Self-Inverse Elements that Commute

Theorem

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

Let $x, y \in \left({G, \circ}\right)$, such that $x$ and $y$ are self-inverse.

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

Proof

Let the identity element of $\left({G, \circ}\right)$ be $e_G$.

• Let $x$ and $y$ commute.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \circ y}\right) \circ \left({x \circ y}\right)\)	\(=\)	\(\displaystyle x \circ \left({y \circ x}\right) \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Associativity
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x \circ \left({x \circ y}\right) \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$x$ and $y$ commute
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x \circ x}\right) \circ \left({y \circ y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Associativity
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e_G \circ e_G\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$x$ and $y$ are self-inverse
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e_G\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Behaviour of Identity

Thus $\left({x \circ y}\right) \circ \left({x \circ y}\right)$, proving that $x \circ y$ is self-inverse.

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

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

Thus $\left({x \circ x}\right) \circ \left({y \circ y}\right) = e_G \circ e_G = e_G$.

Because $x \circ y$ is self-inverse, we have $\left({x \circ y}\right) \circ \left({x \circ y}\right) = e_G$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \circ y}\right) \circ \left({x \circ y}\right)\)	\(=\)	\(\displaystyle \left({x \circ x}\right) \circ \left({y \circ y}\right) = e_G\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \circ \left({y \circ x}\right) \circ y\)	\(=\)	\(\displaystyle x \circ \left({x \circ y}\right) \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Associativity
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y \circ x\)	\(=\)	\(\displaystyle x \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cancellation Laws

So $x$ and $y$ commute.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): Exercise $4.14$