Number of subgroups of order $4$ in the direct product $S_2\times S_4$. I have the following problem:

Find the number of subgroups of 4 elements in the direct product of
permutation groups $S_2 \times S_4$

I started with writing down all the elements in $S_4$ group. Here we have $4!=24$ elements:
$$S_4=\{ (1)(2)(3)(4),\\ (1)(2)(34),(1)(3)(24), (12)(34), (13)(24), (1)(4)(23), (14)(23), (12)(3)(4), (13)(2)(4), (14)(2)(3), \\ (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(132),\\
(1243), (1234), (1342), (1324), (1423), (1432)  \}$$
The group $S_2$ is simple: $\{(1)(2), (12)\}$
In our subgroup we must have the neutral element $\{ (1)(2), (1)(2)(3)(4)\}$. But how can I find out how another $3$ elements can look like?
I understand, that we have to fulfill here 2 conditions:

*

*The product of any two elements must be in the subgroup

*The inverse of each element must be in the subgroup as well


Within the $S_4$ factor of $G$ there are three $C_4$ subgroups and four $C_2\times C_2$ subgroups.
The remaining subgroups of order $4$ all intersect the $S_4$ factor in a group of order $2$.
There are just three $C_4$ groups outside of $S_4$, generated by $(1,2)(3,4,5,6)$, $(1,2)(3,4,6,5)$ and $(1,2)(3,5,4,6)$.
There are three $C_2 \times C_2$ subgroup that intersect $S_4$ in $\langle(3,4)(5,6)\rangle$, namely $\langle (1,2),(3,4)(5,6) \rangle$, $\langle (1,2)(3,4),(3,4)(5,6) \rangle$, and $\langle (1,2)(3,5)(4,6),(3,4)(5,6) \rangle$, and similarly for the other two intersections of that type.
Finally, there are two $C_2 \times C_2$ subgroup that intersect $S_4$ in $\langle(3,4)\rangle$,namely $\langle (1,2),(3,4) \rangle$ and $\langle (1,2)(5,6),(3,4) \rangle$, and simialrly for the other $5$ intersections of that type.
So the total number of subgroups of order $4$ is
$$3+4+3 + (3\times 3) + (2 \times 6) = 31.$$
