Inverse in Group is Unique

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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

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