If $A \subseteq \mathbb{Z}/p\mathbb{Z}$, and $|A| > \frac{p}{3}$, then are there any nontrivial lower bounds for $|AA|$?

Question

If $A \subseteq \mathbb{Z}/p\mathbb{Z}$, and $|A| > \frac{p}{3}$, then are there any nontrivial lower bounds for $|AA|$?
Where $AA=\{a_{1} \cdot a_2:a_1,a_2 \in A\}$, and $p$ is prime.

Writing out manually some multiplication results for some low $p$ and big enough $A$ gives the impression that if $|A| > \frac{p}{3}$, or so, then it should be the case that $|AA| \geq p-1$.

It seems to me that I'm missing some basic knowledge about the multiplication in the $\mathbb{Z}/p\mathbb{Z}$, since there are clear patterns in the respective multiplication table for $AA$, as above, like the fact that it is symmetric over both diagonals, or that each row and column consist of the different numbers, etc.

Answer 1

You seem to be interested in analogues of the Cauchy-Davenport theorem for the multiplicative group. Alas, I'm mostly ignorant about what else is known for groups other than $\Bbb{Z}_p$.

I do want to make a few somewhat trivial observations showing that $p/3$ is not high enough:

1. If $A$ consists of only quadratic residues modulo $p$, then so will $AA$. There are $(p-1)/2$ non-zero quadratic residues, so it is possible for $A$ to contain nearly half of the elements without $AA$ filling in the whole lot (add $0$ to $A$, if you feel like it). In the context of general Cauchy-Davenport problem this is an example of the use of a (large) subgroup.
2. Modulo a prime $p$ there always exists a primitive root $g$ with the property that its powers $1,g,g^2,\ldots,g^{p-2},g^{p-1}=1$ cycle over the non-zero residue classes modulo $p$. If we then fix an integer $m<p/2$, and choose $$A=\{1,g,g^2,\ldots,g^{m-1}\}$$ we immediately see that $|A|=m$ and $$AA=\{1,g,g^2,\ldots,g^{2m-2}\},$$ a set with $2m-1<p-1$ elements. This is "the equality case" in Cauchy-Davenport's inequality.

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The above observations do suggest that (assuming $0\notin A$, so we are working in a group):

• Assuming $|A|>p/3$ we have $|AA|\ge\min\{(p-1)/2, 2|A|-1\}$. For relatively large subsets $A$ like here, the subgroup of quadratic residues seems to be the only obstruction to the general mechanism with $2|A|-1$ as the lower bound for $|AA|$.
• If $p\equiv1\pmod3$, then by selecting $A$ to be the subgroup of $(p-1)/3$ cubic residues we get $AA=A$, so it is possible to have $$|A|=|AA|=(p-1)/3.$$