I would like to verify whether or not the following equation holds:

$e(a,c)^{c1\cdot c2\cdot c3}e(b,c)^{c1\cdot c2\cdot c4}==e(a,c)^{c2\cdot c3}e(b,c)^{c1^2\cdot c2\cdot c4}$ for appropriately defined bilinear map $e$ with different input groups $G_1$ and $G_2$ for $a,b\in G_1$ and $c\in G_2$


1 Answer 1


No it does not. To see it, take $b=1$:

$e(a,c)^{c1\cdot c2\cdot c3}e(b,c)^{c1\cdot c2\cdot c4}= e(a,c)^{c1\cdot c2\cdot c3}e(1,c)^{c1\cdot c2\cdot c4} = e(a,c)^{c1\cdot c2\cdot c3}1^{c1\cdot c2\cdot c4} = e(a,c)^{c1\cdot c2\cdot c3}$


$e(a,c)^{c2\cdot c3}e(b,c)^{c1^2\cdot c2\cdot c4} = e(a,c)^{c2\cdot c3}$

So you would get

$e(a,c)^{c1\cdot c2\cdot c3} = e(a,c)^{c2\cdot c3}$

which is clearly not true in general (unless $c_1 = 1$ or $c_1 = 0$ or $e(a,c) = 1$).

  • $\begingroup$ What if $b \ne 1$? Can we say that it always holds unless $b=1$ and $c_1 = 1$ or $c_1 = 0$ or $e(a,c) = 1$ $\endgroup$
    – curious
    May 27, 2019 at 13:18
  • $\begingroup$ No, it still does never hold in general. I had just taken a specific examples to make it obvious, but there is no general form of your equation that works - you just cannot play around with exponents like that, it would be like asking whether $a^n * b^m = a^m * b^n$ in general -- that might hold for very specific choices of value, but is simply false in general. $\endgroup$ May 27, 2019 at 13:21
  • $\begingroup$ Sorry, stupid miscalculation from me. I removed the comment - the previous, more general comment stands. Basically, what you ask is whether $A^{c_1}*B = A*^B^{c_1}$ is true, for appropriate values of $A,B$. It should be clear for you, with this notation, that this is only true for very specific values of $A,B$ unless $c_1 = 1$. $\endgroup$ May 27, 2019 at 15:23

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