추상대수학, 그 스물네 번째 이야기 | 쉴로브 정리 ( Sylow Theorems )  By 초코맛 도비

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In this post, I would like to talk about Sylow theorems. The Sylow theorems are a collection of theorems named after the Norwegian mathematician Sylow 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. Before we get to know about Sylow theorems, let's find out what Sylow subgroups are.

 

Definition 1. The p-Groups & The Sylow p-Subgroups

Let p be a prime number. By a p-Group, we mean a finite group whose order is a power of p i.e., a finite group whose order is pn for some integer n0. Let G be a finite group and H a subgroup. We call H a p-Subgroup of G if H is a p-group. We call H a p-Sylow Subgroup if H is a maximal p-subgroup of G.

 

Now, let's deal with the Sylow theorems. As mentioned above, the Sylow theorems consist of three theorems, so we will divide the text into three parts. Click on the table of contents below to see those parts.

 

Table of Contents

 

Part 1.

Before proving the 1st theorem of the Sylow theorems, let's show a lemma to prove it.

 

Lemma 1.

Let G be a finite abelian group of order m, let p be a prime number dividing m. Then G has a subgroup of order p.

 

Proof.

We first prove by induction that if a finite abelian group G has exponent n then the order of G divides some power of n. Let bG, b1, and let H be the cyclic sugroup generated by b. (At this point, 1 means an identity element of G.) Then the order of H divides n since bn=1, and n is a multiple of exponent for G/H. Hence the order of G/H divides a power of n by induction hypothesis, and consequently so does the order of G because (G:1)=(G:H)(H:1).

Let G have order divisible by p. By what we have just seen, there exists xG whose order is divisible by p. Let this order be ps for some integer s. Then xs1 and obviously xs has order p, and generates a subgroup of order p, as was to be shown.

 

Theorem 1. The 1st Theorem of Sylow Theorems

Let G be a finite group and p be a prime factor with multiplicity n of the order of G. Then there exists a Sylow p-subgroup of G, of order pn.

 

Proof.

We will prove by induction on the order of G. If the order of G is a prime, our assertion is obvious. We now assume given a finite group G, and assume that the theorem proved for all groups of order smaller than that of G. If there exists a proper subgroup H of G whose index is relatively prime to p, then a Sylow p-subgroup of H will also be one of G, and our assertion follows by induction. We may therefore assume that every proper subgroup of G has an index divisible by p. We now let G act on itself by conjugation. From the fact that the set of the orbits under G is a partition of G, we obtain (G:1)=(Z:1)+(G:Gx).

Here, Z is then center of G, and the term (Z:1) corresponds to the orbits having one element, namely the element of Z. The sum of the right is taken over the other orbits, and each index (G:Gx) is then >1, hence (G:Gx) is divisible by p. Since (G:1) is divisible by p, (Z:1) is also divisible by p, hence G has a non-trivial center.

Since Z is a finite abelian group, by the Lemma 1, Z has an element aZ whose order is p. Let H be the cyclic subgroup generated by a. Since H is a subgroup of the center of G, H is normal in G. Let f:GG/H be a canonical map and let pn be the highest power of p dividing the order of G. Then pn1 divides (G:H). Let K be a Sylow p-subgroup of G/H. And let K=f1(K). Then HK and f maps K onto K. Hence we have an isomorphism K/HK by the first isomorphism theorem for groups. Hence K has order pn1p=pn, as desired and this proves the theorem.

 

Part 2.

Before proving the 2nd theorem of the Sylow theorems, let's show a lemma to prove it.

 

Lemma 2.

Let H be a finite p-group, let Ω be a finite set acted on by H, and let Ω0 denote the set of points of Ω that are fixed under the action of H. Then |Ω||Ω0|(modp).

 

Proof.

Any element xΩ not fixed by H will lie in an orbit of order |H|/|Hx| (where Hx denotes the stabilizer), which is a multiple of p by assumption. The result follows immediately by writing |Ω| as the sum of |Hx| over all distinct orbits Hx and reducing mod p.

 

Theorem 2. The 2nd Theorem of Sylow Theorems

If H is a p-subgroup of G and P is a Sylow p-subgroup of G, then there exists an element g in G such that g1HgP. In particular, all Sylow p-subgroups of G are conjugate to each other, that is, if H and K are Sylow p-subgroups of G, then there exists an element g in G with g1Hg=K.

 

Proof.

Let Ω be the set of left cosets of P in G and let H act on Ω by left multiplication. Applying the Lemma 2 to H on Ω, we see |Ω0||Ω|(G:P)(modp). Now p(G:P) by definition so p|Ω0|, hence in particular |Ω0|0 so there exists some gPΩ0. With this gP, we have hgP=gP for all hH, so HgP=gP and therefore g1HgP. Furthermore, if H is a Sylow p-subgroup, then |g1Hg|=|H|=|P| so that g1Hg=P.

 

 

Part 3. The 3rd Theorem of Sylow Theorems

As the above two theorems give information about the order of the Sylow p-subgroup, the last theorem gives information about the number of the Sylow p-subgroup.

 

Theorem 3.

Let q denote the order of any Sylow p-subgroup P of a finite group G. Let np denote the number of Sylow p-subgroups of G. Then:
(a) np=(G:NG(P)),
(b) np divides |G|/q, and
(c) np1(modp).

 

Proof.

Let Ω be the set of all Sylow p-subgroups of G and let G act on Ω by conjugation. Let PΩ be a Sylow p-subgroup. By Theorem 2, the orbit of P has size np, so by the orbit-stabilizer theorem np=(G:GP). For this group action, the stabilizer GP is given by {gG|gPg1=P}=NG(P), the normalizer of P in G. Thus, np=(G:NG(P)), and it follows that this number is a divisor of (G:P)=|G|/q.

Now let P act on Ω by conjugation, and again let Ω0 denote the set of fixed points of this action. Let QΩ0 and observe that then Q=xQx1 for all xP so that PNG(Q). By Theorem 2, P and Q are conjugate in NG(Q) in particular, and Q is normal in NG(Q), so then P=Q. It follows that Ω0={P} so that by the Lemma 2, |Ω||Ω0|=1(modp).

 

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