Is it possible that $N_G(H)=H$ and $N_G(K)=K$ where $K\subsetneq H$?

Question

Is it possible that $N_G(H)=H$ and $N_G(K)=K$ where $K \subsetneq H$ and $H,K$ are proper subgroups of $G$ ?

Answer 1

$\newcommand{\Span}[1]{\langle #1 \rangle}$Consider for instance the group $G$ of order $18$ which is the semidirect product of an elementary abelian group $\Span{x,y}$ of order $9$ by a group $H = \Span{z}$ of order $2$, acting like $$x^z = x^{-1}, \qquad y^{z} = y^{-1}.$$

Now take $K = \Span{x, z}$, so that $H \subsetneq K$.

We have $N_{G}(H) = H$ and $N_{G}(K) = K$.

Answer 2

Let $G$ be a finite group and $P$ any Sylow $p$-subgroup of $G$. Then if $N_G(P) \leq K \leq G$, we have $N_G(K) = K$.

Thus to find an example, it would be enough to find a finite $G$ with a Sylow $p$-subgroup $P$ such that $N_G(P)$ is not maximal and $N_G(P) \neq G$. One example like this is given in the answer by Andreas.

Answer 3

Same basic answer as Andreas and Mikko:

Parabolic subgroups of groups with $(B,N)$-pairs are self-normalizing, so one gets entire buildings of self-normalizing subgroups.

So to get a chain of length $n-2$, take $G=\operatorname{GL}_n(q)$ for any field $q$, and let $H_i$ be the subgroup of $G$ consisting of matrices $\begin{bmatrix} A & B \\ 0 & T \end{bmatrix}$ with $A \in \operatorname{GL}_i(q)$ an arbitrary invertible $i \times i$ matrix, $B \in q^{[i,(n-i)]}$ an arbitrary $i \times (n-i)$ matrix, and $T$ an upper triangular invertible $(n-i) \times (n-i)$ matrix.

Then $H_1 < H_2 < \ldots < H_{n-1} < H_n = G$ are all self-normalizing, so $H_1 < H_2 < \ldots < H_{n-1}$ is a chain of length $n-2$ of proper, self-normalizing subgroups. Notice that when $q$ is a finite field of characteristic $p$, this is just the Sylow example of Mikko.