Are the normalizers subnormal?

Question

Let $A$ and $B$ be two subgroups of group $G$.

If $A$ is normal in $B$, is $N_G(B)$ normal or subnormal in $N_G(A)$?

If not, what if $A$ is a $p$-subgroup of $G$ and $|B|=\frac{|A|}{p}$?

Answer 1

The answer is no. Take $G=S_3$, $A$ the trivial subgroup of $G$ and $B$ any subgroup of order $2$. Then $N_G(A)=G$ and $N_G(B)=B$, thus $B$ is not subnormal in $G$.

Similar examples with $A \neq 1$ exist. Take for instance $G=S_4$. A Sylow $2$-subgroup $B$ of $G$ is self-normalising (i.e. it is its own normaliser), yet the core of $B$ in $G$ has order $4$ and is isomorphic to $C_2 \times C_2$. You can take that normal subgroup as $A$.

Note the following (it is reasonably important): if $P$ is a Sylow $p$-subgroup of the finite group $G$ then $N_G(P)$ is self-normalising. So if $P$ is not normal in $G$ then $N_G(P)$ cannot possibly be subnormal in $G$. This fact is an application of the Frattini argument.