Timeline for When does $G\times G\times G$ admit a faithful group action on a set of size $|G|$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2022 at 22:52 | comment | added | Geoff Robinson | Also probably worth making explicit that the only non- cyclic exceptional groups of odd order are those of type $(5,5)$ and $(3,3^k)$. | |
| Jan 9, 2022 at 23:05 | comment | added | Sean Eberhard | It may be useful to add that this list in fact includes all (nontrivial) abelian subgroups of all groups $G$ not acting faithfully on $|G|/3$ points. (Proof: Suppose $H \leq G$ has a faithful permutation action on $m$ points. Then the induced representation of $G$ is a faithful permutation action on $[G:H]m$ points.) This and a couple further arguments show that all exceptional $p$-groups for odd $p$ are abelian. | |
| Jan 8, 2022 at 21:26 | history | edited | Sean Eberhard | CC BY-SA 4.0 |
deleted 11 characters in body
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| Jan 8, 2022 at 21:25 | comment | added | Sean Eberhard | @TerryTao Thanks, you're right. | |
| Jan 8, 2022 at 18:20 | comment | added | Terry Tao | As just observed by Geoff, I think the case $(2,3,5)$ should be deleted (it barely fails the strict inequality). | |
| Jan 5, 2022 at 15:28 | comment | added | Sean Eberhard | @YCor Yes, because the claim that the minimal faithful action of $G$ has $m = q_1 + q_2 + \cdots + q_k$ points, applied to $G^3$, shows that the minimal faithful action of $G^3$ has $3m$ points. | |
| Jan 5, 2022 at 15:24 | comment | added | YCor | Does this also solve the original question for abelian groups. Namely, is it true that no $G$ in this list is such that $G^3$ acts faithfully on $|G|$ elements? (It seems so but I haven't entirely checked) | |
| Jan 5, 2022 at 15:04 | history | answered | Sean Eberhard | CC BY-SA 4.0 |