Inverse in Group is Unique/Proof 2

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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 $\struct {G, \circ}$.


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 $a$.


\(\ds b\) \(=\) \(\ds b \circ e\) 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.