Monoid where every element is left-invertible equals group

Statement

Left-invertibility version

If $(M,*)$ is a monoid with identity element (neutral element) $e$, such that for every $a\in M$, there exists $b\in M$ such that $b*a=e$, then $M$ is a group under $*$.

Right-invertibility version

If $(M,*)$ is a monoid with identity element (neutral element) $e$, such that for every $a\in M$, there exists $b\in M$ such that $a*b=e$, then $M$ is a group under $*$.

Related facts

Stronger facts

• Semigroup with left neutral element where every element is left-invertible equals group

Facts used

1. Equality of left and right inverses in monoid

Proof

Proof idea (left-invertibility version)

We need to show that every element of the group has a two-sided inverse. To do this, we first find a left inverse to the element, then find a left inverse to the left inverse. Thus, the left inverse of the element we started with has both a left and a right inverse, so they must be equal, and our original element has a two-sided inverse.

Proof details (left-invertibility version)

Given: A monoid $(M,*)$ with identity element $e$ such that every element is left invertible. An element $a\in M$

To prove: $a$ has a two-sided inverse.

Proof: Suppose $b$ is a left inverse for $a$. Let $c$ be a left inverse for $b$. Then, $b$ has $a$ as a right inverse and $c$ as a left inverse, so by Fact (1), $a=c$. Thus, $a*b=b*a=e$, so $a$ has a two-sided inverse $b$.

Proof details (right-invertibility version)

Given: A monoid $(M,*)$ with identity element $e$ such that every element is right invertible. An element $a\in M$

To prove: $a$ has a two-sided inverse.

Proof: Suppose $b$ is a right inverse for $a$. Let $c$ be a right inverse for $b$. Then, $b$ has $a$ as a left inverse and $c$ as a right inverse, so by Fact (1), $a=c$. Thus, $a*b=b*a=e$, so $a$ has a two-sided inverse $b$.