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?