Prove that $(G,*)$ is a group.

Question

Let $G$ be a non-empty set, $*:G\times{G}\to{G}$ a binary operation that satisfies:

1. $*$ is associative.
2. Exist $e\in{G}$ such that $a*e=a$ $\forall{a\in{G}}$
3. For all $a\in{G}$ there is $i(a)\in{G}$ such that $a*i(a)=e$

Prove that $(G,*)$ is a group.

I have tried the following (without reaching any results):

$e*(a*e)=e*a\Rightarrow (e*a)*e=e*a\Rightarrow (e*a)=(e*a)$

And:

$i(a)*(a*i(a))=i(a)*e\Rightarrow (i(a)*a)*i(a)=i(a)\Rightarrow(i(a)*a)*(i(a)*a)=i(a)*a$

Answer 1

I assume you want to show that $e$ is also a left identity and that $i(a)$ is also a left inverse for $a$.

Let $a\in G$. \begin{align*} i(a)\ast a &= (i(a)\ast a)\ast e\\ &= (i(a)\ast a)\ast(i(a)\ast i(i(a)))\\ &= ((i(a)\ast a)\ast i(a))\ast i(i(a))\\ &= (i(a)\ast(a\ast i(a)))\ast i(i(a))\\ &= (i(a)\ast e)\ast i(i(a))\\ &= i(a)\ast i(i(a))\\ &= e \end{align*}

So we may now use that $a\ast i(a) = i(a)\ast a = e$.

\begin{align*} e\ast a&= (a\ast i(a))\ast a\\ &= a\ast (i(a)\ast a)\\ &= a\ast e\\ &= a. \end{align*}

Answer 2

If I understand your question it remains to show:

$e*a = a$ $\ $ & $\ $ $i(a)*a = e$

$e*a = a$ if and only if $e*a*i(a) = e$ $\implies$ $e*e = e$ since$\ $ $a*i(a) = e$ by assumption.