[proofplan] We show that every element $g \in G$ belongs to the product $K \cdot N_G(P)$. Given $g \in G$, the conjugate $gPg^{-1}$ is a Sylow $p$-subgroup of $K$ (since $K$ is normal in $G$). By [Sylow's Second Theorem](/theorems/3250), $gPg^{-1}$ and $P$ are conjugate inside $K$, so there exists $k \in K$ with $gPg^{-1} = kPk^{-1}$. Then $k^{-1}g \in N_G(P)$, which gives $g = k(k^{-1}g) \in K \cdot N_G(P)$. [/proofplan] [step:Show that $gPg^{-1}$ is a Sylow $p$-subgroup of $K$ for every $g \in G$] Let $g \in G$. Since $K \trianglelefteq G$, we have $gKg^{-1} = K$. The map \begin{align*} \gamma_g: K &\to K \\ x &\mapsto gxg^{-1} \end{align*} is a group automorphism of $K$ (it is a homomorphism with inverse $\gamma_{g^{-1}}$). Since $P \le K$, the image $gPg^{-1} = \gamma_g(P)$ is a subgroup of $K$ with $|gPg^{-1}| = |P|$ (automorphisms preserve cardinality). Because $P$ is a Sylow $p$-subgroup of $K$, its order $|P| = p^a$ where $p^a$ is the largest power of $p$ dividing $|K|$. Therefore $gPg^{-1}$ is a subgroup of $K$ of order $p^a$, hence a Sylow $p$-subgroup of $K$. [guided] Why does conjugation by $g \in G$ send subgroups of $K$ to subgroups of $K$? This is precisely because $K$ is normal in $G$: normality means $gKg^{-1} = K$ for all $g \in G$. So if $P \le K$, then $gPg^{-1} \le gKg^{-1} = K$. This containment is the only place in the entire proof where normality of $K$ is used — it is the key hypothesis that makes the argument work. We now formalise conjugation by $g$ as an automorphism of $K$. Define \begin{align*} \gamma_g: K &\to K \\ x &\mapsto gxg^{-1}. \end{align*} This is well-defined as a map $K \to K$ (not merely $G \to G$) because $K \trianglelefteq G$ ensures $gKg^{-1} = K$. To see it is a homomorphism: $\gamma_g(xy) = gxyg^{-1} = (gxg^{-1})(gyg^{-1}) = \gamma_g(x)\gamma_g(y)$. It is bijective with two-sided inverse $\gamma_{g^{-1}}$, since $\gamma_{g^{-1}}(\gamma_g(x)) = g^{-1}(gxg^{-1})g = x$ and symmetrically. So $\gamma_g \in \operatorname{Aut}(K)$. Since $\gamma_g$ is an automorphism of $K$, it maps subgroups to subgroups and preserves their orders. Therefore $\gamma_g(P) = gPg^{-1}$ is a subgroup of $K$ and $|gPg^{-1}| = |P|$. Finally, write $|K| = p^a m$ with $p \nmid m$, so that $|P| = p^a$ (as $P$ is a Sylow $p$-subgroup of $K$). Then $gPg^{-1}$ is a subgroup of $K$ of order $p^a$, which is a Sylow $p$-subgroup of $K$ by definition. If $K$ were not normal in $G$, the conjugate $gPg^{-1}$ need not lie inside $K$ at all, and the argument would collapse at this first step. [/guided] [/step] [step:Conjugate $gPg^{-1}$ back to $P$ using an element of $K$] By [Sylow's Second Theorem](/theorems/3250) applied to the finite group $K$, any two Sylow $p$-subgroups of $K$ are conjugate in $K$. Since $P$ and $gPg^{-1}$ are both Sylow $p$-subgroups of $K$, there exists $k \in K$ such that \begin{align*} gPg^{-1} = kPk^{-1}. \end{align*} [guided] We now invoke [Sylow's Second Theorem](/theorems/3250), which states: in a finite group, every two Sylow $p$-subgroups are conjugate by an element of that group. We apply this theorem to the ambient group $K$. First, let us verify the hypotheses. Sylow's Second Theorem requires a finite group: $K$ is finite because $K \le G$ and $G$ is a finite group by hypothesis. We need two Sylow $p$-subgroups of $K$: these are $P$ (which is a Sylow $p$-subgroup of $K$ by hypothesis) and $gPg^{-1}$ (which is a Sylow $p$-subgroup of $K$ by the previous step). Applying the theorem to $K$, we obtain an element $k \in K$ such that $gPg^{-1} = kPk^{-1}$. The crucial point is that the conjugating element $k$ belongs to $K$, not merely to $G$. Sylow's Second Theorem produces a conjugating element inside the ambient group in which the theorem is applied, and here we have deliberately chosen $K$ as that ambient group. This is what will allow us, in the next step, to factor $g$ as a product $g = k \cdot (k^{-1}g)$ where the first factor $k$ lies in $K$ and the second factor $k^{-1}g$ turns out to normalise $P$. Why could we not simply apply Sylow's theorem to $G$? We could, but that would only give a conjugating element in $G$, not in $K$, and the decomposition $g \in K \cdot N_G(P)$ requires the first factor to come from $K$. [/guided] [/step] [step:Deduce that $g \in K \cdot N_G(P)$] From $gPg^{-1} = kPk^{-1}$, we obtain $(k^{-1}g)P(k^{-1}g)^{-1} = k^{-1}(gPg^{-1})k = k^{-1}(kPk^{-1})k = P$. Define $n := k^{-1}g$. Then $nPn^{-1} = P$, so $n \in N_G(P)$ (by definition, $N_G(P) = \{g \in G : gPg^{-1} = P\}$). Since $k \in K$ and $n \in N_G(P)$, we write \begin{align*} g = k \cdot n \in K \cdot N_G(P). \end{align*} Since $g \in G$ was arbitrary, $G \subseteq K \cdot N_G(P)$. The reverse inclusion $K \cdot N_G(P) \subseteq G$ holds because both $K$ and $N_G(P)$ are subsets of $G$. Therefore $G = K \cdot N_G(P)$. [guided] We want to show $g \in K \cdot N_G(P)$, i.e., we want to write $g = kn$ with $k \in K$ and $n \in N_G(P)$. We already have $k \in K$ with $gPg^{-1} = kPk^{-1}$. The natural candidate for $n$ is $k^{-1}g$: if this element normalises $P$, we are done. Let us verify. Set $n = k^{-1}g$. Then \begin{align*} nPn^{-1} = (k^{-1}g)P(k^{-1}g)^{-1} = k^{-1}(gPg^{-1})k = k^{-1}(kPk^{-1})k = P. \end{align*} So $nPn^{-1} = P$, which means $n \in N_G(P)$ by definition of the normaliser. Now $g = k \cdot (k^{-1}g) = k \cdot n$ with $k \in K$ and $n \in N_G(P)$, so $g \in K \cdot N_G(P)$. Since $g$ was an arbitrary element of $G$, we have shown $G \subseteq K \cdot N_G(P)$. The reverse inclusion is immediate: $K \subseteq G$ since $K \trianglelefteq G$, and $N_G(P) \subseteq G$ by definition, so every product $kn$ with $k \in K$, $n \in N_G(P)$ is an element of $G$. Therefore $G = K \cdot N_G(P)$. [/guided] [/step]