# Calculation of subgroups of $(Z_{12}, +)$

How calculate all subgroups of $(Z_{12}, +)$? I know that the order of subgroups divide the order of the group, but there is such a smart way to calculate the subgroups of order 6?

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 This is a cyclic group, right? – Hurkyl Oct 6 '12 at 3:48 Why the low accept rate ? – Belgi Oct 7 '12 at 20:34

$\mathbb Z_{12}$ is cyclic, which means all of its subgroups are cyclic as well.

$\mathbb Z_{12}$ has $\phi (12)=4$ generators: $1, 5, 7$ and $11$, $Z_{12}=\langle1 \rangle=\langle 5 \rangle=\langle 7 \rangle=\langle 11 \rangle$.

Now pick an element of $\mathbb Z_{12}$ that is not a generator, say $2$. Calculate all of the elements in $\langle2 \rangle$. This is a subgroup. Repeat this for a different non-generating element. You should find $6$ subgroups.

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Hint: If a subgroup contains an element $n$, then it also contains $n+n, n+n+n, \ldots$

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Adding to above theoretical nice approaches; you can use GAP to find all subgroups of $\mathbb Z_{12}$ as well:

Regarding Subgroups in GAP: please see my comment here and this F.A.Q. – Alexander Konovalov Apr 22 at 21:15