# Inverse in Group is Unique

## Theorem

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

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

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

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:

\(\ds b\) | \(=\) | \(\ds b \circ e\) | as $e$ is the identity element | |||||||||||

\(\ds \) | \(=\) | \(\ds b \circ \paren {a \circ c}\) | as $c$ is an inverse of $a$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {b \circ a} \circ c\) | Group Axiom $\text G 1$: Associativity | |||||||||||

\(\ds \) | \(=\) | \(\ds e \circ c\) | as $b$ is an inverse of $a$ | |||||||||||

\(\ds \) | \(=\) | \(\ds c\) | 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.

$\blacksquare$

## Also see

## Sources

- 1970: B. Hartley and T.O. Hawkes:
*Rings, Modules and Linear Algebra*... (previous) ... (next): Chapter $1$: Rings - Definitions and Examples: $1$: The definition of a ring: Definitions $1.1 \ \text{(b)}$ - 2008: Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*... (previous) ... (next): $\S 1$: Exercise $5$