Timeline for In what sense is the classification of all finite groups "impossible"?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2021 at 16:48 | comment | added | Joshua Grochow | @YCor Actually I wasn't so much referring to your comment as Sergeichuk's result. I am confused about how to reconcile that result with the classification that one gets from Kronecker normal form for pencils. Either way you are right that codimension 3 is wild; and unrestricted codimension is TI-complete (when formulated appropriately) or what one might call "tensor wild" (eg doi.org/10.1016/j.laa.2018.12.022). | |
| Nov 19, 2021 at 9:31 | comment | added | YCor | OK, actually I had no serious clue to say "I think this is wild", and should rather have said "I guess" — this is close to classifying spaces with a pair of alternating forms. Possibly one needs codimension 3 (that's roughly classifying spaces with a triple of alternating forms)? | |
| Nov 19, 2021 at 9:09 | comment | added | Joshua Grochow | @YCor: Ah, but now I am confused. When the center is $\mathbb{Z}/p\mathbb{Z})^2$ I believe there is a tame classification because of Kronecker normal form for matrix pencils (eg the corresponding classification for genus 2 p-groups can be seen in Brooksbank-Maglione-Wilson). I will have to dig/think some more to reconcile these... | |
| Nov 19, 2021 at 9:06 | comment | added | YCor | Ah thanks. I have much less intuition about these ones, say of exponent $p^2$, since this boils down to Lie algebras over no longer fields but $\mathbf{Z}/p^2\mathbf{Z}$. | |
| Nov 19, 2021 at 9:05 | comment | added | Joshua Grochow | @YCor: Ah, sorry, misread! Sergeichuk also proves it for groups whose centers are $\mathbb{Z}/p^2\mathbb{Z}$ if I recall correctly. | |
| Nov 19, 2021 at 9:03 | comment | added | YCor | I said "$V$ 2-codimensional", not 2-dimensional. | |
| Nov 19, 2021 at 8:54 | comment | added | YCor | To specify the first assertion, for $p$ odd prime, a 2-step-nilpotent group of exponent $p$ whose abelianization has rank $n$ is the same as the quotient of the free $n$-generated exponent $p$ 2-step-nilpotent group $L_n$ by a subspace $V$ of its center. The latter can be identified to $\bigwedge^2\mathbf{F}_p^n$ (which has dimension $n(n-1)/2$), and two resulting groups $L_n/V$, $L_n/V'$ are isomorphic iff $V,V'$ are in the same $\mathrm{GL}_n(\mathbf{F}_p)$-orbit. Even for 2-codimensional $V$ (i.e. such groups of order $p^{n+2}$), I think this is wild (say, when $p$ is fixed and $n$ grows). | |
| Nov 19, 2021 at 8:21 | history | edited | Joshua Grochow | CC BY-SA 4.0 |
Updated complexity of p-group iso to be more precise and added refs
|
| Mar 31, 2015 at 13:51 | vote | accept | Keivan Karai | ||
| Mar 4, 2015 at 6:05 | history | answered | Joshua Grochow | CC BY-SA 3.0 |