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A group $G$ is called an FC-group if $|x^G|$ is finite for each $x \in G$ (here $x^G$ is the conjugacy class of $x$ in $G$). Equivalently, $|G:C_G(x)|$ is finite for each $x \in G$.

I think it should be completely obvious that a homomorphic image of an FC-group is an FC-group & maybe I'm just tired, but I've been trying to think of a proof of this for a few minutes now and am coming up blank. It's one of these things that is left unproved whenever FC-groups are mentioned, presumably because the proof is straightforward.

Could somebody throw a proof my direction to ease my troubled mind?

Thank you.

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up vote 4 down vote accepted

Applying the First isomorphism theorem, this boils down to proving that the quotient of an FC group by a normal subgroup is again FC.

This is true, because if two classes in the quotient are conjugate, you can find representatives in the group that were already conjugate. The conjugacy class can therefore only get smaller after dividing out.

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Another way to say it is too look at the (preimage in G of the) centralizer K of xN in G/N. It contains all of N and all of C=C_G(x), so the size of the conjugacy class of x in G is [G:C]=[G:K][K:CN][CN:C] and so is quite a bit bigger than the size [G:K] of the conjugacy class of xN in G/N. –  Jack Schmidt Feb 13 '11 at 19:30
    
Thank you both for your help. I arrived at Jack's answer (more or less) since posting the question. I think my brain was switched off entirely yesterday. –  Bob Heffernan Feb 14 '11 at 5:38
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