Show that all subgroups of $Q_8 \times E_{2^n}$ are normal I want to prove the following statement
All subgroups of $Q_8 \times E_{2^n}$ are normal 
Here $E_{p^n} = \mathbb{Z}_p \times \mathbb{Z}_p \times \cdots \times \mathbb{Z}_p$ (n times)

From some comments below, i made up some informal justification. 
My strategy are following. 
Since $\mathbb{Z}_p$ is cyclic thus abelian its subgroup is normal. 
What i am left is check all subgroups of $Q_8$ are normal. 
There are 4 subgroups $<i>, <j>, <k>, <-1>$, for the first three the index is 2, thus normal and for the last $<-1>$, since $-1 \in Z(Q_8)$ it is normal. 
Thus all subgrouops of $Q_8$ are normal
 A: Let $G\leq Q_8\times E_{2^n}$. Let $(x,y)\in G$. Then $(x,y)^{-1}=(x^{-1},y)\in G$ because $|y|=2$.
Now let $(a,b)\in Q_8\times E_{2^n}$. Then
\begin{equation}
(a,b)(x,y)(a,b)^{-1}=(axa^{-1},y).
\end{equation}
If $a$ and $x$ commute, then $axa^{-1}=x$. So $(a,b)(x,y)(a,b)^{-1}=(x,y)\in G$.
If $a$ and $x$ don't commute, then $a,x\in\{\pm i,\pm j,\pm k\}$ with $x\neq \pm a$. For example, if $x=i$ and $a=j$ then $axa^{-1}=ji(-j)=-i=x^{-1}$. So $(a,b)(x,y)(a,b)^{-1}=(x^{-1},y)\in G$. Other cases are similar. Hence $G\unlhd Q_8\times E_{2^n}$.
A: Note: as mentioned in the comments, it is not true in general that if all the subgroups of $G$ are normal and all the subgroups of $H$ are normal, then all the subgroups of $G \times H$ are normal. A counterexample is $Q_8 \times \mathbb{Z}_4$.
The reason for this is that not all subgroups of a direct product are the direct product of subgroups of the individual factors.
Instead, one way to attack the problem is to use Goursat's lemma, which classifies all subgroups of a direct product. I have outlined how to do this for $Q_8 \times \mathbb{Z}_4$ in this answer. The method given there is easily adapted to $Q_8 \times H$ for any abelian group $H$, as you have in this case. If you work through that method, you should find the result you are looking for.
