anonymous
  • anonymous
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).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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).
anonymous
  • anonymous
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)
anonymous
  • anonymous
If i am not mistaken though, ab^-1 IS in the kernel?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
only if ø(a)= ø(b) --- I have a hard time with proofs, because it seems obvious, right?
anonymous
  • anonymous
if G is a homomorphism, then ø(ab) = ø(a)ø(b)
anonymous
  • anonymous
...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
  • anonymous
yes that is correct
anonymous
  • anonymous
...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
  • anonymous
where did ø(a) go?
anonymous
  • anonymous
= ø(b)
anonymous
  • anonymous
okay so then is that equal to e'?
anonymous
  • anonymous
Is the identity mapped to the identity?
anonymous
  • anonymous
I was not given that information, butI thought that was given in a homomorphism
anonymous
  • anonymous
However, I know that if ø is an isomorphism, then Ker ø={e}. and every isomorphism is a homomorphism

Looking for something else?

Not the answer you are looking for? Search for more explanations.