Homomorphic properties of Paillier 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 $\alpha\cdot\alpha^{-1}$, 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 $\textsf{Dec}(\textsf{sk},\textsf{Enc}(\textsf{pk}, \alpha)^k\bmod N^2)=k\cdot\alpha\bmod N$. 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 $\alpha\bmod N$.



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 $\alpha+\alpha^{-1}$.