I'm asked to prove the following:
Let $N$ be a minimal normal subgroup of $G$ and $E$ a minimal normal subgroup of $N$, and assume that the set $\mathcal{M}=\{E^g\mid g\in G\}$ is finite. Then $E$ is simple, and there exist $E_1,...,E_n$ in $\mathcal{M}$ such that $$N=E_1\times\cdots\times E_n.$$
My attempt:
It's rather easy to show $\prod_{g\in G} E^g$ is a normal subgroup of $G$, then $N=\prod_{g\in G} E^g$ since $N$ is minimal normal in $G$. I've learned previously that $\prod_{g\in G} E^g=E_1\times\cdots\times E_n$, where $E_i\in\mathcal{M}$, but then I'm stuck, what should I do to push it forward?
: ) It'll be really appreciated if any help is provided!
Any $g\in E$ can be written as $g=g_1\cdots g_n$, $g_i\in E_i$. If $E_i$ has a normal subgroup, say $H$, we’ll get $H^g=H^{g_1\cdots g_n}=H^{g_i}$, since $[E_i, E_j]=1$ for any $i\neq j\in\{1,...,n\}$. Thus, $H^g=H$ and $H$ is normal in $G$, which is a contradiction.