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.
Best Answer
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: $(12)(345)$.