0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Matt G
    Mar 2 '16 at 23:49
  • $\begingroup$ The map $H\to HN/N$, $h\mapsto hN $ is subjective. So... $\endgroup$
    – David Hill
    Mar 3 '16 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.