# Inverse in Group is Unique

## Theorem

Let $\struct {G, \circ}$ be a group.

Then every element $x \in G$ has exactly one inverse:

$\forall x \in G: \exists_1 x^{-1} \in G: x \circ x^{-1} = e^{-1} = x \circ x$

where $e$ is the identity element of $\struct {G, \circ}$.

## Proof 1

By the definition of a group, $\struct {G, \circ}$ is a monoid each of whose elements has an inverse.

The result follows directly from Inverse in Monoid is Unique.

$\blacksquare$

## Proof 2

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

By Group Axiom $\text G 3$: Existence of Inverse Element, every element of $G$ has at least one inverse.

Suppose that:

$\exists b, c \in G: a \circ b = e, a \circ c = e$

that is, that $b$ and $c$ are both inverse elements of $a$.

Then:

 $\displaystyle b$ $=$ $\displaystyle b \circ e$ as $e$ is the identity element $\displaystyle$ $=$ $\displaystyle b \circ \paren {a \circ c}$ as $c$ is an inverse of $a$ $\displaystyle$ $=$ $\displaystyle \paren {b \circ a} \circ c$ Group Axiom $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle e \circ c$ as $b$ is an inverse of $a$ $\displaystyle$ $=$ $\displaystyle c$ as $e$ is the identity element

So $b = c$ and hence the result.

$\blacksquare$

## Proof 3

Let $x, y \in G$.

We already have, from the definition of inverse element, that:

$\forall x \in G: \exists x^{-1} \in G: x \circ x^{-1} = e = x^{-1} \circ x$

By Group has Latin Square Property, there exists exactly one $a \in G$ such that $a \circ x = y$.

Similarly, there exists exactly one $b \in G$ such that $x \circ b = y$.

Substituting $e$ for $y$, it follows that $x^{-1}$ as defined above is unique.

$\blacksquare$