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Let $G$ be a group with relation ~ defined as:

If $a,b \in G\,, \,\text{write}\,\, a$~$b$ to mean that $ \exists\, g \in G $ such that

$$ga = bg$$

(a) Prove that ~ is an equivalence relation. (b) Let $x \in G$. Prove that if $[x] = \text{{$x$}}$ then $x$ commutes with every element of $G$. That is, show that for any $y\in G$, we have $xy = yx$.

I've attempted part (a) and managed to prove Reflexivity by arguing that if one chooses $g$ to be $e$ (the identity element) then $a$~$a$. However, I have no idea how to tackle Symmetricity, Transitivity or part (b).

Any guidance would be appreciated.

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    for symmetry use g inverse – 2017-02-18
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    $a\sim b$ means $b=gag^{-1}$, and $b\sim c$ means $c=hbh^{-1}$, right? So $c=h(gag^{-1})h^{-1}$, and you can take it from there. – 2017-02-18
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    @saravanan Of course... Thank you! – 2017-02-18
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    @saravanan I did as you suggested so I get $a = g^{-1}bg \rightarrow ag^{-1} = g^{-1}b$ which I assume is the intended result...? – 2017-02-18

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Suppose that the equivalence class of $x$ is {$x$}

Choose some $g\in G$ and denote $$y:=g^{-1}xg$$

This implies $x$~$y$ , but the equivalence class of $x$ only contains $x$, which implies $y=x$ and therefore $x=g^{-1}xg$ and therefore $gx=xg$

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    I'm having difficulty understanding how this proves that x is commutative with all $y \in G$ since it just implies that if $x \sim y \rightarrow y = x$. – 2017-02-18
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    This works for EVERY $g\in G.$ It would be better wording to say "For each $g\in G$ let $y_g=g^{-1}xg.$ Then $x\sim y_g$ ( etc.),.... so $x$ commutes with each $g$." – 2017-02-18
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    @user254665 The formulation "choose some" indicates that $g$ is arbitary, – 2017-02-18
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    @Peter. Yes, But the comment from the OP indicated that he/she didn't follow it so I attempted to clarify it. – 2017-02-19
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Partial answer :

Symmetry : Suppose $$ga=bg$$ this implies $$gag^{-1}=b$$ (multiply with $g^{-1}$ from right)

and this implies $$ag^{-1}=g^{-1}b$$ (multiply with $g^{-1}$ from left)

This shows symmetry.

For the transitivity, assume $ga=bg$ and $hb=ch$. Then, we have $$c=hbh^{-1}$$ and $$b=gag^{-1}$$

So, we have $$c=hgag^{-1}h^{-1}$$ implying $$g^{-1}h^{-1}c=ag^{-1}h^{-1}$$

showing transitivity.

(This time find out which multiplications are necessary)