Thanks for the link, but the proof assumes an understanding of homomorphisms which we haven't covered yet (it's actually the next chapter in the book). Could you possibly show me an alternate approach?
Yes! I just deleted my link because it was a diaster as an answer. I was just being lazy. I type the proof up real soon right now. Be Patient.
I will use the following really useful theorem.
Theorem: Let G be a group and H be a subgroup of G. Then H is a normal subgroup of G if and only if ghg^{-1} in H for all g in G and h in H.
I will prove only one direction, the one I think is harder. And I leave the second direction which will be very similar as an excersice. I recommend to post your proof of the other direction for me to check it.
In the proof I shall use a triangle symbol. It means "normal subgroup of".
Thanks for the help. I think I've got the other direction down. Let me know if I did something wrong.
Claim: If H is a normal subgroup of G, then H/N is a normal subgroup of G/N.
Since N is a normal subgroup of G and H is a subset of G, we know hNh^(-1) is an element of N, which means H/N exists. Since H is a normal subgroup of G, ghg^(-1) is an element of H for all g in G. So, ghg^(-1)N = gN(hN)g^(-1)N is an element of H/N. Thus, H/N is a subgroup of G/N. Q.E.D.
Sorry for the lack of LaTeX, I still need to learn how to use it.
That is basically it. It is just stated in a very strange way.
Nothing wrong here. But if it was on exam I would state the initial hypothesis such as, N is normal subgroup of G and H is a subgroup of G.Claim: If H is a normal subgroup of G, then H/N is a normal subgroup of G/N.
No need to say "subset of" if it is a subgroup it is certainly a subset .Since N is a normal subgroup of G and H is a subset of G,
Perhaps you want to say "we know that hNh^(-1)=N for all h in H.we know hNh^(-1) is an element of N
You mean well-defined. No need to say that. We already know that N is normal subgroup of G so definitely H.which means H/N exists.
And for all h in H.Since H is a normal subgroup of G, ghg^(-1) is an element of H for all g in G.
That is the most important part of the proof. That is all good.So, ghg^(-1)N = gN(hN)g^(-1)N is an element of H/N. Thus, H/N is a subgroup of G/N. Q.E.D.
It is very readable. Group theory looks easy.Sorry for the lack of LaTeX, I still need to learn how to use it.