[proofplan] We prove both implications by expanding the definition of the commutator. If $G$ is abelian, then the middle factors in $ghg^{-1}h^{-1}$ can be rearranged so that each element cancels with its inverse. Conversely, if every commutator is the identity, then the equation $ghg^{-1}h^{-1}=e$ can be multiplied on the right by $h$ and then by $g$ to obtain $gh=hg$ for arbitrary $g,h \in G$. [/proofplan] [step:Use commutativity to show every commutator is the identity] Assume that $G$ is abelian. Let $g,h \in G$ be arbitrary. Since $G$ is abelian, $hg^{-1}=g^{-1}h$. Therefore \begin{align*} [g,h]=ghg^{-1}h^{-1}=gg^{-1}hh^{-1}=ee=e. \end{align*} Since $g,h \in G$ were arbitrary, $[g,h]=e$ for every $g,h \in G$. [/step] [step:Use vanishing commutators to force every pair to commute] Assume that $[g,h]=e$ for every $g,h \in G$. Let $g,h \in G$ be arbitrary. By the definition of the commutator, \begin{align*} ghg^{-1}h^{-1}=e. \end{align*} Multiplying this equality on the right by $h$ and using associativity gives \begin{align*} ghg^{-1}=h. \end{align*} Multiplying this equality on the right by $g$ and using associativity again gives \begin{align*} gh=hg. \end{align*} Since $g,h \in G$ were arbitrary, every pair of elements of $G$ commutes. Hence $G$ is abelian. [guided] Assume that $[g,h]=e$ for every $g,h \in G$. To prove that $G$ is abelian, we must prove that any two elements commute. Let $g,h \in G$ be arbitrary. The hypothesis applies to this particular pair $g,h$, so expanding the definition of the commutator gives \begin{align*} ghg^{-1}h^{-1}=e. \end{align*} We want to extract the relation $gh=hg$ from this equation. First multiply both sides on the right by $h$. By associativity of the group operation and by the inverse identity $h^{-1}h=e$, the left-hand side becomes \begin{align*} (ghg^{-1}h^{-1})h=ghg^{-1}(h^{-1}h)=ghg^{-1}e=ghg^{-1}. \end{align*} The right-hand side becomes $eh=h$. Hence \begin{align*} ghg^{-1}=h. \end{align*} Now multiply both sides on the right by $g$. Again using associativity and $g^{-1}g=e$, we obtain \begin{align*} (ghg^{-1})g=gh(g^{-1}g)=ghe=gh. \end{align*} The right-hand side becomes $hg$. Therefore \begin{align*} gh=hg. \end{align*} Because the elements $g,h \in G$ were arbitrary, this proves that every pair of elements of $G$ commutes. By the definition of an [abelian group](/page/Abelian%20Group), $G$ is abelian. [/guided] [/step] [step:Combine the two implications] The first step proves that if $G$ is abelian, then $[g,h]=e$ for every $g,h \in G$. The second step proves that if $[g,h]=e$ for every $g,h \in G$, then $G$ is abelian. Therefore $G$ is abelian if and only if every commutator in $G$ is equal to the identity element $e$. [/step]