mjhjmtu

I'm curious about the homomorphic properties of Paillier. So, basically if I have $textsfDec(textsfsk, textsfEnc(textsfpk, alpha) cdot textsfEnc(textsfpk, alpha^-1))$, I will get as result $alpha + alpha^-1$. But, does it also mean that if I have $textsfDec(textsfsk, textsfEnc(textsfpk, alpha)^textsfEnc(textsfpk, alpha^-1))$, then the result will be $alpha cdot alpha^-1$, which will basically cancel each other, and will be left with 1?

No, there is no reason that $textsfDec(textsfsk, textsfEnc(textsfpk,alpha)^textsfEnc(textsfpk, alpha^-1))$ would be $alphacdotalpha^-1$, including when we spread $bmod N$ or $bmod N^2$ here and there.

What does apply is: for overwhelmingly most $alpha$ and $k$ in $Bbb Z$, it holds that $textsfDec(textsfsk,textsfEnc(textsfpk, alpha)^kbmod N^2)=kcdotalphabmod N$. We could take $k=alpha^-1bmod N$ and get $textsfDec(textsfsk,textsfEnc(textsfpk,alpha)^(alpha^-1bmod N)bmod N^2)=1$, but that's not useful anyway, since $alpha^-1bmod N$ reveals $alphabmod N$.

Criticism of the question:

• It is not defined in which group it is computed $alpha^-1$, and that matters.


• 
  $textsfDec(textsfsk, textsfEnc(textsfpk, alpha) cdot textsfEnc(textsfpk, alpha^-1))$ will be $alpha+alpha^-1bmod N$, which may or may not be $alpha+alpha^-1$.

Given two plaintexts $alpha$ and $beta$, Pailler cryptosystem $mathcalE$ homomotphic property is: $mathcalE(alpha)times mathcalE(beta)=mathcalE(alpha+beta)$. So, $mathcalE(alpha)^n=mathcalE(nalpha)$. In your example, $n=mathcalE(alpha^-1)$ and thus after decryption you will have $mathcalE(alpha^-1)times alpha$ ans not $alpha^-1 times alpha$. This is a high level answer, but you need to define the modulo $N$ of your system.