In pairing based cryptography (PBC) we restrict the pairing domains to be subgroups of the $r$-torsion group $E[r]$. This arises two questions to me:

  1. Why do we restrict them to subgroups of $E[r]$? Couldn't we use, for instance, the whole group of points $E$? Or a mixture between various torsion groups? (As long as they are sufficiently large to ensure security)
  2. If so, what happens if I evaluate pairings ouside the two subgroups?

1 Answer 1


First, for cryptographic purposes we will always want to work in groups of finite order. Now suppose that we have a pairing $e(.,.)$ that bilinearly maps pairs of elements of an abelian groups $G_1$ and $G_2$ to $r$th roots of unity in some field. Considering cosets of the subgroup $rG_1$, we can write any element of $G_1$ in the form $rP+R$ where the coset representative $R$ is in the $r$-torsion subgroup. Now pick any other element $S\in G_2$

By bilnearity, we have $$e(rP+R,S) = e(rP,S)e(R,S)=e(P,S)^re(R,S)=e(R,S)$$ so that the pairing depends only on $R$. A similar argument applies to $G_2$. Thus even if we can adapt the construction of $e(.,.)$ to the whole group, there is no security benefit and the additional information in $P$ could be a malicious channel.

Theres no reason that $r$ cannot be composite, but the cryptographic strength of the pairing is no greater than the difficulty of a generic group or order $r$ which (by Pohlig-Hellman) is no greater than the square-root of the largest prime factor of $r$, with partial information being available for smaller subgroups with less work. Thus taking $r$ to be prime is the most efficient construction.

  • $\begingroup$ Here, you assume that $e(\cdot,\cdot)$ maps to some group of $r$-th roots of unity. Why can we assume that? What if this is not the case? $\endgroup$
    – Bean Guy
    Apr 11 at 6:30
  • 1
    $\begingroup$ The same analysis applies when the output is in any cyclic group of order $r$. I'm not familiar with any non-commutative pairing constructions and almost all cryptography uses cyclic groups. $\endgroup$
    – Daniel S
    Apr 11 at 10:44

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