Let $V=\{(1),(1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}$ a subgroup of $S_4$, and let $H$ the subgroup generated by $V \cup \{(1,2,3)\}$ in $S_4$.
I think that we have $$H=V_4 \cup \{(1,2,3), (1,3,2), (1,3,4),(1,4,3),(1,4,2,3),(1,3,2,4),(2,4,3),(2,3,4)\}=:X$$ and $H$ has order $12$. But I don't know how to prove it without doing too many computations. For instance, I could check that $X$ is indeed a subgroup of $S_4$, and I'd be done. But I'd like to avoid the $\sim {12 \choose 2}$ computations. It is also sufficient to show that $H \neq S_4$, but I don't know how to do it.
(As a bonus question, how would you prove that $\langle V \cup \{(1,2)\}\rangle$ has order $8$?)
Thank you for your help.
First of all note that $V$ is the Klein group $\mathbb{Z}_2 \times \mathbb{Z}_2$, and that it acts transitively on the numbers $\{1, 2, 3, 4\}$. Consider the group generated by $V$ and $(1,2)$ and note that it also acts transitively on this set. What is the size of the orbit of $3$, what is the size of the stabilizer group of $3$, conclude the size of the group (using the orbit-stabilizer theorem). The group generated by $V$ and $(1,2,3)$ also acts transitively on $\{1, 2, 3, 4\}$, use the orbit-stabilizer theorem again to find the order of this group.