I'm trying to prove that there are exactly 5 groups of order 42. My approach was to show that there is always a subgroup of order 6, let's say $H$, and a normal group of order 7, let's say $K$. It follows that $G=H\rtimes K$. $K \cong \mathbb{Z_7}$ and $H\cong\mathbb{Z}_6$ or $\cong S_3$. I was able found that there are two non-isomorphic, non-abelian, groups when $H\cong S_3$, namely $S_3\times\mathbb{Z}_7$ and $S_3\rtimes_\sigma \mathbb{Z}_7 $, $\sigma:S_3\rightarrow \text{Aut}(\mathbb{Z}_7)$ such that $\ker\sigma = A_3$. So far so good.
My problem starts when $H\cong\mathbb{Z}_6$. I showed that there are $6$ distinct homomorphisms from $\mathbb{Z}_6$ to $\text{Aut}(\mathbb{Z}_7)$, namely $\sigma_n (\bar{1})(\bar{1})=\bar{n}$. This gives $6$ groups, with presentation:
$$ G=\left<a,b \right>\\ a^7=b^6=e\\ bab^{-1}=a^n $$
one of them ($n=1$) abelian and the others non-abelian. However, this gives us a total of $7$ non-abelian groups of order 42, and I know it must be five. My first instict was to construct isomorphisms between the groups with $H\cong\mathbb{Z}_6$ (the ones with the presentation above), but in vain. Could anyone assist me?
The group given by $bab^{-1}=a^2$ is isomorphic to the group given by $bab^{-1}=a^4$ and the group given by $bab^{-1}=a^3$ is isomorphic to the group given by $bab^{-1}=a^5$. In both cases, the isomorphism is induced by the automorphism of $\mathbb Z_6$: $b\mapsto b^{-1}=b^5$. (Say, if $bab^{-1}=a^2$ then $b^{-1}ab=b^5ab^{-5}=a^{2^5}=a^4$ etc.)