it is motivated by [Density of congruence classes covered by a set](https://mathoverflow.net/questions/46249/density-of-congruence-classes-covered-by-a-set/46277#46277)


Let say just "$n$-gon" for the set of vertices of a regular $n$-gon inscribed in a unit circle, "2-gon" for the set of two opposite points.


is it true that for given positive integers $1 < b\_1 < b\_2 < \dots < b\_k$ the union of $b\_i$-gons has the minimal cardinality when they all have a common vertex?


In other words, if $G\_i$ are subgroups of the finite cyclic group $G$, is it true that 
$$|\cup\_{i=1}^k x\_iG\_i|\ge |\cup\_{i=1}^k G\_i|$$
for arbitrary cosets $x\_iG\_i$? 



union of regular polygons
Let me attempt a proof using the group-theoretic formulation. I will use the additive notation for the group operation.


The proof is by induction on $n=|G|$, with the base being trivial. Let $n=p^rm$ for some prime $p$ with $\gcd(p,m)=1$.
Consider $G' = pG$. Our goal is to reduce the problem for $G$ to its instance for $G'$. Let $R\_j:=G' + jm$. Then $\{R\_0,R\_1,\ldots, R\_{p-1}\}$ is a partition of $G$ into $G'$-cosets. 


Let $G\_i=d\_iG$ be a subgroup of $G$ with $d\_i | n$. If $p|d\_i$ then $G\_i \subseteq G'$ and every $G\_i$ coset belongs to some $R\_j$. In this case
we say that $G\_i$ is *of the first kind*. Otherwise, $d\_i | m$, and translating $G\_i$ by $m$ does not change $G\_i$. We say that such $G\_i$ is *of the second kind*.


Let $S = \cup\_{i=1}^k (G\_i+x\_i)$ be the union under consideration, let $S\_1$ be the union of cosets of the first kind among cosets comprising $S$, and $S\_2$ -- of the second. Let $T\_j=S\_1 \cap R\_j$ for $0\leq j \leq p-1$. Then $T\_j$ is actually union of some of our cosets, and the sets $T\_0,T\_1,\ldots,T\_{p-1}$ are disjoint. Let $T\_j'=T\_j-jm$: we shift all the cosets in $T\_j$ from $R\_j$ to $R\_0=G'$. Note that $S\_2-jm=S\_2$ and therefore the intersection of $T\_j$ and $S\_2$ shifts with $T\_j$. In particular, $|T\_j'-S\_2|=|T\_j-S\_2|$.


The set $S'=S\_2 \cup (\cup\_{j=0}^{p-1}T'\_j) $ is still a union of cosets of $G\_i$'s. We have 


$|S'| \leq |S\_2|+ \sum\_{j=1}^{p-1}|T'\_{j}-S\_2| = |S\_2|+ \sum\_{j=1}^{p-1}|T\_{j}-S\_2|=|S|$.


Thus it suffices to consider $S'$, which means that we may assume that $S\_1 \subseteq G'$. Let $G\_i'=G\_i \cap G'$ and let
$x\_i'$ be chosen so that $G\_i'+x\_i'= (G\_i+x\_i) \cap G'$. By the induction hypothesis applied to $G'$, we get


$a\_1:=| \cup\_{i=1}^k G\_i' | \leq | \cup\_{i=1}^k (G\_i' + x\_i')|=: b\_1$


and also


$a\_2:=|\cup\_{i: G\_i \not \subseteq G'} G\_i'| \leq |\cup\_{i: G\_i \not \subseteq G'} (G\_i' + x\_i')|=:b\_2$,


where in this second inequality we restrict our attention to $G\_i$'s of the second kind. The set $S\_2$ is the disjoint union of $p$ translates of its intersection with $G'$, which intersection is present on the right side of the inequality directly above. It follows that 


$|S|=|S \cap G'|+|S\_2 - G'|=b\_1 + (p-1)b\_2,$ 


while similarly we have


$|\cup\_{i=1}^k G\_i|=a\_1 + (p-1)a\_2$. 


It follows that $|S| \geq |\cup\_{i=1}^k G\_i|$, as desired.




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Finally, let me note that the inequality does not hold for non-cyclic groups. Already for $G = \mathbf{Z}\_2 \times \mathbf{Z}\_2$ the union of three distinct subgroups of $G$ of size $2$ is $G$, while it is possible to choose their cosets with the union of size $3$.


