# Inverse in Group is Unique/Proof 2

## 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

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

By Group Axioms: $G3$: Inverses, 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 Axioms: $G1$: 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.

$\blacksquare$