Sylow and Cauchy theorems

Lagrange's Theorem tells us that if G has a subgroup of order n, then n must divide |G|. Would still this statement be true if we turned it around: If n is a factor of |G|, then must G have a subgroup of order n? This is called the converse of Lagrange's Theorem, and we saw that it is not true (For example, A₄ (of order 12) has no subgroup of order 6); just because Lagrange's Theorem permits a subgroup of a certain size does not guarantee that such a subgroup will exist. Sylow Theory digs deeper, finding which sizes of subgroups are guaranteed to exist.
We shall consider three fundamental theorems of the Norwegian mathematician Ludwig Sylow (1832–1918).

Cauchy's Theorem is a partial converse to Lagrange's Theorem: Partial because it guarantees the existence of a subgroup only if its order is prime. In small groups this can be very helpful because the subgroups of prime order make up a large portion of the group. But in large groups, knowing whether there are larger subgroups would be more helpful. The Sylow Theorems include a more general converse to Lagrange's Theorem, using p-groups.

7.17.1 (Cauchy's Theorem) If G is a finite group whose order is divisible by a prime p, then G contains an element (thereby a subgroup) of order p.

Proof. Clearly, the statement of Cauchy's theorem is true for the case in which G has ­order 2. We must find an element g ≠ e, g^p = e. We introduce a new action of the group ℤ/pℤ (the integers with addition modulo p), to X. Let

that is, X is the set of all ordered p-tuples of elements of G whose product is the neutral element. We prove the theorem by applying an action of the group ℤ/pℤ, and counting orbits. For all t ∈ ℤ_p

which corresponds to a translation of t positions of the indexes and their reduction modulo p.

1. X has a cardinality which is a multiple of p: indeed g₁, g₂, ..., g_p: g_p−1 can be chosen at will in G, i.e. one of the n = |G|; g_p is determined by the relation g₁g₂ ⋅⋅⋅ g_p = e. Thus there are n^p−1 possible strings, which is a multiple of p, since n is.

2. Owing to the orbit stabilizer theorem:

   O|(x)| ⋅ |Stab(x)| = |ℤ_p|

the cardinality of each orbit divides p, then every orbit has either one or p strings. If all orbits with the exception (e, e,..., e) were made by p strings, made by p, the order of X could not be a multiple of p contradicting point 1). Thus there must exist an orbit ( not reduced to e, e,..., e) which contains a single element, hence containing a string (g₁, g₂, ..., g_p) ≠ (e, e,..., e) which is left fixed by every element of ℤ_p. This implies (given the action of ℤ_p on X) that g₁ = g₂ = ⋅⋅⋅ = g_p, that is x₁ is an element of period p.   □

We've seen that given a group of order n, is not said that for every divisor of n there exists a subgroup having that as order. The theorem just proved guarantees the existence that for every prime divisor of the order of a group, there exists a subgroup having that divisor as order: the existence of an element with period p guarantees that the subgroup generated by it has order p.

Example 7.17.2. Let G a group order 200; The number 200 factors into 2³ ⋅ 5², so the only primes dividing 200 are 2 and 5. Cauchy's Theorem therefore says that G contains elements a and b of orders 2 and 5. ■

7.17.3. Definition. A p-group is a group whose order is a power of a prime p. A p-group that is a subgroup of a group G is called, for short, a p-subgroup of G. □

7.17.4 Definition. A p-subgroup P of a finite group, |G| with the property that its cardinality is the largest power of p dividing |G| is known as Sylow p-subgroup.  □

7.17.5 First Sylow theorems. Let G a finite group whose order is p^a ⋅ m, with p a prime number and m not divisible by p. Then G contains a subgroup of order p^a, that is a Sylow p-subgroup.

Proof. Let X the set of all subsets of G having p^a elements. Consider the action of G on X via left translation: given U ∈ X and a g ∈ G, we let g * U := gU. We know that the cardinality of X is the number of combinations C_n,k (i.e. the numer of k-subsests that can be formed out a n-set)

which is not divisible by p (see exercise 3), and since X can be partitioned into orbits, there must exist an element A ∈ X such that O(A) has a number of elements which is not a multiple of p.

