Let us have a set $G$ together with some binary operation $*$. We will use multipicative notation where $ab = a * b$. Let $x, y, z \in G$. If $\langle G, *\rangle$ has the following properties:
- $(xy)z = x(yz)$
- $ex = x$
- $x^{-1}x = e$
for some fixed $e \in G$, then we say that $\langle G, *\rangle$ is a group. In my class, we were also told to show that $xe = x$ and $xx^{-1} = e$. However, these can be derived by the prior three properties.
Prove $xx^{-1} = e$
$$ \begin{align*} e &= (xx^{-1})^{-1}(xx^{-1}) \\ &= (xx^{-1})^{-1}(x(ex^{-1})) \\ &= (xx^{-1})^{-1}(x((x^{-1}x)x^{-1})) \\ &= (xx^{-1})^{-1}(x(x^{-1}x)x^{-1}) \\ &= (xx^{-1})^{-1}((xx^{-1})(xx^{-1})) \\ &= ((xx^{-1})^{-1}(xx^{-1}))(xx^{-1}) \\ &= e(xx^{-1}) \\ &= xx^{-1} \\ \end{align*} $$
Prove $xe = x$
We can use the last proof to solve this faster.
$$ \begin{align*} x &= ex \\ &= (xx^{-1})x \\ &= x(x^{-1}x) \\ &= xe \end{align*} $$