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Let $(G,\cdot)$ be a group. We define a relation on $G$ as follows: if $a,b\in G$ we write $a\sim b$ to mean that there exists $g\in G$ such that $ga=bg$. Let $x\in G$. Prove that if $[x]=\{x\}$ then $x$ commutes with every element of $G$.

Notes:

$\cdot$The relation is an equivalence relation, fulfilling the following conditions:

1)$\forall x, x\sim x$

2)$\forall x\forall y$,$x\sim y \Rightarrow y \sim x$

3)$\forall x \forall y \forall z$, $(x \sim y )\wedge (y \sim z)\Rightarrow x \sim z$

$\cdot$ $[x]$ denotes the equivalence class of $x$, which is the set of all elements $y$ in the domain for which $x \sim y$

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    Note that $[a] = \{ g a g^{-1} : g \in G\}$ for all $a \in G$. – 2017-02-19
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    How did you approach this question? – 2017-02-19
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    [x]={x} means that if xg=gy then y=x. for all g, let xg=q=gg'q so g'q=x. So gx=gg'q=q=xg. – 2017-02-19

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Let $[x]=\{x\} $

Let $g \in G$. Let $xg=q=gg^{-1}q $. Then $x \sim g^{-1}q$. So $x=g^{-1}q $.

So $gx=gg^{-1}q=q=xg$.

So $x$ commutes with $g $.

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Another way of putting it:

For any $x $ and any $g $, $xg=gg^{-1}xg $ so $x \sim g^{-1}xg $ for all $x $ and $g $.

So if $[x]=\{x\}$ then $x = (g^{-1}xg)$ for all $g$. So $gx=g (g^{-1}xg)=xg $ for all $g$.

$\sim $

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    What does $\cong$ mean? – 2017-02-19
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    It means I don't know how to make the equivalent symbol in latex. – 2017-02-19
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    Oh, apparently it's $\sim $.... – 2017-02-19
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    Yes I have figured it out later, thank you – 2017-02-20