Let $K'<K<G$ be a chain of groups s.t. $K\subset N_G(K')$. Is that true that $N_G(K)=N_G(K')$?

Question

Let $K'<K<G$ be a chain of groups s.t. $K'$ is a normal subgroup of $K$. Is that true that $N_G(K)=N_G(K')$? Obviously it is true that $N_G(K)\subset N_G(K')$, but is the inverse true? If this is not true, can I guarantee that they have same topological dimension?

Answer 1

No. For example the Klein four subgroup $K$ of $S_4$ is abelian and has a subgroup $K'$ of order 2. But the normalizer of $K$ is the whole $S_4$ while the normalizer of $K'$ is smaller.