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?
Calculation of subgroups of $(Z_{12}, +)$
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$\begingroup$
abstract-algebra
group-theory
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0This is a cyclic group, right? – 2012-10-06
3 Answers
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$\mathbb Z_{12}$ 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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0$\phi(n)$ is the Euler totient function - the number of coprime numbers less than n. 1,5,7,11 are all coprime with 12 and less than 12 and are the only such numbers, thus $\phi(12)=4$. Because they are all coprime with 12 then they will generate $Z_{12}$. – 2014-05-06
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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:
> LoadPackage("sonata"); Z12:=CyclicGroup(12); A:=Subgroups(Z12); List([1..Size(A)],k->(A[k]));
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1Regarding `Subgroups` in GAP: please see [my comment here](http://math.stackexchange.com/questions/173565/group-of-order-60/173759#comment792635_173759) and [this F.A.Q.](http://www.gap-system.org/Faq/faq.html#7.7) – 2013-04-22