Proving Third Isomorphism Theorem: Normal Subgroups and Homomorphisms

[moo5003]

Problem:

" Prove (Third Isomorphism THeorem) If M and N are normal subgroups of G and N < or = to M, that (G/N)/(M/N) is isomorphic to G/M."

Work done so far:

Using simply definitions I have simplified (G/N)/(M/N) to (GM/N). Now using the first Isomorphism theorem I want to show that a homomorphism Phi from GM to G/M exists. Such that the Kernal of Phi is N.

I constructed phi such that GM -> G/M
where it sends all x |----> xN.

My problem is as follows: How do I know xN is actually in the set G/M. It may just be that I'm going about the proof in a way that is very complicated then it should be. Any help would be greatly appreciated.

 

[moo5003]

Alright I've been looking at some online proofs and I can see were I went wrong. I should have constructed a phi from G/N to G/M.

My only question is how to show that phi from a gN to a gM is onto G/M. I was looking at the proofs online and they didnt seem to make any sense on this part.

 

[matt grime]

The map is I presume the on induced by sending g to [g] its coset in G/M. This is surjective. N is in the kernel so it factors as G-->G/N-->G/M. And the second map must also be surjective.

THinking more concretely, each and every coset of M is a union of cosets of N, so your map from G/N to G/M just identifies these cosets of N.