Timeline for Action on a normal subgroup where each coset acts freely
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2017 at 3:00 | review | Close votes | |||
| Oct 15, 2017 at 3:02 | |||||
| Oct 8, 2017 at 22:02 | vote | accept | Michael Cotton | ||
| Oct 7, 2017 at 0:58 | history | edited | Michael Cotton | CC BY-SA 3.0 |
added 190 characters in body
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| Oct 7, 2017 at 0:52 | comment | added | Michael Cotton | So you can't just arbitrarily biject N with something else. If it's not constructive, then it can't really be of any use, unfortunately. | |
| Oct 7, 2017 at 0:40 | comment | added | Michael Cotton | However, the action needs to be described or defined somehow. | |
| Oct 7, 2017 at 0:39 | comment | added | Michael Cotton | The action doesn't really have to be related much to the structure of N. Also, the only groups of interest here are either countably infinite or of size continuum. | |
| Oct 6, 2017 at 16:56 | comment | added | YCor | "An action of $G$ on $N$": how is the action related to the group structure of $N$? of its embedding in $G$? I understand the question as equivalent to "is there an action on $G$ a set $X$ of the same cardinality as $N$ that is free on $N$"? If $N$ is finite, this is equivalent to require that $N$ is part of a semidirect decomposition. But if both $N,G$ are infinite countable, this is always true. | |
| Oct 6, 2017 at 15:09 | answer | added | Aaron Meyerowitz | timeline score: 1 | |
| Oct 6, 2017 at 10:17 | review | Close votes | |||
| Oct 6, 2017 at 14:08 | |||||
| Oct 6, 2017 at 9:28 | comment | added | François Brunault | In the case $G=\mathbf{Z}/4\mathbf{Z}$ and $N=\{0,2\}$, by your assumption the element $2 \in N$ must act as a transposition of $N$. Then the action of $1$ on $N$ is a permutation $\sigma$ which satisfies $\sigma^2 = (0 2)$, which is impossible. | |
| Oct 6, 2017 at 9:25 | comment | added | Derek Holt | Isn't the cyclic group $C_4$ of order 4 with $|N|=2$ a counterexample? There is only one nontrivial action, and that does not have the specified property. | |
| Oct 6, 2017 at 8:46 | comment | added | Dirk | You surely thought about it for some time. Could you maybe give an example of an action that has the desired property in a non-trivial case (i.e. $N \neq 1$ and $N \neq G$)? I'm having a hard time coming up with an operation that leaves $N$ fixed as a set but every coset "acts" regularly on $N$... | |
| Oct 6, 2017 at 8:41 | comment | added | Dirk | @FrançoisBrunault I think it can, as $g,h$ are supposed to lie in the same coset - finding two distinct elements in the same coset when $N=1$ seems difficult... | |
| Oct 6, 2017 at 4:13 | history | asked | Michael Cotton | CC BY-SA 3.0 |