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I am working through Charles Pinter's A Book on Abstract Algebra.

One of the problems is to prove the equivalence relation iff there is an $x \in G$ such that $a = xbx^{-1}$

To show the reflexive is easy.
If $x = a$ then $a = xax^{-1} = aaa^{-1} =a$
Also if $x = e$ then $a = xax^{-1} = eae^{-1}=a$

The symmetric proof is where I am confused:
Suppose $a = xbx^{-1}$
$x^{-1}a = x^{-1}xbx^{-1}$
$x^{-1}ax = x^{-1}xbx^{-1}x$
$x^{-1}ax = b$ But this is obviously not symmetric.
What am I missing here or conflating?

Transitivity Seems easy. Suppose $a,b,c, x, y \in G$
and $a= xbx^{-1}$ and $b= ycy^{-1}$.
Then we can have $a = xycy^{-1}x^{-1}$ ...For substituting for b
$a = (xy)c(xy)^{-1}$... Because (y^{-1}x^{-1} = (xy)^{-1}
Since $xy,(xy)^{-1} \in G$ We met transitivity check.

Thus $a$ $b$ equivalence relation proved.

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    $\begingroup$ If $x \in G$ then $x^{-1} \in G.$ When you show $b = x^{-1}a x$ then $b$ is indeed related to $a$ $\endgroup$
    – user317176
    Oct 22, 2021 at 0:26
  • $\begingroup$ Oh so we are saying any x. That makes total sense then. Is that correct? $a = xbx^{1}$ then $b = yby^{1}$ is what we are saying? Where in this case $y=x^{-1}$? $\endgroup$ Oct 22, 2021 at 0:36
  • $\begingroup$ $a$ is related to $b$ if there is a member of the group (any $x\in G$) such that $a = xbx^{-1}.$ You could also say that if we take every member of $G$ and calculate $gag^{-1},$ it will generate a subset of $G$ (called the conjugacy class) and all of the members of this subset are related to one another. $\endgroup$
    – user317176
    Oct 22, 2021 at 0:43
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    $\begingroup$ You appear to be misunderstanding what "$b= xax^{-1}$ MEANS. It means that, for this specific pair, a, b, there EXIST some x so that is true. It does NOT mean that the same "x" must work for every a and b. If $b= xax^{-1}$ then $x^{-1}bx= a$. We are just using "$x^{-1}$" as "x" now. $\endgroup$
    – user247327
    Oct 22, 2021 at 12:09

1 Answer 1

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\begin{alignat}{1} &a\sim b \iff \\ &\exists x\in G\mid a=xbx^{-1} \iff \\ &\exists x\in G\mid b=x^{-1}ax \iff \\ &\exists x'(:=x^{-1})\in G\mid b=x'ax'^{-1} &\iff\\ &b\sim a \end{alignat}

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  • $\begingroup$ In English can you say what the second to last line is saying? $\endgroup$ Oct 22, 2021 at 16:14
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    $\begingroup$ If $a$ is equivalent to $b$, then $a=xbx^{-1}$ for some $x\in G$. But then $b=x'ax'^{-1}$ for some $x'\in G$ (take $x'=x^{-1}$). Therefore $b$ is equivalent to $a$. $\endgroup$
    – user943729
    Oct 22, 2021 at 16:26

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