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

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October 14th 2011, 10:57 AM
MSUMathStdnt

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

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

I'm told the proof is by multiplying the two terms, like this (I guess)
$f(g^{-1}) f(g)^{-1} = ??$
but still cannot see what to do.

I also have that $e_G, e_H$ , are the identities in $G$ and $H$ , respectively.
October 14th 2011, 12:12 PM
HallsofIvy

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?
October 14th 2011, 12:53 PM
MSUMathStdnt

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?

$g$ in $G$ implies $g^{-1}$ in $G$ .
$f(g)*f(g^{-1})=f(g*g^{-1})=f(e1)=e2$

Then just left multiply the first and last element by $f(g)^{-1}$ .
April 19th 2013, 02:47 PM
mohammedlabeeb

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

How can I post things by using this beautiful notation?
April 19th 2013, 03:05 PM
emakarov

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.

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