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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?2012-10-22
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    C'est tout! You are done. It is, in fact, rather simple.2012-10-22
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    @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.2012-10-22
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    Relax and enjoy, you are done. Go have an ice cream cone.2012-10-23

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Ok. In fact, this is very trivial.