Subgroups of $S_4$ isomorpic to $S_2$

Question

I am trying t prove that there are only 9 subgroups of $S_4$ isomorphic to $S_2$ but I only find 7 different ones( fix 12, 13, 14, 23, 24 34 and swap the remaining two + identity). Which cases am I missing? And also is there a subgroup of $S_4$ isomorphic to $Z_6$ ?

Answer 1

Consider $\{id, (12)(34)\},$ $\{id, (13)(24)\}$, and $\{id, (14)(23)\}$.

There is no cyclic subgroup of order $6$ in $S_4$, hence no subgroup of order $6$ in $S_4$ isomormorphic to $\mathbb Z_6$.

Answer 2

No, every subgroup of order $6$ in $S_4$ is isomorphic to $S_3$. Since $S_4$ has no elements of order $6$, the group $\mathbb{Z}/6$ cannot accur as subgroup.

Answer 3

Recall that $S_2$ is a cyclic group of order 2. A subgroup of $S_4$ has order 2 if and only if it is generated by a permutation of order 2. A permutation of order 2 in $S_4$ is either of type $(i,j)$ (a transposition) or of type $(i,j)(k,\ell)$ (a product of two disjoint transpositions). The number of such generators is exactly 9: the six transpositions you mention, and the permutations $(12)(34), (13)(24), (14)(23)$. Thus $S_4$ has exactly 9 subgroups of order 2.

If $S_4$ contained a cyclic subgroup of order 6, then this subgroup is generated by some permutation of order 6. However, there does not exist a permutation in $S_4$ of order 6. The order of a permutation is the least common multiple of the lengths of its cycles, and it is not possible to obtain 6 as the least common multiple of the lengths of the cycles of a permutation if the lengths of the cycles sum to 4. Hence $S_4$ does not contain a cyclic group of order 6.