Abstract Nonsense

Crushing one theorem at a time

Sylow’s Theorems Revisited


Point of Post: In this post we give a more refined proof of Sylow’s Theorems.

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Motivation

On this blog I have given a proof of Sylow’s theorems and an alternate proof of Sylow’s first theorem. A year or so later, with more time, and more finesse I’d like to give the simplest, most coherent proof of Sylow’s theorems.

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Sylow’s Theorems

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We assume all the notations such as \text{Syl}_p(G) being the set of Sylow p-subgroups of G and n_p=\#(\text{Syl}_p(G)),etc. are known to the reader:

Theorem (Sylow): Let G be a finite group with |G|=p^kn where (n,p)=1, then:

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\begin{aligned}&\;\;\;(I)\quad \textit{If }p^m\mid |G|\textit{ for some prime }p\textit{ then }G\textit{ contains a subgroup }H\textit{ of order }p^m\\ &\;(II)\quad \textit{If }Q\textit{ is a }p\textit{-subgroup of }G\textit{ and }P\in\text{Syl}_p(G)\textit{ then }Q\leqslant gPg^{-1}\textit{ for some }P\in\text{Syl}_p(G),\\ &\quad\quad\quad\textit{in particular, any two elements of }\text{Syl}_p(G)\textit{ are conjugate}\\ &(III)\quad n_p=\left[G:N_G(P)\right]\textit{ for any }P\in\text{Syl}_p(G)\textit{ and }n_p\equiv 1\text{ mod }p\end{aligned}

Proof:

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(I): We construct such a group inductively. Evidently if m=0 then we may take the trivial group. Suppose then that we have found a subgroup H of |G| with |H|=p^m for m<k. We know then that p\mid \left[G:H\right] and so by a previous theorem we have that p\mid \left[N_G(H):H\right]. Thus, by Cauchy’s Theorem there exists xH\in N_G(H)/H with |\langle xH\rangle|=p. By the fourth isomorphism theorem we know that \langle xH\rangle=K/H for some H\leqslant K\leqslant N_G(H) and so, in particular, we have that |K|=|\langle xH\rangle||H|=p^{m+1}. As stated, from this we may construct subgroups H_0,\cdots,H_k of G with |H_k|=p^k.

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(II): Let Q act on X=G/P by left multiplication. Since Q is a p-group we have by the ‘fundamental theorem’ that \#(X)\equiv \#(X^Q)\text{ mod }p. That said, since p\nmid \#(X) we have in particular that p\nmid \#(X^Q) and so X^Q is non-empty. So, let xP\in X^Q then qxP=xP for all q\in Q and so x^{-1}qx\in P for all q\in Q, or equivalently x^{-1}Qx\subseteq P. Thus, Q\subseteq xPx^{-1}. In particular, if Q\in\text{Syl}_p(G) we have that |Q|=|xPx^{-1}| and so Q=xPx^{-1}.

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(III): Let G act on \text{Syl}_p(G) by conjugation. By (II) we know that this action is transitive. In particular, we have that n_p=\left[G:\text{stab}(P)\right] for any P\in\text{Syl}_p(G). That said, we note by definition that \text{stab}(P)=N_G(P) and so n_p=\left[G:N_G(P)\right].

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For the second part of the statement we let P\in\text{Syl}_p(G) act on \text{Syl}_p(G)=X by conjugation. By the ‘fundamental theorem’ we know that \#(X)\equiv \#(X^P)\text{ mod }P. What we claim though is that X^P=\{P\}. Indeed, it’s evident that P\in X^P, and so now assume that Q\in X^P. We have then that P\subseteq N_G(Q) and so PQ\leqslant G, but we know that

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\displaystyle |PQ|=\frac{|P||Q|}{|P\cap Q|}=\frac{p^{2k}}{|P\cap Q|}

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and since |PQ|\mid |G| and p^k is the maximal power of p dividing |G| we may conclude that |P\cap Q|=p^k and so P=Q. Thus, \#(X)\equiv \#(X^Q)=1\text{ mod }p. Noting then that \#(X)=n_p completes the proof.

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References:

1. Isaacs, I. Martin. Finite Group Theory. Providence, RI: American Mathematical Society, 2008. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

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September 20, 2011 - Posted by | Algebra, Group Theory | , , , , , , ,

4 Comments »

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