# elements in the product of subgroups in $S_4$

Let $N:=\{e,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\}$ be the normal subgroup of $S_4$ and $H:=\langle(1\,2\,3\,4)\rangle$ be the cyclic subgroup of $S_4$ generated by $(1\,2\,3\,4)$. Using the Second Isomporphism Theorem, how do I easily find the elements of $HN$? I know that $|HN|=\frac{|H||N|}{|H\cap N|}=\frac{4\cdot4}{2}=8$, but how can I actually find those 8 elements without brute force?

I don't know how you expect to find the eight elements without making a computation, but the second isomorphism theorem does give you a way to be efficient. Using the isomorphism $HN/N\cong H/H\cap N$ you get that the distinct cosets in $HN/N$ are $N$ and $(1234)N$. Now, compute the elements in these cosets.
• I found the elements of $H/(N\cap H)$ but those aren't necessarily the elements of $HN/N$, they are just isomorphic. How do I find the isomorphism mapping between the two groups? Mar 2 '16 at 23:49
• The map $H\to HN/N$, $h\mapsto hN$ is subjective. So... Mar 3 '16 at 0:29