Timeline for When does $\langle grg^{-1}|g\in G\rangle = \langle gsg^{-1}|g\in G\rangle$ imply $\langle r\rangle=\langle xsx^{-1}\rangle$?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2015 at 17:44 | answer | added | Geoff Robinson | timeline score: 1 | |
| Dec 14, 2015 at 16:23 | comment | added | Ashot Minasyan | An interesting example of a normal group is any free group. Indeed, a classical theorem of W. Magnus states that if two elements $r,s$, of a free group, have the same normal closures then $r$ is conjugate to $s^{\pm 1}$. | |
| Dec 11, 2015 at 11:25 | answer | added | Derek Holt | timeline score: 5 | |
| Dec 11, 2015 at 1:26 | answer | added | Will Sawin | timeline score: 8 | |
| Dec 10, 2015 at 23:16 | answer | added | YCor | timeline score: 7 | |
| Dec 10, 2015 at 22:18 | comment | added | YCor | "What is the largest family of normal groups?" The answer is "the family of all normal groups" :) | |
| Dec 10, 2015 at 22:17 | comment | added | Farid Aliniaeifard | @SamHopkins The reason I call such groups normal was it seams that they have a lot of normal subgroups. | |
| Dec 10, 2015 at 21:57 | comment | added | YCor | With no symbols, the definition of "normal" is: whenever two cyclic subgroups generate the same normal subgroup, they are conjugate. | |
| Dec 10, 2015 at 21:56 | comment | added | Sam Hopkins | "Normal" is surely a bad choice of terminology... | |
| Dec 10, 2015 at 21:51 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
edited title
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| Dec 10, 2015 at 21:45 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
edited title
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| Dec 10, 2015 at 21:09 | comment | added | Fan Zheng | @FaridAliniaeifard Sorry I didn't see you mean the subgroup generated by these elements. | |
| Dec 10, 2015 at 20:53 | history | edited | Farid Aliniaeifard | CC BY-SA 3.0 |
added 9 characters in body
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| Dec 10, 2015 at 20:44 | comment | added | Farid Aliniaeifard | Not necessarily $r=xsx^{-1}$ since r could be a product of the elements of $\langle gsg^{-1}: g\in G \rangle $. | |
| Dec 10, 2015 at 20:34 | comment | added | Fan Zheng | I'm afraid there is possibly some serious confusion here: $r=ere^{-1}\in <grg^{-1}:g\in G>=<gsg^{-1}:g\in G>$, so there is always some $x\in G$ such that $r=xsx^{-1}$. | |
| Dec 10, 2015 at 20:28 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
rephase, clarify.
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| Dec 10, 2015 at 18:59 | history | asked | Farid Aliniaeifard | CC BY-SA 3.0 |