Prove that $Q_{8} \cong H \rtimes G \Rightarrow H=\{e\}$ or $G=\{e\}$.
My proof (is it correct?):
We know that (it's a fact from the lecture): If $M \cong H\rtimes G$, then
- $H \cong K \unlhd M$ and $G\cong L \leq M$.
- $K \cap L =\{e\}$.
- For finite groups: $|K||L|=|M|$.
As every subgroup of $Q_{8}$ is normal (I think I can prove that, but is there a fast way to do it?), the possibilties, for K are:
$$\{-1,1\}, \{1,-1,i,-i\}, \{1,-1,j,-j\}, \{1,-1,k,-k\}, \{1\}, Q_{8}$$ Let $\{\{1,-1,i,-i\}, \{1,-1,j,-j\}, \{1,-1,k,-k\}\}=N$
- If $K=\{-1,1\}$, then from 3. $L$ would have to consist of 4 elements. So $L\in N$. But then $K\cap L \neq\{e\}$.
- If $K \in N$, then $L=\{-1,1\}$ and it can't happen, because of the same reason as in 1.
- If $K=\{1\} $ or $K=Q_8$ ($L=\{1\})$, then everything is ok.
So the only possibilities for $K$ are $\{1\}$ or the whole group (and then $L=\{1\}.$ And $1$ is $e$ in $Q_8$.
Your proof seems fine. Although I think it's enough to write $H \unlhd Q_8$ and $G \leq Q_8$ without using $K$ and $L$. Here's another solution:
As you noted, every subgroup of $Q_8$ is normal. Thus, if $Q_8 \cong H \rtimes G$, then also $G \unlhd Q_8$, hence $Q_8 \cong H \times G$. If $|H|, |G| \leq 4$, then $H$ and $G$ are abelian, and thus $M$ is abelian too, a contradiction.