# 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.

• The second one you're looking for seems to be the dihedral group of size 42, which you get by letting ${\Bbb Z}_2$ act on ${\Bbb Z}_{21}$ by inverting it. – Hew Wolff Jan 1 '15 at 23:00
• Oh, and don't forget that the action can be trivial, so ${\Bbb Z}_{42}$ and other abelian groups should probably be on your list. – Hew Wolff Jan 1 '15 at 23:03
• @HewWolff: I think I see your point now as when $\phi$ acts non-trivially, we obtain precisely $D_{42}$ but when $\phi$ acts trivially, then $\mathbb{Z}_{21} \rtimes \mathbb{Z}_2 = \mathbb{Z}_{21} \times \mathbb{Z}_2 \cong \mathbb{Z}_{42}$. Well, at least that covers three of the six groups! – Libertron Jan 2 '15 at 0:12
• There are just 6 groups of order 42 up to isomorphism. – Alexander Konovalov Jan 3 '15 at 23:15
• Just to point out the (perhaps) obvious: You also need to prove that these groups are not isomorphic. A simple example would be $C_{21}\times C_2\cong C_{14}\times C_3$. Non-trivial examples also exist. – user1729 Jan 3 '15 at 23:50

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.