non-Abelian subgroup of order 6 in $S_5$?

Question

First off, yes, this is a homework question. However, I've been trying everything and I just feel like I must be missing something obvious.

In $S_5$, the only way to get a subgroup of order 6 would be to do a disjoint cycle like $(abc)(de)$, so we can have an order of $lcm(3,2)=6$. However, any subgroup in $S_5$ with distinct elements for $(abc)(de)$ just ends up with the set of $e, (abc)(de), (acb), (de), (abc), (acb)(de)$, and it seems to always be Abelian.

What am I doing wrong? Anything at all would be appreciated.

Answer 1

Think about the set of all permutations that act only on the first $3$ elements -- namely the subgroup $S_3 \leq S_5$. But $|S_3|=6$ and it is nonabelian, so it's exactly what you're looking for!

You do not need an element of order $6$ -- that would be required to get a cyclic group of order $6$, but that is Abelian anyways, so we don't want that.

Answer 2

You've correctly noted that the only elements of order $6$ in $S_5$ are of cycle type $(2,3)$. However, it's not the case that a group of order $6$ needs to contain an element of order $6$. For example, $S_3$ is a non-abelian group of order $6$, and none of its elements have order $6$. Can you find a copy of $S_3$ inside $S_5$?