How do I prove symmetry of a relation given a function?

Question

Let G be a group. For all $g\in G$ , define the function f: G → G that sends x to $gxg^{-1}$. Define the relation ~ on G by a~b if $a = f(b)$ for some $g\in G$. Prove that ~ is an equivalence relation.

How do I do this? Even for Symmetry only. Thanks!

Answer 1

~ is an equivalence relation if:

a~a

a~b $\iff$ b~a

a~b and b~c $\implies$ a~c

Answer 2

You are given that $a \mathrel{R} b$ if and only if there is $g \in G$ such that $a = g b g^{-1}$.

Reflexivity: $a \mathrel{R} a$, take $g = 1$

Transitivity: If $a \mathrel{R} b$ and $b \mathrel{R} c$, then there are $g_1, g_2 \in G$ such that $a = g_1 b g_1^{-1}$ and $b = g_2 c g_2^{-1}$, and thus $a = (g_1 g_2) c (g_2^{-1} g_1^{-1}) = (g_1 g_2) c (g_1 g_2)^{-1}$, and $g_1 g_2 \in G$, thus $a \mathrel{R} c$

Symmetry: If $a \mathrel{R} b$ then $a = g b g^{-1}$, but then (by solving for $b$) it is also $b = g^{-1} a g = (g^{-1}) a (g^{-1})^{-1}$, and $g^{-1} \in G$, thus $b \mathrel{R} a$