# Number of given order subgroups of the torsion subgroup (Elliptic Curve)

I'm studing the paper "ON THE COST OF COMPUTING ISOGENIES BETWEEN SUPERSINGULAR ELLIPTIC CURVES" (link) and, at some point, autors say that (assuming $$e$$ even):

the number of order-$$\ell^{e/2}$$ subgroups of $$E[\ell^e]$$ is $$N=(\ell+1)\ell^{e/2 - 1}.$$

Here $$E$$ is an elliptic curve on $$\mathbb{F}_q,$$ $$q = p^n$$ with $$p$$ prime and $$\ell$$ is another prime diffrent from $$p$$. $$E[\ell^{e}]$$ is the set of $$\ell^e-$$torsion points of the elliptic curve $$E$$.

Can someone help me to understand why the number of order-$$\ell^{e/2}$$ subgroups of $$E[\ell^e]$$ is $$N=(\ell+1)\ell^{e/2 - 1}$$?

I know that $$E[\ell^{e}] \simeq \mathbb{Z}/\ell^e \mathbb{Z}\times \mathbb{Z}/\ell^e \mathbb{Z}$$ so i think that the problem is equivalent to counting order-$$\ell^{e/2}$$ subgroups of $$\mathbb{Z}/\ell^e \mathbb{Z}\times \mathbb{Z}/\ell^e \mathbb{Z}.$$ But

• I'm unable to do it;
• I'm not convinced I'm on the right way.

Any help will be appreciated

If $$c\le e$$ then any subgroup of order $$\ell^c$$ in $$E[\ell^e]$$ will be contained in $$E[\ell^c]$$. The number of elements of order $$\ell^c$$ in $$\Bbb{Z}/\ell^c\Bbb{Z}\times\Bbb{Z}/\ell^c\Bbb{Z}$$ is $$\ell^{2c}-\ell^{2(c-1)}$$ (elements $$(a,b)$$ with $$\ell\nmid a$$ or $$\ell \nmid b$$). Then divide by $$\ell^{c-1}(\ell-1)$$ (the number of elements generating the same subgroup) to get $$\ell^{c-1}(\ell+1)$$ for the number of cyclic subgroups of order $$\ell^c$$.
• thank you for the answer. I have last question for you. Here (link), p. 37/59, C. Martindale and L. Panny give an example: consider two elliptic curves E and E' over $\mathbb{F}_{431}$ (explicit) which are 4-isogenous. Obviously they write that an 4-isogeny $f:E\to E'$ has ker generated by $P\in E[\overline{\mathbb{F}_p}]$ of order $4.$ Why do they say "try all nine possible order 4 kernels..."? With respect with your argouments i think that possible 4 order Kernels are six. I'm wrong? Commented Feb 6, 2023 at 10:24
• @ManuelBravi Not hard to check by hand that there are $6$ cyclic subgroups of order $4$. The only non-cyclic one is $C_2\times C_2$ but it is the kernel of $P\to [2]P$ so it would need that the two curves have the same $j$-invariant (and be in fact isomorphic over $\overline{\Bbb{F}}_p$) Commented Feb 6, 2023 at 12:10