# Why the full $r$-torsion group contains $r^2$ many elements and consists of $r+1$ subgroups

let $$F$$ be a finite field, $$E(F)$$ an elliptic curve of order $$n$$, $$r$$ a factor of $$n$$, $$k(r)$$ for the embedding degree of $$E(F)$$ with respect to $$r$$.

Then why the full $$r$$-torsion group contains $$r^2$$ many elements and consists of $$r+1$$ subgroups?

Only if $$r$$ is prime.
Let $$E[r] = \langle P,Q \rangle$$ be the $$r$$-torsion subgroup. Its cyclic subgroup is defined by a generator of the form $$[a]P + [b]Q$$, but up to a (non-zero) scalar multiple: because $$G$$ and $$[k]G$$ generate the same subgroups.
How to count the subgroups? In the expression $$G = [a]P + [b]Q$$ we can always take $$[a^{-1}]G = [1]P + [ba^{-1}]Q$$ if $$a$$ is nonzero, or $$[b^{-1}]G = [ab^{-1}]P + [1]Q$$ if $$b$$ is nonzero (and one of them has to be non-zero).
Therefore, we can count $$2r$$ generators of the form $$[1]P + [a]Q$$ and $$[a]P + [1]Q$$ for any $$a$$ (incl. 0). But we counted the subgroups where both coefficients are invertible twice, which is when $$a$$ is invertible (non-zero): $$\langle[1]P+[a][Q]\rangle=\langle[a^{-1}]P+[1][Q]\rangle.$$ So we have to correct it and get $$2r-(r-1)=r+1.$$