Does this equation hold in a bilinear map?

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 $$c\in G_2$$

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

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

similarly,

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

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 $$e(a,c) = 1$$).

• What if $b \ne 1$? Can we say that it always holds unless $b=1$ and $c_1 = 1$ or $c_1 = 0$ or $e(a,c) = 1$ May 27 '19 at 13:18
• 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. May 27 '19 at 13:21
• 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 $c_1 = 1$. May 27 '19 at 15:23