Prove that the following relation $\rho$ is an equivalence relation on a group $G$

Question

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:

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

Answer 1

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.

Answer 2

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.