Smallest subgroup of $S_4$ containing a set of permutations. So the question asks to find the order of the smallest subgroup of $S_4$ (call this subgroup $G$) which contains the following set of permutations:
$$\{ (12)(34),(14)(32),(24),(31) \}$$
I know that since the order of $S_4$ is 24, the order of $G$ must divide 24.
I also have that the identity element is included in $G$ so the order of $G$ must be at least $5$ which limits the possible orders to: $6$, $8$, $12$, and $24$.
In a similar question I had that all the permutations were even so could use that it was in $A_4$ but that won't work here. Any tips or tricks I can use to solve this would be appreciated.
 A: Since $(1\ 2)(3\ 4)(1\ 4)(2\ 3) = (1\ 3)(2\ 4)$, it's clear your group contains:
$V = \{e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$, which eliminates $6$ as a possible order.
Since your generated group contains odd permutations, it cannot be $A_4$, which eliminates $12$ as a possibility (as this is the only subgroup of $S_4$ of order $12$).
So we are left with just $8$ or $24$.
Finally, note that:
$(1\ 3)(1\ 2)(3\ 4) = (1\ 2\ 3\ 4)$, so that your generated group contains:
$D_8 = \{e, (1\ 3), (2\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3), (1\ 2)(3\ 4), (1\ 2\ 3\ 4), (1\ 4\ 3\ 2)\}$ (check that this is a group of symmetries of a square with vertices $\{1,2,3,4\}$, and $r = (1\ 2\ 3\ 4), s = (1\ 2)(3\ 4)$).
Since your set is completely contained in this subgroup, and $8 < 24$, we have a winner.
A: It will be a Sylow 2 subgroup of $S_4$
$S_4$ has a Sylow 2 subgroup of order 8. (12)(34) and (24) generate a 4 cycle. So 4 divides the order of G. so we have (1,2)(3,4)(2,4)=(4,1,2,3) =(1,2,3,4).
But we have $(1,2)$ is not in $<(1,2,3,4)>$ thus order of the subgroup is 8,12 or 24. But we know (2,4) and (1,2,3,4) generate $D_8$ so the group generated is $D_8$ since I can express the other elements using these two generators.
Note
(1,2,3,4)(2,4)=(1,2)(3,4)
(4,3,2,1)(2,4)=(1,4)(2,3)
(1,2,3,4)²(2,4)=(1,3)
