, all nonidentity elements have order p. Conversely, any element of order p generates a subgroup of size p. By Lagrange, distinct subgroups of size p must intersect trivially, so their nonidentity elements are disjoint from each other. Therefore each subgroup of size p has its own p − 1 elements of order p, not shared by any other subgroup of size p. The number of elements of order p is therefore k(p − 1). ■

Corollary 0.7. If #G = 6, then G has a normal 3-Sylow subgroup.

Proof. Since we already know groups of size 6 are either cyclic of size 6 (if abelian) or isomorphic to S_{3} (if nonabelian), we could prove this by checking the conclusion holds on such explicitly known groups.

Instead, using Sylow III, we have n_{3}|2 and n_{3} ≡ 1 mod 3. Thus n_{3} = 1, so the 3-Sylow subgroup is normal. ■

Corollary 0.8. If #G = 12, then G has a normal 2-Sylow or 3-Sylow subgroup.

Proof. By Sylow III, n_{2}|3 and n_{2} ≡ 1 mod 2, so n_{2} = 1 or 3. Also n_{3}|4 and n_{3} ≡ 1 mod 3, so n_{3} = 1 or 4.

We want to show n_{2} = 1 or n_{3} = 1.

Assume n_{3} ≠ 1, so n_{3} = 4. Since the 3-Sylows have size 3, Lemma 0.6 says G has n_{3} · 2 = 8 elements of order 3. The number of remaining elements is 12 − 8 = 4. A 2-Sylow subgroup has size 4, and is a subset of this complementary 4 element set, and thus fills it up. Therefore n_{2} = 1.

Since n_{3} ≠ 1 implies n_{2} = 1, either n_{2} or n_{3} = 1. ■

That we were unable to show both n_{2} and n_{3} equal 1 is not an accident. Only one of the choices needs to arise, depending on the group with size 12. For example, A_{4} has n_{2} = 1 but n_{3} ≠ 1, while the mod 6 ax + b group has n_{3} = 1 but n_{2} ≠ 1.

Corollary 0.9. If #G = 30, then G has a normal 3-Sylow subgroup and a normal 5-Sylow subgroup.

Proof. By Sylow III,

n_{3}|10, n_{3} ≡ 1 mod 3 ⟹ n_{3} = 1 or 10.

and

n_{5}|6, n_{5} ≡ 1 mod 5 ⟹ n_{5} = 1 or 6.

The 3-Sylows have size 3 and the 5-Sylows have size 5. If n_{3} = 10 and n_{5} = 6, then Lemma 0.6 says G has n_{3} · 2 = 20 elements of order 3 and n_{5} · 4 = 24 elements of order 5. This adds up to 20 + 24 = 44 elements, which exceeds #G = 30. Thus n_{3} = 1 or n_{5} = 1. We ultimately want to show both numbers equal 1.

Let P be a 3-Sylow subgroup and Q be a 5-Sylow subgroup. By the previous paragraph, at least one of these subgroups is a normal subgroup of G, so their product set P Q is a subgroup of G. Easily #P Q ≤ 15. Since P and Q are subgroups of P Q, Lagrange’s theorem implies #P Q is divisible by 3 and 5, and thus by 15. Therefore #P Q = 15, so P Q is a cyclic group (using structure theorem for pq-groups from class).

Since [G : P Q] = 2, P Q ◁ G. For any g ∈ G, gP g^{−1} and gQg^{−1} are subgroups of g(P Q)g^{−1} = P Q. Since P Q is cyclic, it has only one subgroup of each size, and in particular we must have

gP g^{−1} = P, gQg^{−1} = Q.

As this holds for all g ∈ G, we see P and Q are normal subgroups of G. Thus n_{3} = 1 and n_{5} = 1. ■

Recall that normality is not usually transitive: if N_{1} ◁ N_{2} and N_{2} ◁ G, it need not follow that N_{1} ◁ G (the counterexample from class is G = D_{4}). While it is false that a normal subgroup of a normal subgroup is a normal subgroup, it turns out that a normal Sylow subgroup of a normal subgroup is a normal subgroup, as follows.

