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anonymous
 5 years ago
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).
anonymous
 5 years ago
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).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If i am not mistaken though, ab^1 IS in the kernel?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only if ø(a)= ø(b)  I have a hard time with proofs, because it seems obvious, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if G is a homomorphism, then ø(ab) = ø(a)ø(b)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...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'?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...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) ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so then is that equal to e'?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is the identity mapped to the identity?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was not given that information, butI thought that was given in a homomorphism

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0However, I know that if ø is an isomorphism, then Ker ø={e}. and every isomorphism is a homomorphism
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