0
$\begingroup$

Let $N=pq$, where $p$ and $q$ are prime numbers, and order of $G_1$ and $G_2$ is equal to $p$. Suppose that $e(G_1, G_2)=G_t$ is a pairing of composite order. I know that, usually the order of target group $G_t$ is $p$.

Is it possible to set the order of target group to $pq$? In other words, is it possible to set order of target group be bigger that the order domain groups?

$\endgroup$
2
$\begingroup$

(I write $G_n$ for an abelian group of order $n$.)

Well, you can always embed $G_p$ into a group of order $pq$ as a subgroup, but that is probably not what you want.

Besides such trivial cases, this is impossible: The universal property of tensor products uniquely factors any bilinear map $e\colon\ G_p\times G_p\to G_{pq}$ as $$e\colon\;G_p\times G_p\twoheadrightarrow G_p\otimes_{\mathbb Z}G_p\xrightarrow{\bar e}G_{pq} \text;$$ note in particular that $\bar e$ has the same image as $e$. However, since $p$ is prime, $G_p$ is cyclic and $G_p\otimes_{\mathbb Z}G_p\cong G_p$ via the map $a\otimes b\mapsto ab$; thus $\operatorname{im}e=\operatorname{im}\bar e$ has order dividing $p$.

This essentially shows that any such pairing $e$ is either degenerate or given by a pairing $G_p\times G_p\to G_p$ followed by an embedding.


Here is a more elementary proof, using only basic group theory and bilinearity: Let $e\colon\ G_p\times G_p\to G_{pq}$ be a bilinear pairing and $g$ a generator of $G_p$. Since $e(g^n,g^m)=e(g,g)^{nm}$ for any $n,m\in\mathbb Z$, the element $e(g,g)$ is a generator of the image of $e$. But $$e(g,g)^p=e(g,g)^p=e(g^p,g)=e(1,g)=e(g^0,g)=e(g,g)^0=1\text,$$ thus the order of $e(g,g)$ divides $p$. This implies (as $p$ is prime) that the image of $e$ is either trivial or isomorphic to $G_p$; in particular $e$ is never surjective.

$\endgroup$
  • $\begingroup$ Explain it, please. What is $\bar e$? I have no idea about $\operatorname{im}e$ and difference of $\mapsto $ and $\twoheadrightarrow $. $\endgroup$ – Majid Dec 25 '16 at 12:43
  • $\begingroup$ @Majid I added a simpler proof. For reference: $\bar e$ is the induced map from the tensor product as indicated in the displayed diagram; $\operatorname{im}$ denotes the image set/group of a map; and the arrow $\twoheadrightarrow$ symbolizes a surjective map. $\endgroup$ – yyyyyyy Dec 25 '16 at 14:57
  • $\begingroup$ ok, this map is not surjective, but degenerate? $\endgroup$ – Majid Dec 26 '16 at 6:50
  • $\begingroup$ You said that $G_p \times G_p \rightarrow G_{pq}$ can be given by a pairing $e:G_p \times G_p \rightarrow G_p$ followed by an embedding. Please explain it. $\endgroup$ – Majid Dec 26 '16 at 6:57
  • $\begingroup$ would you please explain this "any such pairing $e$ is either degenerate or given by a pairing $G_p×G_p→G_p$ followed by an embedding" $\endgroup$ – Majid Dec 28 '16 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.