Product of Commuting Elements in Monoid is Unit iff Each Element is Unit/Proof 2

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:

$x \in \map G A$ if and only if $x_i \in \map G A$ for each $1 \le i \le n$.

Proof

If $x_1, \ldots, x_n \in \map G A$, then:

$\ds \prod_{i \mathop = 1}^k x_i \in \map G A$

by Inverse of Product: Monoid: General Result.

Conversely, suppose:

$\ds \prod_{i \mathop = 1}^k x_i \in \map G A$

That is, there is a $z \in A$ such that:

$(1):\quad \ds z \paren {\prod_{i \mathop = 1}^k x_i} = \paren {\prod_{i \mathop = 1}^k x_i} z = e$

We shall show:

$\forall i \in \set {1, \ldots, n} : x_i \in \map G A$

It suffices to show this for $i=1$, since $x_1, \ldots, x_n$ are commuting.

Define:

$\ds z_1 := \paren {\prod_{i \mathop = 2}^k x_i} z$

Then:

$x_1 z_1 = z_1 x_1 = e$

so that $x_1 \in \map G A$.

Indeed:

\(\ds x_1 z_1\)

\(=\)

\(\ds x_1 \paren {\prod_{i \mathop = 2}^k x_i} z\)

Definition of $z_1$

\(\ds \)

\(=\)

\(\ds \paren {\prod_{i \mathop = 1}^k x_i} z\)

\(\ds \)

\(=\)

\(\ds e\)

Moreover:

\(\ds z_1 x_1\)

\(=\)

\(\ds \paren {\prod_{i \mathop = 2}^k x_i} z x_1\)

Definition of $z_1$

\(\ds \)

\(=\)

\(\ds e \paren {\prod_{i \mathop = 2}^k x_i} z x_1\)

\(\ds \)

\(=\)

\(\ds z \paren {\prod_{i \mathop = 1}^k x_i} \paren {\prod_{i \mathop = 2}^k x_i} z x_1\)

by $(1)$

\(\ds \)

\(=\)

\(\ds z \paren {\prod_{i \mathop = 2}^k x_i} \paren {\prod_{i \mathop = 1}^k x_i} z x_1\)

$x_1$ commutes with $x_2, \ldots, x_n$

\(\ds \)

\(=\)

\(\ds z \paren {\prod_{i \mathop = 2}^k x_i} e x_1\)

by $(1)$

\(\ds \)

\(=\)

\(\ds z \paren {\prod_{i \mathop = 2}^k x_i} x_1\)

\(\ds \)

\(=\)

\(\ds z \paren {\prod_{i \mathop = 1}^k x_i}\)

$x_1$ commutes with $x_2, \ldots, x_n$

\(\ds \)

\(=\)

\(\ds e\)

by $(1)$

$\blacksquare$