Let $A =$ the symmetric group in $S_4$. Consider the following subgroup of $A$: $$J=\langle(1234),(12)(34)\rangle$$ Find $|J|$.
I believe that the subgroup must contain the identity and that $(1234)$ is a cyclic subgroup of order $4$ so $4$ must divide the order of $A$ (Cayley's and Lagrange). That gives us $3$ elements and that $|J|$ must be $4$, $8$, $12$, or $24$.
I then multiplied $(1234)*(12)(34)$ to get $(13)$ and that's another element. Also, I know the inverse of $(1234)$ must be in the subgroup as well. I've narrowed the possibilities down to $|J|$ just be $8$, $12$, or $24$. My guess is the answer is $8$ but I'm not completely sure of my calculations and I don't know how to find more elements to prove this.
Let $H$ be the subgroup generated by $r =(1234)$ and $K$ be the subgroup generated by $s = (12)(34)$. Since $srs^{-1} = (2143) = r^{-1}$, $K$ lies in the normalizer $N_G(H)$. Thus $HK$ is a subgroup of order $\frac{|H||K|}{|H \cap K|} = 8$.