2

I'm having trouble with finding a specific isomorphism for the following exercise, and I cannot think of any theorems that will be helpful here. Any help is much appreciated.

For $G$ an abelian group, $N\lhd G$, $K=\{(u,v)\in G\times G:uv^{-1}\in N\}$, prove $K\cong G\times N.$

What I tried: I first showed that $K\subset G\times G$ is a subgroup, and in a next part that for $G/N$ abelian (which is definitely the case here), $K\lhd G\times G$ and $(G\times G)/K\cong G/N$. I don't know how to apply these results here, however, and I also cannot think of any theorems that will be useful here. Therefore I tried to find an explicit isomorphism $\phi:K\to G\times N: (u,v)\mapsto(\cdot,uv^{-1})$, since by definition of $K$, $uv^{-1}\in N$. The problem is what I have to map to the first element. Well-defined, injective and a homomorphism is not difficult but I didn't manage to find a surjective map. I tried mapping $u$,$v$ or $uv$ to the first element, but these maps where not surjective.

Václav Mordvinov
  • 3,268

2 Answers2

1

Hint: prove that

$$G\times G/G\times N\cong G/N$$

marwalix
  • 16,773
  • Thank you. First isomorphism theorem with $\phi:G\times G\to G/N:(g_1,g_2)\mapsto g_2N$, right? However can I conclude directly from $(G\times G)/(G\times N)\cong (G\times G)/K$ that $G\times N\cong K$? – Václav Mordvinov May 23 '18 at 21:27
  • I'm in doubt because of, for example, this post https://math.stackexchange.com/questions/1584568/isomorphic-quotient-groups-fracgh-cong-fracgk-imply-h-cong-k – Václav Mordvinov May 23 '18 at 21:37
  • I accepted the other answer because I'm not sure if your approach works; also see the link I posted. If there is not some extra condition that allows us to make the conclusion $G\times N\cong K$, your approach does not seem to work – Václav Mordvinov May 24 '18 at 08:31
  • You’re right to accept the constructive answer that exhibits an isomorphism. It is true that $G/H\cong G/K$ does not imply $H\cong K$. Here we are in a special case. I have no time right now. I will write down the details this evening. – marwalix May 24 '18 at 08:55
  • @Legoman the linked post considers the other direction: if the quotients $G/H,G/K$ are isomorphic, then we cannot conclude $H\cong K$. However, like you say there might be an argument using the kernels, but I don't see it. @ marwalix, that is allright, take all the time you need. – Václav Mordvinov May 24 '18 at 09:14
  • @VáclavMordvinov I delete my post since I read the wrong stuff. So sorry for this! – InsideOut May 24 '18 at 09:17
1

Barring some computational error, the maps $$ \varphi\colon K\to G\times N:(u,v)\mapsto (u^{-1},uv^{-1}) $$ and $$ \psi\colon G\times N\to K:(g,n)\mapsto (g^{-1},(gn)^{-1}) $$ should give an explicit isomorphism.

Ben West
  • 12,366
  • 1
    Thanks for your answer! I tried the same map, but with $u$ as the first element, and since that one didn't work out, I assumed the case for $u^{-1}$ would be the same. – Václav Mordvinov May 24 '18 at 08:30
  • By the way, now I look at it the case where the first element is $u$ also works... I don't know why I didn't see this yesterday – Václav Mordvinov May 24 '18 at 08:37
  • @VáclavMordvinov exactly, let $(x,n)\in G\times N$, then $f(x,n^{-1}x)=(x,xx^{-1}n)=(x,n)$ and we can guarantee that $(x,n^{-1}x)\in K$ since $x(n^{-1}x)^{-1}=n\in N$. :) I just solved it too! – rae306 May 24 '18 at 16:17
  • @VáclavMordvinov Ok, good, I didn't try $u,v,uv$ based on the question, but it sounds like the isos could be made a little simpler. – Ben West May 24 '18 at 16:19
  • Yes, but either $u$ or $u^{-1}$ are equally straightforward to me, so your answer was exactly what I was looking for. Thank you, really appreciate the effort :) – Václav Mordvinov May 24 '18 at 16:21