# Proofs of Sylow theorems

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.

• 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

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.

• +1 I also like these proofs,formulated by explicitly using group actions. I first learned about Sylow's theorems and their proofs this way, my lecture notes supplemented the material in Herstein's TOPICS IN ALGEBRA, which didn't use this method. – Mathemagician1234 Jul 4 '16 at 18:20

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.

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

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.

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