Cyclic property of an elliptic group In page 32 of the following link [ http://www.cse.iitd.ernet.in/~ssen/csl863/cycgrp.pdf ], it's stated that for an elliptic curve defined over $\mathbb{F}_q$ we have that:
($1$) $E(\mathbb{F}_q)\cong\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}$ with $n_2|n_1$ and $n_2|q-1$.
($2$) $E(\mathbb{F}_q)$ is cyclic $\iff n_2=1$.
A nice proof for the first statement is given by Jyrki Lahtonen to the following question: Isomorphism of Elliptic Curves:.
I was wondering about how to go about proving the second statement for elliptic curves of order $p+1$ for a large enough $p$:
For an elliptic curve $E(\mathbb{F}_p)$ with $|E(\mathbb{F}_p)|=p+1$ we see that in this case for $E(\mathbb{F}_p)\cong\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}$ we require that $n_2|p+1$ and $n_2|p-1$. Clearly, for $p$ large enough, we have either $n_2=1$ (in which casae our work is done) or $n_2=2$ and $n_1=\frac{p+1}{2}$. For statement ($2$) to hold the latter case must not be possible, but I am unsure as to why.
Any help is appreciated.
 A: For an elliptic curve $E$ over $\Bbb{F}_p$ with $|E(\Bbb{F}_p)|=p+1$ it is perfectly possible that $n_2=2$.
Assume that


*

*$p\equiv-1\pmod4$ (so $-1$ is not a quadratic residue modulo $p$), and

*$a$ is a non-zero quadratic residue modulo $p$.


I claim that in this case the curve
$$E:y^2=x^3-ax$$
has $p+1$ rational points as well as $n_2=2$.
The first claim follows from the fact that the polynomial $f(x)=x^3-ax$ is odd. For any $x\in\Bbb{F}_p$ 
either $f(x)=f(-x)=0$ or exactly one of $f(x), f(-x)=-f(x)$ is a quadratic residue (and the other is a quadratic non-residue). This immediately implies that $|E(\Bbb{F}_p)|=p+1$, this argument has been covered many times on this site already.
The second claim follows from the fact that $f(x)=x(x^2-a)$ has three distinct solutions $x=0, x=\pm\sqrt{a}$ in $\Bbb{F}_p$. On an elliptic curve in short Weierstrass form the points with $y=0$ have order two. Therefore on $E(\Bbb{F}_p)$ we have three points of order two. This means that the group is not cyclic, and hence $n_2>1$. The OP already explained why $n_2\in\{1,2\}$ so we can conclude that $n_2=2$.

If, instead, we select $a$ to be a quadratic non-residue, it follows that $E$ has only a single $\Bbb{F}_p$-rational point of order two. So in that case $2\nmid n_2$, and the group $E(\Bbb{F}_p)$ must, indeed, be cyclic of order $p+1$.
