Proof that homomorphism of an inverse equals the inverse of the homomorphism

is an element of group . is a homomorphism from to . Prove that .

I'm told the proof is by multiplying the two terms, like this (I guess)

but still cannot see what to do.

I also have that , are the identities in and , respectively.

Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

Oops, sorry. It is not "inverse of a homomorphism" but "homomorphism of an inverse".

That is, if ab= e1, the identity in the first group (ring, field?), then you want to show that f(a)f(b)= e2. Well, the whole point of a homomorphism is that f(ab)= f(a)f(b). What does that tell you if ab= e1?

Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

How can I post things by using this beautiful notation?

Re: Proof that homomorphism of an inverse equals the inverse of the homomorphism

Quote:

Originally Posted by

**mohammedlabeeb** How can I post things by using this beautiful notation?

See this post.