Monoid where every element is left-invertible equals group

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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

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.