But then, from the relation

it follows that |Stab(A)| must be a multiple of p^a. By definition of stabilizer and given the definition of action, letting an element a ∈ A and a g ∈ St_A, it results ga ∈ A, which means that the right lateral Stab(A) is contained in A for all a ∈ A. This implies that |Stab(A)| ≤ |A|, that is p^a. Since Stab(A)a has the same cardinality of Stab(A) it follows that Stab(A) is a subgroup of G of exactly p^a elements.   □

Continuation of Example 7.17.2.. Let's return to the group G fin Example 7.17.2, of which we assumed only that its order is 200. Cauchy's Theorem told us that there must therefore be two subgroups of orders 2 and 5, but the First Sylow Theorem tells us much more. Those small, prime-order subgroups are inside groups of orders 4 and 25 respectively, and the subgroup of order 4 is in one of order 8.  ■

7.17.6 Lemma. If K is a Sylow p-subgroup of G, then so is xKx⁻¹.

Proof. Let G be a group and x ∈ G. Example 7.8.2 shows that the map f : G ⟶ G given by f(a) = xax⁻¹ is an isomorphism. If K is a subgroup of G, then the image of K under f is xKx⁻¹ = {xkx⁻¹ |k ∈ K}. Hence, xKx⁻¹ is a subgroup of G that is isomorphic to K.   □

Let P the set of all Sylow p-subgroups of G. P is not empty by the first Sylow theorem. If S ∈ P, then P contains also all conjugates of S, through the elements of G, since they are all subgroup of the same largest order p^a. Given the orbit O(S), of S with respect to the action of G on P by conjugation, we have O(S) ⊆ P. In addition the cardinality of O(S) divides m. Indeed, since (see Proposition 7.12.13):

it's clear that Stab(S) ⊇ S, then |Stab(S)| = p^ar. The cardinality of O(S), then divides m, based on the relation |O(S)| |Stab(S)| = |G| and is relatively prime to p.

2^nd Proof of Cauchy Theorem. If p is a prime that divides the order of a group G, then G contains a subgroup K of order p by the First Sylow theorem. Since K is cyclic by Corollary 7.8.6 of Lagranges's Theorem, its generator is an element of order p in G.  □

The orbit-stabilizer theorem O|(x)| ⋅ |Stab(x)| = |G|, introduced in the last paragraph has different interesting applications, among these a way of providing an alternative proof of Cauchy's theorem.

7.17.7 Theorem. Every nontrivial p-group has a nontrivial center.

Proof. Let G a p-group and consider its partition in conjugate classes. The cardinality of every class will be either 1 or a power of p. The center G is made by all elements that commute with every element of G and thus contains all classes which have only one element. If the center were trivial the order of G would be congruent to 1 modulo p, which contradicts the hypothesis that |G| = p^k. □

7.17.8 Theorem. A group G of order p² is abelian.

Proof. For what stated in the previous theorem, the center Z(G) is nontrivial; it possesses either p or p² elements. We suppose it possesses p elements, letting a an element not in Z(G). Then the centralizer of a, C(a) = {g ∈ G | ga = ag} is a subgroup of G containing properly Z(G). Then owing to Lagrange theorem, it coincides with all G, which is absurd because a should be contained in Z(G). It remains the only possibility that Z(G) has p² elements, that is to coincide with all G. G is abelian. □

7.17.9 (Second Sylow Theorem). Let G be a finite group with order p^a ⋅ m and p a prime such that m is not divisible by p. Then any p-subgroup H of G is contained in a p-Sylow subgroup.

Proof. Let P be the set of p-Sylow subgroups of G, and let H be a p-subgroup of G, and let |H| = p^s, s ≤ a since a p-subgroup of G cannot have order greater by p^a by Lagrange's Theorem. Let S a fixed p-Sylow subgroup and O(S) its orbit under the conjugation action. Let H act on O(S): hT = hTh⁻¹ ∀T ∈ O(S): it's an action on on O(S) because the conjugate of an element is still in O(S). O(S) is thus partitioned into suborbits, each one having a cardinality which divides p^s. The cardinality O(S) is the sum of the cardinalities of all these suborbits. Since |O(S)| divides m and is *therefore* relatively prime to p (for all divisors of |G|, being coprime to p is the same as dividing m: "divisor of m, with m coprime to p" ⇒ "divisor coprime to p"), there exists at least one suborbit having just one element (consider 1 + p + p + p² + p³, it's NOT a multiple of p); that is ∃T ∈ O(S) such that hTh⁻¹ = T, ∀h ∈ H.
Then H ⊆ St_T = {g ∈ G | gTg⁻¹ = T} = N_G(T) (normalizer of T in G). Clearly T ⊲̲ N_G(T) (See Proposition 7.12.3). We can then construct the quotient group N_G(T)/T and consider the subgroup HT/T ⊲̲ N_G(T)/T. By the First Isomorphism Theorem, HT/T ≃ H/H ∩ T.

