Group is Abelian iff Opposite Group is Itself

Theorem
Let $({G, \circ})$ be a group.

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

$({G, \circ})$ is an Abelian group if and only if $({G, \circ})=({G, *})$.

Proof
From the Definition of Opposite Group:

Suppose $(G, \circ)$ is Abelien.

By Equality of Algebraic Structures, $({G, \circ}) = ({G, *})$.

Suppose $({G, \circ}) = ({G, *})$.

Combining (1) and (2) gives us:


 * $\displaystyle \forall a,b \in G: a * b = b * a $

$({G, * })$ is Abelian, and $({G, \circ}) = ({G, *})$, therefore $({G, \circ})$ is Abelian