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I am working on computing the Davenport constant $D(G)$ for $S_n$ and $A_n$, i.e., the minimal number $d$ such that every sequence (multiset) of $d$ elements contains some subsequence giving identity as a product, in some order. The small Davenport constant $d(G)$ is the maximal length of a one-free sequence, i.e., $d(G)+1=D(G)$.

In particular, for $A_n$ I want to get a lower bound by computing a general family of one-product free sequences for each $n$. For $A_3$ we know $d(A_3)=2$ with maximal one-product free sequence {(123),(123)}. For $A_4$, we have $d(A_4)=4$ with maximal one product free sequence {(123), (123), (134), (134)} and for $A_5$ we know $d(A_5)=8$ with maximal one-product free {(12345) x 4, (12453) x4}. No other value is known. Moreover, in these three cases $d(A_n)=2^{n-2}$, is this coincidence or could be proved to be some lower/upper bound?

I am looking for patterns to get maximal one-product free sequences for $A_n$ in general and $n=6$ and $n=7$ in particular. One of them, looking at the examples, could be for $n$ odd getting $n-1$ copies of some set of $n$-cycles. For $n$ even, one possible idea could be taking maximal sequence of $A_{n-1}$ and taking another copy conjugating all elements by some $g$ not fixing $n$.

Does anyone come up with some idea to get one-free sequences in $A_n$? Apart from the obvious one of taking $m-1$ times an element of order $m$. Similarly, any idea for getting upper bounds for $D(A_n)$ are also very welcome and helpful ;)

Thank you very much in advance!

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  • $\begingroup$ Minor point: the statements of known values of $D$ look to actually be known values of $d$. More usefully, the Sage code I posted earlier elsewhere can't find a set of three elements of $A_6$ which give a one-product-free multiset of 12 elements. The best is 11 with e.g. $(1,5,4,3,2)^3, (1,6,5,2,3)^4, (1,3,6,2,4)^4$. That rather spoils the idea of finding four $5$-cycles to give a multiset $c_1^4, c_2^4, c_3^4, c_4^4$. $\endgroup$
    – Peter Taylor
    Commented Jul 11 at 15:37
  • $\begingroup$ @PeterTaylor Thanks a lot once again! I am definitely going to use your Sage code from now on since it is very useful for the calculations I need! I really appreciate your help ;) $\endgroup$
    – Mikel Martinez Puente
    Commented Jul 12 at 14:30
  • $\begingroup$ @PeterTaylor If I understood well, the code runs thrugh all possible subsequence products, with all possible ordering right? Because that is so important point, for a fixed subsequence taking the different orders for the product into account! $\endgroup$
    – Mikel Martinez Puente
    Commented Jul 12 at 15:02
  • $\begingroup$ Yes, that's the bottleneck. I can't think of a way to represent the partial products in such a way that you can add one element and generate all of the partial products with it in arbitrary positions, so it recalculates a lot. $\endgroup$
    – Peter Taylor
    Commented Jul 13 at 8:44
  • $\begingroup$ @PeterTaylor Yes, I agree! I asked you this since there is some contradiction I do not understand why it fails, maybe you could help me or my computer does not compute it correctly. As you previously mentioned in one comment above there is some one-free sequence of length 11 in $A_6$ with three distict elements, thus adding any odd permutation in $S_6$ it should be one-free sequence of length 12 in $S_6$ with 4 distinct elements. However, in the Sage code it seems not to exist such a sequence, when I look for it. Do you know why could that be? $\endgroup$
    – Mikel Martinez Puente
    Commented Jul 13 at 12:38

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