# 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 $*$.

## 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 $b$.