[proofplan] We let $P$ act by left multiplication on the set of left cosets $G/Q$ and analyse the fixed points of this action. A counting argument modulo $p$ shows that at least one coset $gQ$ is fixed by every element of $P$, which forces $P \le gQg^{-1}$. Since $P$ and $gQg^{-1}$ are both Sylow $p$-subgroups and therefore have the same order, the inclusion is an equality, giving $P = gQg^{-1}$. [/proofplan] [step:Set up the action of $P$ on the left coset space $G/Q$] Let $P$ and $Q$ be Sylow $p$-subgroups of $G$. Write $|G| = p^a m$ where $a \ge 1$ and $p \nmid m$, so that $|P| = |Q| = p^a$. Define the left coset space \begin{align*} G/Q := \{gQ : g \in G\}. \end{align*} By [Lagrange's Theorem](/theorems/841), the number of left cosets is $|G/Q| = [G : Q] = |G|/|Q| = p^a m / p^a = m$. Since $p \nmid m$, we have $p \nmid |G/Q|$. Define an action of $P$ on $G/Q$ by left multiplication: \begin{align*} P \times G/Q &\to G/Q \\ (x, gQ) &\mapsto x \cdot gQ := (xg)Q. \end{align*} This is well-defined: if $gQ = g'Q$, then $g' = gq$ for some $q \in Q$, so $(xg')Q = (xgq)Q = (xg)Q$. The map satisfies $e \cdot gQ = gQ$ and $(x_1 x_2) \cdot gQ = x_1 \cdot (x_2 \cdot gQ)$ for all $x_1, x_2 \in P$, so it is a group action. [guided] The strategy for proving that two Sylow $p$-subgroups are conjugate is to find a group action whose orbit structure forces the conjugacy relation. The key choice is to let $P$ act on the coset space $G/Q$ by left multiplication, rather than on $G$ itself or on some other set. Why cosets of $Q$? Because a fixed point of this action will directly encode the conjugacy relation $P \le gQg^{-1}$. Let $P$ and $Q$ be Sylow $p$-subgroups of $G$. Write $|G| = p^a m$ where $a \ge 1$ and $\gcd(p, m) = 1$, so that $|P| = |Q| = p^a$ (both are subgroups of order $p^a$, which is the largest power of $p$ dividing $|G|$). Define the left coset space \begin{align*} G/Q := \{gQ : g \in G\}. \end{align*} By [Lagrange's Theorem](/theorems/841), the number of left cosets is \begin{align*} |G/Q| = [G : Q] = \frac{|G|}{|Q|} = \frac{p^a m}{p^a} = m. \end{align*} Since $p \nmid m$, the size of the set being acted upon is not divisible by $p$. This will be the crucial arithmetic fact in the counting argument. Now define an action of $P$ on $G/Q$ by left multiplication: \begin{align*} P \times G/Q &\to G/Q \\ (x, gQ) &\mapsto x \cdot gQ := (xg)Q. \end{align*} We verify this is well-defined: if $gQ = g'Q$, then $g^{-1}g' \in Q$, so $g' = gq$ for some $q \in Q$, and $(xg')Q = (xgq)Q = (xg)Q$ since $q \in Q$. The identity acts as required: $e \cdot gQ = (eg)Q = gQ$. Associativity holds: $(x_1 x_2) \cdot gQ = (x_1 x_2 g)Q = x_1 \cdot (x_2 g Q) = x_1 \cdot (x_2 \cdot gQ)$. So this is indeed a group action. [/guided] [/step] [step:Count fixed points modulo $p$ to find a coset stabilised by all of $P$] Let $(G/Q)^P := \{gQ \in G/Q : x \cdot gQ = gQ \text{ for all } x \in P\}$ denote the set of fixed points of the $P$-action. By the [Orbit-Stabiliser Theorem](/theorems/845), the action of $P$ on $G/Q$ partitions $G/Q$ into orbits, and each orbit has size $|P|/|P_{gQ}|$ where $P_{gQ} = \{x \in P : xgQ = gQ\}$ is the stabiliser of $gQ$ in $P$. Since $|P| = p^a$, every orbit size divides $p^a$ and is therefore either $1$ (a fixed point) or a positive power of $p$. Summing over all orbits: \begin{align*} |G/Q| = |(G/Q)^P| + \sum_{\substack{\text{orbits } \mathcal{O} \\ |\mathcal{O}| > 1}} |\mathcal{O}|. \end{align*} Each term in the sum is divisible by $p$. Since $|G/Q| = m$ and $p \nmid m$, reducing modulo $p$ gives \begin{align*} |(G/Q)^P| \equiv m \not\equiv 0 \pmod{p}. \end{align*} In particular, $(G/Q)^P \neq \varnothing$: there exists at least one coset $gQ$ fixed by every element of $P$. [guided] We now exploit the fact that $P$ is a $p$-group acting on a set whose size is coprime to $p$. This is a standard technique: when a $p$-group acts on a finite set, the number of fixed points is congruent to the size of the