Group is Abelian iff Opposite Group is Itself

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Theorem

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

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


$\struct {G, \circ}$ is an Abelian group if and only if:

$\struct {G, \circ} = \struct {G, *}$


Proof

By definition of opposite group:

$(1): \quad \forall a, b \in G : a \circ b = b * a$


Necessary Condition

Let $\struct {G, \circ}$ be Abelian.

Then:

\(\ds \forall a, b \in G: \, \) \(\ds a \circ b\) \(=\) \(\ds b \circ a\) Definition of Abelian Group
\(\ds \) \(=\) \(\ds a * b\) Definition of Opposite Group
\(\ds \leadsto \ \ \) \(\ds \struct {G, \circ}\) \(=\) \(\ds \struct {G, *}\) Equality of Algebraic Structures

$\blacksquare$


Sufficient Condition

Let $\struct {G, \circ} = \struct {G, *}$.

Then:

\(\ds \forall a, b \in G: \, \) \(\ds a \circ b\) \(=\) \(\ds a * b\) Equality of Algebraic Structures
\(\ds \) \(=\) \(\ds b \circ a\) Definition of Opposite Group

Thus by definition $\struct {G, \circ}$ is an Abelian group.

$\blacksquare$