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I have to: "Find all abelian subgroups of a $4$-element permutation group $\Sigma_4$"

I don't know what is $\Sigma_4$. Don't know how to bite this topic. The exercise seems too general to me. Any help?

luka5z
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  • should I use the fact that cyclic implies abelian? But how can I find ALL of them? – luka5z Sep 13 '14 at 17:37
  • Look here which subgroups are abelian: http://math.stackexchange.com/questions/566997/number-of-subgroups-of-s-4. See also here: http://groupprops.subwiki.org/wiki/Subgroup_structure_of_symmetric_group:S4 – Dietrich Burde Sep 13 '14 at 17:39
  • Ok, so all cyclic groups are abelian. So there are 24 abelian subgroups (there are 24 elements in this group, and a group generated by one element is cyclic). So how do I find all of them? – luka5z Sep 13 '14 at 17:44
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    I'd advice you first to learn about the group $;S_n;$ (this is the most used notation for what you wrote $;\Sigma_n;$). This group also has abelian subgroups that are not cyclic. – Timbuc Sep 13 '14 at 17:44
  • I know $S_4$ is the group of permutations of set ${1,2,3,4}$ – luka5z Sep 13 '14 at 17:45
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    Well @luka5z, in fact $;\Sigma_4=S_4;$ – Timbuc Sep 13 '14 at 17:46
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    There is a simple algorithm for gaining intuition for a problem that is too difficult, which is to make it simpler. For example, can you solve the question for $S_3$? If not, how about $S_2$? $S_1$? For example, by looking at $S_3$, you can see that there are $5$ abelian subgroups, not $6$, so your idea of counting elements must have a flaw somewhere. – Andrew Dudzik Sep 13 '14 at 18:28

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All subgroups of order $n\le 4$ of $S_4$ are abelian, and all subgroups of order $n\ge 5$ are non-abelian. The latter are $S_3, D_8,A_4$ and $S_4$, up to isomorphism. The abelian subgroups up to isomorphism then are the trivial group, the cyclic group of order $2$, $3$ and $4$, and the Klein four-group $V_4$. Here $V_4$ is abelian but not cyclic. If you have to find all abelian subgroups, then we must list all possible subgroups of order $n\le 4$.

Dietrich Burde
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