Timeline for Bound on the order of a finite group generated by elements $a$ and $b$ of order 2 and $n \geq 3$ such that the sum of the images of $a$, $b$ and $b^{-1}$ under any ordinary representation has only rational eigenvalues
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2013 at 4:01 | answer | added | Alireza Abdollahi | timeline score: 2 | |
| Jun 9, 2013 at 7:34 | answer | added | katie | timeline score: 0 | |
| Jun 7, 2013 at 9:10 | comment | added | S. Carnahan♦ | @katie, thank you. That is the sort of information that can be helpful when inserted into the text of the question. | |
| Jun 7, 2013 at 6:34 | comment | added | katie | I don't know what you mean by background and motivation. I was came across this problem when I was studying Q-groups in the book "Structure and Representations of Q-Groups" by Dennis Kletzing. I needed to know, what can happen if we put more restriction. Is that produce an infinite class of groups like Q-groups, or not. This was the most obvious extension I could have think of. Characters turning away many other information about representations. Problem get a lot challenging when we look representation. I know the answer for this, if I was just taking characters rather than representations. | |
| Jun 7, 2013 at 3:31 | comment | added | Yemon Choi | @katie: seeing as your question has already attracted votes to close, I think you should edit your question to provide some of the background and motivation. | |
| Jun 6, 2013 at 20:33 | comment | added | katie | Update! This result is valid for all groups of order less than 460 up to now (verified by computers). | |
| Jun 6, 2013 at 20:28 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Fixed order of b also in the title.
|
| Jun 6, 2013 at 19:22 | comment | added | katie | $D_6, D_8, D_{12}, A_4, S_4, A_4\times \mathbb{Z}_2, D_6\times \mathbb{Z}_4, D_8\times \mathbb{Z}_3$ (where $D_n$ is the dihedral group of order $n$) are the only groups I have found satisfying the conditions stated in the problem. | |
| Jun 6, 2013 at 19:09 | comment | added | katie | @Stefan Kohl, b can be anything other than an involution (and obviously identity) so I edited again. thanks | |
| Jun 6, 2013 at 19:07 | history | edited | katie | CC BY-SA 3.0 |
added 14 characters in body; added 12 characters in body
|
| Jun 6, 2013 at 18:53 | comment | added | Derek Holt | @stefan: she did not specify that the order of $b$ is 3, only that it is greater than 2. | |
| Jun 6, 2013 at 17:44 | comment | added | Stefan Kohl♦ | @katie: I have tried to formulate your question in a more appropriate way -- please check! | |
| Jun 6, 2013 at 17:43 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Tried to formulate the question in a more appropriate way.
|
| Jun 6, 2013 at 11:50 | comment | added | Lior Silberman | Need specifically $a,b$ not to commute. Otherwise let $b\in C_10$ be of order $5$ and take the character such that $\chi(ab) = \exp(2\pi i/10)$. Then the given eigenvalue is (up to rational arithmetic) $\cos(2\pi/5)$ which is irrational. | |
| Jun 6, 2013 at 10:23 | comment | added | Derek Holt | I assumed that the addition was in the group algebra. The condition holds with $a$ and $b$ generating $S_4$ with $b$ of order 3. | |
| Jun 6, 2013 at 8:43 | comment | added | S. Carnahan♦ | Do you mean $\tau(a) + \tau(b) + \tau(b^{-1})$? I don't know what you mean by addition in a non-abelian group (although addition in the group ring would yield the above formula). | |
| Jun 6, 2013 at 8:39 | comment | added | user6976 | @Derek: I think we of course can assume that $G$ is generated by $a,b$, and $25$ is equivalent to $24$. | |
| Jun 6, 2013 at 8:35 | comment | added | Derek Holt | Why 25? $|\langle a,b \rangle|$ must be even, so why not 24? Why are assuming that $G$ is non-abelian? Can't we just assume that $G = \langle a,b \rangle$? | |
| Jun 6, 2013 at 5:46 | history | edited | katie | CC BY-SA 3.0 |
added 7 characters in body
|
| Jun 6, 2013 at 4:47 | comment | added | user6976 | @Misha: $G$ is probably assumed finite. I too think it is a homework, so I voted to close. To keep it open, the OP needs to provide some motivation. | |
| Jun 6, 2013 at 4:29 | comment | added | Misha | @Yemon: 25 probably came from some incompletely reproduced homework problem, as one can take $G$ to be an infinite finitely generated simple group, with $a$ and $b$ generating a subgroup isomorphic to, say $Z_2 *Z_3$. | |
| Jun 6, 2013 at 4:11 | comment | added | Yemon Choi | Where do you get the number 25 from? | |
| Jun 6, 2013 at 3:54 | history | asked | katie | CC BY-SA 3.0 |