[proofplan] We prove the two conclusions separately. For the divisibility $n_p \mid m$: by [Sylow's Second Theorem](/theorems/3250), the conjugation action of $G$ on the set $\operatorname{Syl}_p(G)$ of all Sylow $p$-subgroups is transitive, so $n_p = [G : N_G(P)]$ for any fixed Sylow $p$-subgroup $P$. Since $P \le N_G(P)$, Lagrange gives $n_p \mid m$. For the congruence $n_p \equiv 1 \pmod{p}$: we let a fixed Sylow $p$-subgroup $P$ act on $\operatorname{Syl}_p(G)$ by conjugation and show that $P$ itself is the unique fixed point of this action, so the fixed-point count is $1$ and the $p$-group orbit-counting argument gives $n_p \equiv 1 \pmod{p}$. [/proofplan] [step:Express $n_p$ as the index $[G : N_G(P)]$ via transitivity of the conjugation action] Let $\operatorname{Syl}_p(G) := \{P \le G : |P| = p^a\}$ denote the set of all Sylow $p$-subgroups of $G$, so that $n_p = |\operatorname{Syl}_p(G)|$. The group $G$ acts on $\operatorname{Syl}_p(G)$ by conjugation: \begin{align*} G \times \operatorname{Syl}_p(G) &\to \operatorname{Syl}_p(G) \\ (g, Q) &\mapsto gQg^{-1}. \end{align*} This is well-defined since conjugation preserves subgroup order: $|gQg^{-1}| = |Q| = p^a$. By [Sylow's Second Theorem](/theorems/3250), any two Sylow $p$-subgroups are conjugate, so this action is transitive: there is a single orbit, namely $\operatorname{Syl}_p(G)$ itself. Fix a Sylow $p$-subgroup $P \in \operatorname{Syl}_p(G)$. The stabiliser of $P$ under the conjugation action is the normaliser $N_G(P) = \{g \in G : gPg^{-1} = P\}$. By the [Orbit-Stabiliser Theorem](/theorems/845), the size of the orbit of $P$ equals the index of its stabiliser: \begin{align*} n_p = |\operatorname{Syl}_p(G)| = |G \cdot P| = [G : N_G(P)] = \frac{|G|}{|N_G(P)|}. \end{align*} [guided] The first goal is to express $n_p$ as a group-theoretic index, which will make the divisibility $n_p \mid m$ transparent. The natural action to consider is $G$ acting on $\operatorname{Syl}_p(G)$ by conjugation, since [Sylow's Second Theorem](/theorems/3250) tells us precisely that this action is transitive. Define $\operatorname{Syl}_p(G) := \{P \le G : |P| = p^a\}$, the set of all Sylow $p$-subgroups. The conjugation action \begin{align*} G \times \operatorname{Syl}_p(G) &\to \operatorname{Syl}_p(G) \\ (g, Q) &\mapsto gQg^{-1} \end{align*} is well-defined because conjugation is an automorphism of $G$, and automorphisms preserve subgroup order: $|gQg^{-1}| = |Q| = p^a$, so $gQg^{-1} \in \operatorname{Syl}_p(G)$. It is a group action: $eQe^{-1} = Q$ and $(g_1 g_2)Q(g_1 g_2)^{-1} = g_1(g_2 Q g_2^{-1})g_1^{-1}$. By [Sylow's Second Theorem](/theorems/3250), for any $P, Q \in \operatorname{Syl}_p(G)$, there exists $g \in G$ with $Q = gPg^{-1}$. This is exactly the statement that the conjugation action is transitive: $\operatorname{Syl}_p(G)$ consists of a single orbit. Now fix any $P \in \operatorname{Syl}_p(G)$. The stabiliser of $P$ under conjugation is \begin{align*} \operatorname{Stab}_G(P) = \{g \in G : gPg^{-1} = P\} = N_G(P), \end{align*} which is precisely the normaliser of $P$ in $G$. By the [Orbit-Stabiliser Theorem](/theorems/845), the orbit size equals the index of the stabiliser: \begin{align*} n_p = |G \cdot P| = [G : N_G(P)] = \frac{|G|}{|N_G(P)|}. \end{align*} Why is this useful? Because we can now read off divisibility properties of $n_p$ from the position of $N_G(P)$ in the subgroup lattice of $G$. [/guided] [/step] [step:Deduce the divisibility $n_p \mid m$ from the inclusion $P \le N_G(P)$] Since $P$ normalises itself (every subgroup is contained in its own normaliser), we have $P \le N_G(P)$. By [Lagrange's Theorem](/theorems/841), $|P|$ divides $|N_G(P)|$, so $p^a \mid |N_G(P)|$. Write $|N_G(P)| = p^a \ell$ for some positive integer $\ell$. Then \begin{align*} n_p = \frac{|G|}{|N_G(P)|} = \frac{p^a m}{p^a \ell} = \frac{m}{\ell}. \end{align*} Since $n_p = m / \ell$ is an integer, $\ell \mid