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Prove that if $g$ and $\bar{g}$ commute then so do $\alpha(g)$ and $\alpha(\bar{g})$.

Let $\alpha : G \to H$ is a homomorphism. Let $g, \bar{g} \in G$.

Here's what I have done -

$g\bar{g} = \bar{g}g$

$\alpha(g\bar{g}) = \alpha(\bar{g}g)$

$\alpha(g)\alpha(\bar{g}) = \alpha(\bar{g})\alpha(g)$

Is that it, it seems a bit too trivial?

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Why do you expect this not to be trivial? – Chris Eagle Oct 22 '12 at 23:44
C'est tout! You are done. It is, in fact, rather simple. – ncmathsadist Oct 22 '12 at 23:44
@ChrisEagle: Probably because I had a bad start with group theory and I always expect the worst :) ...hopefully that will change this weekend when I put in 3 days straight studying it. – Jim_CS Oct 22 '12 at 23:47
Relax and enjoy, you are done. Go have an ice cream cone. – ncmathsadist Oct 23 '12 at 0:03
up vote 2 down vote accepted

Ok. In fact, this is very trivial.

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