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

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.

• You're assuming the conclusion.
– user403337
Feb 24, 2020 at 0:48
• @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 $uv\sim xy$. Feb 24, 2020 at 0:53
• You don't want to show that $uv\sim uv$ for reflexivity, but that $u\sim u$. Similarly, for symmetry and transitivity. Feb 24, 2020 at 1:09

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