TITLE: The group structure of elliptic curve over $\mathbb F_p$
QUESTION [5 upvotes]: I want to find the group of the elliptic curve $y^2=x^3-x$ over $\mathbb F_p$ for all primes $p \le 29$. But I know only 1 fact about the structure of this group: $E(\mathbb F_p)=\mathbb Z/m \mathbb Z \times \mathbb Z/nm\mathbb Z$ for $\gcd (m,p)=1, p =1 \mod m$. It doesn't really help. Is there any method to find the group without huge calculation?

REPLY [2 votes]: Note that the point at infinity, together with the points $(-1,0),(0,0),(1,0)$ form a subgroup of $ E(\Bbb Q)$ isomorphic to $(\Bbb Z/2\Bbb Z)^2$, you will find this subgroup in $E(\Bbb F_p)$ for any $p > 2$.
Moreover, your curve has complex multiplication (by $i : (x,y) \mapsto (-x,iy)$), so a lot can be said via the explicit class field theory for $\Bbb Q(i)$.
If $p \equiv 3 \pmod 4$, the number of points on $E(\Bbb F_p)$ is $p+1$. Since there is no prime factor in common between $p+1$ and $p-1$ except possibly $2$, and we know that $\Bbb (\Bbb Z/2\Bbb Z)^2$ is a subgroup, but not $(\Bbb Z/4\Bbb Z)^2$ (because $p-1 \equiv 2 \pmod 4$), we must have $E(\Bbb F_p) \approx (\Bbb Z/2\Bbb Z) \times (\Bbb Z/\frac{p+1}2\Bbb Z)$.
The points $(i,1-i),(i,-1+i),(-i,1+i),(-i,-1-i)$ with the previous four form a subgroup of order $8$ in $E(\Bbb Q(i))$ (isomorphic to $(\Bbb Z/2\Bbb Z) \times (\Bbb Z/4\Bbb Z)$), so this subgroup will always be present in $E(\Bbb F_p)$ with $p \equiv 1 \pmod 4$
If $p \equiv 1 \pmod 4$, then there is a unique way to write $p = (a+ib)(a-ib)$ with $a+ib \equiv 1 \pmod {2+2i}$. Then, $\# E(\Bbb F_p) = p+1-2a$, and $E(\Bbb F_p)$ contains a factor $(\Bbb Z/m\Bbb Z)^2$ if and only if $(a,b) \equiv (1,0) \pmod m$ (the Frobenius automorphism $Frob_p$ acts on $E(\overline{\Bbb F_p})$ a lot like multiplication by $a+ib$ does on the torus $\Bbb Q[i] / \Bbb Z[i]$).