# Invertible Elements of Monoid form Subgroup

## Theorem

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.

Let $U \subseteq S$ be the subset of $S$ consisting of the invertible elements of $S$.

Then $\struct {U, \circ}$ forms a subgroup of $S$.

## Proof

We have from Inverse of Identity Element is Itself that $e$ is invertible.

Hence $e \in U$ and so $U \ne \O$.

Let $x, y \in U$.

As $x$ and $y$ are invertible, it follows that $x^{-1}$ and $y^{-1}$ both exist in $S$.

Both $x^{-1}$ and $y^{-1}$ also have inverses $x$ and $y$ respectively, and so themselves are invertible.

Hence both $x^{-1} \in U$ and $y^{-1} \in U$.

Then we have:

\(\displaystyle \paren {x \circ y} \circ \paren {y^{-1} \circ x^{-1} }\) | \(=\) | \(\displaystyle x \circ \paren {y \circ y^{-1} } \circ x^{-1}\) | $\circ$ is associative in $S$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \circ e \circ x^{-1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \circ x^{-1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e\) |

and similarly for $\paren {y^{-1} \circ x^{-1} } \circ \paren {x \circ y}$

Hence $x \circ y$ is invertible.

The result follows by the Two-Step Subgroup Test.

$\blacksquare$

## Sources

- 1978: John S. Rose:
*A Course on Group Theory*... (previous) ... (next): $2$: Examples of Groups and Homomorphisms: $2.3$