# All Elements Self-Inverse then Abelian

## Theorem

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

Suppose that every element of $G$ is self-inverse.

Then $G$ is abelian.

## Proof

Every element of $G$ is self-inverse, that is:

$\forall x \in G: x \circ x = e$

In particular, for all $x, y \in G$:

$\paren {x \circ y} \circ \paren {x \circ y} = e$

that is, $x \circ y$ is also self-inverse.

From Self-Inverse Elements Commute iff Product is Self-Inverse, it follows that:

$\forall x, y \in G: y \circ x = x \circ y$

that is: $x$ and $y$ commute for all $x, y \in G$.

Hence $G$ is an abelian group.

$\blacksquare$