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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

  1. $(\mathbb{Z}_5\times \mathbb{Z}_{13})\rtimes \mathbb{Z}_9$
  2. $(\mathbb{Z}_5\times \mathbb{Z}_{13})\rtimes (\mathbb{Z}_3\times\mathbb{Z}_3)$.

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?

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  • $\begingroup$ What do you mean by "by cheating, shamefully"? Do you mean you have found out, not by your own effort, that there exists only one nontrivial semidirect product $(\mathbb{Z}_5\times \mathbb{Z}_{13})\rtimes (\mathbb{Z}_3\times\mathbb{Z}_3)$, up to isomorphism? $\endgroup$ – Batominovski Aug 2 '18 at 10:09
  • $\begingroup$ I know that (by referring to another source) that there are only 4 distinct groups of order 585. $\endgroup$ – user134070 Aug 2 '18 at 10:13
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    $\begingroup$ To show that the two groups are isomorphic, consider $c' = c d$ in the first group, and note that $a, b, c', d$ satisfy the relations of the second group. $\endgroup$ – Andreas Caranti Aug 2 '18 at 11:20
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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.

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