# Group as product of subsets

There's a fairly simple result that states that

for a finite group $G$ and two subsets $A, B$ with $$|A| + |B| > |G|,$$ any $g \in G$ has a representation $g = a*b$ with $a \in A$, $b \in B$.

To prove this, just consider that the sets $A$ and $gB^{-1} = \{gb^{-1} : b \in B\}$ can't be disjoint.

This immediately implies that every number modulo a prime $p$ is the sum of two squares:

let $G = \mathbb{Z}/p\mathbb{Z}$ and $A = B = \{x^2 : x \in \mathbb{Z}/p\mathbb{Z}\}$.

My question: can this result be extended meaningfully to products of three or more subsets? I'm interested in representations mod $p$ as sums of cubes and higher powers.

The most obvious (to me) generalization

for a finite group $G$ and three subsets $A, B, C$ with $$|A| + |B| + |C| > |G|,$$ any $g \in G$ has a representation $g = a*b*c$ with $a \in A$, $b \in B$, $c \in C$.

is clearly false: just take $G = \mathbb{Z}/2\mathbb{Z}$ and $A = B = C = 0$.

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I have the distinct impression that I've seen something along these lines somewhere else in this site before... – Arturo Magidin Jun 19 '12 at 1:37

For every $k$, every number modulo a prime $p$ is a sum of two $k$th powers, if $p$ is large enough (as a function of $k$). There is a long list of names associated with this result, and with either finding simpler proofs or with getting better bounds for $p$. Some of the history is given in Ribenboim's 13 Lectures on Fermat's Last Theorem (to be precise, in Lecture XII, Fermat's Congreunce). Ribenboim is only interested in expressing a $k$th power as a sum of two $k$th powers, but many of the arguments make no essential use of the assumption that the sum is to be a $k$th power.

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