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Let $(G,*)$ be a group. For $a, b \in G$, define $a R b$ if and only if there is an element $x \in G$ with $a = xbx^{-1}$. Prove that $R$ is an equivalence relation.

It seems I am really bad at abstract algebra please help. It seems obvious to me that $x^{-1}a(x^{-1})^{-1} = b$ so its obviously symmetric, can I use the same fact to show $b = ycy^{-1}$ and thus $x^{-1}ax = ycy^{-1}$ to show transitivity somehow?

Reflexive seems ok oddly $a= eae^{-1}$ where $e$ is then identity element? Am I on the right track?

I hate to ask so many questions here but after I get through something on this site I feel like I actually understand what I am doing as opposed to, ok I have a solution that means nothing to me...

Thanks for helping me study for midterms :)

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    $\begingroup$ Mind that $x^{-1}ax=b$ does not show it is symmetric (in fact the order in which the inverse appears does matter!). But a little tweak makes your idea work: $x^{-1}ax=x^{-1}a(x^{-1})^{-1}=b$. $\endgroup$
    – user39280
    Feb 3, 2013 at 19:53

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You're right about reflexive: $a\sim a$ since $a=eae^{-1}$.

For symmetric, we want to show $a\sim b$ implies $b\sim a$. This is clear though, since $a\sim b$ implies $a=xbx^{-1}$. Since $G$ is a group, we have $x^{-1}ax=b$. Now let $y=x^{-1}$, and we have $b=yay^{-1}$, i.e., $b\sim a$ as desired.

For transitivity, a similar idea applies: Suppose $a\sim b$ and $b\sim c$. Then $a=xbx^{-1}$ and $b=ycy^{-1}$. Thus, $a=x(ycy^{-1})x^{-1}=(xy)c(xy)^{-1}$, so $a\sim c$.

This proves that $\sim$ is an equivalence relation.

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  • $\begingroup$ Wow that makes perfect sense! i am physics student so alot of the time i don't have a problem with the ideas its just how to implement the proofs wherein the difficulty lies! Thank both of you so much! $\endgroup$
    – Faust
    Feb 3, 2013 at 20:07
  • $\begingroup$ One stupid question...is the reason we can "let $y=x^{-1}$" simply because we know $(G,*)$ is a group and by definition there must be an inverse for every element? $\endgroup$ Apr 16, 2017 at 15:50

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