set modulo $p$. Let $(G/Q)^P := \{gQ \in G/Q : x \cdot gQ = gQ \text{ for all } x \in P\}$ denote the set of fixed points. The action of $P$ partitions $G/Q$ into disjoint orbits. By the [Orbit-Stabiliser Theorem](/theorems/845), the orbit of $gQ$ has size \begin{align*} |P \cdot gQ| = \frac{|P|}{|P_{gQ}|} = \frac{p^a}{|P_{gQ}|} \end{align*} where $P_{gQ} = \{x \in P : xgQ = gQ\}$ is the stabiliser of $gQ$ in $P$. The stabiliser $P_{gQ}$ is a subgroup of $P$, so by [Lagrange's Theorem](/theorems/841), $|P_{gQ}|$ divides $|P| = p^a$, which means $|P_{gQ}| = p^j$ for some $0 \le j \le a$. The orbit size is therefore $p^{a-j}$, which is either $1$ (when $j = a$, i.e., $P_{gQ} = P$, meaning $gQ$ is a fixed point) or a positive power of $p$ (when $j < a$). Summing over all orbits gives the orbit-counting decomposition: \begin{align*} |G/Q| = |(G/Q)^P| + \sum_{\substack{\text{orbits } \mathcal{O} \\ |\mathcal{O}| > 1}} |\mathcal{O}|. \end{align*} Every non-singleton orbit has size divisible by $p$, so the entire sum on the right is divisible by $p$. Since $|G/Q| = m$ and $p \nmid m$, reducing modulo $p$: \begin{align*} |(G/Q)^P| \equiv m \pmod{p}. \end{align*} Since $p \nmid m$, we have $|(G/Q)^P| \not\equiv 0 \pmod{p}$, so in particular $(G/Q)^P \neq \varnothing$. This is the key point: the arithmetic forces at least one fixed point to exist. What does a fixed point mean concretely? It means there exists some $g \in G$ such that $xgQ = gQ$ for every $x \in P$. [/guided] [/step] [step:Translate the fixed-point condition into the inclusion $P \le gQg^{-1}$] Let $gQ$ be a fixed point of the $P$-action, so $xgQ = gQ$ for all $x \in P$. The condition $xgQ = gQ$ means that $xg$ and $g$ lie in the same left coset of $Q$, which holds if and only if $g^{-1}(xg) \in Q$, i.e., $g^{-1}xg \in Q$. Since this holds for every $x \in P$, we have \begin{align*} g^{-1}Pg \subseteq Q. \end{align*} Equivalently, $P \subseteq gQg^{-1}$. Now $gQg^{-1}$ is a subgroup of $G$ conjugate to $Q$, so $|gQg^{-1}| = |Q| = p^a = |P|$. Since $P \subseteq gQg^{-1}$ and both sets have the same finite cardinality $p^a$, we conclude \begin{align*} P = gQg^{-1}. \end{align*} This shows $P$ and $Q$ are conjugate in $G$. For the equivalent formulation: $G$ acts on $\operatorname{Syl}_p(G)$ by conjugation, $(g, R) \mapsto gRg^{-1}$. Since any two Sylow $p$-subgroups $P$ and $Q$ satisfy $Q = gPg^{-1}$ for some $g \in G$, every pair lies in the same orbit, so the action is transitive. [guided] We now decode what a fixed point of the action tells us about the relationship between $P$ and $Q$. Let $gQ$ be a fixed point, so $x \cdot gQ = gQ$ for all $x \in P$. By definition of the action, $x \cdot gQ = (xg)Q$, so the condition $xgQ = gQ$ means that $xg$ and $g$ lie in the same left coset of $Q$. Two elements lie in the same left coset of $Q$ if and only if their "ratio" belongs to $Q$: $xgQ = gQ$ iff $g^{-1}(xg) \in Q$, i.e., $g^{-1}xg \in Q$. Since $gQ$ is fixed by every element of $P$, this holds for all $x \in P$: \begin{align*} g^{-1}xg \in Q \quad \text{for all } x \in P. \end{align*} In other words, $g^{-1}Pg \subseteq Q$, or equivalently $P \subseteq gQg^{-1}$. Now we use the fact that both $P$ and $Q$ are Sylow $p$-subgroups. The subgroup $gQg^{-1}$ is conjugate to $Q$, so $|gQg^{-1}| = |Q| = p^a$. We also have $|P| = p^a$. The inclusion $P \subseteq gQg^{-1}$ between two finite sets of the same cardinality forces equality: \begin{align*} P = gQg^{-1}. \end{align*} This is precisely the statement that $P$ and $Q$ are conjugate in $G$. For the equivalent formulation: the group $G$ acts on the set $\operatorname{Syl}_p(G)$ of all Sylow $p$-subgroups by conjugation, $g \cdot P = gPg^{-1}$. We have shown that for any two Sylow $p$-subgroups $P$ and $Q$, there exists $g \in G$ with $Q = gPg^{-1}$, i.e., $Q$ lies in the orbit of $P$. Since $P$ and $Q$ were arbitrary, there is only one orbit, which means the action is transitive. [/guided] [/step]