Left inverse, right inverse, and invertible element of a monoid

Let $(M,\ast ,e)$ be a monoid (Semigroup and monoid) and let $g\in M$. An element $h\in M$ is

• a left inverse of $g$ when $h\ast g=e$;
• a right inverse of $g$ when $g\ast h=e$;
• a two-sided inverse, or simply an inverse, of $g$ when it is both, that is when $h\ast g=e=g\ast h$.

The element $g$ is invertible, or a unit of $M$, when it has a two-sided inverse. The set of units of $M$ is written $M^{\times }$.

The identity is always invertible, since $e\ast e=e$ exhibits $e$ as a two-sided inverse of itself.

• One-sided inverses are not enough on their own. The definition keeps the three notions apart because the sharp uniqueness statement, and hence the right to write $g^{-1}$, is about their interaction: in a monoid a left inverse of $g$ and a right inverse of $g$ are automatically equal (In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided). Until that is proved, "the inverse of $g$" is not legitimate language.

• $M^{\times }$ is a group under the restricted operation (The invertible elements of a monoid form a group under the restricted operation), and a monoid is a group exactly when $M^{\times }=M$ (Group and abelian group).