Computing all semidirect products $\mathbb{Z}/3\mathbb{Z}\rtimes D_8.$

Question

Let $D_8 = \langle a,b \mid a^2 = e, b^4 = e, aba^{-1} = b^{-1}\rangle.$ So I begin by considering $\text{Aut} (\mathbb{Z}/3\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}.$ Next, consider a homomorphism $\varphi:D_8\to \mathbb{Z}/2\mathbb{Z}.$ There is obviously the trivial homomorphism of mapping everything to $0,$ which results in the direct product $\mathbb{Z}/3\mathbb{Z}\times D_8.$ Next, let us map the reflection $a$ to $1,$ which means $\varphi(a)=\varphi(ab)=\varphi(ab^2)=\varphi(ab^3)=1$ and everything else is $0.$ How would we present the resulting group? Further, I believe that there is only one other possibility but I am unable to figure out this possibility (without accidentally introducing a third possibility...? if we let only $\varphi(b)=1$ then we could also have $\varphi(a)=\varphi(b)=1$ as our third possibility). I think I'm just a bit confused about this whole idea, so help is appreciated.

Answer 1

Expanding the comment by @Spin:

There is an isomorphism $D_8\to D_8$ mapping \begin{eqnarray*} a&\mapsto& ab,\\ b&\mapsto& b, \end{eqnarray*} which allows us to identify solutions 3. and 4. from my comment. That is why you and I got four solutions, when in fact there are only three distinct semi-direct products.

As you said 1. is just $\mathbb{Z}/3\mathbb{Z}\times D_8$.

In 2. the elements $b$ and $t$ commute, so $bt$ generates a cycle of order $12$. Conjugation by $a$ is inversion on this cycle, so 2. is $D_{24}$.

In 3. the elements, $b,t$ generate a subgroup of order $12$. This subgroup contains an element $x=tb^2$ of order $6$, satisfying $x^3=b^2$ and $bxb^{-1}=x^5=x^{-1}$. Thus this subgroup is generated by $x,b$ and is isomorphic to $Q_{12}$.

To see that 1., 2., 3. are non-isomorphic, note that 1. is the only one with a central subgroup of order $3$, so different to the other two. Finally all subgroups of $D_{24}$ of order $12$ are either isomorphic to $C_{12}$ or $D_{12}$, hence not $Q_{12}$. Thus 2. is different to 3..