0
$\begingroup$

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 $x \sim y$

$\endgroup$
  • 3
    $\begingroup$ 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
2
$\begingroup$

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 $

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.