I'm trying to prove that an image of a self-normalizing subgroup of a finite group under a surjective homomorphism is again self-normalizing in the target group.
I proved that if the following is true then the statement above is also true.
If H is a self-normalizing subgroup of a finite group G and N is a normal subgroup of G, then NH is also a self-normalizing subgroup of G.
However, I couldn't figure out why this holds.
Also, these two statement is used in some papers such as "Nilpotent self-normalizing subgroups of soluble groups", but I couldn't find the reference that contains the proof of them.
Can anyone help?
This is false.
To look for a counterexample, we could look for semidirect products $S = N \rtimes H$ with $N$ elementary abelian such that $C_N(H)=1$, for which $N$ has more than one conjugacy class of complements in $S$.
In this situation, $S$ has outer automorphisms $\alpha$ that permute these complements, and so we might hope to find an example $G = S \rtimes \langle \alpha \rangle$, in which the subgroup $H$ is self-normlaizing, but its image in $G/N$ is not.
For one such an example, we can start with $S = N \rtimes H$ with $|N|=2^4$ and $H \cong A_5$, where the action of $A_5$ on $N$ is that of ${\rm SL}(2,4) \cong A_5$ on its natural module. Then $|H^1(S,N)|= 4$, and we can construct a group $S \rtimes \langle \alpha \rangle$, where $\alpha$ is an automorphism of order $2$ that interchanges two non-conjugate complements of $N$ in $S$.
The group $G$ is $\mathtt{SmallGroup}(1920,240995)$ in the GAP/Magma databases. The subgroup $H$ is isomorphic to $A_5$ and $N$ is the unique normal subgroup of order $2^4$, so you can easily check the claimed properties.