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: 
**Hint**: can you prove that every ideal is generated by one element in $\mathbb{Z}$? Can you prove that if $ \langle x\rangle \subset \langle y \rangle $, then $ x $ is divisible by $ y$?

And yes, your approach is correct
> 
> I am trying to prove $\langle a \rangle$ is a maximal ideal in $\mathbb{Z}$, if and only if $a$ is prime number.
> 
> 
> ... assume $a \in \mathbb{Z}$, ... [if] it's not [a] prime number we can write as $a = xy$, for some integers $x$ and $y$. Then $\langle a \rangle \subset \langle x \rangle$ and $\langle a \rangle \subset \langle y \rangle$.
> 
> 
> 

Looks good so far.

> 
> while if $\langle a \rangle$ is maximal ideal, [there does not] exist $\langle k \rangle$ such that $\langle a \rangle \subset \langle k \rangle\subset \mathbb{Z}$.
> 
> 
> 

I think the unspoken assumption here is that $k \in \mathbb{Z}$, so that $\langle k \rangle$ is a principal ideal. Then I think you either need to prove that all ideals in $\mathbb Z$ are principal, or refer to a proof of that fact by someone else.

Because without that result, how can you be sure $\langle a \rangle \subset \langle a, k \rangle \subset \mathbb{Z}$ is not a possibility?

Rewrite the previous set comparison as $\langle a \rangle \subseteq \langle a, k \rangle \subseteq \mathbb{Z}$. If $\gcd(a, k) = 1$, then $\langle a, k \rangle = \mathbb{Z}$. But if $\gcd(a, k) = |a|$, then $\langle a \rangle = \langle a, k \rangle$.

But that's just me talking without proving all ideals are principal in this domain.

> 
> so $\langle a \rangle$ cannot be maximal ideal.
>  we can say by contradictory if $a$ is prime, $\langle a \rangle$ is maximal ideal.
> 
> 
> is this correct? 
>  and how to prove the other way. 
> 
> 
> 

With the missing assumption filled in, then yes, this is correct. I'm not sure what you mean by "the other way."