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Here is my question :

Find $n$ such that $(\mathbb Z_6,+)$ is isomorphic to a group of permutations that is a subgroup of $S_n$.

I thought the answer would be $S_3$ but it turns out that $S_3$ is not abelian. Now I'm completely lost.

Any help would be appreciated.

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  • $\begingroup$ Note that they are asking you to show an isomorphism with a subgroup of a symmetric group. Hint: $\mathbb{Z}_6$ is sometimes called a cyclic group. $\endgroup$ – angryavian Nov 9 '15 at 0:04
  • $\begingroup$ Do you know Cayley's theorem, that every finite group is isomorphic to a permutation group? Do you know the proof of Cayley's theorem? Do you kinow how to find the order of a permutation? Can you find an element of order $6$ in $S_6$ In $S_5$? $\endgroup$ – bof Nov 9 '15 at 0:06
  • $\begingroup$ The question has been answered - look here. $\endgroup$ – Dietrich Burde Jan 18 '16 at 18:52
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Since $(\mathbb Z_6,+)$ is a cyclic group of order $6$, you need to find an element of order $6$ in $S_n$.

$S_3$ does not have an element of order $6$. Nor does $S_4$.

$S_6$ clearly does: the $6$-cycle $(123456)$.

Perhaps surprisingly, so does $S_5$, even though it's not a $6$-cycle: $(12)(345)$.

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