Given a finite group $G$, $N \trianglelefteq G$, and $H \leq G$, such that $G$ is the inner semidirect product of $N$ and $H$, and knowing that the conjugacy action of $H$ on $N$ divides $N$ in two orbits, prove that:
- There exists a prime $p$ such that any nontrivial element of $N$ has order $p$
- $N$ is abelian
- It doesn't exist $M \trianglelefteq G$ nontrivial and properly contained in $N$.
My solution for $(1)$: I know that $N$ is divided in two orbits, and since there's one orbit that contains only $1_G$, all the elements $n\in N:n \neq 1_G$ are in the same orbit, meaning that they all have the same order, since the action is the conjungancy one. It follows that the order of $N$ must be $p^\alpha$, $p$ prime. I've been stuck for some time on $(2)$ and $(3)$ though, so any hint would be really appreciated. Thanks.
Thanks to the hints given in the comment section, I've been able to solve the problem, and I'm posting here the solution for any future reference.
For $(3)$, if there was $M \leq N\,,\,M\neq N$ normal in $G$, it would mean that $h^{-1}mh \in M$ for every $h \in H$ and $m \in M$. But since $N$ is divided in only two orbits, it exists $x \in H$ such that $x^{-1}mx=n\,,\,n\in N - M$.
For $(2)$, I know that $|N|=p^\alpha$, and that there exists a normal subgroup of $X \leq N$ that has order $p^{\alpha-1}$, meaning that $N/X$ is abelian and $N' \leq X$. But $N'$ is characteristic in $N$, so for $(3)$ it must be trivial $\Rightarrow N$ is abelian.