Timeline for Length *vs* table of characters
Current License: CC BY-SA 3.0
12 events
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| Feb 27, 2017 at 23:23 | comment | added | Richard Lyons | Answer to a related question: the length is uniquely determined by the character table, as a consequence of the classification of finite simple groups. This is because if $G$ is the direct product of isomorphic finite simple groups, the number of factors in the direct product is determined by $|G|$. This appears in W. Kimmerle et al., Proc. London Math. Soc. (3) 60 (1990), no. 1, 89–122, and uses CFSG. | |
| Feb 2, 2017 at 21:56 | comment | added | Kevin Buzzard | David's point is that to answer your question in the negative you need two groups with the same character table but different lengths; this and the observation about the order shows that one of these groups must be non-solvable. | |
| Feb 2, 2017 at 20:05 | comment | added | Denis Serre | @David. Of course I know the relation $|G|=\sum\chi(e)^2$, but because it gives only the order of the group, which does not determine the length, one needs something else in the table. | |
| Feb 2, 2017 at 17:35 | comment | added | David E Speyer | Note that the length of any solvable group is the number of prime factors in its order (counted with multiplicity). Moreover, the order of a group is recoverable from its character table, as $\sum \chi(1)^2$. Thus, to have a hope of building a counter-example, we need to use non-solvable groups. | |
| Feb 2, 2017 at 17:06 | comment | added | Jim Humphreys | On second thought, I doubt that the character table determines the length (though I don't have good counterexamples at hand). While it's easy to identify normal subgroups, it's much harder to pass from the character table of $G$ to that of a proper subgroup (as seen in going from $S_n$ to $A_n$). | |
| Feb 2, 2017 at 16:28 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 290 characters in body
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| Feb 2, 2017 at 16:01 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 10 characters in body
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| Feb 2, 2017 at 16:01 | comment | added | Denis Serre | @benblumsmith. Yes of course ! I fix it immediately. | |
| Feb 2, 2017 at 15:59 | comment | added | benblumsmith | Do you mean that $\ell(G) = 1$ if and only if there does not exist a character $\chi\neq 1$ such that...? | |
| Feb 2, 2017 at 15:17 | comment | added | Denis Serre | @Jim. I agree with your first sentence (this is basically what I meant when I considered the case $\ell(G)=1$). However, once you know the list of normal subgroups, how do you reconstruct the Joradn-Hölder factors ? | |
| Feb 2, 2017 at 15:01 | comment | added | Jim Humphreys | If I recall correctly, the character table does allow one to locate the normal subgroups (as unions of some conjugacy classes). From this you can probably recover the length, but I'd have to check the details. | |
| Feb 2, 2017 at 14:57 | history | asked | Denis Serre | CC BY-SA 3.0 |