Counting the number of semidirect products I am having some issue understanding how to count the number of semidirect products we can obtain. Take for example $\mathbb{Z}_{7} \rtimes \mathbb{Z}_6$. I understand that this would be counting the number of homomorphisms $\phi: \mathbb{Z}_6 \rightarrow Aut(\mathbb{Z}_{7}) \cong \mathbb{Z}_6$, but how would I go about counting these homomorphisms, or actually generating the semidirect products.
 A: Basically, the way you do this is by listing the homomorphisms, then thinking about the result you got.
For instance, note that $\mathbb Z_6$ is generated by a single element $g$ of order $6$. In particular, a homomorphism $f:\mathbb Z_6\rightarrow G$ is determined by $f(g)$ subject to the condition that $f(g)^6=e$. More specifically, since every element in $\mathbb Z_6$ has order diving $6$, for each $\alpha\in \mathbb Z_6$ there is a unique homomorphism $f_{\alpha}:\mathbb Z_6\rightarrow \mathbb Z_6$ with $f_{\alpha}(g)=\alpha$. Explicitly, in additive notation, this is just $f_{\alpha}(x)=\alpha x$. So, there are six homomorphisms.
If you want to actually write down the semi-direct products that this yields, recall that the $G\rtimes_{\phi} H$ where $\phi:H\rightarrow \operatorname{Aut}(G)$ is basically defined to have $G$ as a normal subgroup on which an element $h\in H$ acts by conjugation as $\phi(h)$. Observing that $\operatorname{Aut}(\mathbb Z_7)$ is just maps of the form $x\mapsto x^c$ for $c\in \{1,\ldots,6\}$ and we have shown that a generator of $\mathbb Z_6$ could be mapped to any such element by some homomorphism $f_c$, we can actually write down a generic presentation for a semidirect product, where $c\in\{1,\ldots,6\}$:
$$\mathbb Z_7\rtimes_{f_c} \mathbb Z_6\cong \langle x,g |x^7=e,g^6=e,gxg^{-1}=x^c\rangle.$$
Then, you can use ad hoc reasoning to figure out which of these six groups are isomorphic.
(Of course, there are some more sophisticated ways to deal with the question of isomorphism, but for this case it's just as easy to classify the six groups by hand.)
