Show that $N(H)$ is the largest subgroup of $G$ in which $ H$ is normal.

Question

Let $G$ be a group and $H \leq G$. The normalizer of $H$ in $G$ is $$N(H) = \{g \in G|gHg^{−1} = H\}$$ If $H$ is a normal subgroup of $K \leq G$ then $K \leq N(H)$. Show that $N(H)$ is the largest subgroup of $G$ in which $ H$ is normal.

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Thanks

Answer 1

Let $x\in K$. Since $H\lhd K$, $xHx^{-1}=H$ which implies $x\in N(H)$. Therefore $K\subseteq N(H)$.

Answer 2

H is normal in K and K $\leq G $ Let $a \in K$ . Hence, $aH = Ha $ implies $ H = aHa^{-1} $ which implies $ a \in N(H)$. Hence, $K\subseteq N(H) $