m$, and therefore $n_p \mid m$. [guided] We now prove the first conclusion: $n_p \mid m$. From the previous step, $n_p = [G : N_G(P)] = |G| / |N_G(P)|$. The key observation is that $P$ is a subgroup of its own normaliser. This is immediate from the definition: for any $x \in P$, the conjugate $xPx^{-1} = P$ (since $P$ is closed under its own group operation), so $x \in N_G(P)$. Thus $P \le N_G(P)$. By [Lagrange's Theorem](/theorems/841), the order of a subgroup divides the order of the group, so $|P| = p^a$ divides $|N_G(P)|$. Write $|N_G(P)| = p^a \ell$ for some positive integer $\ell$. Substituting into the formula for $n_p$: \begin{align*} n_p = \frac{|G|}{|N_G(P)|} = \frac{p^a m}{p^a \ell} = \frac{m}{\ell}. \end{align*} Since $n_p$ is a positive integer (it counts the Sylow $p$-subgroups), $\ell$ divides $m$. Therefore $n_p = m / \ell$ divides $m$. Note what this argument uses: the transitivity from [Sylow's Second Theorem](/theorems/3250) (to get $n_p = [G : N_G(P)]$) and the containment $P \le N_G(P)$ (to extract the factor $p^a$ from both numerator and denominator). The divisibility $n_p \mid m$ says that the number of Sylow $p$-subgroups is bounded by and divides the "$p'$-part" of $|G|$. [/guided] [/step] [step:Let $P$ act on $\operatorname{Syl}_p(G)$ by conjugation and identify the fixed points] We now prove the congruence $n_p \equiv 1 \pmod{p}$. Define an action of the fixed Sylow $p$-subgroup $P$ on $\operatorname{Syl}_p(G)$ by conjugation: \begin{align*} P \times \operatorname{Syl}_p(G) &\to \operatorname{Syl}_p(G) \\ (x, Q) &\mapsto xQx^{-1}. \end{align*} This is the restriction of the $G$-action to the subgroup $P \le G$. Let $\operatorname{Syl}_p(G)^P := \{Q \in \operatorname{Syl}_p(G) : xQx^{-1} = Q \text{ for all } x \in P\}$ denote the set of fixed points. [claim:$P$ is the unique fixed point of the $P$-action on $\operatorname{Syl}_p(G)$] Let $Q \in \operatorname{Syl}_p(G)^P$, so $xQx^{-1} = Q$ for all $x \in P$. This means $P \le N_G(Q)$. Since $Q \trianglelefteq N_G(Q)$ (every subgroup is normal in its normaliser), the product $PQ$ is a subgroup of $N_G(Q)$: we have $PQ = QP$ because $Q \trianglelefteq N_G(Q)$ and $P \le N_G(Q)$, so $PQ$ is closed under multiplication and inversion. By the [Second Isomorphism Theorem](/theorems/843), $|PQ| = |P| \cdot |Q| / |P \cap Q|$. Since $P$ and $Q$ are both $p$-groups, $P \cap Q$ is a $p$-group, so $|P \cap Q| = p^b$ for some $0 \le b \le a$. Therefore \begin{align*} |PQ| = \frac{p^a \cdot p^a}{p^b} = p^{2a - b}. \end{align*} Since $PQ \le N_G(Q) \le G$ and $|G| = p^a m$ with $p \nmid m$, by [Lagrange's Theorem](/theorems/841) we have $|PQ| \mid |G|$. The highest power of $p$ dividing $|G|$ is $p^a$, so $p^{2a-b} \mid p^a m$. Since $p \nmid m$, we need $2a - b \le a$, i.e., $b \ge a$. Combined with $b \le a$, this gives $b = a$, so $|P \cap Q| = p^a = |P| = |Q|$. Since $P \cap Q \subseteq P$ and $|P \cap Q| = |P|$, we conclude $P \cap Q = P$, i.e., $P \subseteq Q$. By the same cardinality argument $P = Q$. [/claim] [proof] Proved within the claim statement above. [/proof] Therefore $\operatorname{Syl}_p(G)^P = \{P\}$, and in particular $|\operatorname{Syl}_p(G)^P| = 1$. [guided] To prove $n_p \equiv 1 \pmod{p}$, we use the standard $p$-group fixed-point technique: restrict the conjugation action from $G$ to a Sylow $p$-subgroup $P$, and count fixed points modulo $p$. Define the action of $P$ on $\operatorname{Syl}_p(G)$ by conjugation: \begin{align*} P \times \operatorname{Syl}_p(G) &\to \operatorname{Syl}_p(G) \\ (x, Q) &\mapsto xQx^{-1}. \end{align*} This is well-defined (it is the restriction of the transitive $G$-action to $P$). The fixed points are those $Q \in \operatorname{Syl}_p(G)$ satisfying $xQx^{-1} = Q$ for all $x \in P$, i.e., those $Q$ normalised by $P$: $P \le N_G(Q)$. Which Sylow $p$-subgroups does $P$ normalise? Certainly $P$ normalises itself, so $P \in \operatorname{Syl}_p(G)^P$. We