**Proof plan.** Write $|G| = p^a m$ with $p \nmid m$. We prove existence (Sylow 1) by acting on subsets of size $p^a$, conjugacy (Sylow 2) by having a $p$-subgroup act on cosets, and the congruence and divisibility of $n_p$ (Sylow 3) via a $P$-conjugation orbit argument. **Step 1: Existence (Sylow 1).** Let $\Omega$ be the collection of all subsets of $G$ of size $p^a$. Then \begin{align*} |\Omega| = \binom{p^a m}{p^a}. \end{align*} A combinatorial argument shows that for $0 \leq k < p^a$, the numbers $p^a m - k$ and $p^a - k$ are divisible by the same power of $p$. Consequently, $|\Omega|$ is coprime to $p$. The [group](/page/Group) $G$ acts on $\Omega$ by $g \cdot X = \{gx : x \in X\}$. By the [Orbit-Stabilizer Theorem](/theorems/845), the orbit sizes divide $|G| = p^a m$. Since $|\Omega|$ is coprime to $p$, some orbit must have size coprime to $p$; call the corresponding set $X$. Then $|G_X| \cdot |\operatorname{Orb}(X)| = p^a m$, and since $|\operatorname{Orb}(X)|$ is coprime to $p$, we have $p^a \mid |G_X|$. Conversely, for any $g \in G$ and $x \in X$, the inclusion $g \in (gx^{-1}) \cdot X$ shows $G = \bigcup_{Y \in \operatorname{Orb}(X)} Y$. As this union has at most $|\operatorname{Orb}(X)| \cdot |X| = |\operatorname{Orb}(X)| \cdot p^a$ elements, we get $|G_X| = p^a m / |\operatorname{Orb}(X)| \leq p^a$. Combining with $p^a \mid |G_X|$ gives $|G_X| = p^a$, so $G_X$ is a Sylow $p$-subgroup. **Step 2: Conjugacy (Sylow 2).** We prove the stronger statement: if $Q \leq G$ is any $p$-subgroup and $P \in \operatorname{Syl}_p(G)$, then $g^{-1}Qg \leq P$ for some $g \in G$. Let $Q$ act on $G/P$ by $q \cdot gP = qgP$. The orbit sizes divide $|Q| = p^b$, so are powers of $p$. Since $|G/P| = m$ is coprime to $p$, there must be an orbit of size $1$, say $\{gP\}$. This means $qgP = gP$ for all $q \in Q$, so $g^{-1}qg \in P$ for all $q \in Q$, i.e. $g^{-1}Qg \leq P$. Applying this when $Q$ is another Sylow $p$-subgroup: $g^{-1}Qg \leq P$, and since $|g^{-1}Qg| = |Q| = p^a = |P|$, equality holds: all Sylow $p$-subgroups are conjugate. **Step 3: Congruence $n_p \equiv 1 \pmod{p}$ (Sylow 3, first part).** Let $P \in \operatorname{Syl}_p(G)$. The group $P$ acts on $\operatorname{Syl}_p(G)$ by conjugation; orbit sizes divide $|P| = p^a$, so are either $1$ or divisible by $p$. The orbit $\{P\}$ has size $1$. Suppose $\{Q\}$ is another orbit of size $1$, i.e. $p^{-1}Qp = Q$ for all $p \in P$, so $P \leq N_G(Q)$. Both $P$ and $Q$ are Sylow $p$-subgroups of $N_G(Q)$, so by Step 2 they are conjugate in $N_G(Q)$. But conjugation in $N_G(Q)$ fixes $Q$ by definition, so $P = Q$. Thus $\{P\}$ is the only orbit of size $1$, and \begin{align*} n_p = |\operatorname{Syl}_p(G)| \equiv 1 \pmod{p}. \end{align*} **Step 4: Divisibility $n_p \mid m$ (Sylow 3, second part).** By Step 2, $G$ acts transitively on $\operatorname{Syl}_p(G)$ by conjugation. By the [Orbit-Stabilizer Theorem](/theorems/845), $n_p = |\operatorname{Syl}_p(G)|$ divides $|G| = p^a m$. Since $n_p \equiv 1 \pmod{p}$, $p \nmid n_p$, so $n_p \mid m$. $\square$