Note that |HT/T| divides m and therefore relatively prime to p. HT/T = HT = {hT = Th | h in H}, HT is a subgroup of G containing T, so its order is p^d k for some k coprime to p and such that k|m with d ≤ a. Now note that |H / H ∩ T| = |H| / H ∩ T| = [H : H ∩ T] = p^r for some r ≥ 0 and ≤ s (we applied Lagrange's theorem that is |G| / |H| = (G:H) where (G:H) is the number of cosets, but this number is nothing else than |G/H|). Thus it must necessarily be |H/H ∩ T| = 1, that is H ⊆ T. We have so proved that every p-subgroup H is contained in a Sylow p-subgroup of G.   □

The next Corollary shows that every Sylow p-subgroup of G can be obtained from a conjugate in the fashion of Lemma 7.17.7.

7.17.10 Corollary. Any two Sylow p-subgroups of a finite group G are conjugate in G.

Proof. We have to prove that O(S) = P, that is the oribit O(S) defined as in the first Sylow theorem, contains all Sylow p-subgroup. Let Q and R be two p–Sylow subgroups of G. According to the statement of the corollary, there is an a ∈ G such that aQa⁻¹ ⊆ R. We proved in the last theorem that every p-subgroup H is contained in a conjugate T that is H ⊆ T. In particular in the case in which H is a Sylow p-subgroup, it must necessarily be H = T since every Sylow p-subgroup has p^a elements.   □

Corollary 7.17.11. Let G be a finite group and K a Sylow p-subgroup for some prime p. Then K is normal in G if and only If K is the only Sylow p-subgroup in G.

Proof. We know that x⁻¹Kx is a Sylow p-subgroup for every x ∈ G. If K is the only Sylow p-subgroup of G, then we must have x⁻¹Kx = K for every x ∈ G. Therefore, K is normal by Proposition 7.12.7. Conversely, suppose K is normal and let P be any Sylow p-subgroup. By the Second Sylow Theorem there exists x ∈ G such that P = x⁻¹Kx. Since K is normal, P = x⁻¹Kx = K. Therefore K is the unique Sylow p·subgroup. □

The preceding theorems establish the existence of Sylow p-subgroups and the relationship between any two such subgroups. The next theorem tells us how many Sylow p-subgro ps a given group may have.

7.17.12 Third Sylow Theorem. The number n, of Sylow p-subgroups of a finite group G divides |G| = p^a ⋅ m with (p,m) = 1 and is congruent to 1 modulo p. So

Proof. Let P be the set of all Sylow p-subgroups of G and S ∈ P. Consider the action of the Sylow p-subgroup H on O(S) by conjugation; there is only one suborbit of O(S) with one element; indeed if there were another suborbit with order 1, H would be identical to another Sylow p-subgroup. Being the cardinality of all other suborbits a power p^k (with k > 0), the cardinality of O(S) is thus conruent to 1 modulo p.   □

This theorem helps us narrow down, just based on a group's order, the possible number of Sylow p-subgroups that the group can have.

