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Jul 10, 2018 at 7:27 answer added M. Farrokhi D. G. timeline score: 7
Jul 9, 2018 at 20:17 answer added Seva timeline score: 1
Jul 9, 2018 at 14:03 comment added Dirk Intuitively I would say that you will find such a set $S$ for most groups, yes. You might be interested in this conjecture here ( en.wikipedia.org/wiki/Diameter_(group_theory) ), stating that $N(S,G)$ remains small no matter what set $S$ you take, at least as long as $G$ is simple.
Jul 9, 2018 at 12:32 comment added Johannes Hahn One example that comes to mind are the Coxeter groups. For example for $G=S_n$ with S=the neighbour-transpositions the length of the longest element (the permutation $(1,n)(2,n-1)...$) is $\frac{n(n-1)}{2}=O(n^2)$, but the order of the group is of course $n! \gg n^2$
Jul 9, 2018 at 12:32 comment added LeechLattice Computing the quantity $N(S,G)$ is hard in general. The GAP package may be useful for your purpose.
Jul 9, 2018 at 12:20 comment added LSpice I can imagine an asymptotic meaning, but how do I decide for a single group $G$ and generating set $S$ whether $\lvert S\rvert, N(S, G) \ll \lvert G\rvert$ is true?
Jul 9, 2018 at 12:18 comment added LeechLattice Isn't that the diameter of the Cayley graph?
S Jul 9, 2018 at 11:46 history suggested Ali Taghavi
I add a tag.
Jul 9, 2018 at 11:43 review Suggested edits
S Jul 9, 2018 at 11:46
Jul 9, 2018 at 11:35 review First posts
Jul 9, 2018 at 11:41
Jul 9, 2018 at 11:34 history asked Logikus CC BY-SA 4.0