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).

To prove that ø(a) = ø(b) if and only if ab^(-1) exists in Ker(ø), we must show two separate implications:

1. If ø(a) = ø(b), then ab^(-1) exists in Ker(ø).
2. If ab^(-1) exists in Ker(ø), then ø(a) = ø(b).

Let's prove each implication separately:

1. If ø(a) = ø(b), then ab^(-1) exists in Ker(ø):

To prove this, we need to show two things: (i) ab^(-1) is well-defined and (ii) ab^(-1) is in the Kernel of ø.

(i) Well-definedness:

Suppose that for some a and b in G, we have ø(a) = ø(b). We want to show that ab^(-1) is well-defined, meaning that it does not depend on the specific choices of a and b.

Let's consider another element c in G such that ø(c) = ø(a) = ø(b). We need to show that ac^(-1) = ab^(-1).

Now, since ø(a) = ø(b), we have ø(a) * ø(b)^(-1) = e', where e' is the identity element in G'. This implies ø(a) * ø(b)^(-1) = ø(b) * ø(b)^(-1) (since ø(b) * ø(b)^(-1) = e' for any element in G'). By applying the homomorphism property, we get ø(ab^(-1)) = e'. Similarly, we have ø(ac^(-1)) = e'.

Since ø(ab^(-1)) = ø(ac^(-1)) = e', we can conclude that ab^(-1) = ac^(-1), proving the well-definedness of ab^(-1).

(ii) Kernel of ø:

We now need to show that ab^(-1) is in the Kernel of ø, denoted as Ker(ø). Recall that Ker(ø) consists of all elements g in G such that ø(g) = e', where e' is the identity element in G'.

Since ø(a) = ø(b), we have ø(a) * ø(b)^(-1) = ø(b) * ø(b)^(-1), which simplifies to ø(ab^(-1)) = e'.

This means that ab^(-1) is in the Kernel of ø, satisfying the condition.

Therefore, if ø(a) = ø(b), then ab^(-1) exists in Ker(ø).

2. If ab^(-1) exists in Ker(ø), then ø(a) = ø(b):

To prove this, we assume that ab^(-1) exists in Ker(ø) and show that ø(a) = ø(b).

Since ab^(-1) is in the Kernel of ø, we have ø(ab^(-1)) = e', where e' is the identity element in G'.

By applying the homomorphism property, we get ø(a) * ø(b)^(-1) = e'.

If we multiply both sides of this equation by ø(b), we get ø(a) = ø(b), as desired.

Therefore, if ab^(-1) exists in Ker(ø), then ø(a) = ø(b).

By proving both implications, we have shown that ø(a) = ø(b) if and only if ab^(-1) exists in Ker(ø).