# Symmetric property of $x \in G$ such that $a = xbx^{-1}$

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.

• If $x \in G$ then $x^{-1} \in G.$ When you show $b = x^{-1}a x$ then $b$ is indeed related to $a$ Oct 22, 2021 at 0:26
• 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 $y=x^{-1}$? Oct 22, 2021 at 0:36
• $a$ 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. Oct 22, 2021 at 0:43
• 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. Oct 22, 2021 at 12:09

\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}
• 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 $a$.