# Self-Inverse Elements Commute iff Product is Self-Inverse

## Theorem

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

Let $x, y \in \struct {G, \circ}$, such that $x$ and $y$ are self-inverse.

Then $x$ and $y$ commute if and only if $x \circ y$ is also self-inverse.

## Proof

Let the identity element of $\struct {G, \circ}$ be $e_G$.

### Necessary Condition

Let $x$ and $y$ commute.

Then:

 $\ds \paren {x \circ y} \circ \paren {x \circ y}$ $=$ $\ds x \circ \paren {y \circ x} \circ y$ $\circ$ is associative $\ds$ $=$ $\ds x \circ \paren {x \circ y} \circ y$ $x$ and $y$ commute $\ds$ $=$ $\ds \paren {x \circ x} \circ \paren {y \circ y}$ $\circ$ is associative $\ds$ $=$ $\ds e_G \circ e_G$ $x$ and $y$ are self-inverse $\ds$ $=$ $\ds e_G$ Definition of Identity Element

Thus $\paren {x \circ y} \circ \paren {x \circ y} = e_G$, proving that $x \circ y$ is self-inverse.

$\Box$

### Sufficient Condition

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

We already have that $x$ and $y$ are self-inverse.

Thus:

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

Because $x \circ y$ is self-inverse, we have:

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

Thus:

 $\ds \paren {x \circ y} \circ \paren {x \circ y}$ $=$ $\ds \paren {x \circ x} \circ \paren {y \circ y} = e_G$ $\ds \leadsto \ \$ $\ds x \circ \paren {y \circ x} \circ y$ $=$ $\ds x \circ \paren {x \circ y} \circ y$ $\circ$ is associative $\ds \leadsto \ \$ $\ds y \circ x$ $=$ $\ds x \circ y$ Cancellation Laws

So $x$ and $y$ commute.

$\blacksquare$