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It seems that there are many ways to prove the Sylow theorems. I'd like to see a collection of them. Please write down or share links to any you know.

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Check out the early chapters of Dixon & Mortimer for a nice selection of proofs of basic facts about group theory. –  Jyrki Lahtonen Jun 22 '12 at 16:49
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5 Answers 5

I love Wielandt's proof for the existence of Sylow subgroups (Sylow I). Isaacs uses this proof in his books Finite Group Theory and Algebra: A Graduate Course. As Isaacs mentions, the idea of the proof is not very natural and does not generalize to other situations well but it is simply beautiful. First a lemma:

Lemma: Let $p,a,b$ be natural numbers where $p$ is prime and $a \geq b$. Then $$\binom{pa}{pb} \equiv \binom{a}{b} \pmod{p}$$

Proof. Consider the polynomial $(x + 1)^{pa} = (x^p+1)^a \in \mathbb{F}_p[x]$. Computing the coefficient of the $x^{pb}$ term in two different ways yields the result.

Proof of Sylow I: Let $|G| = p^nm$ such that $p \nmid m$. Let $$\Omega = \{ X \subseteq G: |X| = p^n\} $$ (Note that we are taking every subset of $G$ with $p^n$ elements).
$G$ acts on $\Omega$ by left multiplication. Observe that $$|\Omega| = \binom{p^nm}{p^n} \equiv m \pmod{p}$$ by repeated usage of the lemma. Hence $p \nmid |\Omega|$, therefore $\Omega$ has an orbit $\mathcal{O}$ such that $p \nmid |\mathcal{O}|$. Now let $X \in \mathcal{O}$ and let $H$ be the stabilizer subgroup of $X$. Since $|G:H| = |\mathcal{O}|$ (orbit-stabilizer theorem), we deduce that $p^n$ divides $|H|$; in particular $p^n \leq |H|$. On the other hand, for $x \in X$ by definition of stabilizing $Hx \subseteq X$ and hence $$|H| = |Hx| \leq |X| = p^n$$ Thus $H$ is a Sylow $p$-subgroup.

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Here is a proof by Keith Conrad. In fact, just googling "sylow theorem proof" produces many examples.

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Let $p$ be a prime dividing the order of the finite group $G$. The existence of a Sylow $p$-subgroup can be proved in a standard manner by induction on $|G|$. Let $x \in G$ be a non-central element (i.e. $x \not \in Z(G)$). If the index of the centralizer $|G:C_G(x)|$ is not divisible by $p$ then we may apply induction and find a Sylow $p$-subgroup of $G$ inside $C_G(x)$. So we may assume that $p$ divides the indeces of the centralizers of the non-central elements. By the orbit-stabilizer equation $|G| = |Z(G)| + \sum_i |G:C_G(x_i)|$ where the $x_i$ are representatives of non-central conjugacy classes. In particular $p$ divides $|Z(G)|$. By Cauchy theorem there exists $x \in Z(G)$ of order $p$. By induction there exists a Sylow $p$-subgroup $H/\langle x \rangle$ of $G/\langle x \rangle$, implying that $H$ is a Sylow $p$-subgroup of $G$.

Now let $P$ be a Sylow $p$-subgroup of $G$, and let $\Omega$ be the set of conjugates of $P$ in $G$, $\Omega = \{P^g\ |\ g \in G\}$. $P$ acts on $\Omega$ by conjugation, with $P$ as unique fixed point. Indeed, if $P$ fixes $R \in \Omega$ then $P$ normalizes $R$, so that $PR$ is a $p$-subgroup of $G$ containing $P$ ($|PR| = |P| \cdot |R|/|P \cap R|$), hence by maximality $P=R$. Note that each orbit of an action of a $p$-group has size a power of $p$. In particular partitioning $\Omega$ into $P$-orbits gives $|\Omega| \equiv 1 \mod(p)$. We are left to prove that every Sylow $p$-subgroup of $G$ belongs to $\Omega$. Suppose by contradiction that there exists a Sylow $p$-subgroup $Q$ of $G$ such that $Q \not \in \Omega$. $Q$ acts on $\Omega$, and by the previous argument there are no fixed points under this action, thus $p$ divides $|\Omega|$, contradiction.

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Check out Artin's Algebra book. He proves all three of them. The book is just called Algebra. The Artin I'm talking about is Michael, not Emil. It's a great book, and covers a lot of material.

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Look Hungerford proofs, he uses the same idea for prove all Sylow's theorems. Which uses the same Lemma of $p$-groups actions.

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