Timeline for Finite groups with integral character table
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Aug 18, 2022 at 4:33 | vote | accept | Sebastien Palcoux | ||
| Aug 17, 2022 at 21:39 | comment | added | Sebastien Palcoux | oeis.org/A064527 Numbers k such that there exists a finite group G of order k such that all entries in its character table are integers. | |
| Aug 17, 2022 at 14:24 | history | became hot network question | |||
| S Aug 17, 2022 at 14:04 | history | suggested | kabenyuk | CC BY-SA 4.0 |
improved formatting
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| Aug 17, 2022 at 13:56 | review | Suggested edits | |||
| S Aug 17, 2022 at 14:04 | |||||
| Aug 17, 2022 at 9:44 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
remove about perfect group, covered by simple example PSp(6,2) metioned in comment
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| Aug 17, 2022 at 9:32 | comment | added | Sebastien Palcoux | My first guess is false, see Corollary B.1 in On finite rational groups and related topics by Feit and Seitz: Let $G$ be a noncyclic simple group. Then $G$ has an integral character table iff $G =Sp_6(2)$ or $O_8^+(2)’$. | |
| Aug 17, 2022 at 9:16 | comment | added | Sebastien Palcoux | @YCor Yes all the Weyl groups are $\mathbb{Q}$-groups, according to the third line of the introduction of the book of D. Kletzing (cited in Alex answer). The Weyl group of $E_6$ also has a nonabelian simple normal subgroup that is not an alternating group. en.wikipedia.org/wiki/E6_(mathematics)#Weyl_group | |
| Aug 17, 2022 at 9:10 | answer | added | Geoff Robinson | timeline score: 7 | |
| Aug 17, 2022 at 8:35 | comment | added | Sebastien Palcoux | @YCor: Then the order of a $\mathbb{Q}$-group must be in oeis.org/A124240 | |
| Aug 17, 2022 at 8:34 | comment | added | Sebastien Palcoux | @DerekHolt: ok, done! | |
| Aug 17, 2022 at 8:33 | history | edited | Sebastien Palcoux | CC BY-SA 4.0 |
edit suggested by Derek Holt in comment
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| Aug 17, 2022 at 8:23 | comment | added | Derek Holt | You should be aware that the GAP structure description of a group does not always identify a group up to isomorphism. It would be more informative to print the numbers of the groups of each order. | |
| Aug 17, 2022 at 8:23 | comment | added | YCor | Answer to your third question: yes $1$ is the only example of odd order. If $G$ has this property and has odd order $\ge 1$, let $C$ be a cyclic subgroup of prime order $p$. Then the normalizer of $C$ is transitive on $C-\{1\}$, so its order is divisible by $p-1$, hence is even. (This also proves that whenever a prime $p$ divides $|G|$, then $p-1$ divides $|G|$.) | |
| Aug 17, 2022 at 8:20 | comment | added | YCor | I think Weyl groups are known to be examples. Using the Weyl group of $E_8$ you get an example with a nonabelian simple normal subgroup that is not an alternating group. | |
| Aug 17, 2022 at 7:11 | answer | added | Alex B. | timeline score: 16 | |
| Aug 17, 2022 at 6:24 | history | asked | Sebastien Palcoux | CC BY-SA 4.0 |