proof question? help please?

Question

let f: G---> H be a homomorphism between groups G and H and let a belongs to G. prove that {x belongs to G | f(x) = f(a)} = (ker(f))a

Answer 1

Easy HOMEWORK.

Hint: Suppose f(x)=f(a), then f(xa^(-1))=e, so xa^(-1) is in ker(f). Conversely, suppose xa^(-1) is in ker(f), show f(x)=f(a).

Answer 2

As you should know, the way to prove two sets are equal is to show that each one is contained in the other. Let x be in the right hand side. Thus x=ya, where y is in the kernel of f. Thus f(x)=f(y)f(a)=ef(a)=f(a), so x is in the left hand set. Conversely, suppose f(x)=f(a), and consider the quantity y=xa^(-1). Then f(y)=f(x)f(a^(-1))=f(x)f(a)^(-1)=e since f(x)=f(a) by hypothesis. Thus y is in ker(f) and by multiplying the definition of y on the right by a we obtain x=ya. Thus x is in the right hand set.

Answer 3

Let S = {x belongs to G | f(x) = f(a)}.

Let y belong to ker(f). Then f(y)=eH, the identity element of H. Since f is a homomorphism, f(y*a) = f(y)f(a) = eH*f(a) = f(a), so ya belongs to S. Therefore (ker(f))a is a subset of S.

Now, let z belong to S, and let b be the inverse of a. Then f(z)*f(b) = f(a)*f(b) = f(a*b) = f(eG) = eH. Therefore zb belongs to ker(f), and zba = z belongs to ker(f)a, so S is a subset of (ker(f))a.

Answer 4

When I see this, I am glad that I didn't do well in math! The only place that I have seen kerf before this is in wood shop.
TRIZ is the method that I would use to solve this problem. When you have an insoluble problem, you solve a subset and apply that to the whole. Can you construct some sets and run your kerf on them?