Show $f \in S_4$ does not generate the whole $S_4$

Question

I want to show that $f \in S_4$ does not generate the whole $S_4$, that is $\{f^k\mid k\in \mathbb{N}\}$ is not the whole $S_4$. I think of the possible disjoint cycle structures $f$ can have: i) one 4-cycle such as $(1 2 3 4)$, ii) one 3-cycle and one 1-cycle eg. $(1 3 2)(4)$, iii) one 2-cycle and two 1-cycles eg. $(1 2)(3)(4)$, iv) two 2-cyles eg. $(1 2)(3 4)$, v) 4 1-cyles eg. the identity. When I work out the powers I see that the possible generators are (except for the structure of $f$ itself) are in i) 2 2-cycles and in ii-v) 4 1-cycles. Hence every element in $S_4$ only produces two different cycle structure and can therefore not generate the whole set $S_4$. I am unsure if I have included all the possibilities in the proof and if there is a neater approach.

Answer 1

Here's a neater approach: for any integers $j,k$, $f^j$ and $f^k$ commute. However, $S_4$ is not abelian.