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 1

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.

$\blacksquare$

Proof 2

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

$\ds a * e$

$=$

$\ds e \circ a = a$

$\ds e * a$

$=$

$\ds a \circ e = a$

So: $e \in \struct {G, *}$

Hence:

$\struct {G, *}$ is non-empty.

$\Box$

As $\struct {G, \circ}$ is a group:

$\struct {G, \circ}$ is closed

every element of $\struct {G, \circ}$ has an inverse.

Therefore:

$\quad \forall a, b \in \struct {G, \circ}: b^{-1} \circ a \in \struct {G, \circ}$

Hence, by definition of $*$:

$\quad \forall a, b \in \struct {G, *}: a * b^{-1} \in \struct {G, *}$

$\Box$

The result follows from the One-Step Subgroup Test and Group is Subgroup of Itself.

$\blacksquare$

Sources

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