# Abstract Nonsense

## 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. 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.

September 20, 2011 -

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