Continuation of example 7.17.2. Before seeing an extended application of this theorem, let's see how it impacts our running example of the group G of order 200. The Third Sylow Theorem requires that the number of Sylow 5-subgroups be a factor of 200 that's congruent to 1 mod 5. The factors of 200 are

and you can verify that the only one congruent to 1 mod 5 is the number 1 itself. Thus there is only one Sylow 5-subgroup, and it must therefore be normal. However, the Third Sylow Theorem does not narrow down the possible number of Sylow 2-subgroups as much. That number must be a factor of 200 that's congruent to 1 mod 2, which leaves three options, 1, 5, and 25. So the Third Sylow Theorem does not give us full information about the number of Sylow 2-subgroups. Perhaps some additional reasoning would narrow the number down further, or perhaps not.  ■

7.17.13 Example Find all Sylow 2 and Sylow 3-subgroups of A₄.

Solution: |A₄| = 12 = 2² ⋅ 3. Hence by Sylow First Theorem, A₄ has Sylow 2-subgroups and Sylow 3- Subgroups of order 4 and 3 respectively. Let n₂ and n₃ denote respecively the number of distinct Sylow 2 and Sylow 3-subgroups of A₄. Then by Sylow Third Theorem n₂ | 12 and n₂ = 2k + 1 for some integer k ≤ 0. Similary, n₃| 12 and n₃ = 3l + 1 for some integer l ≥ 0. The only possibilities are n₂ = 1 and n₃ = 3. Both can occur. ■

7.17.14 Example Every subgroup of order 20 contains a normal subgroup.

Proof. 20 = 2² ⋅ 5. The number of 5-Sylow subgroups must be a divisor of 4 (=m) that is it can be 1, 2, or 4, but it also needs to be congruent to 1 modulo 5. The only choice is therefore 1. If a Sylow p-subgroup in unique, this is necessarily a normal subgroup (see exercise 1).  ■

A group G is simple if it is not the identity group but its only normal subgroups are the trivial ones, the whole group and the identity.

We just proved that a group with order 20 is not simple.

Another consequence of Sylow's theorems is the following:

7.17.14 Theorem Let G a group with order p^a ⋅ m, with m non divisible by p. Then G possesses a chain of subgroups

such that for 0 ≤ i ≤ a, H_i is a normal subgroup of H_i+1 and |H_i+1/H_i| = p.

Proof.We will proceed by induction on |G|. For |G| = 1 there is nothing to prove. We suppose the theorem true for every grop with order less that the order of G and we'll prove it for G. G contains a subgroup of order p^a. We know now that such a group has a non-trivial center Z(H). Then for Cauchy's theorem, there exists a subgroup H₁ with order p contained in Z(H). H₁ is a normal subgroup of H (being contained in the center), thus the quotient H/H₁ is a group with order less than the order of G. By the inductive hypothesis, there exists a chain of subgroups in H/H₁

such that

and

By the 2nd isomorphism theorems from eq 7.12.1 we get that H_i ⊲̲ H_i+1 and from eq 7.12.2 that

7.17.16 Corollary Every group G with order p^am, (m,p) = 1, has a subgroup H_i of order pⁱ, for every i = 0, 1, ..., a.

Proof. The result is implicit in the theorem.  □

7.17.17 Corollary Every p-group of G contains a subgroup with index p in G.

Proof. Just exchange H_a in the theorem with G.  □

7.17.18 Lemma. A permutation group G (recall this is any subgroup of S_n) which contains an odd permutation has a normal subgroup of index 2.

Proof. Since G contains an odd permutation, G is not contained in A_n. Also since A_n is normal in S_n the set A_nG is a subgroup of S_n by Lemma 7.12.4. Thus A_n ⊂ A_nG properly. But [S_n: A_n] = 2 by Proposition 7.13.6 and thus A_nG = S_n. By the second isomorphism theorem we have

Thus A_n ∩ G is subgroup of index 2 in G and hence a normal subgroup of G completing the proof of the lemma.   □

7.17.19 Theorem. An abelian group G is simple if and only if it is a cyclic group of prime order.

Proof. If G is cyclic of prime order, then Lagrange's theorem implies that G has not subgroups of order different fro 1 or |G| and so G is simple. Conversely, if G is simple, then every nonidentity element g ∈ G must generate G, that is G = ⟨g⟩ is cyclic. In fact, if the abelian group G is simple, then it does not contain non-trivial subgroup, and because there must be cyclic subgroups in it, it will itself be cyclic. If G were infinite, then ⟨g²⟩ would be a nontrivial proper subgroup of G, a contradiction. Hence, G is finite and cyclic. But if p | |G| where p is prime, the Cauchy's theorem implies that G has a subgroup H of order p and so G = H has prime order.  □