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?
 A: I'm not sure if this a more "compact" solution, but this follows from a more general result that the $\langle (123\ldots n),(12) \rangle = S_{n}$, which I think would be cleaner and more instructive to show. Hint: see if you can show that this subgroup generates all transpositions in $S_{n}$, and therefore generates $S_{n}$. 
A: Conjugate $(12)$ by powers of $(1234)$ to get $(12), (23), (34)$. Then, you can conjugate further to get all transpositions.
A: I hope you are familiar to $S_4$ presentation:
$$S_4=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$$ Considering a nice Corollary of Von Dyck's Theorem we can set $a=(1,2),~~~b=(1,2,3,4)$ and see that $(ab)^3=1,a^2=b^4=1$. Also, an small program in GAP's environment could show this fast and easy:
gap> f:=FreeGroup("a","b");
gap> a:=f.1;; b:=f.2;; 
     s:=f/[a^4,b^2,(a*b)^3];;
gap> StructureDescription(s);
                                         "S4"

A: The two methods I would use are either


*

*show all transpositions appear. If $c$ is the cycle and $t$ the transposition, start by considering $c^{-1}tc$.

*note that we have an element of order 4 already, and multiplying the ones we have together gives an element of order 3. Hence the group generated has order divisible by 12. But the only subgroup of size 12 is $A_4$ which this clearly is not.

A: 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.
