Question

prove that the relation of conjugacy in a group is an equivalence relation

Answer 1

before proving this statement ,we have to understand both given terms ‘ conjugacy ’ and ‘ equivalence relation ’ .

conjugacy :- Two elements a and b are said to be conjugate when $b=x^{-1}ax$.
Where x is arbitrary element.

now, equivalence relation :- two elements are said to be in equivalence relation when both follow $\textbf{reflexivity}$, $\textbf{symmetry}$ and $\textbf{transitivity}$.

let's start !
1 . does it follow reflexivity ? of course yes , every element is conjugate to itself. if a be the element then it possible to find other element x such that $a=x^{-1}ax$.

2. does it follow symmetry ?.
can we write a and b such that $a=y^{-1}by$?, of course yes, because there is some elements x where $b=x^{-1}ax$ then, there must be possible to get some elements y such that $a=y^{-1}by$

3. does it follow transitivity?.
transitive means, if a is conjugate of b and b is conjugate of c for transitive a will conjugate of c. yes of course, we can easily find some elements z such that $b=x^{-1}ax$ <=> $c=y^{-1}by$ <=> $a=z^{-1}cz$

hence, here it is clear that the relation in conjugacy in a group is an equivalnce relation.