# Monoid where every element is left-invertible equals group

## Contents

## Statement

### Left-invertibility version

If is a monoid with identity element (neutral element) , such that for every , there exists such that , then is a group under .

### Right-invertibility version

If is a monoid with identity element (neutral element) , such that for every , there exists such that , then is a group under .

## Related facts

### Stronger facts

## Facts used

## 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 with identity element such that every element is left invertible. An element

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

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

### Proof details (right-invertibility version)

*Given*: A monoid with identity element such that every element is right invertible. An element

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

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