Prove that the group of order 3 is cyclic. 
Question: Prove that the group of order 3 is cyclic.

Attempt:
Let H be a group of order 3. By definition of group, there can be only one identity element in the group H.
So, $H=\left \{ e,x,y \right \}$.
By definition of cyclic group,
we have that the elements x and y
$x=g^{n} \exists n \in \mathbb{Z}$
$y=g^{n} \exists n \in \mathbb{Z}$
In particular, n is positive for if it were not, a contradiction would arise from having more than one identity element.
Any hints or assistance is appreciated.
Thank in advance.
 A: You're trying to PROVE that $G$ is cyclic, so you cannot (yet) assert that $x = g^n$. Instead, consider $x \cdot x$. It must be either $x, y,$ or $e$. If it's $x$, then you have
$$
x^2 = x\\
x^2 (x^{-1}) = x x^{-1}\\
x = e
$$
which is a contradiction, because $x$ and $e$ are distinct elements of the group. 
If $x^2 = e$, then $x$ has order 2, but 2 does not divide 3, so this contradicts Lagrange's theorem. 
Finally, we conclude that $x^2 = y$, and thus the group is cyclic, generated by the element $g = x$. 
{Alternative if you don't like Lagrange yet:}
In the case where we suppose that $x^2 = e$: 
The elements $xe, xx,$ and $xy$ must all be distinct for if two were the same, then multiplying by $x^{-1}$ on the left would show that two of $e, x, y$ were the same, which is impossible. 
Since $xe = x$ and we're assuming $x^2 = e$, we must have 
$$
xy = y.
$$
multiplying on the right by $y^{-1}$ gives $x = e$, a contradiction. So $x^2 = e$ is also impossible. 
A: Actually, any group of prime order is cyclic.
Indeed, let $p=\lvert G\rvert$ and $a\in G$, $a\ne e$. By Lagrange's theorem, the subgroup $\langle\mkern1.5mu a\mkern1.5mu \rangle$ generated by $a$ has order a divisor of $\lvert G\rvert=p$, and is $>1$ since $a\ne e$, whence $\lvert\langle\mkern1.5mu  a\mkern1.5mu \rangle\rvert=p$, i.e. $\;\langle\mkern1.5mu  a\mkern1.5mu \rangle=G$.
