Classifying groups of order 585 I am trying to classify the groups of order 585. (It is known that there are 4 of distinct non-isomorphic groups, but I am not assuming it.)
The question further asks to show that any group of this size has a cyclic subgroup of prime index, and the center is non-trivial with a composite order.
My attempt so far:
Note that $|{G}|=585=3^2\cdot 5\cdot 13$. By Sylow, $n_3\in \{1,13\}$, $n_5=1$, and $n_{13}=1$. Denote by $S_{3}$, $S_5$, $S_{13}$, the Sylow subgroups of respective order. It then follows that $S_5\cong \mathbb{Z}_5$, $S_{13}\cong \mathbb{Z}_{13}$, and $S_3$ is isomorphic to either $\mathbb{Z}_9$ or $\mathbb{Z}_3\times\mathbb{Z}_3$.
Take $\mathbb{Z}_5\times \mathbb{Z}_{13}$. This is a normal subgroup that is cyclic (but not with prime index), hence all groups of order 585 are either of the form


*

*$(\mathbb{Z}_5\times \mathbb{Z}_{13})\rtimes \mathbb{Z}_9$

*$(\mathbb{Z}_5\times \mathbb{Z}_{13})\rtimes (\mathbb{Z}_3\times\mathbb{Z}_3)$.


The immediate ones are:
(i) $\mathbb{Z}_5\times \mathbb{Z}_{13}\times \mathbb{Z}_9$
(ii) $\mathbb{Z}_5\times \mathbb{Z}_{13}\times \mathbb{Z}_3\times \mathbb{Z}_3$
These clearly have cyclic subgroups of prime index, namely, $\mathbb{Z}_{13}\times \mathbb{Z}_9$ (or $\mathbb{Z}_5\times \mathbb{Z}_9$) and $\mathbb{Z}_5\times \mathbb{Z}_{13}\times \mathbb{Z}_3$. Also, these are abelian groups, so they obviously have non-trivial center of composite order (namely, 585).
Okay, let's find then the non-trivial semi-direct products. To investigate the Case 1, we need to identify the homomorphisms $\varphi:\mathbb{Z}_9\to\mathrm{Aut}(\mathbb{Z}_5\times \mathbb{Z}_{13})\cong\mathbb{Z}_5^*\times\mathbb{Z}_{13}^*$. Since $|\varphi(1)|\mid 9$, it follows that the only non-trivial homomorphism is given by $1\mapsto (1,3)$. This gives rise to the presentation
$$
<a,b,c,d:a^5=b^{13}=c^9=e,ab=ba,ac=ca,cbc^{-1}=b^3>.
$$
This has a cyclic subgroup generated by $ac$ whose index is 13. This group has a non-trivial center since $a\in Z(G)$. Moreover, the order of $Z(G)$ is composite since it also contains $c^3$: $c^3bc^{-3}=b^{27}=b$. Hence the order of $Z(G)$ is divisible by 15.
Let's move on to the Case 2 where $G$ is given by $(\mathbb{Z}_5\times \mathbb{Z}_{13})\rtimes (\mathbb{Z}_3\times\mathbb{Z}_3)$. To define the homomorphism, we need to determine where such a homomorphism sends $(1,0)$ and $(0,1)$ of $\mathbb{Z}_3\times\mathbb{Z}_3$. By the same logic, we determine that either $(1,0)\mapsto (1,3);\,(0,1)\mapsto (1,1)$ OR $(1,0)\mapsto (1,3);\,(0,1)\mapsto (1,3)$. The map $(1,0)\mapsto (1,1);(0,1)\mapsto (1,3)$ gives rise to an isomorphic group.
The corresponding presentations are:
$$
\begin{split}
<a,b,c,d
&:a^5=b^{13}=c^3=d^3=e,ab=ba,cd=dc,ac=ca,bc=cb,ad=da\\
&;dbd^{-1}=b^3>,
\end{split}
$$
(namely, $a$ commutes with everything, but $b$ only commutes with $a$ and $c$) and
$$
\begin{split}
<a,b,c,d
&:a^5=b^{13}=c^3=d^3=e,ab=ba,ac=ca,ad=da,cd=dc\\
&;cbc^{-1}=b^3,dbd^{-1}=b^3>.
\end{split}
$$
Here is where I have the problem (by cheating, shamefully): These two better be isomorphic! But I don't see how they can be isomorphic... Any help?
 A: Use (or first prove) the following fact: If
$$\varphi,\psi:\ G\ \longrightarrow\ \operatorname{Aut}(H),$$
are two group homomorphisms and $\chi\in\operatorname{Aut}(H)$ is such that $\varphi(g)=\psi(g)\chi$ for all $g\in G$, then
$$G\rtimes_{\varphi}H\cong G\rtimes_{\psi}H.$$

For the first case, you have overlooked the homomorphism
$$\mathbb{Z}_9\ \longrightarrow\ \mathbb{Z}_5^*\times\mathbb{Z}_{13}^*:\ 1\ \longmapsto\ (1,9),$$
which is not a problem, as it yields a semidirect product isomorphic to the one you found by the result above. For the second case, you have overlooked the nontrivial homomorphisms
$$\mathbb{Z}_3\times\Bbb{Z}_3\ \longrightarrow\ \mathbb{Z}_5^*\times\mathbb{Z}_{13}^*:\ 
\begin{array}{c}
(0,1)\ \longmapsto\ (1,1)\\
(1,0)\ \longmapsto\ (1,9)
\end{array},$$
$$\mathbb{Z}_3\times\Bbb{Z}_3\ \longrightarrow\ \mathbb{Z}_5^*\times\mathbb{Z}_{13}^*:\ 
\begin{array}{c}
(0,1)\ \longmapsto\ (1,3)\\
(1,0)\ \longmapsto\ (1,9)
\end{array},$$
$$\mathbb{Z}_3\times\Bbb{Z}_3\ \longrightarrow\ \mathbb{Z}_5^*\times\mathbb{Z}_{13}^*:\ 
\begin{array}{c}
(0,1)\ \longmapsto\ (1,9)\\
(1,0)\ \longmapsto\ (1,1)
\end{array},$$
$$\mathbb{Z}_3\times\Bbb{Z}_3\ \longrightarrow\ \mathbb{Z}_5^*\times\mathbb{Z}_{13}^*:\ 
\begin{array}{c}
(0,1)\ \longmapsto\ (1,9)\\
(1,0)\ \longmapsto\ (1,3)
\end{array},$$
$$\mathbb{Z}_3\times\Bbb{Z}_3\ \longrightarrow\ \mathbb{Z}_5^*\times\mathbb{Z}_{13}^*:\ 
\begin{array}{c}
(0,1)\ \longmapsto\ (1,9)\\
(1,0)\ \longmapsto\ (1,9)
\end{array},$$
which is again not a problem, as they yield semidirect products isomorphic to the one you found, again by the result above.

I'm also a bit surprised by your abstract approach of semidirect products. In stead of working with presentations, why not work with concrete sets with concrete operations? For example, the semidirect product $(\Bbb{Z}_5\times\Bbb{Z}_{13})\rtimes\Bbb{Z}_9$ can be constructed as the set $(\Bbb{Z}_5\times\Bbb{Z}_{13})\times\Bbb{Z}_9$ with operation
$$((a_1,b_1),c_1)\cdot((a_2,b_2),c_2)=((a_1+a_2,b_1+3^{c_1}b_2),c_1+c_2).$$
This makes it much easier to see whether two given semidirect products are isomorphic.
