Product of Commuting Elements in Monoid is Unit iff Each Element is Unit/Proof 1

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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

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:

\(\ds y z x\) \(=\) \(\ds y z x x y z\) since $x y z = e$
\(\ds \) \(=\) \(\ds y z x y x z\) using $x y = y z$ and Monoid Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds y \paren {z x y} x z\) Monoid Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds y x z\) since $z x y = e$
\(\ds \) \(=\) \(\ds x y z\) since $y x = x y$
\(\ds \) \(=\) \(\ds \paren {x y} z\) Monoid Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds e\)

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.

$\Box$


Sufficient Condition

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

$\blacksquare$

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