I am trying to classify groups of order 18. So far, I have shown that a group $G$ of order 18 is given by $G\cong C_9 \rtimes_{\varphi} C_2$ or $G\cong (C_3 \times C_3)\rtimes_{\varphi} C_2$.
If $G\cong C_9 \rtimes_{\varphi} C_2$, then $\mid \varphi(1) \mid$ divides $2$, so that $\varphi(1)$ is trivial or inverts a generator of $C_9$. I concluded that $G \cong C_{18}$ in the former case and $G \cong D_{18}$ in the latter case.
I am now considering the case $G\cong (C_3 \times C_3)\rtimes_{\varphi} C_2$, but I am stuck. I have found this article, but they do not do it the same way as me (finding the image of $\varphi(1)$ in $Aut(C_3 \times C_3)$).
Is there a way to do this using my method? My goal is to be able to do questions like this on my algebra qual without having to fiddle around. I was messing around with the fact that $Aut(C_3 \times C_3)\cong GL_2(C_3)$.
Best Answer
$\varphi(1)$ is an automorphism of $C_3 \times C_3$ of order $1$ or $2$. If it has order $1$, then you get the direct product $C_3 \times C_3 \times C_2 \cong C_3 \times C_6$, so let's suppose it has order $2$.
It helps to think of ${\rm Aut}(C_3 \times C_3)$ as the group ${\rm GL}(2,3)$ of invertible $2 \times 2$ matrices over the field of order $3$. An element of order $2$ has minimal polynomial $x^2-1$ or $x+1$ , so it has eigenvalues $1,-1$ or $-1,-1$. In either case, the matrix is diagonalizable.
So, we can find generators $t,u$ of $C_3 \times C_3$ (corresponding to eigenvectors of the matrix) such that, in the first case, $\phi(1)$maps $t \mapsto t$, $u \mapsto u^{-1}$ and, in the second case $t \mapsto t^{-1}$, $u \mapsto u^{-1}$. So there are two isomorphism classes of nontrivial semidirect products.