5

There is a problem in Lang's book that I don't quite understand how to proceed. It is problem #5, pg 75. I have already shown that the subgroups N and N' can be identified as normal in G, G'. But I don't know how to show that the image of H in G/N$\times$G'/N' is the graph of an isomorphism.

Problem statement:

Let $G, G'$ be groups, and let $H$ be a subgroup of $G \times G'$ such that the two projections $p_1 : H \to G$ and $p_2 : H \to G'$ are surjective. Let $N$ be the kernel of $p_2$ and $N'$ be the kernel of $p_1$.

One can identify $N$ as a normal subgroup of $G$, and $N'$ as a normal subgroup of $G'$. Then the image of $H$ in $G/N \times G'/N'$ is the graph of an isomorphism $ G/N \approx G'/N' $.

Community
  • 1
Enigma
  • 3,909
  • 3
    For convenience, could you include the statement from Lang of the problem? – coffeemath Sep 14 '14 at 06:10
  • 1
    Seconding coffeemath's plea. A large number of people who could answer your question either don't own a copy of Lang, or have it in their office or somewhere else out of reach. Basically you are making it difficult for them to help you. – Jyrki Lahtonen Sep 14 '14 at 06:15
  • 2
    Here is the problem, sorry for the inconvenience. Let G, G' be groups, and let H be a subgroup of G$\times$ G' such that the two projections $p_1$: H$\rightarrow$ G and $p_2$: H$\rightarrow$ G' are surjective. Let N be the kernel of $p_2$ and N' the kernel of $p_1$. One can identify N as a normal subgroup of G, and N' as a normal subgroup of G'. Then the image of H in G/N$\times$ G'/N' is the graph of an isomorphism G/N$\approx$ G'/N'. – Enigma Sep 14 '14 at 06:18

1 Answers1

5

It seems to me that you're at a stage where you don't know what you need to try proving, so instead of giving a full answer, I'll give a hint that will hopefully clarify things.

Prove the following lemma: Given groups $H, K, K'$ and surjective morphisms $H \to K$ and $H \to K'$ with the same kernel, the image of the induced morphism $H \to K \times K'$ is the graph of an isomorphism between $K$ and $K'$. (Note that having the same kernel is actually a necessary condition, since no elements $(1,k')$ or $(k,1)$ are permitted to appear in the image.) To prove the lemma, it may be helpful to factor first through the common kernel $M$, at which point it should become obvious by identifying each factor with $H/M$.

Now return to the original situation and see what needs to be proved.

Dave
  • 2,087