# Alternate Definition of a Group

A set $G$ with binary composition denoted multiplicatively is a group if

(i) The composition is associative.
(ii) For every pair of elements $a,b \in G$, the equations $ax = b$ and $ya = b$ have unique solutions in $G$.

Proof: Binary operation implies that the set $G$ under consideration is closed under the operation. Now to prove that the set $G$ is a group, we have to show the left identity exists and each element of $G$ possesses a left inverse with respect to the operation under consideration.
It is given that for every pair of elements $a,b \in G$ the equation $ya = b$ has a solution in $G$. Therefore, say $e \in G$ such that $e \cdot a = a$.
Now, let us suppose that $b$ is any arbitrary element of $G$. Therefore, there exists $e \in G$ such that $ax = b$.
Thus

$b = ax$
$\Rightarrow eb = e\left( {ax} \right)$
$\Rightarrow eb = \left( {ea} \right)x$
$\, \Rightarrow eb = ax$
$\Rightarrow eb = b$

Therefore there exists $e \in G$ such that $eb = b, \forall b \in G$
$\therefore e$ is the left identity.

Now, let $b = e \in G$ be an element. $ya = b \Rightarrow ya = e$ $\Rightarrow y$ is the inverse of $a$ in $G$.
Let $y = {a^{ – 1}}$ such that ${a^{ – 1}}a = e$, then ${a^{ – 1}} \in G$ as $ya = e$ has a solution in $G$.

Thus ${a^{ – 1}}$ is the left inverse of $a$ in $G$. Therefore each element of possesses a left inverse. Hence $G$ is a group for the given composition if the postulate and (ii) are satisfied.