4 Element abelian subgroup of S5. I have a homework question from my intro to group theory class. 
Question: 
Find a 4 element abelian subgroup of $S_5$.  Write it's table.
This is where I've gotten so far, but I don't even know if I'm on the right mental track.
We know $S_5 = \{1,2,3,4,5\}$.
Abelian means the subgroup of $S_5$ is commutative. 
I am having trouble figuring out how to start this.  Do I pick any 4 elements of $S_5$?  Or are there particular elements that must be chosen?
Do I perform permutations on the group first?
Or do I chose my 4 elements, and then perform the permutations?
Thank you!
 A: $S_5 \neq \{1, \,2,\, 3,\, 4, \,5\}$, but rather, $S_5$ is the group of permutations of the 5 elements in $\{1,\, 2,\, 3,\, 4,\, 5\}.$
Hint: Pick an abelian subgroup of order $4$ (hence not just any four elements, but four elements in $S_4$ that together meet the criteria of a commutative subgroup of $S_5$.) 
Such a group can easily be found by choosing a permutation, call it $\sigma$, of order $4$, and then determining the cyclic subgroup $H$ of $S_5$ that is generated by $\sigma$: $\;\; H =\{id,\, \sigma,\, \sigma^2, \,\sigma 3\}$. And so we have a cyclic, and therefore abelian, subgroup $H$ of $S_5$. 
Additional hint: Let $\sigma$ be a $4$-cycle in $S_5$ (A $4$-cycle is cycle of length $4)$.
Take, for example, the $4$-cycle $\sigma = (1\,2\,3\,4)(5) = (1\,2\,3\,4)\in S_5.\,$ Then $\,\sigma^2 = (1\,3)(2\,4) \in S_5,\,$ and $\,\sigma^3 = (1\,4\,3\,2)\in S_5.\,$ Finally, $\sigma^4 = id$. Hence we have that $\langle \sigma\rangle = \{\sigma, \sigma^2\, \sigma^3, id\}$ is a cyclic (hence commutative) subgroup of order $4$ in $S_5$.
