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

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.
