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I am trying to describe the kernel of the homomorphism $f:G \rightarrow G$ s.t. $G$ is abelian and $f(x) = x^5$.

So far I have that

$$ ker(f) = \{x : x^5 = e \} = \{x : x = xe = x(x^5) = x^{6 + 5k} \} $$

But this isn't making use of the abelianness of $G$ nor does it seem that interesting (which I assume the problem is getting at something interesting). Is there something I'm missing?

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    It makes use of $G$ being abelian in your first line. $f$ will in general be no homomorphism, but as $G$ is abelian, $f(xy) = (xy)^5 =\text{ ($G$ abelian) } x^5y^5 = f(x)f(y)$.2012-10-20
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    You already have the answer and wrote it down: $\,\ker f=\{x\in G\;;\;x^5=1\}\,$ . Nothing more can be said in such a general case and of course, as the other commenters noted, you have to use *abelian* in order to make sure the map $\,f\,$ is a homomorphism.2012-10-20

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