[proofplan] We verify the three defining properties of an [equivalence relation](/page/Equivalence%20Relation). Reflexivity follows by conjugating an element by the identity element. Symmetry follows by inverting the conjugating element. Transitivity follows by multiplying the two conjugating elements in the correct order. [/proofplan] [step:Conjugate by the identity element to prove reflexivity] Let $e \in G$ denote the identity element of the group $G$. For every $g \in G$, the identity laws give \begin{align*} ege^{-1} = ege = g. \end{align*} Thus, taking $h := e$, there exists $h \in G$ such that $g = hgh^{-1}$. Therefore $g \sim g$ for every $g \in G$, so $\sim$ is reflexive. [/step] [step:Invert the conjugating element to prove symmetry] Let $g,k \in G$ and suppose $g \sim k$. By definition of $\sim$, there exists an element $h \in G$ such that \begin{align*} k = hgh^{-1}. \end{align*} Since $G$ is a group, $h^{-1} \in G$. Multiplying this equality on the left by $h^{-1}$ and on the right by $h$ gives \begin{align*} h^{-1}kh = h^{-1}(hgh^{-1})h = (h^{-1}h)g(h^{-1}h) = ege = g. \end{align*} Since $(h^{-1})^{-1} = h$, this can be written as \begin{align*} g = h^{-1}k(h^{-1})^{-1}. \end{align*} Thus there exists an element $h^{-1} \in G$ conjugating $k$ to $g$, so $k \sim g$. Hence $\sim$ is symmetric. [guided] We must prove symmetry: assuming $g \sim k$, we need to prove $k \sim g$. The assumption $g \sim k$ means that $k$ is obtained from $g$ by conjugation. More precisely, there exists an element $h \in G$ such that \begin{align*} k = hgh^{-1}. \end{align*} To reverse this operation, use the inverse element $h^{-1} \in G$. We solve the displayed equality for $g$ by multiplying on the left by $h^{-1}$ and on the right by $h$. Associativity of the group operation and the inverse laws give \begin{align*} h^{-1}kh &= h^{-1}(hgh^{-1})h \\ &= (h^{-1}h)g(h^{-1}h) \\ &= ege \\ &= g. \end{align*} Here $e \in G$ denotes the identity element. Now rewrite this equality in the exact form required by the definition of conjugacy. Since $(h^{-1})^{-1} = h$, we have \begin{align*} g = h^{-1}k(h^{-1})^{-1}. \end{align*} Thus the element $h^{-1} \in G$ conjugates $k$ to $g$. Therefore $k \sim g$, which proves symmetry. [/guided] [/step] [step:Compose the conjugating elements to prove transitivity] Let $g,k,\ell \in G$ and suppose $g \sim k$ and $k \sim \ell$. By definition of $\sim$, there exist elements $a,b \in G$ such that \begin{align*} k = aga^{-1} \end{align*} and \begin{align*} \ell = bkb^{-1}. \end{align*} Substituting the first equality into the second gives \begin{align*} \ell = b(aga^{-1})b^{-1}. \end{align*} By associativity and the inverse rule $(ba)^{-1} = a^{-1}b^{-1}$, this becomes \begin{align*} \ell = (ba)g(a^{-1}b^{-1}) = (ba)g(ba)^{-1}. \end{align*} Since $ba \in G$, there exists an element of $G$ conjugating $g$ to $\ell$. Hence $g \sim \ell$, so $\sim$ is transitive. [/step] [step:Combine the three properties to conclude equivalence] We have shown that $\sim$ is reflexive, symmetric, and transitive on $G$. Therefore $\sim$ is an equivalence relation on $G$. [/step]