Timeline for Almost squared finite groups
Current License: CC BY-SA 4.0
38 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2021 at 5:56 | answer | added | kabenyuk | timeline score: 1 | |
| Apr 6, 2021 at 20:18 | comment | added | Taras Banakh | @StevenStadnicki On the other hand, $24=2^3\cdot 3$ and $48=2^4\cdot 3$ are not exceptional, but $80=2^4\cdot 5$ is exceptional. | |
| Apr 6, 2021 at 19:59 | comment | added | Steven Stadnicki | The two orders, in addition to being $n^2-1$, have very similar 'shape' — $2^4\cdot5$ and $2^5\cdot 7$ — so if there isn't a group of order 224 perhaps there's an explanation lurking there too. | |
| Apr 6, 2021 at 19:42 | comment | added | Taras Banakh | @StevenStadnicki An hour ago I have written exactly the same question to Kabenyuk (who can compute all those large groups without GAP, because GAP is too slow). I do not know what happens with 224 and also why there are no examples of almost squared groups of order 80 (how to prove this without brute force). | |
| Apr 6, 2021 at 18:25 | comment | added | Steven Stadnicki | Random observation: the examples here (and in the answers) all have $|G|=p^2-1$ for prime $p$; the non-example $|G|=80$ corresponds to a squaring set of size $9$. It'd be fascinating to know if there are any examples with $|G|=224$. | |
| Apr 6, 2021 at 15:47 | answer | added | Taras Banakh | timeline score: 0 | |
| Apr 4, 2021 at 17:20 | answer | added | kabenyuk | timeline score: 2 | |
| Mar 21, 2021 at 14:49 | answer | added | kabenyuk | timeline score: 5 | |
| Mar 5, 2021 at 19:50 | history | edited | Taras Banakh | CC BY-SA 4.0 |
edited body
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| Mar 5, 2021 at 19:44 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added info about 120-element groups
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| Feb 26, 2021 at 11:07 | comment | added | Geoff Robinson | By the way, finding an almost square root where the identity shows up twice in $A.A$ (and every other element shows up once) necessitates finding a $(0,1)$-matrix $M$ such that $M^{2} = I + J$ where $J$ is the all-ones matrix. This is reminiscent of the problem of finding an incidence matrix for a projective plane, which is notoriously difficult in exceptional dimensions ( ie dimension $n^{2}+n+1$ with $n$ not a prime power). | |
| Feb 25, 2021 at 21:36 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Reworded a Remark
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| Feb 23, 2021 at 22:17 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added info about groups of order 80.
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| Feb 22, 2021 at 16:31 | history | edited | Taras Banakh | CC BY-SA 4.0 |
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| Feb 22, 2021 at 16:03 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added Problem 3'
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| Feb 22, 2021 at 15:38 | history | edited | Taras Banakh | CC BY-SA 4.0 |
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| Feb 21, 2021 at 22:40 | comment | added | Taras Banakh | @LSpice You asked about the almost square root of $S_4$: the set of permutations $A={ (12), (123), (13)(24), (1324), (243) }$ is such a root (a personal communication by V.Gavrylkiv). | |
| Feb 21, 2021 at 22:09 | comment | added | Taras Banakh | @LSpice This was a private communication from Ravsky and it concerned squared groups but the idea worked also for almost squared groups. In my partial answer (below) this Ravsky's observation is improved just a bit. | |
| Feb 21, 2021 at 22:03 | comment | added | LSpice | Where does Alex Ravsky's observation about the centre appear? | |
| Feb 21, 2021 at 14:11 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Corrected SL to GL
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| Feb 21, 2021 at 13:13 | answer | added | Geoff Robinson | timeline score: 8 | |
| Feb 21, 2021 at 13:01 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added some info.
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| Feb 21, 2021 at 12:19 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added one example
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| Feb 21, 2021 at 11:27 | answer | added | Taras Banakh | timeline score: 6 | |
| Feb 21, 2021 at 8:21 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added one more problem.
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| Feb 21, 2021 at 7:27 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Added a subquestion to Problem 4.
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| Feb 21, 2021 at 7:25 | comment | added | Taras Banakh | @LSpice I have added your question about order of central elements to Problem 4. | |
| Feb 21, 2021 at 6:46 | comment | added | Taras Banakh | @LSpice For the almost squared groups $D_8$ and $S_4$ the center indeed has exponent 2. GAP calculations show that $S_4$ has 96 almost roots. The simplest is $\{ a, b, c, ac, bc \}$ where $a,b,c$ are elements of $S_4$ such that $a^2=b^3=c^2=1$, $ba=ab^2$, $(aca)^2=1$, $cb=baca$... GAP shows some strange presentation of $S_4$ (if the group (24,12) in GAP is indeed $S_4$). Maybe I will try to find the almost square root of $S_4$ by hands (using some more convenient presentations). | |
| Feb 21, 2021 at 5:26 | comment | added | LSpice | Trivial observation: if $A$ is an almost square root of $G$, then so is $z A$ for every $z \in \operatorname Z(G)$; so perhaps it makes sense to ask in Problem 4 if some central translate of $A$ satisfies your condition—unless you are making the additional hypothesis that the centre of an almost squared group has exponent $2$? (One could perhaps also ask about how many almost square roots there are; surely it would be too much to expect that they are all central translates of one another.) | |
| Feb 21, 2021 at 5:16 | comment | added | LSpice | What is an almost square root of $\mathrm S_4$? | |
| Feb 21, 2021 at 4:46 | comment | added | Taras Banakh | @LSpice Right. Thank you. I corrected to "every". | |
| Feb 21, 2021 at 4:44 | history | edited | Taras Banakh | CC BY-SA 4.0 |
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| Feb 20, 2021 at 21:06 | comment | added | Will Sawin | which by orthogonality of characters implies that $A$ is reasonably evenly distributed among the conjugacy classes of $G$ - if I calculated right that the sum over conjugacy classes $C$ of $( | A \cap C| - |A| |C| /|G|)^2 / |C|$ is at most $(n^2-2)/(n^2-1)$, but this doesn't seem strong enough to lead to a classification. | |
| Feb 20, 2021 at 21:05 | comment | added | Will Sawin | I tried to extend the arguments in the representation ring. The eigenvalues of $\sum_{a \in A} a$ on every nontrivial irreducible representation of $a$ are roots of unity, since the eigenvalues of its square are roots of unity on these representations. Thus $|\operatorname{tr} (\sum_{a \in A} a, \dim V)| \leq \dim V$ for all nontrivial representations, so $\sum_v |\operatorname{tr} (\sum_{a \in A} a, \dim V)| ^2 \leq |A|^2 + |G|-1= 2 (n^2-1)$ | |
| Feb 20, 2021 at 19:53 | history | edited | Taras Banakh | CC BY-SA 4.0 |
Changed title
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| Feb 20, 2021 at 18:41 | history | edited | Taras Banakh | CC BY-SA 4.0 |
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| Feb 20, 2021 at 18:36 | history | edited | Taras Banakh | CC BY-SA 4.0 |
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| Feb 20, 2021 at 18:20 | history | asked | Taras Banakh | CC BY-SA 4.0 |