Commutators are Identity iff Group is Abelian

Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.

For $g, h \in G$, let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.

Then $\struct {G, \circ}$ is abelian :
 * $\forall g, h \in G: \sqbrk {g, h} = e$

Necessary Condition
Let $\struct {G, \circ}$ be such that:


 * $\forall g, h \in G: \sqbrk {g, h} = e$

From Commutator is Identity iff Elements Commute:


 * $\forall g, h \in G: g \circ h = h \circ g$

Hence $\struct {G, \circ}$ is abelian by definition.

Sufficient Condition
Let $\struct {G, \circ}$ be an abelian group.

Then by definition:
 * $\forall g, h \in G: g \circ h = h \circ g$

From Commutator is Identity iff Elements Commute:
 * $\forall g, h \in G: \sqbrk {g, h} = e$

Hence the result.