Show that $\langle(1234),(12)\rangle = S_4$ I am trying to show that $\langle(1234),(12)\rangle = S_4$. I can multiply and get all $24$ permutations manually but isn't there a more compact solution?

Write $H$ for the subgroup generated by those two permutations. Then $(1234)(12)=(234)$, so $H$ contains certainly the elements of $\langle(1234)\rangle$, $\langle (234)\rangle$ and $(12)$, hence $\vert H\vert \geq 7$ and therefore $\vert H\vert \geq 8$ ($\vert H\vert$ must divide $24=\vert S_{4}\vert$). Since $(1234)\in H$ but such a permutation is not even, i.e it doesn't belong to $A_{4}$, $H$ is not a subgroup of $A_{4}$. By Lagrange's theorem, $\vert H\vert$ must equal $24$ (the only subgroup of order $12$ in $S_{4}$ is $A_{4}$) and we're done.
