Timeline for Finite groups such that every irrep can be induced from trivial irrep of a subgroup ?
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2021 at 18:28 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Clarified introduction.
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| Nov 13, 2016 at 16:17 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
cleaned up typos
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| Sep 6, 2012 at 16:54 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Changed $|G|$ to $|K|$.
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| Sep 6, 2012 at 16:53 | comment | added | Geoff Robinson | @Alexander: Of course, it should be $|K| \equiv 1$ (mod $|H|$). I will correct. Thanks! | |
| Sep 6, 2012 at 15:49 | comment | added | Alexander Chervov | @Geoff you write: " |G|≡1 (mod |H|) " but order of subgroup divides order of group, so |G|≡0 mod |H|... I feel it is stupid quest, but still... | |
| Sep 5, 2012 at 14:22 | vote | accept | Alexander Chervov | ||
| Sep 4, 2012 at 21:53 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
added 751 characters in body
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| Sep 4, 2012 at 20:56 | comment | added | Frieder Ladisch | Perhaps it is a good idea to edit this answer so that it contains the result of David's comment in the first paragraph? Your answer contains the proof anyway, and this characterization is, imho, the definitive answer to this question. For example, from the known properties of Frobenius complements it follows at once that lots of groups have the property of the question (all simple groups, all perfect groups but one, all groups containing a non-cyclic group of order $pq$...). | |
| Sep 4, 2012 at 19:57 | comment | added | Geoff Robinson | Yes, see my comment below David's for the first question. Yes, my answer is compatible with Will's but a little more precise. | |
| Sep 4, 2012 at 7:44 | comment | added | Alexander Chervov | @Geoff Robinson let me thank you again, I am traveling now, so cannot fully absorb your answer. May I ask some brief questions. 1) Is it correct that "The Frobenius complements are the only groups which fail to have this property" David wrote this with "?" it seems to me his argument is correct, but I'm not expert so not quite sure. 2) It seems to me that yours answer is comptible with Will Sawin's, is it correct ? I mean Will wrote "if there is NON-cyclic abelian subgroup - than property holds" you wrote that all Sylow subgroups are abelian cyclic (p\ne2) -this implies Will's property ? | |
| Aug 28, 2012 at 12:18 | comment | added | Alexander Chervov | @Geoff Robinson thank you very much for yours kind answer. I always learn a lot from them. Let me take some time to absorb it. | |
| Aug 28, 2012 at 7:27 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
simplified argument at end
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| Aug 28, 2012 at 4:45 | history | edited | Will Sawin | CC BY-SA 3.0 |
fixed paragraphs
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| Aug 27, 2012 at 9:36 | comment | added | Geoff Robinson |
@David: Yes, it's a characterization, I had ben worried abut faithfulness, but of course an irreducible character which misses ${\rm Ind}_{H}{G}(1)$ for each npn-trivial subgroup $H$ of $G$ is necessarily faithful.
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| Aug 27, 2012 at 4:56 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Included proof of characterization of Frobenius complements.
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| Aug 27, 2012 at 0:44 | comment | added | David E Speyer | I didn't know this characterization of Frob. complements before reading Geoff's answer, but it seems to be exactly what we need. | |
| Aug 27, 2012 at 0:43 | comment | added | David E Speyer | Isn't the answer precisely "The Frobenius complements are the only groups which fail to have this property?" Suppose $G$ is a group and $V$ is a representation which does not occur in $Ind_H^G 1$ for any $H$. Then, for any $g \in G$, the restriction $V|_{\langle g \rangle}$ must have no trivial component, so $g$ acts on $V$ without eigenvalue $1$ and the group is a Frob. complement. Conversely, suppose that there is some $H \subset G$ such that $V|_H$ has a trivial component. Then choose $h \in H$, and $V|_{\langle h \rangle}$ has a trivial component, hence $h$ has eigenvalue $1$. | |
| Aug 26, 2012 at 23:51 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Showed that everything works if trivial is replaced by linear
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| Aug 26, 2012 at 23:39 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
gave other examples
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| Aug 26, 2012 at 23:28 | comment | added | Geoff Robinson | @Will: Yes, that is true. | |
| Aug 26, 2012 at 23:25 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typos corrected
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| Aug 26, 2012 at 23:24 | comment | added | Will Sawin | Due to the classification of Hamiltonian groups, only the quaternions times an odd cyclic subgroup have a faithful irreducible representation. | |
| Aug 26, 2012 at 22:08 | history | answered | Geoff Robinson | CC BY-SA 3.0 |