Timeline for Non-simple groups $G$ with only non-trivial quotient isomorphic to $G$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2023 at 7:57 | comment | added | Dominic van der Zypen | Thanks for this suggestion @DaveBenson and also thanks to YCor for making me aware of this. | |
| Nov 29, 2023 at 22:41 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
added non-simple
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| Nov 29, 2023 at 10:38 | comment | added | YCor | For every infinite cardinal $\kappa$, the group of even finitely supported permutations of $\kappa$ is a simple group of cardinal $\kappa$. Hence, as @DaveBenson suggested, you should modify the question to For which infinite cardinals $\kappa$, is there a non-simple group with Property Q having cardinal $\kappa$? | |
| Nov 28, 2023 at 15:19 | answer | added | Keith Kearnes | timeline score: 12 | |
| Nov 28, 2023 at 15:11 | vote | accept | Dominic van der Zypen | ||
| Nov 30, 2023 at 7:57 | |||||
| Nov 28, 2023 at 14:50 | comment | added | HJRW | @EmilJeřábek: since both interpretations of the question are trivial, I’m not sure what the substantive point of disagreement is here. Your reading of the question is just as trivial as mine. | |
| Nov 28, 2023 at 13:41 | answer | added | YCor | timeline score: 13 | |
| Nov 28, 2023 at 11:17 | comment | added | Emil Jeřábek | Though I agree that the question as such, “for which cardinals ...”, is immediately answered by the existence of arbitrarily large infinite simple groups; it would have to be reformulated along Dave Benson’s lines to make it research-level. | |
| Nov 28, 2023 at 11:10 | comment | added | Emil Jeřábek | @HJRW The definition is formulated precisely, and does not require $s$ to be an isomorphism. This is clearly intended, as the motivating dual property in the first paragraph likewise does not require the inclusion map itself to be an isomorphism, lest $\mathbb Z$ would not have the property. I am perfectly aware of the OP’s history of non-research-level questions, but that does not warrant arbitrarily changing the meaning of his question to make it more trivial than it is. | |
| Nov 28, 2023 at 10:32 | comment | added | HJRW | @EmilJeřábek: Whether or not property $Q$ coincides with simplicity depends on whether or not the isomorphism is required to be induced by $s$. The question doesn't specify, but since the OP writes "I don't know whether they [$\mathbb{Z}/p\mathbb{Z}$] are the only finite groups with property $Q$", I don't think we can be confident about what the definition of property $Q$ actually is. I stand by my vote to close. Please note that the OP is clearly an enthusastic mathematician, but has a history of asking naïve questions that aren't "research level". This is another one. | |
| Nov 28, 2023 at 9:24 | comment | added | Dave Benson | If the question were to understand the non-simple groups with property $Q$, I think it might not get closed. | |
| Nov 28, 2023 at 9:21 | comment | added | Emil Jeřábek | By general universal algebra, any group is a subdirect product of subdirectly irreducible groups that are its quotients, thus every group with property Q must be subdirectly irreducible. In particular, this easily implies that the only abelian groups with property Q are $C_p$ and $\mathbb Z(p^\infty)$. | |
| Nov 28, 2023 at 9:21 | comment | added | ADL | Here is a Math.SE question dealing with the non-simple groups in $Q$. (Take home points are: finite generation implies simple, and there are non-abelian countable examples.) Of course, simplicity is enough to answer the stated question (e.g.), so I agree with @HJRW re closure. | |
| Nov 28, 2023 at 9:16 | review | Close votes | |||
| Dec 5, 2023 at 3:01 | |||||
| Nov 28, 2023 at 9:13 | comment | added | Emil Jeřábek | @HJRW This is not true. For example, Prüfer groups $\mathbb Z(p^\infty)$ have property Q even though they are not simple. Note that the definition does not require $s$ itself to be an isomorphism. | |
| Nov 28, 2023 at 8:54 | comment | added | HJRW | Groups with your “property Q” are better known as simple groups. Of course, this is one of the most famous and best-studied notions in group theory. I’m voting to close. | |
| Nov 28, 2023 at 8:45 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |