Group relations: Prove that $\forall u,v\in G$, $uv\sim vu$

Question

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.

Answer 1

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.

Answer 2

You simply need a $g\in G$ such that $guvg^{-1}=vu$. But $g=v$ works: $vuvv^{-1}=vu$.