Equivalence of Definitions of Commutator of Group Elements

Theorem

The following definitions of the concept of Commutator of Group Elements are equivalent:

Definition $1$

The commutator of $g$ and $h$ is the element of $G$ defined and denoted:

$\sqbrk {g, h} := g^{-1} \circ h^{-1} \circ g \circ h$

Definition $2$

The commutator of $g$ and $h$ is the element $c$ of $G$ with the property:

$h \circ g \circ c := g \circ h$

Proof

Throughout the following, Group Axiom $\text G 1$: Associativity is assumed implicitly.

Let $g, h \in G$ be arbitrary elements of $G$.

Let $c$ be the commutator of $g$ and $h$ by definition $2$.

Then we have:

\(\ds h \circ g \circ c\)

\(=\)

\(\ds g \circ h\)

Definition 2 of Commutator of Group Elements

\(\ds \leadstoandfrom \ \ \)

\(\ds h^{-1} \circ h \circ g \circ c\)

\(=\)

\(\ds h^{-1} \circ g \circ h\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \leadstoandfrom \ \ \)

\(\ds g \circ c\)

\(=\)

\(\ds h^{-1} \circ g \circ h\)

Definition of Inverse Element

\(\ds \leadstoandfrom \ \ \)

\(\ds g^{-1} \circ g \circ c\)

\(=\)

\(\ds g^{-1} \circ h^{-1} \circ g \circ h\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \leadstoandfrom \ \ \)

\(\ds c\)

\(=\)

\(\ds g^{-1} \circ h^{-1} \circ g \circ h\)

Definition of Inverse Element

\(\ds \leadstoandfrom \ \ \)

\(\ds c\)

\(=\)

\(\ds \sqbrk {g, h}\)

Definition 1 of Commutator of Group Elements

$\blacksquare$