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

$\displaystyle g$ is an element of group $\displaystyle G$. $\displaystyle f(g)$ is a homomorphism from $\displaystyle G$ to $\displaystyle H$. Prove that $\displaystyle f(g^{-1})=f(g)^{-1}$.

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

$\displaystyle f(g^{-1}) f(g)^{-1} = ??$

but still cannot see what to do.

I also have that $\displaystyle e_G, e_H$, are the identities in $\displaystyle G$ and $\displaystyle H$, 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

Quote:

Originally Posted by

**HallsofIvy** 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?

$\displaystyle g$ in $\displaystyle G$ implies $\displaystyle g^{-1}$ in $\displaystyle G$.

$\displaystyle f(g)*f(g^{-1})=f(g*g^{-1})=f(e1)=e2$

Then just left multiply the first and last element by $\displaystyle f(g)^{-1}$.

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.