Properties of Inverses of Commuting Elements

Let $$\left({S, \circ}\right)$$ be a monoid whose identity is $$e_S$$.

Theorem 1
Let $$x, y \in S$$ such that $$y$$ is invertible.

Then $$x$$ commutes with $$y$$ iff $$x$$ commutes with $$y^{-1}$$.

Proof of Theorem 1

 * Let $$x$$ commute with $$y$$. Then:

$$ $$ $$ $$ $$ $$

So $$x$$ commutes with $$y^{-1}$$.


 * Now let $$x$$ commute with $$y^{-1}$$.

From the above, it follows that $$x$$ commutes with $$\left({y^{-1}}\right)^{-1}$$.

From Inverse of an Inverse, $$\left({y^{-1}}\right)^{-1} = y$$

Thus $$x$$ commutes with $$y$$.

Theorem 2
Let $$x, y \in S$$ such that $$x$$ and $$y$$ are both invertible.

Then $$x$$ commutes with $$y$$ iff $$x^{-1}$$ commutes with $$y^{-1}$$.

Proof of Theorem 2

 * Let $$x$$ commute with $$y$$. Then:

$$ $$ $$

So $$x^{-1}$$ commutes with $$y^{-1}$$.


 * Now let $$x^{-1}$$ commute with $$y^{-1}$$.

From the above, $$\left({x^{-1}}\right)^{-1}$$ commutes with $$\left({y^{-1}}\right)^{-1}$$.

From Inverse of an Inverse, $$\left({x^{-1}}\right)^{-1} = x$$ and $$\left({y^{-1}}\right)^{-1} = y$$.

Thus $$x$$ commutes with $$y$$.

Theorem 3
Let $$x, y \in S$$ such that $$x$$ and $$y$$ are both invertible.

Then $$x$$ commutes with $$y$$ iff $$\left({x \circ y}\right)^{-1} = x^{-1} \circ y^{-1}$$.

Proof of Theorem 3
$$ $$ $$

Theorem 4
Let $$x, y \in S$$ such that $$x$$ and $$y$$ are both invertible.

Then $$x \circ y \circ x^{-1} = y$$ iff $$x$$ and $$y$$ commute.

Proof of Theorem 4
As it is taken for granted that $$\circ$$ is associative, we can dispense with parentheses.

We also take for granted the fact that $$x$$ and $$y$$ are cancellable from Invertible also Cancellable.

Then:

$$ $$ $$

Theorem 5
Let $$x, y \in S$$ such that $$x$$ and $$y$$ are both invertible.

Then $$x \circ y \circ x^{-1} \circ y^{-1} = e_S$$ iff $$x$$ and $$y$$ commute.

Proof of Theorem 5
$$ $$ $$ $$