Why are there 5 subgroups that are generated by double transpositions on four elements in $S_5$?

I am trying to calculate the total number of subgroups for each subgroup in $S_5$. I am working on subgroups of order $4$.

I understand that there are $\frac{{5 \choose 4}4!}{4}$ subgroups for $Z_4$ in $S_5$ (there are ${5 \choose 4}4!$ $4$-cycles in $S_5$ and 4 representations of the same subgroup). I also understand that there are $\frac{{5 \choose 2}{3 \choose 2}}{2}$ subgroups generated by a pair of disjoint transpositions (${5 \choose 2}$ selections for the first transposition, ${3 \choose 2}$ selections for the second transposition, and 2 representations of the same subgroup).

From my teacher, I know that I can create more Klein four-groups subgroups through two double transpositions. For example, $\{(), (12)(34), (13)(24), (14)(23)\}$ is a Klein four-group constructed through two double transpositions in $S_5$.

On further inspection from this website, I know that there are $5$ subgroups generated by double transpositions on four elements in $S_5$.

Why are there 5 subgroups generated by double transpositions on four elements in $S_5$?

Solution 1:

Hint: They're in bijection with the elements of the set $\{1,2,3,4,5\}$ they don't move.