Cfrgtkky

I'm curious about the homomorphic properties of Paillier. So, basically if I have $textsf{Dec}(textsf{sk}, textsf{Enc}(textsf{pk}, alpha) cdot textsf{Enc}(textsf{pk}, alpha^{-1}))$, I will get as result $alpha + alpha^{-1}$. But, does it also mean that if I have $textsf{Dec}(textsf{sk}, textsf{Enc}(textsf{pk}, alpha)^{textsf{Enc}(textsf{pk}, alpha^{-1})})$, 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 $textsf{Dec}(textsf{sk}, textsf{Enc}(textsf{pk},alpha)^{textsf{Enc}(textsf{pk}, alpha^{-1})})$ 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 $textsf{Dec}(textsf{sk},textsf{Enc}(textsf{pk}, alpha)^kbmod N^2)=kcdotalphabmod N$. We could take $k=alpha^{-1}bmod N$ and get $textsf{Dec}(textsf{sk},textsf{Enc}(textsf{pk},alpha)^{(alpha^{-1}bmod N)}bmod N^2)=1$, but that's not useful anyway, since $alpha^{-1}bmod N$ reveals $alphabmod N$.

Criticism of the question:

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


• 
  $textsf{Dec}(textsf{sk}, textsf{Enc}(textsf{pk}, alpha) cdot textsf{Enc}(textsf{pk}, alpha^{-1}))$ will be $alpha+alpha^{-1}bmod N$, which may or may not be $alpha+alpha^{-1}$.

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