Sylow subgroups made easy

8 January 2024

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Here I want to present a quick and easy proof of the first Sylow theorem. It uses basic combinatorics only and yields a super elementary proof of Cauchy’s theorem for free. The goal is to prove the following statement:

Theorem: Let $G$ be a finite group of order $\abs{G} = p^r m$ where $p$ is a prime number and $m$ is coprime to $p$. Then $G$ has a subgroup of order $p^r$.

We need a small number-theoretic lemma which has a beautiful proof using group actions of cyclic groups.

Lemma: Let $p$ be a prime number and $m \in \bN_+$. Then for all $r \in \bN$ we have

$$\binom{p^r m}{p^r} \equiv m \mod p.$$

Proof: Let $C$ be the cyclic group of order $p^r$. This group acts on itself by left multiplication, so the action extends to $S \coloneqq \bigsqcup_{i=1}^m C$. Hence, we have the desired action on

$$X = \{ A \subseteq S : \abs{A} = p^r \}.$$

Notice that $\abs{X} = \binom{p^r m}{p^r}$. Let $A \in X$. Then we have $\abs{\Orb(A)} \cdot \abs{\Stab(A)} = p^r$ and therefore

$$\abs{X} \equiv \abs{\operatorname{Fix}_C(X)} \mod p.$$

Now, $A \in X$ is fixed by $C$ precisely if $A$ is a union of copies of $C$. But since $\abs{A} = \abs{C}$, it is exactly a single copy of $C$. Hence, there are $m$ fixed points. We conclude $\abs{X} \equiv m \mod p$.

With this in mind we can now prove the existence of Sylow subgroups.

Proof (of theorem): Let $G$ act on $X \coloneqq \{ A \subseteq G : \abs{A} = p^r \}$ by left multiplication. Since $m$ and $p$ are coprime, we conclude $p \nmid \abs{X}$ by the previous lemma. Hence, there exists some $A \in X$ with $p \nmid \abs{\Orb(A)}$ and thus $\abs{\Stab(A)} \geq p^r$. Finally, for each $a \in A$, the map

$$\Stab(A) \to A, \quad x \mapsto xa$$

is injective, showing $\abs{\Stab(A)} \leq p^r$. We conclude that $\Stab(A)$ is the desired subgroup of order $p^r$. Neat.

A very nice consequence of this argument is that we get Cauchy’s theorem practically for free.

Corollary (Cauchy’s theorem): Let $G$ be a finite group. If a prime number $p$ divides $\abs{G}$, then $G$ has an element of order $p$.

Proof: By the previous theorem, $G$ has a subgroup $P$ of order $p^r$ for some $r > 0$. Let $x \in P \setminus \{ 1 \}$. Then by Lagrange’s theorem, $x$ has order $p^k$ for some $0 < k \leq r$. Hence, the element

$$y \coloneqq x^{p^{k-1}}$$

has order $p$, since $y \neq 1$ and $y^p = x^{p^k} = 1$.

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