Prove any group with 3 elements is isomorphic to $\mathbb{Z}_3$ Let $G = \left\{e, a, b \right\}$
Since $G$ has $3$ elements, $a \not= e , \,b \not= e, \,a \not= b$
$G$ is closed so $ab \in G$
If $ab = a$, this means that $b = e$
and if $ab = b$, $ a = e $
So we are left with $ab = e$
Define a function
$\varphi (a) = [1]$
$\varphi (b) = [2]$
To show that this function is an isomorphism, 
$\varphi (e) = \varphi (ab) = \varphi (a) + \varphi (b) = [1] + [2] = [0]$
So $\varphi (e) = [0]$
So this function takes the identity to the identity, so it's an isomorphism.
 A: More easier, let $G=\{e,a,b\}$.
$$ab=ba=e$$ and thus, it's commutative. Therefore it's a cyclic group with $3$ elements and thus, it's isomorphic to $\mathbb Z_3$.
A: Consider $G=\{e,a,b\}$ where $e$ is the identity; then the Cayley table is at the beginning
$$
\begin{matrix}
e & a & b \\
a &  &  \\
b &  &
\end{matrix}
$$
In position $(2,2)$ we can have either $e$ or $b$. But if we have $e$, then position $(2,3)$ would have $b$, which is impossible. So we get
$$
\begin{matrix}
e & a & b \\
a & b & e \\
b &  &
\end{matrix}
$$
and now the table is easily completed:
$$
\begin{matrix}
e & a & b \\
a & b & e \\
b & e & a
\end{matrix}
$$
Thus $a^2=b$ and $a^3=ba=e$. Therefore $a$ is a generator of the group. So $G$ is cyclic of order $3$ and so isomorphic to $\mathbb{Z}/3\mathbb{Z}$. You get an explicit isomorphism by comparing the Cayley tables.

In a more abstract way: consider the subgroup $H$ generated by $a$. Since $a\ne e$, we have $|H|>1$. By Lagrange's theorem $|H|$ divides $3$, so $|H|=3$ and therefore $G$ is cyclic with $a$ as a generator.
Consider the homomorphism $\varphi\colon\mathbb{Z}\to G$ defined by $\varphi(n)=a$. Then $\varphi$ is surjective and its kernel must be $3\mathbb{Z}$ because of the order. Thus $\varphi$ induces an isomorphism $\mathbb{Z}/3\mathbb{Z} \to G$.
A: You have shown that $ab=e$. Now $a^2$ cannot equal $e$, because that would give an element of order two, forbidden because 2 does not divide 3. And $a^2$ cannot equal $a$, because then $a$ would equal $e$. So $a^2=b$, and thus $a^3=e$, so the group is cyclic.
Edit: Much simpler, since $a$ cannot be order two, it must be order three, and thus the group is cyclic.
