# Inverse in Group is Unique/Proof 2

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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

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.

$\blacksquare$

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.4$: Lemma $5$ - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): Chapter $\text{II}$: Groups: The Group Property - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Proposition $3.2$