Sylow theorems
Definition (Sylow subgroup)
Let \(p\) be a prime and \(G\) a group. We say that \(H\leq G\) is a Sylow \(p-\)subgroup if \(\lvert H \rvert = p^{e}\) and \(p^{e}\) is the largest power of \(p\) dividing \(\lvert G \rvert\).
Example (Sylow in \(S_{4}\))
Consider \(S_{4}\). Note that \(\lvert S_{4} \rvert=4! = 2^{3} \cdot 3\). A Sylow \(2-\)subgroup has size \(2^{3}\). A Sylow \(3-\)subgroup has size \(3\). A Sylow \(5\)-subgroup is \(\{ e \}\).
Theorem (First Sylow theorem)
If \(p\) is prime and \(G\) is finite, there exists a Sylow \(p-\)subgroup.
Proof: Let \(\lvert G \rvert = p^{e}k\) be such that \(p \not\mid k\). Let \(\Omega:=\{ T \subseteq G: \lvert T \rvert=p^{e} \}\). Consider the action \(\alpha\) of \(G\) on \(\Omega\) such that \(\alpha _g(T) = gT\). (it is left as an exercise to check that it is an action). We have that \(\lvert \Omega \rvert = \binom{p^{e}k}{p^{e}} \equiv k \mod p\). Since \(p\) does not divide \(k\), we have that \(p\ \not\mid \Omega\). We know that
\[\Omega = \dot{\bigcup}_{i \in I} \mathcal{O}_{i},\]where the \(\mathcal{O}_{i}\) are the distinct orbits. Hence \(\lvert \Omega \rvert = \sum_{i \in I} \lvert \mathcal{O}_{i} \rvert\). Therefore, there exists an orbit \(\mathcal{O}_{i}\) such that \(p \not\mid \lvert \mathcal{O}_{i} \rvert\). Hence, there exists \(H \in \Omega\) such that \(p \not\mid \mathcal{O}_{H}\) (taking \(H\) as a representative of the class of \(\mathcal{O}_{i}\)) . We have that \(\lvert \mathcal{O}_{H} \rvert = [G:G_{H}] = \frac{\lvert G \rvert}{\lvert G_{H} \rvert} = \frac{p^{e}k}{\lvert G_{H}. \rvert}\) From this, we have that \(\lvert \mathcal{O}_{H} \rvert \mid p^{e}k \implies \lvert \mathcal{O}_{H} \rvert \mid k\). Let \(A = \bigcup_{T \in \mathcal{O}_{H}} T = \bigcup_{g \in G} gH\). We shall prove that \(G=A\). It is clear that \(A \subseteq G\). We shall first show that \(A\) is stable under left translations. Let \(g \in G\), \(T \in \mathcal{O}_{H}\). We must show that \(gA \subseteq A\). Given \(x \in A\), there exists \(T \in \mathcal{O}_{H}\) such that \(x \in T\) and therefore \(gx \in gT \subseteq A\), from which the result follows. Finally, take \(h_{0} \in H\) (it exists since \(H \neq \emptyset\)); then \(1_{G} \in h_{0}^{-1}H \in \mathcal{O}_{H}\), so that \(1_{G} \in A\). Given any \(g \in G\), stability under translations gives \(g = g \cdot 1_{G} \in gA \subseteq A\). Conclude that \(G \subseteq A\) and therefore \(G=A\). Thus, we have that
\[p^{e}k = \lvert G \rvert = \lvert A \rvert =\left\lvert \bigcup_{T \in \mathcal{O}_{H}}T \right\rvert \leq \sum_{T \in \mathcal{O}_{H}} \lvert T \rvert = \lvert \mathcal{O}_{H} \rvert \cdot \lvert T \rvert = \lvert \mathcal{O}_{H} \rvert \cdot p^{e},\]and therefore \(k\leq \lvert \mathcal{O}_{H} \rvert\). But \(\lvert \mathcal{O}_{H} \rvert \mid k\), whence \(\lvert \mathcal{O}_{H} \rvert = k\). Finally, since
\[\lvert \mathcal{O}_{H} \rvert \cdot \lvert G_{H} \rvert = p^{e} \cdot k \implies \lvert G_{H} \rvert = \frac{p^{e}k}{k} = p^{e}.\]Thus, \(\lvert G_{H} \rvert = p^{e}\) and it is a group.
Theorem (Second Sylow theorem)
Let \(G\) be a finite group and \(p\) prime. If \(P\) and \(Q\) are Sylow \(p-\)groups, then they are conjugate, i.e., there exists \(g \in G\) such that \(gPg^{-1} = Q\).
Proof: Exercise. Hint: Consider the action \(\alpha:P \times G / Q \to G / Q\) such that \(\alpha_{g}(hQ) = ghQ\).
Theorem (Third Sylow theorem)
Let \(G\) be finite and \(p\) prime and \(n_{p}\) the number of Sylow \(p-\)groups. Then \(p \mid n_{p}-1\).
Proof: Hint: Define \(\Lambda_{p}: \{ H \subseteq G: H \text{ is a Sylow }p-\text{subgroup} \}\). Consider the action \(\alpha: G \times\Lambda_{p} \to \Lambda_{p}\) such that \(\alpha_{g}(P) = gPg^{-1}\).