Elliptic Curve: point of prime order $n$ I am attempting to understand ElGamal Encryption on Elliptic Curves (beginnning of this paper)
I have an elliptic curve $E\left( \mathbb{F}_{11} \right)$ defined by $y^2 = x^3 + 2x + 3$
I need to choose a point $P$ on $E\left(\mathbb{F}_{11}\right)$ of prime order $n$ 
I haven't been able to find anything to help me understand what is meant by 'prime order $n$'. Can someone point me in the direction of anything useful or explain to me how to find such a point $P$ please?
I am a computer scientist, so assume limited knowledge of any complex maths/theorems - all my knowledge on elliptic curves is self-taught through Googling...
 A: You were already given a nice answer in comments, but lets add a little detail.
We find all of the points on the curve (Note that $O$ is the point at infinity)
$$(x, y) = O, (0, 5), (0, 6), (2, 2), (2, 9), (3, 5), (3, 6), (4, 3), (4, 
  8), (6, 0), (8, 5), (8, 6), (10, 0)$$
You can verify each of these by checking that
$$y^2 = x^3 + 2x + 3 \pmod{11}$$
If we take a point $P$ and write out $P, P+P, \ldots$, we get 


*

*$1P = (2, 2)$

*$2P = (0, 5)$

*$3P = (3, 5)$

*$4P = (4, 3)$

*$5P = (8, 6)$

*$6P = (10, 0)$

*$7P = (8, 5)$

*$8P = (4, 8)$

*$9P = (3, 6)$

*$10P = (0, 6)$

*$11P = (2, 9)$

*$12P = O$


Compare this list of points with the list above. What do you notice? Every one of the points must be a point on the curve.
Next, how many points did we cycle through, which is the order?
This elliptic curve has order $\#E = |E| = 12$ since it contains $12$ points in its cyclic group.
There is a theorem called Hasse‘s Theorem: Given an elliptic curve module $p$, the number of points on the curve is denoted by $\#E$ and is bounded by
$$p+1-2\sqrt{p} ≤ \#E ≤ p+1+2 \sqrt{p}$$
Interpretation: The number of points is close to the prime $p$. 
A: I do not have enough reputation to comment, so I answer it here.  
The comment says: 

Thanks, this makes sense, but if I choose $P=(4,3)$ then I get $\#E=3$ as $3P=\mathcal{O}$, What made you choose P=(2,2)?   

Because not all points are generators, however, only the generator point can generate all the points on the curve.
So, $P=(2,2)$ is one generator, but  $P=(4,3)$ is not.
you can read this :       https://www.entrust.com/wp-content/uploads/2014/03/WP_Entrust_Zero-to-ECC_March2014.pdf 
