# 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.