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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

phy_math
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    Why would you want to prove such a statement? – Marc van Leeuwen May 08 '17 at 14:20
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    Hint: Try to prove that for any $x,y \in Q_8$ there exists an odd $n$ such that $yxy^{-1}=x^n$. – CJS May 08 '17 at 15:06
  • @MarcvanLeeuwen I think the context is the following: according to wikipedia, a group $G$ such that every subgroup is normal is isomorphic to $Q_8 \times B \times D$ with $Q_8$ the quaternion group, $B$ a group where every nonidentity element has the same odd order and $D$ a $2$-group. See https://en.wikipedia.org/wiki/Dedekind_group – Jef May 08 '17 at 16:10
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    Note that simply because all subgroups of both $Q_8$ and $E_{p^n}$ are normal does not imply that all subgroups of $Q_8 \times E_{p^n}$ are normal. For a counterexample, take $G = Q_8 \times \mathbb{Z}_4$, $H = \langle (i,1) \rangle$ and $g = (j,1)$ and note that it is not, in fact, normal, although all subgroups of $Q_8$ and $\mathbb{Z}_4$ are normal. – Isomorphism Aug 27 '17 at 21:04

2 Answers2

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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 $G\unlhd Q_8\times E_{2^n}$.

Collier Gaiser
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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.

Jonathan Rayner
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