Timeline for Quotienting a virtually cyclic group by an element [closed]
Current License: CC BY-SA 4.0
12 events
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Dec 15, 2021 at 15:18 | history | closed |
Yemon Choi Carl-Fredrik Nyberg Brodda Eric Peterson Bugs Bunny coudy |
Needs more focus | |
Nov 8, 2021 at 10:56 | comment | added | YCor | @piper1967 I don't have a reference, but this seems to be a straightforward verification. In general, if $N$ is this normal subgroup, the quotient $K/\langle\!\langle h\rangle\!\rangle$ is naturally isomorphic to the finite group $G/N$. | |
Nov 8, 2021 at 10:44 | comment | added | piper1967 | @YCor can you give reference to the proof of the fact that the condition holds if and only if the normal subgroup of G generated by $\{g^{-1}u(g) : g \in G\}$ is G itself? | |
Nov 8, 2021 at 10:18 | review | Close votes | |||
Dec 15, 2021 at 15:18 | |||||
Nov 8, 2021 at 10:14 | comment | added | Derek Holt | The problem I always have with this type of question is that I have no idea what sort of answer the poster is looking for. All you can really say is that sometimes it is and sometimes it is not. | |
Nov 8, 2021 at 9:58 | comment | added | YCor | Yes. There are easy examples. (1) Take $G$ simple non-abelian and any non-trivial action for the semidirect product. (2) Take $G$ abelian of odd order and $h$ acting on $G$ by $x\mapsto -x$. Etc. In general let $u$ be the automorphism of $G$ induced by $h$. Then the required property holds iff the normal subgroup of $G$ generated by $\{g^{-1}u(g):g\in G\}$ is $G$ itself. | |
Nov 8, 2021 at 9:36 | comment | added | HJRW | I've taken the liberty of making the title more descriptive. Please feel free to edit as you please, but something much more speicific than the original title would be preferable. | |
Nov 8, 2021 at 9:35 | history | edited | HJRW | CC BY-SA 4.0 |
Made the title more descriptive
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Nov 8, 2021 at 8:51 | history | edited | Mark Wildon | CC BY-SA 4.0 |
added 78 characters in body
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Nov 8, 2021 at 8:43 | comment | added | piper1967 | (h) is a cyclic subgroup of K. I asked when K/H, where H is the normal subgroup generated by (h), is trivial. | |
Nov 8, 2021 at 8:37 | comment | added | LeechLattice | Does "$K/(h)$ trivial" implies K is infinite cyclic? If it's true, how could $G<K$ be normal? | |
Nov 8, 2021 at 8:22 | history | asked | piper1967 | CC BY-SA 4.0 |