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Apr 20 at 12:40 comment added Jarosław Błasiok Well, you could just as well take $Y$ to be a uniform on an arthmetic progression $\{0, 2, \ldots, 2q\}$ where $q$ is some number around $p/4$ and $X = 0$ deterministically, $Z=2$ deterministically. So really the right thing to look at are arithmetic progressions in $G$, as opposed to subgroups of $G$ --- and there is quite a bit more of the former than latter for $\mathbb{Z}/\mathbb{Z}p$.
Apr 20 at 12:36 vote accept alon
Apr 20 at 12:36 comment added alon Very interesting. Though, the Y you gave is 0.5-indistinguishable to uniform on subgroup {0} For a small enough epsilon, I'd like to get such a 0.5 bound for either {0} or G.
Apr 20 at 12:10 comment added kodlu I was about to write the same. In general Pinsker (reverse Pinsker?) style inequalities can be used to try and achieve the best possible but will have penalty terms.
Apr 20 at 12:01 history answered Jarosław Błasiok CC BY-SA 4.0