# Subgroup of $\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}$.

I want to check my solution of this (simple) problem: find all subgroups $H$ of $\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}$, such that $|H|=36$.

My attempt: $|(\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z})/H|=3$, so $$(\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}) \cong \mathbb{Z}/3\mathbb{Z};$$ using the correspondence theorem I can calculate the subgroups of $\mathbb{Z}/3\mathbb{Z}$, that is only $\{0\}$.

In $\mathbb{Z}/6\mathbb{Z}\times \mathbb{Z}/18\mathbb{Z}$ not exixts an element of order $36$, so the unique subgroup is $\mathbb{Z}/9\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$.

Do I do some mistakes?

• There are no elements of order $4$ in the original group. Also, for any given group structure, there may be several different (but isomorphic) subgroups with that structure. Feb 9 '17 at 8:48
• Right, so I suppose the subgroup is isomorphic to \mathbb{Z}/9\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/62\mathbb{Z}$. Or not? – user348628 Feb 9 '17 at 8:52 • Assuming you mean$\Bbb Z/9\Bbb Z\times \Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$, that's one option. What subgroups are isomorphic to that group? But remember that you can also have$\Bbb Z/3\Bbb Z\times \Bbb Z/3\Bbb Z$instead of$\Bbb Z/9\Bbb Z$. And, again, they do not want just a group isomorphic to the subgroups you find, they want the actual subgroup. E.g., if they asked about subgroups of order$2$, instead of$\Bbb Z/2\Bbb Z$, they want$\{(0,9),(0,0)\}$and$\{(3,0),(0,0)\}$and$\{(3,9),(0,0)\}$. For order$36$you may want to describe it more compactly, but that's what they're after. Feb 9 '17 at 8:59 • Ok I get it, but the corrispondece theorem assures that the subgroup is unique? How can I have a lot of subgroups? I don't understand this point. – user348628 Feb 9 '17 at 9:07 • That's not what the correspondence theorem says. It says that the subgroups that contain a given$H$correspond to the subgroups of$\Bbb Z/3\Bbb Z$. But the different possible$H$do not contain one another, so the correspondence theorem is not relevant to this problem. Feb 9 '17 at 9:10 ## 1 Answer The subgroups of$\mathbb{Z}/n\mathbb{Z}$are all of the form$m(\mathbb{Z}/n\mathbb{Z})$where$m \mid n$. For example, the subgroups of$\mathbb{Z}/6\mathbb{Z}= \{\bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5\}are \begin{align} 1(\mathbb Z/6\mathbb Z) &= \{\bar 0, \bar 1, \bar 2, \bar 3, \bar 4, \bar 5\} \\ 2(\mathbb Z/6\mathbb Z) &= \{\bar 0, \bar 2, \bar 4 \} \\ 3(\mathbb Z/6\mathbb Z) &= \{\bar 0, \bar 3 \} \\ 6(\mathbb Z/6\mathbb Z) &= \{\bar 0 \} \\ \end{align} Note also that| m(\mathbb{Z}/n\mathbb{Z}) | = \dfrac nm$. The divisors of$6$are$m \in \{1,2,3,6\}$and the divisors of$18$are$n \in\{1, 2, 3, 6, 9, 18\}$If you want$|m(\mathbb{Z}/6\mathbb{Z}) \times n(\mathbb{Z}/18\mathbb{Z})| = 36$then you need to find all$m$and$n$such that$\dfrac 6m \cdot \dfrac{18}{n} = 36$, which simplifies to$mn = 3$. So your subgroups are •$1(\mathbb{Z}/6\mathbb{Z}) \times 3(\mathbb{Z}/18\mathbb{Z})$•$3(\mathbb{Z}/6\mathbb{Z}) \times 1(\mathbb{Z}/18\mathbb{Z})$This can be "simplified" to •$ \mathbb{Z}/6\mathbb{Z}\times \{\bar 0, \bar 3, \bar 6, \bar 9, \overline{12},\overline{15}\}$•$\{\bar 0, \bar 3 \} \times \mathbb{Z}/18\mathbb{Z}$• You seem to be missing an argument for why all these subgroups should be the direct product of a subgroup from each factor (I don't really see why that should be the case). Feb 9 '17 at 9:25 • Also, you should have$\frac{6}{m}\cdot\frac{18}{n}=36$, which simplifies to$mn=3$. Feb 9 '17 at 9:27 • I suppose are$3(\mathbb{Z}/6\mathbb{Z})\times \mathbb{Z}/18\mathbb{Z}$and$\mathbb{Z}/6\mathbb{Z}\times 3(\mathbb{Z}/18\mathbb{Z})\$.
– user348628
Feb 9 '17 at 9:38
• I am still not seeing any good reasons why the subgroups need to have this form. Feb 9 '17 at 9:49
• @TobiasKildetoft I'm pretty sure it's a consequence of the basis theorem for finite abelian groups. Feb 9 '17 at 10:07