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My friend Wim van Dam asked me the following question:

  • For every finite group $G$, does there exist a subset $S\subset G$ such that $\left|S\right| = O(\sqrt{\left|G\right|})$ and $S\times S = G$? Also, can we describe such an $S$ explicitly?

I believe it's not hard to show that if we choose $S$ uniformly at random, then $\left|S\right| = O(\sqrt{\left|G\right| \log \left|G\right| })$ suffices. But this still leaves the problems of giving an explicit construction and of removing the $\log \left|G\right|$ term.

Of course, if $G$ happens to have a subgroup $H$ of size $\approx \sqrt{\left|G\right|}$, then we just take the union of $H$ with a collection of coset representatives and are done. So, this suggests the possibility of a win-win analysis, where we identify the relatively rare finite groups that don't have a subgroup of the right size and handle them in a different way. Work on approximate subgroups might also be relevant.

In the likely event that this has already been solved, just a link to the paper would be great (a half hour of googling didn't succeed).

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    $\begingroup$ For a finite cyclic group of p elements, you can use 'base sqrt p', and you can extend to Cartesian products, which covers finite abelian groups and finite mostly abelian groups. Maybe the tough part is for subdirectly irreducible groups? Gerhard "Another Use For General Algebra" Paseman, 2017.05.27. $\endgroup$ Commented May 28, 2017 at 0:25
  • $\begingroup$ Yes, I agree abelian groups are easily handled (Wim pointed that out to me). So then what are examples of nonabelian groups that lack subgroups of a suitable size? $\endgroup$ Commented May 28, 2017 at 0:27
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    $\begingroup$ On second thought, we need to be more careful with the Cartesian product property, as e.g. $3\sqrt{s}3\sqrt{t}$ is not bounded above by $3\sqrt{st}$. Maybe the log term is important after all. Gerhard "Master Of The Second Guess" Paseman, 2017.05.27. $\endgroup$ Commented May 28, 2017 at 0:34
  • $\begingroup$ I would suggest seeing how dihedral groups behave. If you have an $O()$ preserving method for handling Cartesian products, then you can handle most small groups. Gerhard "Simplicity Can Be Rather Complicated" Paseman, 2017.05.27. $\endgroup$ Commented May 28, 2017 at 0:38
  • $\begingroup$ Deviating from the question scope, there are counterexamples in semi groups (nonsurjective) and in surjective magmas (binars, groupoids). It would be interesting to see if there is a surjective semigroup counterexample and even a surjective quasigroup example. Gerhard "Looks Where The Light Is" Paseman, 2017.05.27. $\endgroup$ Commented May 28, 2017 at 0:49

3 Answers 3

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It was brought to my attention by Noga Alon that my previous answer (which I keep to avoid any confusion) was in fact incorrect: the Rohrbach conjecture got solved completely by Finkelstein, Kleitman, and Leighton in 1988 ("Applying the classification theorem for finite simple groups to minimize pin count in uniform permutation architectures"), and independently by Kozma and Lev in 1992 ("Bases and decomposition numbers of finite groups").

Explicitly, there exists a subset $S\subseteq G$ of size $|S|\le(4/\sqrt 3)|G|^{1/2}$ such that every element of $G$ is representable as a product of two elements from $S$.

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This problem, first raised in 1937 by H. Rohrbach, has been considered, for instance, in the paper "On $h$-bases and $h$-decompositions of the finite solvable and alternating groups" (J. Number theory 49 (3), 1994) by Gadi Kozma and Arieh Lev. The paper contains historical remarks and further references.

The abstract of the paper reads as follows:

Let $G$ be a finite group such that every composition factor of $G$ is either cyclic or isomorphic to the alternating group on $n$ letters for some integer $n$. Then for every positive integer $h$ there is a subset $A\subseteq G$ such that $|A|\le(2h-1)|G|^{1/h}$ and $A^h=G$. The following generalization for the group $G$ also holds: For every positive integer $h$ and any nonnegative real numbers $\alpha_1,\alpha_2,\dotsc,\alpha_h$ so that $\alpha_1+\alpha_2+\dotsb+\alpha_h=1$ there are subsets $A_1,A_2,\dotsc,A_h\subseteq G$ such that $|A_1|\le|G|^{\alpha_1}$, $|A_i| \le 2|G|^{\alpha_i}$ for $2\le i\le h$ and $A_1A_2\dotsb A_h=G$. In particular, the above conclusions hold if $G$ is a finite group and either $G$ is an alternating group or $G$ is solvable.

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  • $\begingroup$ Thanks so much for the pointer, Seva! So then, should I take it that the problem for completely arbitrary $G$ is still open? $\endgroup$ Commented May 28, 2017 at 19:36
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    $\begingroup$ @ScottAaronson: looks like this, but I am not a big expert in this problem. $\endgroup$
    – Seva
    Commented May 28, 2017 at 19:39
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An update: following a lead from Seva's answer, I discovered in that their 2003 paper Communication Complexity of Simultaneous Messages, Babai, Gal, Kimmel, and Lokam (see Section 7, "Decompositions of Groups") state explicitly that what I want is an open problem. Or strictly speaking, a slight generalization to the $k$-fold rather than $2$-fold Cartesian product---but if more had been known for the $2$-fold case they would've said so. They call what I'm asking for the "Rohrbach Conjecture."

Besides discussing Kozma and Lev's progress on the Rohrbach Conjecture (mentioned in Seva's answer), Babai et al. also prove the following weaker result (their Theorem 2.17, special case relevant for this MO post):

    For every finite group $G$, there exists a subset $S\subset G$ with $\left|S\right| = O(\sqrt{\left|G\right|})$ such that $\left|S\times S\right| \ge (1-1/\sqrt{e})\left|G\right|$.

In an embarrassing postscript, Anna Gal, one of the authors of this paper, occupies the office next to mine!

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    $\begingroup$ I like the postscript... $\endgroup$
    – Seva
    Commented May 29, 2017 at 6:18
  • $\begingroup$ @ScottAaronson do you know if it is enough to solve it to finite simple groups? The formulation of Kozma's and Lev's theorem seems to hint it. $\endgroup$ Commented May 29, 2017 at 9:37

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