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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
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3 Answers

up vote 3 down vote accepted

$\mathbb Z_{12}$ 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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I have got trouble in understanding the line "$\mathbb Z_{12}$ has $\phi (12)=4$ generators: $1, 5, 7$ and $11$, ..".How did I get the generators $1, 5, 7$ and $11$.Can someone explain,please? –  learner May 22 '13 at 11:05
    
$\phi(n)$ 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 $Z_{12}$. –  alex-1729' May 6 at 18:43
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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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The OP wanted a theoretical solution instead of yours. –  Nancy Rutkowskie Oct 7 '12 at 12:56
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Regarding Subgroups in GAP: please see my comment here and this F.A.Q. –  Alexander Konovalov Apr 22 '13 at 21:15
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