# Finding $Aut(\mathbb{C_2} \times \mathbb{D}_8)$

I've tried the following argument to find $Aut(\mathbb{C_2} \times \mathbb{D}_8)$:

The group is generated by $(\rho,1)$, $(\sigma,1)$ and $(\sigma,-1)$ where $\rho$ has order 8 and $\sigma$ has order 2.

Using an order argument, there are 4 possibilities for the image of $(\rho,1)$ (namely those elements of $\mathbb{C_2} \times \mathbb{D}_8$ whose order is 8: $(\rho,1)$, $(\rho,-1)$, $(\rho^{-1},1)$ and $(\rho^{-1},-1)$). Likewise there are 4 possibilities for the image of $(\sigma,1)$, and then 3=4-1 for the image of $(\sigma,-1)$. (I cannot map this to the image of $(\sigma,1)$).

Therefore there can be at most $4^23$ automorphisms.

I doubt if this is exact. And even if it is, what would the structure of the automorphism group be? (It looks tempting to guess $C_4 \times C_4 \times C_3$, but that's abelian...)

EDIT: As someone pointed out in the comments, I've carelessly missed out that there are actually 16 possibilities for the image of (σ,1), two for each reflection. Does that make 15=16-1 possibilities for the image of (σ,-1)?

• The automorphism group of $D_8$ is $D_8$ so the automorphism group of $C_2 \times D_8$ has a subgroup isomorphic to $D_8$. There are, by the way, more than $4$ possibilities of where to send $\sigma$. Just by order and the fact that it can't commute with $\rho$ it looks like you can send it to $(\rho^m\sigma, (-1)^n)$, so $16$ possibilities. – Jim Nov 9 '17 at 3:51
• Oh, why does it matter if it can't commute with $\rho$? You're right though, that just makes everything even more complex since the upper bound on the order of the group is so huge. – SSF Nov 9 '17 at 4:01
• See here – Mikko Korhonen Nov 9 '17 at 12:30
• If $ab \neq ba$ in a group $G$ and $\phi$ is an injective homomorphism then $\phi(ab) \neq \phi(ba)$ so $\phi(a)$ and $\phi(b)$ don't commute. Thus when you're looking at pairs that $\rho$ and $\sigma$ can be sent to, you know you can't send $\sigma$ to $-1$ or $\rho^4$ because even though those do have order $2$ they commute with all of your options for where to send $\rho$ so you have to rule them out. – Jim Nov 9 '17 at 12:41
• As for where to send $(\sigma, -1)$, there aren't $15$ possibilities. Instead of asking where to send $(\sigma, 1)$ and $(\sigma, -1)$ it might better to ask where to send $(\sigma, 1)$ and $(1, -1)$. As $(1, -1)$ is in the center by the same argument it has to get sent to something in the center. The center is generated by $(\rho^4, 1)$ and $(1, -1)$. We can rule out $(\rho^4, 1)$ because all $4$ of your options of where to send $\rho$ yield $\rho^4$ getting sent to itself. So you can send $(1, -1)$ to itself or $(\rho^4, -1)$. That's two options. – Jim Nov 9 '17 at 12:48