Sylow's theorem

Let $𝐺$ be a finite group of order $𝑝 𝑟 𝑚$ where $𝑝$ is prime and $𝑝 ∤ 𝑚$ . Then #m/thm/group

$S y l 𝑝 (𝐺)$ , the set of Sylow $𝑝$ -subgroups, is non-empty;
Every $𝑝$ -subgroup of $𝐺$ is contained in a Sylow p-subgroup;
All Sylow $𝑝$ -subgroups are conjugate in $𝐺$ ;
If $𝑛 𝑝 = ∣ S y l 𝑝 (𝐺) ∣$ then $𝑛 𝑝$ divides $𝑚$ and $𝑛 𝑝 \equiv 𝑝 1$ .

Proof due to Wielandt

In this proof group actions are taken to be right actions.

Let $Ω$ be the set of all subsets of $𝐺$ of size $𝑝 𝑟$ , and $𝐺$ act on $Ω$ by right multiplication. Note

| Ω | = (𝑝 𝑟 𝑚 𝑝 𝑟) = 𝑝 𝑟 𝑚 𝑝 𝑟 (𝑝 𝑟 𝑚 - 1) 𝑝 𝑟 - 1 \dots 𝑝 𝑟 𝑚 - (𝑝 𝑟 - 1) 1

which is not divisible by $𝑝$ , so there must be a $𝐺$ -orbit on $Ω$ of degree not divisible by $𝑝$ , Let $Σ$ be such an orbit and $𝑆 \in Σ$ . Let $𝑃 = (𝐺) 𝜌 𝑆 = {𝑔 \in 𝐺 : 𝑆 𝑔 = 𝑆} \leq 𝐺$ . We claim $| 𝑃 | = 𝑝 𝑟$ , whence follows $𝑃$ is a Sylow $𝑝$ -subgroup. Since $| 𝐺 | = 𝑝 𝑟 𝑚 = | Σ | | 𝑃 |$ , we know $𝑝 𝑟 ∣ | 𝑃 |$ . Now

(𝑃) 𝜌 𝑠 = {𝑔 \in 𝑃 : 𝑠 𝑔 = 𝑠} = 1

and $(𝑃) 𝜌$ acting on $𝑆$ is free, so each $(𝑃) 𝜌$ -orbit of $𝑆$ has size $| 𝑃 |$ . Herefore $| 𝑃 | ∣ | 𝑆 |$ and thus $𝑃 \in S y l 𝑝 (𝐺)$ .

To recap, if $𝑆 \in Ω$ and $∣ 𝑆 (𝐺) 𝜌 ∣$ is not divisible by $𝑝$ , then $(𝐺) 𝜌 𝑆 \in S y l 𝑝 (𝐺)$ , proving ^S1.

Now let $𝐻$ be a $𝑝$ -subgroup of $𝐺$ , and $Σ$ be as above. Then $(𝐻) 𝜌$ acts on $Σ$ . The $(𝐻) 𝜌$ -orbits on $Σ$ have sizes which are powers of $𝑝$ . Since $| Σ |$ does not divide $𝑝$ , it follows there must exist a singleton orbit ${𝑇}$ , so $(𝐻) 𝜌 \leq (𝐺) 𝜌 𝑇$ , the latter being a Sylow $𝑝$ -subgroup, proving ^S2.

Let $𝑃 1, 𝑃 2 \in S y l 𝑝 (𝐺)$ . From directly above, $𝑃 1 = (𝐺) 𝜌 𝑇 1$ and $𝑃 2 = (𝐺) 𝜌 𝑇 2$ for some $𝑇 1, 𝑇 2 \in Σ \subseteq 𝐺$ , thus $𝑇 1 𝑔 = 𝑇 2$ for some $𝑔 \in 𝐺$ . It follows

𝑃 𝑔 1 = (𝐺) 𝜌 𝑇 1 𝑔 = (𝐺) 𝜌 𝑇 2 = 𝑃 2

proving ^S3.

$𝐺$ acts (transitively by ^S3) on $S y l 𝑝 (𝐺)$ by conjugation, and $| 𝐺 | = ∣ S y l 𝑝 (𝐺) ∣ | N 𝐺 (𝑃) |$ where $N 𝐺 (𝑃)$ is the normalizer of $𝑃 \in S y l 𝑝 (𝐺)$ by the Orbit-stabilizer theorem. Since $𝑃 \leq N 𝐺 (𝑃)$ , it follows $𝑝 𝑟$ divides $| N 𝐺 (𝑃) |$ , whence $∣ S y l 𝑝 (𝐺) ∣$ divides $𝑚$ .

$𝑃$ acts by conjugation on $S y l 𝑝 (𝐺)$ , and by ^S2 $𝑃 = (𝐺) 𝜌 𝑆$ for some $𝑆 \in Σ$ . The only orbit of size 1 is ${𝑃}$ : For if $𝑄 \in S y l 𝑝 (𝐺)$ then $𝑄 𝑔 = 𝑄$ for all $𝑔 \in 𝑃$ , so $𝑃 \leq N 𝐺 (𝑄)$ . Since $𝑃 𝑄 \leq N 𝐺 (𝑄)$ , and $𝑃 𝑄$ is a $𝑝$ -group, so $𝑃 = 𝑄$ .

All orbits of $𝑃$ on $S y l 𝑝 (𝐺)$ have size a $𝑝$ -power, therefore all orbits other than ${𝑃}$ have size divisible by $𝑝$ . Therefore $𝑛 𝑝 \equiv 𝑝 1$ , proving ^S4.

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