Sylow's Theorems

We prove each part in turn.

Proof of [1] (Existence via counting): Let $\mathcal{S}$ denote the collection of all subsets of $G$ having exactly $p^n$ elements. Then $|\mathcal{S}| = \binom{|G|}{p^n} = \binom{p^n m}{p^n}$. We analyze this binomial coefficient modulo $p$. Writing $\binom{p^n m}{p^n} = \frac{(p^n m)!}{(p^n)!(p^n m - p^n)!} = \frac{(p^n m)!}{(p^n)!(p^n(m-1))!}$, we use the fact that the highest power of $p$ dividing $k!$ is $\sum_{i=1}^{\infty} \lfloor k/p^i \rfloor$. For the numerator: $v_p((p^n m)!) = \sum_{i=1}^{\infty} \lfloor p^n m / p^i \rfloor = p^{n-1}m + p^{n-2}m + \cdots + m + \lfloor m/p \rfloor + \cdots$ For the denominator: $v_p((p^n)!) = p^{n-1} + p^{n-2} + \cdots + 1$ and $v_p((p^n(m-1))!) = p^{n-1}(m-1) + \cdots$ Computing $v_p(\binom{p^n m}{p^n}) = v_p((p^n m)!) - v_p((p^n)!) - v_p((p^n(m-1))!)$, the $p^{n-1}$ terms give $p^{n-1}m - p^{n-1} - p^{n-1}(m-1) = 0$, and similarly for all higher powers of $p$. The remaining terms involve only $m$ and $m-1$, which are coprime to $p$, so $v_p(\binom{p^n m}{p^n}) = 0$. Thus $p \nmid \binom{p^n m}{p^n}$. Now let $G$ act on $\mathcal{S}$ by left multiplication: $g \cdot S = \{gs : s \in S\}$. Since $p \nmid |\mathcal{S}|$, there exists at least one orbit whose size is not divisible by $p$. Let $S_0$ be a representative of such an orbit, and let $H = \{g \in G : gS_0 = S_0\}$ be its stabilizer. By the orbit-stabilizer theorem, $|G| = |\text{Orb}(S_0)| \cdot |H|$, so $p^n m = |\text{Orb}(S_0)| \cdot |H|$. Since $p \nmid |\text{Orb}(S_0)|$, we have $p^n \mid |H|$. We claim $|H| \leq p^n$. Note that $H \subseteq S_0$ (since if $h \in H$, then $hS_0 = S_0$, and $e \in S_0$, so $h = he \in hS_0 = S_0$). Thus $|H| \leq |S_0| = p^n$. Therefore $|H| = p^n$, and $H$ is a Sylow $p$-subgroup. For the "moreover" clause, let $P$ be a Sylow $p$-subgroup and $p^k \mid |G|$ with $k \leq n$. By Cauchy's theorem applied repeatedly, $P$ contains an element of order $p$, hence a subgroup of order $p$. Inductively, we can show $P$ contains a subgroup of order $p^k$ for each $k \leq n$.

Proof of [2] (Conjugacy): Let $P \in \text{Syl}_p(G)$ and let $Q$ be any $p$-subgroup of $G$ (not necessarily maximal). We show that $Q \subseteq gPg^{-1}$ for some $g \in G$. Let $Q$ act on the set $G/P = \{gP : g \in G\}$ of left cosets by left multiplication. Since $|G/P| = m$ and $\gcd(p,m) = 1$, and $Q$ is a $p$-group, there must exist a fixed point: some coset $gP$ such that $qgP = gP$ for all $q \in Q$. This means $g^{-1}qg \in P$ for all $q \in Q$, so $Q \subseteq gPg^{-1}$. Now if $Q$ is also a Sylow $p$-subgroup, then $|Q| = p^n = |P| = |gPg^{-1}|$, so $Q = gPg^{-1}$.

Proof of [3] (Counting): Let $P \in \text{Syl}_p(G)$. By (2), $G$ acts transitively on $\text{Syl}_p(G)$ by conjugation. By the orbit-stabilizer theorem, $n_p = [G : N_G(P)]$ where $N_G(P) = \{g \in G : gPg^{-1} = P\}$ is the normalizer of $P$. Since $P \subseteq N_G(P)$, we have $|N_G(P)| = p^n \cdot k$ for some integer $k$, so $n_p = \frac{p^n m}{p^n k} = \frac{m}{k}$. Thus $n_p \mid m$. For the congruence, let $P$ act on $\text{Syl}_p(G)$ by conjugation. Since $P$ is a $p$-group, each orbit has size a power of $p$. The orbit of $P$ itself is $\{P\}$ (size 1) since $pPp^{-1} = P$ for all $p \in P$. If $Q \in \text{Syl}_p(G)$ with $Q \neq P$, we claim the orbit of $Q$ under $P$-conjugation has size divisible by $p$. The stabilizer is $\{p \in P : pQp^{-1} = Q\} = P \cap N_G(Q)$. If this equals $P$, then $P \subseteq N_G(Q)$, so $P, Q \subseteq N_G(Q)$. But $P$ and $Q$ are both Sylow $p$-subgroups of $N_G(Q)$, so by conjugacy within $N_G(Q)$, there exists $n \in N_G(Q)$ with $Q = nPn^{-1}$. But $n \in N_G(Q)$ means $nQn^{-1} = Q$, so $nPn^{-1} = Q$ gives $P = Q$, contradiction. Thus $P \cap N_G(Q) \subsetneq P$, so $|P \cap N_G(Q)| \leq p^{n-1}$, and the orbit size $[P : P \cap N_G(Q)] \geq p$. Therefore $n_p = 1 + (\text{sum of orbit sizes} \geq p) \equiv 1 \pmod{p}$. □