Let $G$ be a finite group. Define $\tau(G)$ as the minimal number, such that $\forall X \subset G$ if $|X| > \tau(G)$, then $XXX = \langle X \rangle$.
Is there some sort of formula for $\tau(S\_n)$, for the symmetric group $S\_n$?


Here $XXX$ stands for $\{abc| a, b, c \in X\}$.


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


1) $\tau(\mathbb{C}\_n) = \lceil \frac{n}{3} \rceil + 1$, where $\mathbb{C}\_n$ is cyclic of order $n$;


2) Gowers, Nikolov and Pyber proved the fact that $\tau(\mathrm{SL}(n, p)) \leq 2|\mathrm{SL}(n, p)|^{1-\frac{1}{3(n+1)}}$ for prime $p$.


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



Is there some sort of formula for $\tau(S_n)$?
Here is a lower bound: let $H$ be a subgroup of $S\_{n}$ containing $(12)$ and let $\sigma$ be the $n$-cycle $(12 \ldots n)$. Let $X = H \cup \{\sigma \}$ and note that $\langle X \rangle = S\_{n}$ since we already have $S\_{n} = \langle (12),\sigma \rangle.$ 


Suppose that we have chosen $H$ so that $|X| > \tau(S\_{n}).$ Then we must have $S\_{n} = XXX$. But $XXX = H \cup H\sigma \cup H\sigma H \cup H\sigma^{2} \cup \sigma H \cup \sigma H \sigma \cup \sigma^{2}H \cup \sigma^{3}$ has cardinality at most $|H|^{2} + 6|H| + 1,$ so we must have $|H| + 3 > \sqrt{n!}.$ 


Hence we may choose $k$ minimal so that $k! \geq \tau(S\_{n}).$ Then taking $H$ to be the natural copy of $S\_{k}$ inside $S\_{n}$, we must have $(k! + 3)^{2} > n!$.
 Then we must have $\tau(S\_{n}) > \frac{\sqrt{n!} - 3}{k} \geq \frac{\sqrt{n!} - 3}{n}.$


