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
28 events
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| Jan 12, 2022 at 20:45 | comment | added | Geoff Robinson | I should have made explicit that there are indeed non-Abelian exceptional groups $G$ of all the orders $12,20,24,28, 36$ and $44$. For $q \in \{3,5,7,9,11 \}$, we take $G = \langle x,y : x^{q} = y^{4} = 1, y^{-1}xy = x^{-1} \rangle$. As mentioned in the post, we take $G = C_{3} \times Q_{2^{k}}$ for a non-Abelian example of order $3 \times 2^{k}$ for each $k \geq 3$. We may also take $G = Q_{2^{k}}$ as a non-Abelian exception for each $k \geq 3$, and $G$ a dihedral group of order $2p^{k}$ for each odd prime $p$ and each positive integer $k$ for a non-Abelian exceptional example. | |
| Jan 10, 2022 at 22:42 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Removed last edit
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| Jan 10, 2022 at 14:35 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Discussed $2$-group case.
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| Jan 10, 2022 at 13:39 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
text deleted
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| Jan 9, 2022 at 22:56 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
minor corrections
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| Jan 9, 2022 at 22:15 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
clarification
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| Jan 9, 2022 at 18:04 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
[Edit removed during grace period]
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| Jan 9, 2022 at 17:14 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Minor text changes for clarification and economy
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| Jan 9, 2022 at 12:51 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Reordered small amount of text for clarity
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| Jan 8, 2022 at 23:24 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Added a previously omitted case.
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| Jan 8, 2022 at 17:40 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Added examples
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| Jan 8, 2022 at 13:34 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
typo
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| Jan 8, 2022 at 13:23 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Added note
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| Jan 8, 2022 at 12:33 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Made some clarifications and a summary
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| Jan 7, 2022 at 22:04 | comment | added | Terry Tao | It could be useful to update the answer to list all the remaining unresolved cases. Leaving aside the abelian cases treated in Sean's answer, it seems from the above discussion that the only possible exceptional groups are 2-groups, or of order 12, 20, 24, 28, 36, 44, or of order $3p^k$ for some $p \geq 5$, but I am not 100% sure that this is the correct conclusion of all the above discussion. | |
| Jan 6, 2022 at 13:20 | comment | added | Geoff Robinson | I think your argument can be amended to something which works. It took too long to write it as a comment. | |
| Jan 5, 2022 at 20:36 | vote | accept | Craig | ||
| Jan 5, 2022 at 20:19 | comment | added | Sean Eberhard | You're right, that's a gap. | |
| Jan 5, 2022 at 20:08 | comment | added | Geoff Robinson | @SeanEberhard : I see that $Z_{2}(G)$ is Abelian from the argument above, but why does that imply that $G$ itself is Abelian (maybe I am just missing some point)? | |
| Jan 5, 2022 at 19:35 | comment | added | Sean Eberhard | Let $G$ be an exceptional $p$-group with $p \in \{3, 5\}$. As discussed, all elements of order $p$ are central. Let $H, K$ be subgroups of $Z_2(G)$ of order $p^2$. If $H \cap K = 1$ then we get an action on $2n/p^2$ points. Otherwise, $|H \cap K| \geq p$. Let $h$ and $k$ be generators of $H$ and $K$ respectively such that $h^p k^p = 1$. Then $(hk)^p = h^p k^p [h,k]^{\binom{p}{2}} = [h,k]^{\binom{p}{2}}$, so $(hk / [h,k]^{-(p-1)/2})^p = 1$, so $hk \in Z(G)$. This shows that $Z_2(G)/Z(G)$ has at most one subgroup of order $p$, so it's cyclic, so it's trivial, so $G$ is abelian. | |
| Jan 5, 2022 at 17:38 | comment | added | Sean Eberhard | Well, $C_9 \rtimes C_9$ is not interesting, because the two $C_9$'s have trivial intersection so induce a faithful action on 18 points. | |
| Jan 5, 2022 at 17:15 | comment | added | Sean Eberhard | Good point. As you said in your answer, groups of order $p$ must be normal. Since the automorphism group of a group of order $p$ has order $p-1$, they must in fact be central. So all elements of order $p$ are central. What about something like $C_9 \rtimes C_9$. | |
| Jan 5, 2022 at 16:21 | comment | added | Terry Tao | The arguments in Section 3 of this Babai-Goodman-Pyber paper tandfonline.com/doi/abs/10.1080/00927879308824639 should help treat the 2-group case, though I haven't checked the details. | |
| Jan 5, 2022 at 16:07 | comment | added | Sean Eberhard | Are there exceptional $p$-groups for $p \in \{3, 5\}$ apart from cyclic groups, $C_3^2$, and $C_5^2$? | |
| Jan 5, 2022 at 14:50 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
amended one argument- then corrected typo
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| Jan 5, 2022 at 14:11 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Jan 5, 2022 at 14:00 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
added extra comment
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| Jan 5, 2022 at 13:52 | history | answered | Geoff Robinson | CC BY-SA 4.0 |