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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 classes – WLOG Jan 7 '12 at 20:52
Do you know the relationship between congruences and homomorphisms (or quotients)? – Brian M. Scott Jan 7 '12 at 20:58
What does it mean to raise an element of a group to another element of the group? Typo? – user18063 Jan 7 '12 at 21:07
@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). – Arturo Magidin Jan 7 '12 at 22:37
@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 ${}^y({}^xa) = {}^{yx}a$. – Arturo Magidin Jan 7 '12 at 22:46
up vote 3 down vote accepted

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? – user14174 Jan 8 '12 at 0:12
Hi @Lmn6,I'm saying that if $\sim$ is a congruence relation, then $a\sim b\iff a=b$. Now, look at (1) – kneidell Jan 8 '12 at 7:11
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. – user14174 Jan 8 '12 at 10:35

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