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I am working on the Davenport constant for symmetric groups, $D(G)$ , which is the minimal number $d$ such that every sequence of $d$ elements in the group G is one-product sequence, i.e, we can always get identity as a product of some of these elements. Domokos, Cziszter and Szöllősi computed by GAP this constant for all non-abelian groups of order at most 42 and also for $A_5$ , which is $D(A_5)=9$ , i.e., with any sequence of 9 even elements we get identity as product of some of them. Moreover, we also know that maximal one-product free sequences of length 8 consist only of 5-cycles.

Now, I am trying to prove that any sequence of 9 odd permutations in $S_5$ is also one-product sequence and we can always get identity. I am struggling a lot proving this with the computer due to huge computational cost. Is there any correspondence or relation between odd and even permutations so that $D(A_5)=9$ implies this claim? It would be very nice and helpful if you suggest me all the ideas you come up with. Thanks a lot in advance!

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    $\begingroup$ If you have a one-product free sequence in a subgroup $H$ of a group $G$, and you add to it any $g\in G\setminus H$, you obtain a one-product free sequence in $G$, so $D(S_5)$ is at least $10$. I believe it is easy to see that it is exactly $10$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 29 at 14:43
  • $\begingroup$ I think a sequence of the following $9$ permutations is no one-product sequence: $12$, $23$, $34$, $45$, $13$, $24$, $35$, $14$, $25$ $\endgroup$
    – Maarten Havinga
    Commented Jun 29 at 16:03
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    $\begingroup$ @MaartenHavinga $(12)(34)(45)(13)(24)(35)(14)(25)$ is trivial :) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 29 at 16:27
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    $\begingroup$ Hi @MaartenHavinga, thanks for your comment! However, I think that (34)*(35)*(24)*(13)*(14)*(45)*(23)*(12) = $1_{S_5}$ (with multiplication from right to left). Thus, it doesn't work. $\endgroup$
    – Mikel Martinez Puente
    Commented Jun 29 at 16:28
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    $\begingroup$ Thanks @მამუკაჯიბლაძე for your help! Nice approach, I thought a similar strategy few days ago but I got stuck. I still don't see it 100% clear how getting a subsequence in $a_1^{'}, ..., a_k^{'}$ where all elements are in $A_5$ concludes the proof. Shouldn't we take into account the order of multiplication of these elements? Imagine we have $a_2{'}*a_6^{'}*a_3^{'}*a_1^{'}=1_{S_5}$. with $a_i=a_i^{'}$ for $i=1,2,3$ and $a_4^{'}=(12)a_i(12)$... And also taking products in distinct order for your given sequence above in $S_5$ don't we get distinct $a_i^{'}$ in each case? $\endgroup$
    – Mikel Martinez Puente
    Commented Jun 29 at 18:38

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