**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.6**

Solution: Let $H \leq Q_8 \times E_{2^k}$ be a subgroup, and let $\alpha = (a,x) \in H$ and $\beta = (b,y) \in Q_8 \times E_{2^k}$. Since $E_{2^k}$ is abelian, we have $\beta\alpha\beta^{-1} = (bab^{-1}, x)$.

We saw previously that the conjugacy classes of $Q_8$ are $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, and $\{k,-k\}$. In particular, we have $bab^{-1} \in \{a, a^3\}$.

If $bab^{-1} = a$, then $\beta\alpha\beta^{-1} = (a,x) \in H$. If $bab^{-1} = a^3$, then (since every element of $E_{2^k}$ has order 2) $$\beta\alpha\beta^{-1} = (a^3,x) = (a^3, x^3) = (a,x)^3 \in H.$$ Thus $H$ is normal.