Question

Let ø: G--> G' be a homomorphism from the group G to the group G'. Prove that ø(a)= ø(b) if and only if ab^-1 exists in Ker(ø).

Answer 1

The Kernel of G is the subset that is mapped to the identity of G' by ø. So I think you need to show that if ab^-1 is not in the Kernel, then ø(a) cannot = ø(b).

Answer 2

yes, I know that I have to show two sides: the first side is :
to show ø(a)= ø(b) and only if ab^-1 exists in Ker ( ø)
Then I need to show that:
If ab^-1 exists in Ker ( ø), then ø(a)= ø(b)

Answer 3

If i am not mistaken though, ab^-1 IS in the kernel?

Answer 4

only if ø(a)= ø(b) --- I have a hard time with proofs, because it seems obvious, right?

Answer 5

if G is a homomorphism, then ø(ab) = ø(a)ø(b)

Answer 6

...so ø(ab^-1) = ø(a)ø(b^-1)
Isn't it true that if G is a homomorphism, (injective) that if ø(g) --> g' then there is no other g that is mapped to g'?

Answer 7

yes that is correct

Answer 8

...if ø(ab^-1) = ø(a)ø(b^-1) and ø(a)= ø(b), then ø(ab^-1) = ø(b)ø(b^-1) = ø(bb^-1) ...

Answer 9

where did ø(a) go?

Answer 10

= ø(b)

Answer 11

okay so then is that equal to e'?

Answer 12

Is the identity mapped to the identity?

Answer 13

I was not given that information, butI thought that was given in a homomorphism

Answer 14

However, I know that if ø is an isomorphism, then Ker ø={e}. and every isomorphism is a homomorphism