Timeline for Ascending Chain Condition for finite normalizers
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2011 at 20:53 | comment | added | Qayum Khan | I believe that "$i \geq 0$" should be "$i \geq 1$" in the edit. Note: [ x_3^{-1} x_1 x_3 = x_3^{-1} x_3^4 x_3 = x_3^4 = x_1. ] | |
| Sep 6, 2011 at 10:07 | comment | added | Derek Holt | The normalizer of a Klein 4-group in a dihedral group of order $16$ has order 8, so $H_1$ is not normal in $H_3$. I have added a more precise definition of the groups. | |
| Sep 6, 2011 at 10:03 | history | edited | Derek Holt | CC BY-SA 3.0 |
added 169 characters in body; added 3 characters in body; added 3 characters in body
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| Sep 6, 2011 at 0:43 | comment | added | Qayum Khan | Maybe I'm misunderstanding, but I got $H_3 \subseteq N_G(H_1)$ ... | |
| Sep 5, 2011 at 8:54 | comment | added | Derek Holt | But do you have a problem with the example I gave? | |
| Sep 5, 2011 at 1:49 | comment | added | Qayum Khan | For your idea, it seems one can consider a slightly bigger group, $\hat{G} = O(2) = U(1) \rtimes_{c} C_2$, where $c$ is complex-conjugation, and the subgroup $H = D_2 = O(1) \rtimes_{c} C_2$. However $H \triangleleft \hat{G}$; hence $H \triangleleft G$. | |
| Sep 2, 2011 at 9:11 | history | answered | Derek Holt | CC BY-SA 3.0 |