Corollary 0.10. Let P ◁ N and N ◁ G, where P is a Sylow subgroup of N. Then P ◁ G.

Proof. For any g ∈ G, from P ⊂ N we have

gP g^{−1} ⊂ gNg^{−1} = N,

where the equation follows from normality of N in G. Since P is a Sylow subgroup of N, so is gP g^{−1} (it has the same size as P). However, since P ◁ N, it is the only Sylow subgroup of N with its size, by Sylow II. Therefore gP g^{−1} = P. As this holds for all g ∈ G, P ◁ G. ■

The rest of this handout is devoted to the topic of simple groups. We call a group simple when its only normal subgroups are the trivial subgroup and the whole group. For example, the groups of prime size are simple, for the crude reason that they have no subgroups at all besides the whole group and the trivial subgroup. One special property of simple groups is that a homomorphism on a simple group is either an embedding or sends all points to the identity. Indeed, the kernel of a homomorphism is a normal subgroup, so a homomorphism on a simple group is either one-to-one (if the kernel is trivial) or sends all points to the identity (if the kernel is the whole group).

We have seen how to use the Sylow theorems to prove a group is not simple. For example, groups of size 12 or 30 are not simple, by Corollaries 0.8 and 0.9. Meanwhile, in class we saw that p-groups and groups of size pq are solvable (it is a non-trivial theorem of Burnside that groups of size p^{a}q^{b} are solvable, but his proof used representation theory; a purly group-theoretic proof was only found many decades later by Thompson).

So far we have used the Sylow theorems to prove groups with a certain size have a nontrivial normal subgroup, and therefore are not simple. Now we use the Sylow theorems to prove a certain kind of group is simple.

Theorem 0.11. If #G = 60 and n_{5} > 1, then G is a simple group.

For example, since #A_{5} = (1/2)5! = 60 and A_{5} has more than one 5-Sylow subgroup (cf. Corollary 0.5), A_{5} is a simple group. That is, A_{5} has no normal subgroups besides A_{5} and the trivial subgroup. It can be shown that A_{n} is a simple group for any n ≥ 5, but we do not discuss that proof here.

Proof. The proof will use many of our previous corollaries of the Sylow theorems.

Assume G is not simple, so there is N ◁ G with 1 < #N < 60. We will get a contradiction.

First we show #N is not divisible by 5. Assume 5|#N, so N contains a 5-Sylow subgroup, which is also a 5-Sylow subgroup of G since 60 = 5 · 12. Because N ◁ G, Sylow II shows all the 5-Sylow subgroups of G lie in N.

Let n_{5} be the number of 5-Sylows in G (which we know are all subgroups of N). Since n_{5}|12 and n_{5} ≡ 1 mod 5, n_{5} = 1 or 6. Because n_{5} > 1, n_{5} = 6. Therefore N contains six different subgroups of size 5. By Lemma 0.6,

#N > n_{5} · 4 + 1 = 25.

Since #N is a proper factor of 60, #N = 30. But then, by Corollary 0.9, N has only one 5-Sylow subgroup. This is a contradiction, so #N is not divisible by 5. This means #N can be 2, 3, 4, 6, or 12.

If #N equals 6, then by Corollary 0.7 N contains a normal subgroup of size 3. If #N equals 12, then by Corollary 0.8 N contains a normal subgroup of size 3 or 4. In either case, such a subgroup is a Sylow subgroup of N and therefore is a normal subgroup of G by Corollary 0.10. Because such a normal subgroup of G has size 3 or 4, which is one of the other possibilies 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 \overline{G} = G/N, so \overline{G} is a group with size 30, 20, or 15. But a group of size 15 is cyclic and thus has a normal 5-Sylow subgroup. It is easy to see by the Sylow theorems that a group of size 20 has a normal 5-Sylow subgroup. By Corollary 0.9, a group of size 30 has a normal 5-Sylow subgroup. Therefore in all cases \overline{G} contains a normal 5-Sylow subgroup, say \overline{P}, with #\overline{P} = 5.

