The question is as follows:
Let $G$ be a finite group acting by automorphisms on a finite Abelian group $A$, and assume that $ A$ is completely reducible as a $G$-group.
Then if $N \trianglelefteq G$, show that $A$ is completely reducible when viewed as an $N$-group.
Hint: It suffices to handle the case where $A$ is $G$-simple. Observe that if $B \subset A$ is an $N$-simple $N$-subgroup, then so is $B^g$ for all $g \in G$.
Some definitions:
$G$ acts by automorphisms on a finite Abelian group $A$ means that for every $a_1, a_2 \in A$ and $g \in G$ we have $(a_1 a_2) \cdot g = (a_1 \cdot g)(a_2 \cdot g)$ and each element of $G$ induces an automorphism of $A$. And $A$ is $G-$simple if $A \neq 0$ and $A$ and $\{0\}$ are the only $G-$invariant subgroups of $A$. And $A$ is $G-$completely reducible if it is a direct sum of $G-$simple modules. This is equivalent to being generated by its simple submodules.
Now by knowing all these, how can we show that $A$ is completely reducible when viewed as an $N$-group?
Thanks!
Using the fact that $A$ is $G$-reducible, write $A=A_1\times...\times A_n$ where each $A_i$ is $G$-simple. Now if we can show that each $A_i$ is $N$-reducible, then it follows that $A$ is $N$-reducible. That is because if you write $A_i=A_{i1}\times ...\times A_{im_i}$, then you can also write $A=A_{11}\times ...\times A_{1m_1}\times ...\times A_{n1}\times ...\times A_{nm_n}$. This is the explanation why we can assume that $A$ is $G$-simple.
Now assume that $A$ is $G$-simple, let $B$ be an $N$-invariant, $N$-simple subgroup of $A$ and pick $g\in G$. Consider $B^g=\{b\cdot g\mid b\in B\}$. Now $B^g$ is $N$-invariant since $$(b\cdot g)\cdot n=b\cdot(gn)=b\cdot (gng^{-1}g)=(b\cdot gng^{-1})\cdot g \in B^g$$ ( $gng^{-1}\in N$ since $N$ is normal and $b\cdot gng^{-1}\in B$ since $B$ is $N$-invariant).
Also note that $B^g$ is $N$-simple : Assume that $H^g=\{h\cdot g\mid h\in H\}\subseteq B^g$ is $N$-invariant. For any $n\in N$ one can write $n=gmg^{-1}$ for some $m\in N$ since $N$ is normal. Then $$H^n=H^{gmg^{-1}}=(H^g)^{mg^{-1}}=((H^g)^m)^{g^{-1}}=(H^g)^{g^{-1}}=H$$ (where $(H^g)^m=H^g$ follows by the fact that $H^g$ is $N$-invariant). So $H$ is $N$-invariant. Since $B$ is $N$-simple, we get $H=0$ or $H=B$ which implies that $H^g=0$ or $H^g=B^g$.
Now say $S$ is the subgroup generated by $\bigcup_{g\in G}B^g$. Since the generating set is $G$-invariant, so is $S$. Since $A$ is $G$-simple, we get $A=S$. So $A$ is generated by $N$-invariant, $N$-simple subgroups. Hence $A$ is $N$-reducible.
Edit : If you want to learn more about this argument you may check the Clifford's Theorem and its proof.