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Let $G$ be a group.

  1. Prove that the relation $a\sim b$ if $b=gag^{-1}$ for some $g\in G$, is an equivalence relation on $G$.
  2. Prove that $\forall u,v\in G$, $uv\sim vu$.

So I've proved (1). My confusion lies in the fact that they appear to be the same question. I'm sure I must be wrong, but my approach was to again show that $\sim$ is an equivalence relation. My proof is as follows:

Proof.

  1. Suppose $u,v\in G$. Then $e(uv)e^{-1}=uv$. Therefore $uv\sim uv$ and $\sim$ is reflexive.
  2. Suppose $uv\sim vu$ and that $u,v\in G$. Then $vu=g(uv)g^{-1}$ and \begin{align} g^{-1}(vu)g&=g^{-1}(g(uv)g^{-1})g\\\ &=(g^{-1}g)uv(g^{-1}g)\\\ &=uv \end{align} Therefore, $uv\sim vu$ and $\sim$ is symmetric.
  3. Suppose $uv\sim vu$ and $vu\sim xy$. Then, there exists $g,h\in G$ such that $vu=g(uv)g^{-1}$ and $xy=h(vu)h^{-1}$. Then, \begin{align} xy&=h(vu)h^{-1}\\\ &=h(g(uv)g^{-1}\\\ &=(hg)uv(hg)^{-1}\\\ &=uv \end{align} Therefore $uv\sim xy$ and $\sim$ is transitive.

Thus, thus $uv\sim vu$ for all $u,v\in G$.

And this proof is almost the same as the proof I did for (1), so naturally I'm second guessing my answer for (2). Any help would be greatly appreciated.

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    $\begingroup$ You're assuming the conclusion. $\endgroup$
    – user403337
    Feb 24, 2020 at 0:48
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    $\begingroup$ @ChrisCuster I thought so when I was writing it, and maybe I need to review my relations proofs, but I thought for reflexive we prove that $uv\sim uv$, symmetry we prove that if $uv\sim vu$ then $vu\sim uv$ and then transitivity that if $uv\sim vu$ and $vu\sim xy$ then $uv\sim xy$. $\endgroup$
    – Van-Sama
    Feb 24, 2020 at 0:53
  • $\begingroup$ You don't want to show that $uv\sim uv$ for reflexivity, but that $u\sim u$. Similarly, for symmetry and transitivity. $\endgroup$ Feb 24, 2020 at 1:09

2 Answers 2

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For the second part we need to show $uv\sim vu$ for any $u,v\in G$. So we need to find $ g \in G$ such that $ vu = g (uv)g^{-1}$ ... $g=u^{-1}$ will do.

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    $\begingroup$ Oh.. I see it. I suppose I struggle with knowing what I'm supposed to be doing. This was much simpler than I thought. Thank you. $\endgroup$
    – Van-Sama
    Feb 24, 2020 at 1:44
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You simply need a $g\in G$ such that $guvg^{-1}=vu$. But $g=v$ works: $vuvv^{-1}=vu$.

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