Problem 5.1.6 (All subgroups of $Q_8 \times E_{2^n}$ are normal)

Proof. By definition, $E_{2^n}$ is the elementary abelian group of order $2^n$:

where $Z_2=⟨x⟩,|x|=2$.

Consider some subgroup $H$ of $G=Q_8\times E_{2^n}$. If $h\in H$, then $h=(u,v)$, where $u\in Q_8$ and $v\in E_{2_n}$. Then $u^4=1$ and $v^2=1$, therefore

Let $g=(s,t)$ be any element of $Q_8\times E_{2^n}$. Since $E_{2^n}$ is abelian, $\mathit{𝑔h}{g}^{-1}=(\mathit{𝑠𝑢}{s}^{-1},\mathit{𝑡𝑣}{t}^{-1})=(\mathit{𝑠𝑢}{s}^{-1},v)$, so

Note that in $Q_8$,

so for every $y\in Q_8$

Since there exists an automorphism which maps $i$ on every element of $Q_8-\{1,-1\}$, the same is true if we replace $i$ by $\pm i,\pm j$ or $\pm k$ (and $t(\pm 1)t^{-1}=\pm 1$ for every $t\in Q_8$). This shows that for every $s\in Q_8$ and every $u\in Q_8$,

Therefore, by (1),

Since this is true for every $h\in H$ and every $g\in G$, this proves

All subgroups of $Q_8\times E_{2^n}$ are normal. □