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 $G$.


$\cdot$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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    $\begingroup$ Note that $[a] = \{ g a g^{-1} : g \in G\}$ for all $a \in G$. $\endgroup$ – Andreas Caranti Feb 19 '17 at 16:52
  • $\begingroup$ How did you approach this question? $\endgroup$ – Error 404 Feb 19 '17 at 16:53
  • $\begingroup$ [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. $\endgroup$ – fleablood Feb 19 '17 at 16:55

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 $.


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 $

  • $\begingroup$ What does $\cong$ mean? $\endgroup$ – David Feb 19 '17 at 19:10
  • $\begingroup$ It means I don't know how to make the equivalent symbol in latex. $\endgroup$ – fleablood Feb 19 '17 at 20:06
  • $\begingroup$ Oh, apparently it's $\sim $.... $\endgroup$ – fleablood Feb 19 '17 at 20:11
  • $\begingroup$ Yes I have figured it out later, thank you $\endgroup$ – David Feb 20 '17 at 3:04

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