Timeline for Is this class of groups already in the literature or specified by standard conditions?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 22, 2022 at 0:06 | comment | added | YCor | @SamHopkins oops, you're right, I wasn't careful. I should have said only: a f.g. group in which every proper subgroup is cyclic. | |
| Sep 21, 2022 at 23:47 | comment | added | Sam Hopkins | @YCor: Sorry I must be reading your comment wrong then; I thought you said "a finitely generated group [...] in which every proper subgroup is finite is lossless." That would seem to include all finite groups, no? | |
| Sep 21, 2022 at 23:43 | comment | added | YCor | @SamHopkins So what? There's no contradiction. Lossy means that $P(H,K)$ fails for some $K\le H\le G$. I mentioned that $P(H,H)$ holds whenever $G$ is finite but can fail for $G$ infinite. | |
| Sep 21, 2022 at 23:41 | comment | added | Sam Hopkins | @YCor: I don't understand your last comment; didn't Kyle say he knows examples of lossy finite groups? | |
| Sep 21, 2022 at 21:39 | comment | added | YCor | Call $P(H,K)$ the given condition (so "lossless" means that $P(H,K)$ holds for all $K\le H\le G$). Then $P(H,K)$ is trivial when $H$ is normal. Note that the condition $P(H,H)$ is trivial when $H$ is finite but fails when there exists $g\in G$ such that $gHg^{-1}<H$. The existence of a subgroup conjugate to a proper subgroup of itself is quite frequent for infinite groups (e.g. every group with a nonabelian free subgroup, and many others). Anyway, a f.g. group in which every proper subgroup is cyclic, or in which every proper subgroup is finite (cf. Tarski monsters) is lossless. | |
| Sep 21, 2022 at 20:13 | comment | added | Geoff Robinson | Often in finite group theory, it is how the fusion of subgroups ( or sometimes just individual elements) of a fixed Sylow $p$-subgroup $S$ of $G$ is effected within $G$ which is important. In the more genral setting of dealing with all subgroups, as you seem to, then paper of people like A. Dress on idempotents in the Burnside ring of a finite group may be relevant to you (and there have been much work on uses of the Burnside ring in group representation theory over the decades since) | |
| Sep 21, 2022 at 19:50 | comment | added | kyleormsby | Thanks, @GeoffRobinson. We will definitely look into the control of fusion (of subgroups) perspective! Do you know of any papers that explicitly use such a condition? | |
| Sep 21, 2022 at 19:43 | comment | added | Geoff Robinson | In finite group theory, ( and transfer, etc), often control of fusion can be setwise. So, here a finite group theorist might say that for each subgroup $H$ of $G$ the fusion of subgroups of $H$ within $G$ is controlled by $N_{G}(H)$. It is true that in work of Alperin and others, and later work on fusion systems, we need control of strong fusion (I've always felt that strong control of fusion would be a better expression), wher we require $h$ and $g$ ( as in your definition) to induce the same conjugation elementwise on $K$ , that is, $h^{-1}g \in C_{G}(K)$). | |
| Sep 21, 2022 at 18:19 | comment | added | kyleormsby | @NeilStrickland - Yes, they do look to be related, but it seems that control of fusion is an elementwise condition whereas our condition is an equality of sets. | |
| Sep 21, 2022 at 17:52 | comment | added | kyleormsby | @SamHopkins - Good catch! I have fixed a typo in the definition of lossless. When $gKg^{-1}\le H$ for some $g\in G$, we in fact require $h\in N_G(H)$ such that $gKg^{-1} = hKh^{-1}$. | |
| Sep 21, 2022 at 17:50 | history | edited | kyleormsby | CC BY-SA 4.0 |
added 6 characters in body
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| Sep 21, 2022 at 17:20 | comment | added | Neil Strickland | This is somewhat related to "control of fusion", I think. | |
| Sep 21, 2022 at 17:16 | history | edited | LSpice | CC BY-SA 4.0 |
Name of paper and collaborators
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| S Sep 21, 2022 at 16:50 | review | First questions | |||
| Sep 21, 2022 at 17:31 | |||||
| S Sep 21, 2022 at 16:50 | history | asked | kyleormsby | CC BY-SA 4.0 |