Every group is naturally isomorphic to its opposite group via the inverse map

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Statement

Let $G$ be a group. Then, consider the opposite group of $G$, which is a group with the same underlying set, and such that the binary operation is defined by:

$x * y := yx$

In other words, products are taken with order reversed. Then, $G$ is isomorphic to the opposite group via the map $g \mapsto g^{-1}$.

This isomorphism is natural in the sense that it gives a natural isomorphism between the identity functor and the functor sending each group to its opposite group.

Facts used

1. Inverse map is involutive: This states that $(xy)^{-1} = y^{-1}x^{-1}$ for all $x,y$ in a group, and $(x^{-1})^{-1} = x$ for all $x$ in a group.