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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).

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  1. anonymous
    • 5 years ago
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    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).

  2. anonymous
    • 5 years ago
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    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)

  3. anonymous
    • 5 years ago
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    If i am not mistaken though, ab^-1 IS in the kernel?

  4. anonymous
    • 5 years ago
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    only if ø(a)= ø(b) --- I have a hard time with proofs, because it seems obvious, right?

  5. anonymous
    • 5 years ago
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    if G is a homomorphism, then ø(ab) = ø(a)ø(b)

  6. anonymous
    • 5 years ago
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    ...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'?

  7. anonymous
    • 5 years ago
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    yes that is correct

  8. anonymous
    • 5 years ago
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    ...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) ...

  9. anonymous
    • 5 years ago
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    where did ø(a) go?

  10. anonymous
    • 5 years ago
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    = ø(b)

  11. anonymous
    • 5 years ago
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    okay so then is that equal to e'?

  12. anonymous
    • 5 years ago
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    Is the identity mapped to the identity?

  13. anonymous
    • 5 years ago
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    I was not given that information, butI thought that was given in a homomorphism

  14. anonymous
    • 5 years ago
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    However, I know that if ø is an isomorphism, then Ker ø={e}. and every isomorphism is a homomorphism

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