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If the conjugacy relation on a group $G$ (i.e. $a \sim b \iff \exists x\in G\colon b=a^x $) is a congruence then $G$ is abelian. How to prove that?

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    @lh$f$: we should link to that from the Ask A Question page... It is tiring to have to repeat it with variations!2012-01-08

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Note the following:

  1. A group $G$ is abelian iff any conjugacy class of it consists of exactly one element.
  2. In any group $G$ the conjugacy class of $1$ is exactly $\lbrace 1\rbrace$.
  3. For any $a,b\in G$ $a\sim a^b$.

Now suppose that $\sim $ is a congruence relation, then $a\sim b\wedge c\sim d\Rightarrow ac\sim bd$ Try using the above 3 statements (after proving them, of course) with $c=a^{-1}$ and $d=(a^{-1})^b$, and see where this gets you :-)

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    Oh, thanks! I already got that result using $c$ just like you say and $d=c$ obtaining the same relation without using (3). But I did not known how to conclude with (1). Thanks again.2012-01-08