yes, for ANY subgroup N, a ~ b if ab^-1 is in N is an equivalence relation. but a congruence on a group must satisfy an additional condition:
g ~ g' and h ~ h' ---> gh ~ g'h' for all g,g'h,h' in G.
to prove THIS, you will need normality.
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
Well I must say that I was wondering about the difference between a congruence relation and an equivalence relation. So far I have not found the gh ~ g'h' condition in either of my texts. (Though they may assume I already know that fact, or perhaps it is buried in another theorem.) I'll give it another go.
Thanks!
-Dan
often the condition gh ~ g'h' is disguised as something along the lines of: a congruence is an equivalence compatible with the group operation.
this is something of an old-fashioned approach, it is more customary to just state everything in terms of homomorphisms, these days.
a normal sub-group IS the kernel of a homomorphism, saying N is normal in G is the same as saying: there is a quotient homomorphism G--->G/N
or more succintly, 1--->N-->G--->G'--->1 is exact.