claim $P$ is the *only* fixed point. Suppose $Q \in \operatorname{Syl}_p(G)^P$, so $P \le N_G(Q)$. Since $Q$ is normal in its normaliser $N_G(Q)$ (by definition of normaliser), and $P$ is also a subgroup of $N_G(Q)$, the product $PQ$ is a subgroup of $N_G(Q)$. Why? Because $Q \trianglelefteq N_G(Q)$ implies $PQ = QP$ (cosets of a normal subgroup commute), so $PQ$ is closed under multiplication: $(p_1 q_1)(p_2 q_2) = p_1 (q_1 p_2) q_2$, and $q_1 p_2 \in QP = PQ$, so the product stays in $PQ$. Similarly for inverses. Now we compute $|PQ|$. By the product formula (a consequence of the [Second Isomorphism Theorem](/theorems/843)): \begin{align*} |PQ| = \frac{|P| \cdot |Q|}{|P \cap Q|}. \end{align*} Both $P$ and $Q$ have order $p^a$. The intersection $P \cap Q$ is a subgroup of the $p$-group $P$, so $|P \cap Q| = p^b$ for some $0 \le b \le a$. Thus $|PQ| = p^{2a-b}$. Since $PQ \le N_G(Q) \le G$, [Lagrange's Theorem](/theorems/841) gives $|PQ| \mid |G| = p^a m$. The $p$-part of $|G|$ is exactly $p^a$, and since $p \nmid m$, the highest power of $p$ that can divide $|PQ|$ is $p^a$. So $2a - b \le a$, giving $b \ge a$. Since $b \le a$ as well, $b = a$. This means $|P \cap Q| = p^a = |P|$. Since $P \cap Q \subseteq P$ and both have the same cardinality, $P \cap Q = P$, so $P \subseteq Q$. By the same reasoning (or by equal cardinalities), $P = Q$. Therefore $\operatorname{Syl}_p(G)^P = \{P\}$: the unique fixed point is $P$ itself. [/guided] [/step] [step:Apply the $p$-group orbit-counting argument to conclude $n_p \equiv 1 \pmod{p}$] The action of $P$ (a $p$-group of order $p^a$) on $\operatorname{Syl}_p(G)$ partitions $\operatorname{Syl}_p(G)$ into orbits. By the [Orbit-Stabiliser Theorem](/theorems/845), each orbit has size $|P| / |P_Q| = p^a / |P_Q|$ where $P_Q = \{x \in P : xQx^{-1} = Q\}$ is the stabiliser of $Q$ in $P$. Since $|P_Q|$ divides $|P| = p^a$, every orbit size is a power of $p$: either $1$ (a fixed point) or $p^k$ for some $k \ge 1$. Decomposing by orbits: \begin{align*} n_p = |\operatorname{Syl}_p(G)^P| + \sum_{\substack{\text{orbits } \mathcal{O} \\ |\mathcal{O}| > 1}} |\mathcal{O}|. \end{align*} Each non-singleton orbit contributes a term divisible by $p$. From the previous step, $|\operatorname{Syl}_p(G)^P| = 1$. Reducing modulo $p$: \begin{align*} n_p \equiv 1 \pmod{p}. \end{align*} This completes the proof: $n_p \mid m$ (from the second step) and $n_p \equiv 1 \pmod{p}$ (from the orbit-counting argument). [guided] We now assemble the conclusion. The $P$-action on $\operatorname{Syl}_p(G)$ partitions the set into disjoint orbits. By the [Orbit-Stabiliser Theorem](/theorems/845), the orbit of any $Q \in \operatorname{Syl}_p(G)$ has size \begin{align*} |P \cdot Q| = \frac{|P|}{|P_Q|} \end{align*} where $P_Q = \{x \in P : xQx^{-1} = Q\}$ is the stabiliser of $Q$ in $P$. Since $P_Q \le P$ and $|P| = p^a$, [Lagrange's Theorem](/theorems/841) gives $|P_Q| = p^j$ for some $0 \le j \le a$, so $|P \cdot Q| = p^{a-j}$. This is $1$ when $j = a$ (meaning $P_Q = P$, i.e., $Q$ is a fixed point) and a positive power of $p$ when $j < a$. Summing over all orbits: \begin{align*} n_p = |\operatorname{Syl}_p(G)^P| + \sum_{\substack{\text{orbits } \mathcal{O} \\ |\mathcal{O}| > 1}} |\mathcal{O}|. \end{align*} The sum on the right is a sum of positive powers of $p$, hence divisible by $p$. We showed in the previous step that $|\operatorname{Syl}_p(G)^P| = 1$, so \begin{align*} n_p \equiv 1 \pmod{p}. \end{align*} To summarise the two results: the divisibility $n_p \mid m$ came from the orbit-stabiliser analysis of the transitive $G$-action on $\operatorname{Syl}_p(G)$, which expressed $n_p = [G : N_G(P)]$ and then used $P \le N_G(P)$ to cancel the $p$-part. The congruence $n_p \equiv 1 \pmod{p}$ came from restricting the action to $P$ and showing that $P$ is the unique fixed point, so the $p$-group fixed-point count is exactly $1$. [/guided] [/step]