Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
there are different approaches. wikipedia… 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:… – user641 Apr 13 '11 at 19:40
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.