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:

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

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:

\(\displaystyle b\) \(=\) \(\displaystyle b \circ e\) $\quad$ as $e$ is the identity element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle b \circ \paren {a \circ c}\) $\quad$ as $c$ is an inverse of $a$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {b \circ a} \circ c\) $\quad$ Group Axioms: $G1$: Associativity $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e \circ c\) $\quad$ as $b$ is an inverse of $a$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle c\) $\quad$ as $e$ is the identity element $\quad$

So $b = c$ and hence the result.

$\blacksquare$


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