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Question: Let $G$ be a group. A relation $\rho$ on $G$ is defined by "$a \rho b$ if and only if $b=g\circ a\circ g^{-1}$ for some $g\in g$; $a,b\in G$". Prove that $\rho$ is an equivalence relation.

Proceed:
I can show that $\rho$ is symmetric by the following way :
Let $a\rho b$ then $b=g\circ a\circ g^{-1}$ for some $g\in G$; $a,b\in G$. Now we have $a=g\circ b\circ g^{-1}\implies \rho$ is symmetric. But I stuck to show reflexive and transitivity. How can I do this?

Updated:

$b=g\circ a\circ g^{-1}\implies a=g^{-1}\circ b\circ g\implies a=g^{-1}\circ b\circ (g^{-1})^{-1}$

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  • $\begingroup$ Check your symmetry argument...$gbg^{-1}=g^2ag^{-2}$. Hint: for reflexive, try simple value of $g$. For transitive...just write it out. $\endgroup$
    – lulu
    Commented Mar 12, 2017 at 19:02

2 Answers 2

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In a group $G$ we have $a=e\circ a\circ e^{-1}$ and so $a\rho a$ holds $\implies \rho$ is reflexive.

Let $a\rho b$. Then $b=g\circ a\circ g^{-1}\implies a=g^{-1}\circ b\circ g\implies a=g^{-1}\circ b\circ (g^{-1})^{-1}\implies \rho$ is symmetric.

Let $a\rho b$ and $b\rho c$. Then $b=g\circ a\circ g^{-1}$ and $c=g\circ b\circ g^{-1}\implies c=g^2\circ a\circ (g^{2})^{-1}\implies \rho$ is transitive and consequently $\rho$ is an equivalence relation.

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  • $\begingroup$ Not right for transitivity. The "$g$" that works for $a \rho b$ may not be the same one that works for $b \rho c$. $\endgroup$ Commented Mar 12, 2017 at 19:47
  • $\begingroup$ @EthanBolker Thanks for point out my mistake. Please tell me the transitivity in your answer. $\endgroup$
    – Primo
    Commented Mar 12, 2017 at 19:54
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    $\begingroup$ If $b = g_1\circ a \circ g_1^{-1}$ and $c = g_2\circ b \circ g_2^{-1}$, then $$\begin{aligned}c &= g_2\circ b \circ g_2^{-1} = g_2 \circ (g_1 \circ a \circ g_1^{-1}) \circ g_2^{-1}\\ &= (g_2 \circ g_1) \circ a \circ (g_1^{-1} \circ g_2^{-1}) = (g_2\circ g_1)\circ a \circ (g_2\circ g_1)^{-1}.\end{aligned}$$ $\endgroup$ Commented Mar 12, 2017 at 20:06
  • $\begingroup$ @DanielFischer Thank you so much. (+1) $\endgroup$
    – Primo
    Commented Mar 12, 2017 at 20:48
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Too long for a comment. Hope it helps.

You have to write more clearly to write a proof. The sentence

Let $a \rho b$ then $b=g∘a∘g^{-1}$ for some $g∈G; a,b∈G$

is not right. You started with $a$ and $b$, so you should not then say "for some $a,b∈G$". Then to show that the relation is symmetric you have to find some element $x$ in $G$ such that $a=x∘b∘x^{-1}$. You didn't do that - in your question you seem to think that $x=g$ will do the job.

For the other two properties, write down what you know and then what you have to prove.

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  • $\begingroup$ @EB here the relation holds for some $g\in G$, but $a,b\in G$ fixed $\endgroup$
    – Primo
    Commented Mar 12, 2017 at 19:15
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    $\begingroup$ Your update is good. Now find an $x$ that will tell you about reflexivity, and one for transitivity. Your comment isn't quite right. $a$ and $b$ aren't "fixed", they are a pair of elements for which there happens to be a $g$. $\endgroup$ Commented Mar 12, 2017 at 19:15

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