Finding $6$ pairwise non-isomorphic semi-direct products of order $42$ I'm stuck on an old algebra prelim question concerning semi-direct products. I am to find $6$ pairwise non-isomorphic groups of order $42$ that are semi-direct products. I think my main issue is how to determine appropriate maps $\phi: K \rightarrow \text{Aut}(H)$ with $H \unlhd G$, $H \cap K= 1$, $G=HK$ so that $G \cong H \rtimes_{\phi} K$.  
My attempt: A group $G$ of order $42= 2 \cdot 3 \cdot 7$ has a normal Sylow $7$-subgroup. So a possible choice of one of these groups is given by the presentation $$\langle x,y: x^6= y^7= 1, xyx^{-1}= y^3 \rangle,$$ where we have a normal Sylow $7$-subgroup $\langle y \rangle$ inverted by $x$, an element of order $6$. This describes the semi-direct product $\mathbb{Z}_7 \rtimes \mathbb{Z}_6$. Another possibility, I think is that since we know that a group of order $42$ has a subgroup of order $21$ (index $2$), then that subgroup is normal. So in this case, we could have $H= \mathbb{Z}_{21}$. Then we can let $K= \mathbb{Z}_2$, and think about $\phi: \mathbb{Z}_2 \rightarrow \text{Aut}(\mathbb{Z}_{21})$. Clearly, $H \cap K=1$ by Lagrange's Theorem but what should I use for $\phi$? From this, I think that my second possibility would probably have to be $\mathbb{Z}_{21} \rtimes_{\phi} \mathbb{Z}_2$. Still not sure what the other four semi-direct products are or how to find them.
 A: Let us recall the list given by the comments above:
$1)$ $\mathbb{Z}_{21}\times \mathbb{Z}_{2}=\mathbb{Z}_{42}$
$2)$ $\mathbb{Z}_{21}\rtimes \mathbb{Z}_2=D_{42}$ (Diedral group, action on $\mathbb{Z}_{21}$ sends $x$ to $-x$)
$3)$ $\mathbb{Z}_{21}\rtimes \mathbb{Z}/2=\mathbb{Z}_7\times D_{6}$ (the action of $\mathbb{Z}_2$ on $\mathbb{Z}_{21}=\mathbb{Z}_7\times \mathbb{Z}_3$ is trivial on $\mathbb{Z}_7$ but sends $x$ to $-x$ on $\mathbb{Z}_3$ )
$4)$ $\mathbb{Z}_{21}\rtimes \mathbb{Z}/2=\mathbb{Z}_3\times D_{14}$ (the action of $\mathbb{Z}_2$ on $\mathbb{Z}_{21}=\mathbb{Z}_7\times \mathbb{Z}_3$ is trivial on $\mathbb{Z}_3$ but sends $x$ to $-x$ on $\mathbb{Z}_7$ )
$5)$ $\mathbb{Z}_{2}\times (\mathbb{Z}_7\rtimes \mathbb{Z}_3)$ (the action of $\mathbb{Z}_3$ on $\mathbb{Z}_7$ sends $x$ onto $2x$ and is of order $3$ since $2^3=8=1$ in $\mathbb{Z}_7$)
It remains to show that these give really five non-isomorphic examples of groups. In each case, the group $G$ contains a unique subgroup $H$ of order $21$, which is moreover normal. The group $H$ is not abelian only in the last case, so we can forget $5)$. You then look at the action of $G$ on $H$ and consider the subgroup of fixed points. This gives a subgroup of order $21$, $1$, $7$, $3$ respectively. This achieves the proof.
