Suppose we have two independent random variables $X$ (with distribution $p\_X$) and $Y$ (with distribution $p\_Y$) which take values in the cyclic group $\mathbb{Z}\_n$. Let $Z = X +Y$, where the addition is done modulo $n$. Distribution of $Z$ is given by 
$$p\_Z = p\_X \circledast p\_Y$$ where $\circledast$ stands for cyclic convolution. Suppose we were interested in minimising $H(Z)$ while keeping $H(X)$ and $H(Y)$ fixed. I also impose an additional constraint that $p\_X(0) > 0$ and $p\_Y(0) > 0$. The conjecture is as follows- 



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> The $p\_X$ and $p\_Y$ which minimise $H(Z)$ have to be supported on the smallest possible subgroup $G$ of $Z\_n$ that satisfies $$\log\_2(|G|) \geq \max(H(X),H(Y))$$
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*Some Remarks:* 


1. The constraint that $p\_X(0) > 0$ and $p\_Y(0) > 0$ is essentially to rule out $X$ and $Y$ being supported on some coset of $G$. One can even assume $p\_X(0)$ and $p\_Y(0)$ are the maximum probabilities in the distributions $p\_X$ and $p\_Y$ since we can always "rotate" the distributions and still keep $H(Z)$ the same.
2. The intuition behind this is that we have the trivial lower bound
$$H(Z) \geq \max(H(X),H(Y)$$
and when $H(Y) \leq H(X) = \log\_2 |G|$, this lower bound is actually attained. Also intuitively the more concentrated a distribution, the lesser its entropy. So forcing $X$ and $Y$ to be supported in $G$ also ensures $Z$ is supported only on $G$, which is "small".
3. I'm not sure but I think one can also pose this as a norm-extremisation problem instead of entropy by using the fact $$H(p\_X) = \lim\_{p \to 1}\frac{1}{1-p}\log\_2 \sum p\_X(i)^p$$
4. I unfortunately have very little empirical evidence. An exhaustive search of all possible "entropy fixing" distributions of $X$ and $Y$ goes out of hand very fast. My codes run only till $n=5$ in reasonable time, and $4$ being the only composite number till then, I've verified it only for $4$. Since things are much more cleaner for $2^n$ than for an arbitrary composite number, maybe its true only for $\mathbb{Z}\_{2^n}$.



Entropy conjecture for distributions over $\mathbb{Z}_n$
The conjecture is wrong! It wasn't as complicated as I thought it was. 


A simple counter example is over $\mathbb{Z}\_6$. Consider $H(X) = 1$ and $H(y) = 1 +\epsilon$ where $\epsilon$ is very small. Now the conjecture would imply $X$ and $Y$ should both be supported on $\{0,2,4\}$ giving $H(X+Y) \approx 1.3326$. But consider $$p\_X = [0.5, 0, 0, 0.5, 0, 0]$$ and $$p\_Y = [0.5 - \delta, \frac{\delta}{2}, \frac{\delta}{2},0.5 - \delta, \frac{\delta}{2}, \frac{\delta}{2}]$$ for a suitable $\delta$ which adjusts the entropy to be $1+ \epsilon$. 
$$p\_X \circledast p\_Y = p\_Y $$
so we have $H(X+Y) = 1 + \epsilon < 1.3326$. Thus disproved.


