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 if and only if:

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

Proof

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.

$\Box$

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.

$\blacksquare$

Sources

• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): commutator