Inverse in Group is Unique/Proof 2

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 $\struct {G, \circ}$.

Let $\struct {G, \circ}$ be a group whose identity element is $e$.

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:

$\ds b$

$=$

$\ds b \circ e$

$\ds $

$=$

$\ds b \circ \paren {a \circ c}$

as $c$ is an inverse of $a$

$\ds $

$=$

$\ds \paren {b \circ a} \circ c$

$\ds $

$=$

$\ds e \circ c$

as $b$ is an inverse of $a$

$\ds $

$=$

$\ds c$

So $b = c$ and hence the result.

$\blacksquare$