35
$\begingroup$

Suppose G is a finite group. Define $\tau(G)$ as the minimal number, such that $\forall X \subset G$ if $|X| > \tau(G)$, then $XXX = \langle X \rangle$. What is $\tau(A_n)$?

Similar problems for some different classes of groups are already answered:

1) $\tau(\mathbb{Z}_n) = \lceil \frac{n}{3} \rceil + 1$ (this is a number-theoretic fact proved via arithmetic progressions)

2) Gowers, Nikolov and Pyber proved the fact that $\tau(SL_n(\mathbb{Z}_p)) = 2|SL_n(\mathbb{Z}_p)|^{1-\frac{1}{3(n+1)}}$ (this fact is proved with linear algebra)

However, I have never seen anything like that for $A_n$. It will be interesting to know if there is something...

$\endgroup$
  • $\begingroup$ I don't even understand the question. What do you mean when you say $XXX=\langle{X\rangle}$? $\endgroup$ – crskhr Aug 6 '18 at 13:44
  • $\begingroup$ @crskhr, $XXX$ stands for the group subset product: $XXX = \{abc| a, b, c \in X\}$; $\langle X \rangle$ stands for group subset closure: $\langle X \rangle$ is the minimal subgroup that contains $X$. $\endgroup$ – Yanior Weg Aug 6 '18 at 13:52
  • $\begingroup$ Does anyone have the results for small values of $n$? $\endgroup$ – mathworker21 Aug 21 '18 at 8:57
  • $\begingroup$ I expect that its pretty doable for 1 through 4. But then, for 5 it seems quite hard. $\endgroup$ – user24142 Aug 23 '18 at 1:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.