Timeline for Is a group determined by the number of ways its elements multiply to the identity under some ordering?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2020 at 4:17 | comment | added | R. van Dobben de Bruyn | An example of what can happen is $(G,\mu_1) = H_+ \times H_-$ and $(G,\mu_2) = H_+ \times H_-^{\operatorname{op}}$ (which is abstractly isomorphic using $(g_+,g_-) \mapsto (g_+,g_-^{-1})$). A possible stronger question could be if it is always of this form. A more moderate strengthening is whether there exists an isomorphism that respects the $f_n$. | |
| Jun 18, 2020 at 4:14 | comment | added | R. van Dobben de Bruyn | Some more things you can recover from the $f_n$: orders of elements, subgroups, when two elements commute, centralisers, centre, conjugacy classes, normal subgroups, .... I don't know enough group theory to know if a subset of these properties recovers the group (maybe assuming $G$ is finite), but this is all pretty strong evidence that at least $(G,\mu_1)$ and $(G,\mu_2)$ are abstractly isomorphic. | |
| Jun 17, 2020 at 14:38 | comment | added | Will Sawin | It's equivalent to know, for each $g$ and $h$, the multiset $\{gh, hg\}$. (And knowing that multiset is equivalent to knowing the set.) You'e shown that counting gives us this information. Conversely, if we have $n$ elements, and count all the ways to choose two of them, multiply them in any order, then choose another, and multiply it in any order, and so on, until all are chosen, we will count each ordering the same number $2^{n-2}$ of times. So to recover your information, we can sum over the possible pairwise products and then divide by $2^{n-2}$. | |
| Jun 17, 2020 at 13:58 | history | edited | YCor |
edited tags
|
|
| Jun 17, 2020 at 13:52 | history | edited | Chris H | CC BY-SA 4.0 |
title change
|
| Jun 17, 2020 at 13:51 | comment | added | YCor | Actually what you count is more precise, since you retain the number of permutations. | |
| Jun 17, 2020 at 13:49 | comment | added | Chris H | Edited for clarity, I wasn't quite sure how to phrase the title, if you can think of better wording that would be very helpful. | |
| Jun 17, 2020 at 13:47 | history | edited | Chris H | CC BY-SA 4.0 |
clarifying the question
|
| Jun 17, 2020 at 13:32 | comment | added | YCor | In the title "multiset" sounds weird, as for a multiset in a non-abelian group, to "multiply to 1" doesn't make sense. The first question is a bit unclear: is $n$ fixed? it seems not, after reading the sequel. Also in the second question you mean "unique up to isomorphism", or maybe up to permutation of $X$ preserving $f$. | |
| Jun 17, 2020 at 13:27 | history | asked | Chris H | CC BY-SA 4.0 |