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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?

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  • $\begingroup$ 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. $\endgroup$ – Arthur Feb 9 '17 at 8:48
  • $\begingroup$ 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? $\endgroup$ – user348628 Feb 9 '17 at 8:52
  • $\begingroup$ 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. $\endgroup$ – Arthur Feb 9 '17 at 8:59
  • $\begingroup$ 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. $\endgroup$ – user348628 Feb 9 '17 at 9:07
  • $\begingroup$ 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. $\endgroup$ – Arthur Feb 9 '17 at 9:10
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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}\}$
  • $\{\bar 0, \bar 3 \} \times \mathbb{Z}/18\mathbb{Z}$

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      $\begingroup$ 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). $\endgroup$ – Tobias Kildetoft Feb 9 '17 at 9:25
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      $\begingroup$ Also, you should have $\frac{6}{m}\cdot\frac{18}{n}=36$, which simplifies to $mn=3$. $\endgroup$ – Servaes Feb 9 '17 at 9:27
    • $\begingroup$ I suppose are $3(\mathbb{Z}/6\mathbb{Z})\times \mathbb{Z}/18\mathbb{Z}$ and $\mathbb{Z}/6\mathbb{Z}\times 3(\mathbb{Z}/18\mathbb{Z})$. $\endgroup$ – user348628 Feb 9 '17 at 9:38
    • $\begingroup$ I am still not seeing any good reasons why the subgroups need to have this form. $\endgroup$ – Tobias Kildetoft Feb 9 '17 at 9:49
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      $\begingroup$ @TobiasKildetoft I'm pretty sure it's a consequence of the basis theorem for finite abelian groups. $\endgroup$ – steven gregory Feb 9 '17 at 10:07

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