Timeline for Correspondence between even and odd permutations in $S_5$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 30 at 8:20 | comment | added | Gerry Myerson | Similar question from same user, mathoverflow.net/questions/474181/… (also one on math.stack, but I'll let OP put in the link). | |
| Jun 29 at 18:38 | comment | added | Mikel Martinez Puente | 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? | |
| Jun 29 at 17:27 | comment | added | მამუკა ჯიბლაძე | Some argument like this: each element of $S_5$ is either in $A_5$ or $a\cdot(12)$ with $a\in A_5$, so if you have a sequence $a_1(12)^{\varepsilon_1},...,a_k(12)^{\varepsilon_k}$ with each $\varepsilon_i$ either $0$ or $1$ then its product is $a_1'\cdots a_k'(12)^\varepsilon$ where each $a_i'$ is either $a_i$ or $(12)a_i(12)$ and $\varepsilon$ is the sum of all $\varepsilon_i$ modulo $2$. If $k\geqslant10$ then for $\varepsilon=0$ you can find a subsequence in $a_1',\cdots,a_k'$ with product $1$, while for $\varepsilon=1$ you can find it in $a_1',...,a_{k-1}'$ | |
| Jun 29 at 16:28 | comment | added | Mikel Martinez Puente | 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. | |
| Jun 29 at 16:27 | comment | added | მამუკა ჯიბლაძე | @MaartenHavinga $(12)(34)(45)(13)(24)(35)(14)(25)$ is trivial :) | |
| Jun 29 at 16:10 | comment | added | Mikel Martinez Puente | Yes, I agree @მამუკაჯიბლაძე. You are absolutely right, in fact there is a formula so that for a normal subgroup H we know that $D(H) + D(G/H) -1 \leq D(G)$. In this particular case, $D(A_5)=9$ and $D(S_5/A_5)=D(C_2)=2$, hence as you said $D(S_5)$ is at least 10. Any maximal one-product free sequence of length 8 in $A_5$ along with an odd permutation is one-product free in $S_5$. However, how could I prove that it is indeed $D(S_5)=10$?? I don't see how it is that easy to prove it. I'd need to prove that any sequence of length 11 is one-product, but how? Thanks a lor for your help! | |
| Jun 29 at 16:03 | comment | added | Maarten Havinga | I think a sequence of the following $9$ permutations is no one-product sequence: $12$, $23$, $34$, $45$, $13$, $24$, $35$, $14$, $25$ | |
| Jun 29 at 14:43 | comment | added | მამუკა ჯიბლაძე | 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 $10$. | |
| S Jun 29 at 13:41 | review | First questions | |||
| Jun 29 at 14:05 | |||||
| S Jun 29 at 13:41 | history | asked | Mikel Martinez Puente | CC BY-SA 4.0 |