Here is the question:

I can easily show that right congruence mod N is a congruence relation on G, but normality never enters the proof! In the other direction all I can show is that if right congruence mod N is a congruence relation that N < G. Again, normality doesn't seem to enter.Let N < G. Show that N is normal in G iff right congruence mod N is a congruence relation on G.

As far as proofs go, here is a sketch:

Right congruence mod N means that for all a and b in G we have that ab^{-1} is in N. From this I can easily show reflexivity, symmetry, and transitivity...ie. right congruence mod N is a congruence relation on G. None of this uses the normality of N in G.

In the other direction I can show that e is in N and if a is in N then a^{-1} is also in N. So N < G. Again I can't seem to find an argument for normality.

I did have the thought of showing that left congruence mod N forms the same congruence relation on G, but since the problem statement specifically refers to right congruency I decided that this approach was not the one the text was aiming for.

-Dan