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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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    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. – 2017-02-09
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    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? – 2017-02-09
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    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. – 2017-02-09
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    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. – 2017-02-09
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    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. – 2017-02-09
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    I understand; so in this case, which is the general procedure and which theorem have I to use? – 2017-02-09
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53317/discussion-between-g-cantisani-and-arthur). – 2017-02-09
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    An idea: if I consider subgroup $K$ such that $|K|=18$, it is $K\subseteq H$, can I look at the quotient that is isomorphic to $\mathbb{Z}/6\mathbb{Z}$? Now can I look the subgroup of this group and consider the subgroup of right index? – 2017-02-09

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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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      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). – 2017-02-09
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      Also, you should have $\frac{6}{m}\cdot\frac{18}{n}=36$, which simplifies to $mn=3$. – 2017-02-09
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      I suppose are $3(\mathbb{Z}/6\mathbb{Z})\times \mathbb{Z}/18\mathbb{Z}$ and $\mathbb{Z}/6\mathbb{Z}\times 3(\mathbb{Z}/18\mathbb{Z})$. – 2017-02-09
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      I am still not seeing any good reasons why the subgroups need to have this form. – 2017-02-09
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      Is true that the subgroup of $A\times B$ is in the form of $H\times K$, if $H – 2017-02-09
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      @G.Cantisani No, there are also subgroups not of that form. – 2017-02-09
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      Ok so does exists a general procedure to solve this type of exercises? – 2017-02-09
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      @TobiasKildetoft I'm pretty sure it's a consequence of the basis theorem for finite abelian groups. – 2017-02-09
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      No, there are definitely other subgroups. Whether any of them have order $36$ I didn't actually check, but I don't see why none of them would. – 2017-02-09
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      OP wants the order of subgroups to be $36$ here its $18$ – 2017-02-09
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      @TobiasKildetoft - please show me one. – 2017-02-09
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      Take an element $x$ of order $3$ in the first factor and $y$ of order $3$ in the second factor. Then the subgroup generated by $(x,y)$ is not of the given form. – 2017-02-09
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      @TobiasKildetoft It's also not of order 36. – 2017-02-09
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      Yes, as I said, I didn't check whether any such had order $36$, but I see no reason they should not be (I haven't answered the question precisely because I haven't worked through these things). – 2017-02-09
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      The main issue seems to be whether OP is familiar with the classification of finite abelian groups. If not, the solution needs more work. – 2017-02-10
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      @learnmore - I know. I'm trying to come up with a sufficient answer of manageable size. – 2017-02-10
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      @Leppala Even using that, I don't see any good way for the OP to be able to spot the additional subgroups. For example, if we pick an isomorphism $\varphi$ between the (unique) quotients of order $3$ of each factor, then the set of elements $\{(x,y)\mid \varphi(x) = y\}$ is a subgroup of index $3$ (where we take $\varphi(x) = y$ as a statement when considered on the quotients). I don't see any way a students at this point would be expected to find subgroups of that form. – 2017-02-13