How to show that the normalizer is the largest subgroup of a group in which a set is normal?

Question

Assume we have $H \trianglelefteq K\subseteq G$

We want to prove the $K \subseteq N_G(H)$

However all I have so far is:

Assume $K \subseteq G$ such that $H$ is normal in $K$. Then it must be that $kHk^{-1} \subseteq H$

If $kHk^{-1} = H$ Then trivially $k \in N_G(H)$, but I do not know what to do if that is not the case.

Answer 1

As $H$ is a normal subgroup of $K$, for $k \in K$, you have that $kHk^{-1} \subseteq H$ and $k^{-1}Hk \subseteq H$. So

$$H = (kk^{-1})H(kk^{-1}) = k(k^{-1}Hk)k^{-1} \subseteq kHk^{-1} \subseteq H.$$

This gives you that $kHk^{-1} = H$.

Answer 2

You have your solution. Since $H$ is normal in $K$, every element of $K$ normalizes $H$ (as in your solution) and hence $K\subseteq N_G(H)$.