Consider the projection π: G → \overline{G}. Set H = π^{−1}(\overline{P}). Since \overline{P} ◁ \overline{G}, H ◁ G. Moreover, H ≠ G. Thus H is a proper normal subgroup of G. Since π sends H onto \overline{P}, #H is divisible by 5. But we showed earlier that G contains no proper normal subgroups of size divisible by 5.

Since all choices for #N have been eliminated, there is no such N. Thus G is simple. ■

The next result, which concludes our applications of the Sylow theorems, shows that A_{5} is the only simple group of size 60, up to isomorphism.

Theorem 0.12. Every simple group of size 60 is isomorphic to A_{5}.

Proof. To prove a group is isomorphic to A_{5}, we want to make the group act on a set of 5 objects and then show this action is given only by the even permutations of the 5 objects.

We will seek an action on cosets. Let G be a simple group of size 60. Suppose G has a subgroup H with [G : H] = 5, so the left multiplication action of G on the coset space G/H gives a homomorphism

φ: G → Perm(G/H) ≅ S_{5}.

The kernel of φ is a normal subgroup of G, and therefore is trivial or is G, since G is simple. If g ∈ G is in the kernel of φ, then gH = H, so g ∈ H. In particular, the kernel of φ is a subgroup of H and therefore the kernel can’t be G. The kernel of φ must be trivial, so φ is an embedding of G into S_{5}; G is isomorphic to its image φ(G). In particular, φ(G) is a simple group of size 60. Let’s prove this image is A_{5}.

If φ(G) ⊄ A_{5}, then φ(G) contains an odd permutation. That means the sign homomorphism

sgn: φ(G) → {±1}

is surjective, so its kernel is a normal subgroup of φ(G) with index 2. However, φ(G) doesn’t have such normal subgroups since it is isomorphic to the simple group G and therefore is itself simple. We conclude that all elements of φ(G) have sign 1, so φ(G) ⊂ A_{5}. Both groups have size 60, so φ(G) = A_{5}.

We have shown that if G has a subgroup H with index 5, then the action of G on G/H gives an isomorphism of G with A_{5}. Now we prove G has a subgroup with index 5.

\underline{Step 1}: For any proper subgroup H ⊂ G, [G : H] ≥ 5.

Let t = [G : H]. The left multiplication action of G on G/H gives a homomorphism G → Perm(G/H) ≅ S_{t}. Since H is a proper subgroup and G is simple, this homomorphism has trivial kernel. (The reason follows as before, when we were only concerned with index 5 subgroups: the kernel is a subgroup of H and therefore is a proper normal subgroup of G, which must be trivial since G is simple.) Therefore we have an embedding of G into S_{t}, so 60|t!. This can happen only when t ≥ 5.

\underline{Step 2}: G has a subgroup with index 5.

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

n_{2} ∈ {1, 3, 5, 15}, n_{3} ∈ {1, 4, 10}, n_{5} ∈ {1, 6}.

Since G is simple, n_{2}, n_{3}, n_{5} > 1. Moreover, because Sylow III* says each n_{p} is the index of a subgroup of G, Step 1 tells us n_{2}, n_{3}, n_{5} ≥ 5. Therefore

n_{2} ∈ {5, 15}, n_{3} = 10, n_{5} = 6.

We would love to show n_{2} = 5, since Sylow III* then says there is a subgroup of G with index 5, and that would lead to an isomorphism of G with A_{5} for reasons explained already.

To show n_{2} = 5, we eliminate the possibility that n_{2} = 15. By Lemma 0.6, G has n_{3} · 2 = 20 elements of order 3 and n_{5} · 4 = 24 elements of order 5. This is a total of 44 elements, which leaves at most 60 − 44 = 16 elements that can lie in the 2-Sylow subgroups of G. Each 2-Sylow subgroup of G has size 4, so if n_{2} = 15 then we’d have 15 different groups of size 4 squeezed into a 16 element subset of G. It is left to the reader to show this is impossible. (Warning: unlike groups of prime size, groups of size 4 can overlap nontrivially, but at most in a subgroup of size 2.) Therefore n_{2} ≠ 15, so n_{2} = 5 and we have proved G has a subgroup with index 5 (thanks to Sylow III*). ■