Product of Commuting Elements in Monoid is Unit iff Each Element is Unit

Theorem
Let $A$ be a monoid.

Let $\map G A$ be the group of units of $A$.

Let $n \ge 2$ be an integer.

Let $x_1, \ldots, x_n$ be commuting elements in $A$.

Let:
 * $\ds x = \prod_{i \mathop = 1}^n x_i$

Then:
 * $x \in \map G A$ $x_i \in \map G A$ for each $1 \le i \le n$.

Necessary Condition
We proceed by induction on $n$.

For all $n \ge 2$, let $\map P n$ be the proposition:
 * for every set of commuting elements $x_1, \ldots, x_n$ in $A$
 * if $\ds \prod_{i \mathop = 1}^n x_i \in \map G A$, then $x_i \in \map G A$ for each $1 \le i \le n$.

Basis for the Induction
Let $x, y \in A$ be commuting elements in $A$ such that $x y \in \map G A$.

Let $z \in A$ be such that $\paren {x y} z = z \paren {x y} = e$.

We then have:

So we have:
 * $\paren {y z} x = x \paren {y z} = e$

So we have $x \in \map G A$ with $x^{-1} = y z$.

Then we have $y = x^{-1} z^{-1}$.

From Inverse of Product: Monoid: General Result, it follows that $y \in \map G A$ as well.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * for every set of commuting elements $x_1, \ldots, x_k$ in $A$
 * if $\ds \prod_{i \mathop = 1}^k x_i \in \map G A$, then $x_i \in \map G A$ for each $1 \le i \le k$.

Then we need to show:
 * for every set of commuting elements $x_1, \ldots, x_{k + 1}$ in $A$
 * if $\ds \prod_{i \mathop = 1}^{k + 1} x_i \in \map G A$, then $x_i \in \map G A$ for each $1 \le i \le k + 1$.

Induction Step
This is our induction step.

Let $x_1, \ldots, x_{k + 1}$ be commuting elements of $A$ such that:
 * $\ds \prod_{i \mathop = 1}^{k + 1} x_i \in \map G A$

We can write:
 * $\ds \prod_{i \mathop = 1}^{k + 1} x_i = \paren {\prod_{i \mathop = 1}^k x_i} x_{k + 1}$

Using the basis for the induction, we have that:
 * $\ds \prod_{i \mathop = 1}^k x_i \in \map G A$ and $x_{k + 1} \in \map G A$.

Using the induction hypothesis, we then have:
 * $x_1, \ldots, x_k \in \map G A$ and $x_{k + 1} \in \map G A$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Sufficient Condition
This is precisely Inverse of Product: Monoid: General Result.