# SYLOW’S THEOREM

**Sylow theorems** are a collection of theorems named after the Norwegian mathematician Peter Ludwig Sylow (1872) that give detailed information about the number of subgroups of fixed order that a given finite group contains. The Sylow theorems form a fundamental part of finite group theory and have very important applications in the classification of finite simple groups.

## Sylow’s theorems and their proofs.

**Definitions.** Let G be a group, and let p be a prime number.

• A group of order p^{k} for some k ≥ 1 is called a **p-group**. A subgroup of order p^{k} for some k ≥ 1 is called a **p-subgroup**.

• If |G| = p^{α} m where p does not divide m, then a **subgroup** of order p^{α} is called a **Sylow p-subgroup** of G.

**Notation.** *S _{y}l_{p}(G) = the set of Sylow p-subgroups of G*

*n _{p}(G) = the # of Sylow p-subgroups of G = |S_{y}l_{p}(G)|*

### Diving Into Details

Let G be a group of order p^{α} m, where p is a prime, m ≥ 1, and p does not divide m. Then:

*S*._{y}l_{p}(G) not equal to ∅, i.e. Sylow p-subgroups exist*All Sylow p-subgroups are conjugate in G, i.e., if P*_{1}and P_{2}are both Sylow p-subgroups, then there is some g ∈ G such that P_{1}= g P_{1}g^{−1}. In particular, n_{p}(G) = (G : N_{G}(P)).*Any p-subgroup of G is contained in a Sylow p-subgroup.**n*_{p}(G) ≡ 1 mod p.

**First Sylow Theorem:** There is a subgroup of G whose order is p^{e}.

*Proof:* We let g be the set of all subsets of G of order p^{e}. One of these subsets is the subgroup we are looking for, but instead of finding it directly we will show that one of these subsets has a stabilizer of order p^{e}. The stabilizer will be the required subgroup.

**Lemma.** The number of subsets of order p^{e} in a set of n = p^{e}m elements

(p not dividing m) is the binomial coefficient

Moreover N is not divisible by p.

Proof. It is a standard fact that the number of subsets of order p^{e}: is this binomial coefficient. To see that N is not divisible by p, note that every time p divides a term (n – k) in the numerator of N, it also divides the term (p^{e} – k) of the denominator exactly the same number of times: If we write k in the form k = p^{i} l, where p does not divide l, then i < e. Therefore (n – k) and (p^{e} – k) are both divisible by pi but not divisible by p^{1+i}.

We decompose g into orbits for the operation of left multiplication, obtaining the formula.

Since p does not divide N, some orbit has an order which is not divisible by p, say the orbit of the subset U.

**Second Sylow Theorem:** Let K be a subgroup of G whose order is divisible by p, and let H be a Sylow p-subgroup of G. There is a conjugate subgroup H’ = gHg^{– 1} such that K n H’ is a Sylow subgroup of K.

*Proof:* We are given a subgroup K and a Sylow subgroup H of G, and we are to show that for some conjugate subgroup H I of H, the intersection K n H I is a Sylow subgroup of K. Let S denote the set of left cosets G/H. The facts that we need about this set are that G operates transitively, that is, the set forms a single orbit, and that H is the stabilizer of one of its elements, namely of s = 1H. So the stabilizer of as is the conjugate subgroup aHa^{-1}. We restrict the operation of G to K and decompose S into K-orbits. Since H is a Sylow subgroup, the order of S is prime to p. So there is some K-orbit O whose order is prime to p. Say that O is the K-orbit of the element as. Let H’ denote the stabilizer aHa^{-1} of as for the operation of G. Then the stabilizer of as for the restricted operation of K is obviously H’ n K, and the index [K:H’nK] is |O|, which is prime to p. Also, since it is a conjugate of H, H’ is ap-group. Therefore H’ n K is a p -group. It follows that H’ n K is a Sylow subgroup of K.

**Third Sylow Theorem:** Let |G| = n, and n = p^{e}m. Let s be the number of Sylow p-subgroups. Then s divides m and is congruent 1 (modulo p): s|m, and s = ap + 1 for some integer a ≥ O.

*Proof:* The Sylow subgroups of G are all conjugate to a given one, say to H. So the number of Sylow subgroups is s = [G:N], where N is the normalizer of H. Since H⊂N, [G:N] divides [G:H] = m. To show s ≡ 1 (modulo p), we decompose the set {H1, … , Hs} of Sylow subgroups into orbits for the operation of conjugation by H = H1. An orbit consists of a single group Hi if and only if H is contained in the normalizer M of Hi. If so, then H and Hi are both Sylow subgroups of M, and Hi is normal in M. There is only one H orbit of order 1, namely {H}. The other orbits have orders divisible by p because their orders divide |H|, by the Counting Formula. This shows that s ≡ 1 (modulo p).