Commutation with Inverse in Monoid

Theorem

Let $\struct {S, \circ}$ be a monoid.

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

Then $x$ commutes with $y$ if and only if $x$ commutes with $y^{-1}$.

Proof

Necessary Condition

Let $x$ commute with $y$.

Then:

\(\ds y^{-1} \circ x\)

\(=\)

\(\ds \paren {y^{-1} \circ x} \circ e\)

Monoid Axiom $\text S 2$: Identity

\(\ds \)

\(=\)

\(\ds y^{-1} \circ \paren {x \circ e}\)

Monoid Axiom $\text S 1$: Associativity

\(\ds \)

\(=\)

\(\ds y^{-1} \circ \paren {x \circ \paren {y \circ y^{-1} } }\)

Invertibility of $y$

\(\ds \)

\(=\)

\(\ds y^{-1} \circ \paren {\paren {x \circ y} \circ y^{-1} }\)

Monoid Axiom $\text S 1$: Associativity

\(\ds \)

\(=\)

\(\ds y^{-1} \circ \paren {\paren {y \circ x} \circ y^{-1} }\)

$x$ commutes with $y$

\(\ds \)

\(=\)

\(\ds \paren {y^{-1} \circ \paren {y \circ x} } \circ y^{-1}\)

Monoid Axiom $\text S 1$: Associativity

\(\ds \)

\(=\)

\(\ds \paren {\paren {y^{-1} \circ y} \circ x} \circ y^{-1}\)

Monoid Axiom $\text S 1$: Associativity

\(\ds \)

\(=\)

\(\ds \paren {e \circ x} \circ y^{-1}\)

Invertibility of $y$

\(\ds \)

\(=\)

\(\ds x \circ y^{-1}\)

Monoid Axiom $\text S 2$: Identity

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

$\Box$

Sufficient Condition

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

From the above, it follows that $x$ commutes with $\paren {y^{-1} }^{-1}$.

From Inverse of Inverse in Monoid:

$\paren {y^{-1} }^{-1} = y$

Thus $x$ commutes with $y$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.5: \ 2^\circ$