Elliptic curve group over a prime finite field $F_p$ If $p$ is a big prime, and the elliptic curve $E$ is defined over $F_p$ by the equation $y^2=x^3+ax+b$ where $a,b\in F_p$. The point on $E/F_p$ together with the infinite point $\mathcal{O}$ form a group $G=\{(x,y):x,y\in F_p;(x,y)\in E/F_p\}\cup\{\mathcal{O}\}$. Then  is $G$ a cyclic group? what's the order of $G$?
The cyclic group over the ECDLP problem is posed is a subset of the set of point of the elliptic curve. That is to say, not all the points in the referred curve will be in the cyclic group. What you've called as $G$ confuses with another notation of the generator of a cyclic group $&lt;G&gt;=\{G,[2]G,\cdots,[n]G=\mathcal{O}\}$.

This is the meaning of what you've may heard as the cofactor ($h$). There is a relation between the number of points in the elliptic curve (the cardinal $|E(\mathbb{F}_p)|$) and the order of the mention cyclic group ($org(G)=n$). Following the P1363, the cardinal of a cryptographically good curve should be prime or "almost" prime. 

What this almost mean is to factorize in one big number and another very small. $|E(\mathbb{F}_p)|=n\cdot h$, where $n &gt;&gt;&gt; h$. Back to P1363, $h$ shall not be other than $1$, $2$ or $4$. In a $h=1$ your cyclic group fits with the complete set of points in the elliptic curve. On bigger cofactors, smaller the cyclic group; but don't panic, a $192$ bit curve with a cofactor $2$ can be like a $191$ curve with a cofactor $1$ (despite other issues that suggest to avoid cofactor $1$).