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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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    @Lmn6 Sorry but are you sure about the problem? I know that conjugacy relation is a congruence and conjugacy classes are congruence classes2012-01-07
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    Do you know the relationship between congruences and homomorphisms (or quotients)?2012-01-07
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    What does it mean to raise an element of a group to another element of the group? Typo?2012-01-07
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    @stef: while conjugacy is an equivalence relation, I believe by congruence, they mean that equivalence classes are cosets of a normal subgroup.2012-01-07
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    @lhf: great link! Thanks2012-01-07
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    @user18063: Not a typo, but a standard notation for conjugation: $a^x=x^{-1}ax$. (Some define $a^x$ as $xax^{-1}$ instead.)2012-01-07
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    @Brian: No one that I am aware of does; they *do* write ${}^xa$ for $xax^{-1}$, though (the $x$ goes on the same side as the $x$ goes in the notation).2012-01-07
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    @Arturo: I’m virtually certain that I’ve seen it a time or two, though I can’t recall where. The $^xa$ notation is new to me but makes good sense.2012-01-07
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    @Brian: The idea is that $(a^x)^y = a^{xy}$ under the usual definition; but if you define $a^x$ as $xax^{-1}$, then you get the rather prone-to-errors $(a^x)^y = a^{yx}$. With ${}^xa$, you again get ${}^y({}^xa) = {}^{yx}a$.2012-01-07
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    @lhf: 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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    Please be patient, I was not made for math. From what you say I get $a=b$ for every choice of $a$ and $b$. Where does the commutativity show up?2012-01-08
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    Hi @Lmn6,I'm saying that if $\sim$ is a congruence relation, then $a\sim b\iff a=b$. Now, look at (1)2012-01-08
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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