The symmetric groups $S_4$ and $S_5$

Question

In which symmetric group $S_4$ or $S_5$ is there an element of order 2 and an element of order 3 that commute?

I have worked out the 9 elements of order 2 and 8 elements of order 3 in $S_4$. Is there a more efficient way of working this out or is it a case of going through each element and seeing if they commute?

Answer 1

To add some more context to Mathsingh's comment:

Recall that in the symmetric group $S_n$, two cycles commute if and only if their supports are disjoint. Since you are interested in a 2-cycle and a 3-cycle, this condition means

$(a,b)$ and $(x,y,z)$ commute if and only if

1. $a,b \not \in \{ x,y,z \}$
2. $x,y,z \not \in \{a,b \}$

Now notice that, in $S_4$ there is not enough room! By the pigeonhole principle, we cannot have $5$ different values ($a,b,x,y,z$) all non-equal. So any 2-cycle and 3-cycle you choose will have overlapping supports, and thus cannot commute.

In $S_5$ however, we can choose, say, $(1,2)$ and $(3,4,5)$. These cycles have disjoint supports, and thus commute.

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I hope this helps ^_^