Timeline for Is the small Davenport constant for $S_n$, $d(S_n)=n(n-1)/2$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 11 at 13:07 | comment | added | Noam D. Elkies | @MikelMartinezPuente I don't know, sorry. The lower bound required only coming up with one example; upper bounds require actually proving something in all cases . . . | |
| Jul 11 at 10:53 | comment | added | Mikel Martinez Puente | Thanks a lot! That makes sense. Do you think there is also such an easy upper bound for $A_n$ or $S_n$? The unique idea I come up with is that $D(S_n) \leq n*D(S_{n-1})$ and of course $D(S_n) \leq 2*D(A_n)$, but still is need to know previous values... | |
| Jul 10 at 19:11 | comment | added | Noam D. Elkies | You're welcome. For $A_n$, I don't know whether the maximal order has another name; I suppose you can call it $g_+(n)$. It must be at least as large as $g(n)/2$ (if the maximal order in $S_n$ is attained by an odd permutation then consider its square), and thus also increases faster than any polynomial in $n$. | |
| Jul 10 at 18:58 | comment | added | Mikel Martinez Puente | Thank you so much for the contribution! It is weird I dind't realise to check first such a natural lower bound of $D(S_n)$. By the way, do you know if there is an equivalent function as the Landau function for alternating groups? Thanks :) | |
| Jul 10 at 18:22 | vote | accept | Mikel Martinez Puente | ||
| Jul 10 at 11:34 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
added 4 characters in body
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| Jul 10 at 2:05 | history | answered | Noam D. Elkies | CC BY-SA 4.0 |