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Having proved the Sylow theorem for general linear group over finite field, how to prove it for any finite group?

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there are different approaches. wikipedia en.wikipedia.org/wiki/… has a proof as will any algebra textbook (rotman, dummit and foote, etc) –  yoyo Apr 13 '11 at 11:10
    
@yoyo: The proof you suggested doesn't make use of availability of proof for GL(n,F_p).. –  user8186 Apr 13 '11 at 12:12
    
You'll probably want to first show that a finite group $G$ can be embedded into any sufficiently large general linear group $\text{GL}(n, \mathbb{F}_q)$. Then any $p$-subgroup of $G$ is embedded as a $p$-subgroup of $\text{GL}(n, \mathbb{F}_q)$. –  Zhen Lin Apr 13 '11 at 12:26
    
This is the approach taken in Bogopolski's Group Theory text; see page 13. A google book preview: books.google.com/… –  user641 Apr 13 '11 at 19:40
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You show any group embeds in a GL group by embedding the symmetric groups $S_n$ in a GL group, then using Cayley's theorem. –  user641 Apr 13 '11 at 19:41

1 Answer 1

up vote 3 down vote accepted

Let $G$ be a finite group which has a Sylow $p$-group. (Of course, every $G$ has a Sylow group, but we are assuming we don't know that yet.)

Theorem: If $H$ is a subgroup of $G$, then $H$ has a Sylow subgroup.

Proof: Let $|G|=p^k m$ and $|H| = p^l n$ where $p$ does not divide $m$ or $n$. Let $P$ be a $p$-Sylow of $G$. Let $X$ be the set $G/P$. So $|X| = m$. In particular, $|X| \not \equiv 0 \mod p$. Consider the action of $H$ on $X$; there must be some orbit whose size is not divisible by $p$. Let this orbit be $Y$, and let $Q$ be the stabilizer of a point of $Y$. So $|Y| = |H|/|Q|$, and we see that $p^l$ divides $|Q|$. On the other hand, $Q$ is a subgroup of a conjugate of $P$, so $Q$ is a $p$-group. We thus see that $Q$ is $p$-Sylow in $H$. QED

So, as Zhen Lin says, if you prove that any finite group $H$ embelds in $GL_n(\mathbb{F}_p)$, and you check that $GL_n(\mathbb{F}_p)$ has a $p$-Sylow, then you show that every group has a $p$-Sylow.

You can push this argument a bit further and prove Sylow 2. I seem to recall that I had trouble getting to Sylow 3, though.

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What is $b$? Do you mean $p$? –  Qiaochu Yuan Apr 13 '11 at 19:44
    
Fixed, thanks!! –  David Speyer Apr 13 '11 at 19:49
    
Sylow 3 (taking the numbering of the wikipedia) is a corollary of Sylow 1+2 by taking the action of the $G$ by conjugation on the set of $p$-Sylow subgroups. –  j.p. Apr 14 '11 at 7:33

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