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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

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It is the number of cyclic subgroups.

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 $\ell^c$.

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  • $\begingroup$ 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? $\endgroup$
    – Manuel Bravi
    Commented Feb 6, 2023 at 10:24
  • $\begingroup$ @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 $\overline{\Bbb{F}}_p$) $\endgroup$
    – reuns
    Commented Feb 6, 2023 at 12:10

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