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.
$\blacksquare$