There seems to be some combinatorial fact that every subset $A$ of $G=(\mathbb{Z}/p)^{\times n}$ of cardinality $\frac{p^n-1}{p-1}+1$ containing $\vec{0}$ satisfies $(p-1)A=G$. ($p$ is a prime number.)
Is that true and known in the literature?
I would appreciate a reference or a proof.
Thanks!



Subsets of $(\mathbb{Z}/p)^{\times n}$
There is an alternative elementary way to prove this and see where the number $\frac{p^n-1}{p-1}+1$ comes from.


**Lemma:** If $|A|\geq \frac{p^n-1}{p-1}+1$ then every point in $\mathbb F\_p^n$ lies in a line that passes from two distinct points in $A$.


Proof: If this was false then there would be a point $P\notin A$ such that every line through $P$ contains at most one point of $A$. However the number of distinct lines passing through $P$ is $\frac{p^n-1}{p-1}$, the size of the corresponding projective space. This gives a contradiction, since it implies $|A|\le \frac{p^n-1}{p-1}$.


Returning to the statement in the OP, starting with an arbitrary point $P\in \mathbb F\_p^n$ the lemma tells us that there are points $M,N\in A$ such that $-P$ lies in the line passing through $M$ and $N$. Now consider the line consisting of the points $kM+(p-1-k)N$, for $0\le k\le p-1$. This line contains $(p-1)M$ and $(p-1)N$, therefore it also contains $(p-1)(-P)=P$. This means that $P=kM+(p-1-k)N$ for some $k$, which is in $(p-1)A$, as desired.


