Better Proof of Sylow’s First Theorem
There is a proof of the first Sylow Theorem that I much prefer to the standard one (which involves looking at the class equation and doing induction). I found it in some notes by Peter Cameron, and he says that it is the original proof that Sylow used. I prefer it because it is for me more intuitive, it involves group actions, and it has more mathematical flavor.
Suppose $G$ is a finite group. For any prime $p$, define a Sylow $p$-subgroup of $G$ to be a group of order $p^n$, where $n$ is the highest exponent such that $p^n \mid \# G$. Sylow's First Theorem says A finite group $G$ has Sylow $p$-subgroups for all primes $p$. Note by the definition that the Sylow $p$-subgroup of $G$ is the trivial group for all primes $p$ such that $p \not \mid \#G$. Lemma: A group $G$ has a Sylow p-subgroup iff there is an action by $G$ on a set such that all stabilizers are $p$-subgroups and there exists an orbit of size coprime to $p$. Proof: (==>) The stabilizer of the orbit which is of size coprime to $p$ is a Sylow $p$-subgroup.
(<==) Suppose $P < G$ is the given group. Let $G$ act on the cosets of $P$ by multiplication. It's a transitive action with a single orbit of size coprime to $p$, and the stabilizers are all of the same order as $P$ itself. QED
Lemma: If $G$ has a Sylow $p$-subgroup, then all subgroups of $G$ have Sylow $p$-subgroups. Proof: Suppose $H<G$. Take the action as in the previous lemma, and restrict the action to $H$. Clearly the stabilizers remain $p$-subgroups (they are subgroups of the stabilizers of the $G$ action). To show that there remains an orbit of size coprime to $p$, consider the orbit of the $G$ action that is coprime to $p$. It breaks up into smaller orbits under the action by $H$, and they cannot all be divisible by $p$, or else the original orbit would be divisible by $p$. QED Lemma: $GL_n(\mathbb{F}_p)$ has a Sylow $p$-subgroup. Proof: Recall that $GL_n(\mathbb{F}_p)$ has order $p^{n(n-1)/2}(p^n-1)(p^{n-1}-1)\dotsb(p-1)$. And note that the upper triangular matrices with ones on the diagonal have order $p^{n(n-1)/2}$. QED
Now prove the theorem by noting that every group $G$ is a subgroup of a symmetric group $S_n$ by Cayley's Theorem, and $S_n < GL_n(\mathbb{F}_p)$ by the permutation matrices.