# Subgroups of the Semi-Direct Product $\mathbb{Z}/7\mathbb{Z} \rtimes (\mathbb{Z}/7\mathbb{Z})^{\times}$

I want to list all the subgroups of the semi-direct product $$\mathbb{Z}/7\mathbb{Z} \rtimes (\mathbb{Z}/7\mathbb{Z})^{\times}$$, under the homomorphism $$\theta: (\mathbb{Z}/7\mathbb{Z})^{\times} \rightarrow \mathrm{Aut}(\mathbb{Z}/7\mathbb{Z})$$, $$\theta: a \mapsto \theta_{a}$$ where $$\theta_{a}(i)=ai$$. Until now, I know that the subgroups of $$(\mathbb{Z}/7\mathbb{Z})^{\times}$$ will be of orders $$1, 2, 3$$ or $$6$$ and moreover they will be unique (similarly, the cyclic group with $$7$$ elements only has the trivial subgroups).

I was thinking that the subgroups of the semi-direct product would be semi-direct products of the subgroups of $$\mathbb{Z}/7\mathbb{Z}$$ and $$(\mathbb{Z}/7\mathbb{Z})^{\times}$$. Is my claim correct? If not, what would be a way to compute those subgroups?

• Also, I think this is the first time I have ever seen a $\mathbb{7}$. How disorienting! – Alex Kruckman Apr 14 '16 at 1:42
• It is not true that the subgroups of orders $2$, $3$, and $6$ are unique. There are $7$ of each - although those of the same order are all all conjugate. There are also subgroups of order $14$, $21$ and $42$. In this example, all subgroups are themselves semidirect products, but that would not be true for all semidirect products. – Derek Holt Apr 14 '16 at 2:29
• I see, how would one be able to find all those subgroups? Or a general procedure to find one of them? Thanks for the remark. – hoyast Apr 14 '16 at 3:50

Let $H$ be a subgroup of the group $G$ of order $42$ in question.
• If $7$ divides $|H|$, then, since $G$ contains normal subgroup $P$ of order $7$ which is Sylow-$7$ subgroup, $H$ will contain the normal subgroup $P$. Then $H/P$ is subgroup of $G/P\cong \mathbb{Z}_7^{\times}$; so possible orders of $H/P$ are $1,2,3,6$ and it is unique (in $\mathbb{Z}_7^{\times}$), according to which we will get unique subgroups of $G$ of order $7,14,21,42$.
• If $7$ does not divide $|H|$ then $|H|$ will be $1,2,3$ or $6$, $H$ will be conjugate to subgroup of $\mathbb{Z}_7^{\times}$; this is because of the following: if $|H|=1$ then it is obvious. If $|H|=6$, then $H$ is complement of a normal Hall subgroup $\mathbb{Z}_7$ and by Schur-Zassenhaus, the complements of a normal Hall subgroup are conjugate. If $|H|=2$ or $3$ then $H$ will be Sylow-$2$ or Sylow-$3$ subgroup, and $\mathbb{Z}_7^{\times}$ contains Sylow-$2$ and Sylow-$3$ subgroup, hence $H$ will be conjugate to a subgroup of $\mathbb{Z}_7^{\times}$.
• In general, the subgroups of semi-direct product $N\rtimes H$ will not be of the form $N_1\rtimes H_1$ for $N_1\leq N$ and $H_1\leq H$. Simple example: $S_3=Z_3\rtimes Z_2$. – p Groups Apr 14 '16 at 5:12