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Let G be a group. For $x,y\in G$, define $x\sim y$ if there exists some element $a\in G$ such that $y=axa^{-1}$. Show that ~ defines an equivalence relation on $G$.

rschwieb
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Michelley
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  • aaaand what have you tried :')? – M.B. Jan 31 '14 at 07:52
  • Reflexive: Suppose a∈G where y=axa^-1. Now if a=1, then y=x. Hence x~x. I am pretty sure there is something wrong here. – Michelley Jan 31 '14 at 07:54
  • There you go. Now try to prove that the relation is reflexive. If $y=axa^{-1}$, can you show that for some $b$, you have the equation $x=byb^{-1}$? – 5xum Jan 31 '14 at 07:56
  • Do you mean symmetric? Suppose b∈G where x=bab^-1. Is that what you mean? – Michelley Jan 31 '14 at 07:59
  • I mean $x=byb^{-1}$. And no, you cannot suppose you have such a $b$. If you suppose $x\equiv y$, you must PROVE that $y\equiv x$. This is done by assuming that you have such an $a\in G$ that $y=axa^{-1}$ and then finding such a $b\in G$ that $x=byb^{-1}$. – 5xum Jan 31 '14 at 08:21

1 Answers1

1

$x \sim x$

$a:=e$ satisfies the condition.

$x \sim y \implies y \sim x$

Suppose $x \sim y$. Then there exists an $a$ such that $y=axa^{-1}$. Rearranging the equation: $$a^{-1}ya=x$$ Recall that for an arbitrary group and an element $x$ in the group $(x^{-1})^{-1}=x$. Thus $a^{-1}$ satisfies the condition.

$x \sim y \wedge y \sim z \implies x \sim z$

Suppose $x \sim y$ and $y \sim z$. Then there exist $a,b$ such that: $$x=aya^{-1} \space \space \space (1)$$ $$y=bzb^{-1}$$ From equation (1) we have: $$y=a^{-1}xa$$ $$a^{-1}xa=bzb^{-1}$$ $$b^{-1}a^{-1}xab=z$$ Recall that in an arbitrary group all elements $x,y$ satisfy $(xy)^{-1}=y^{-1}x^{-1}$

So $ab$ indeed satisfies the condition. This concludes the proof.

gndz
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