Opposite Group is Group

Theorem

Let $\struct {G, \circ}$ be a group.

Let $\struct {G, *}$ be the opposite group to $G$.

Then $\struct {G, *}$ is a group.

Proof

Group Axiom $\text G 0$: Closure

$\struct {G, *}$ is closed:

$b \circ a \in G \implies a * b \in G$

$\Box$

Group Axiom $\text G 1$: Associativity

$*$ is associative on $G$:

\(\ds a * \paren {b * c}\)

\(=\)

\(\ds \paren {c \circ b} \circ a\)

Definition of $*$

\(\ds \)

\(=\)

\(\ds c \circ \paren {b \circ a}\)

Associativity of $\circ$

\(\ds \)

\(=\)

\(\ds \paren {a * b} * c\)

Definition of $*$

$\Box$

Group Axiom $\text G 2$: Existence of Identity Element

Let $e$ be the identity of $\struct {G, \circ}$:

\(\ds a * e\)

\(=\)

\(\ds e \circ a = a\)

\(\ds e * a\)

\(=\)

\(\ds a \circ e = a\)

Thus $e$ is the identity of $\struct {G, *}$.

$\Box$

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

Let the inverse of $a \in \struct {G, \circ}$ be $a^{-1}$:

\(\ds a * a^{-1}\)

\(=\)

\(\ds a^{-1} \circ a = e\)

\(\ds a^{-1} * a\)

\(=\)

\(\ds a \circ a^{-1} = e\)

Thus $a^{-1}$ is the inverse of $a \in \struct {G, *}$

$\Box$

So all the group axioms are satisfied, and $\struct {G, *}$ is a group.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \epsilon$