Timeline for Embedding $\mathrm{PGL}(n,q^h)$ in $\mathrm{PGL}(nh,q)$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 11, 2015 at 8:17 | vote | accept | Tom De Medts | ||
| Apr 29, 2015 at 13:57 | history | edited | Tom De Medts | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 29, 2015 at 10:26 | history | edited | Tom De Medts | CC BY-SA 3.0 |
deleted 128 characters in body
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| Apr 29, 2015 at 9:37 | answer | added | peliukas | timeline score: 5 | |
| Apr 29, 2015 at 7:32 | history | edited | Tom De Medts | CC BY-SA 3.0 |
added 189 characters in body
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| Apr 28, 2015 at 18:15 | comment | added | Derek Holt | I have changed it back to $n^h$. I just retyped some of the words to apparently increase the amount of editing. | |
| Apr 28, 2015 at 18:14 | history | edited | Derek Holt | CC BY-SA 3.0 |
edited body
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| Apr 28, 2015 at 18:04 | comment | added | LSpice | @DerekHolt, I'm sorry. I was the one who changed it, because I thought that it was a typo; but I should have asked. I'll un-change it now. EDIT: Darn, it's too few characters to make a new edit. Is there any way that I or someone else could request that it be reverted? | |
| Apr 28, 2015 at 17:46 | comment | added | Derek Holt | Originally it read "${\rm PGL}(n,q^h)$ is isomorphic to a subgroup of ${\rm PGL}(n^h,q)$", which I believe is correct, but that has seems to have been changed! | |
| Apr 28, 2015 at 17:38 | comment | added | LSpice | Should "I believe I read somewhere that $\operatorname{PGL}(n, q^h)$ is isomorphic to a subgroup of $\operatorname{PGL}(n h, q)$" be "… that $\operatorname{PGL}(2, q^2)$ is isomorphic to a subgroup of $\operatorname{PGL}(4, q)$" (as in @DerekHolt's answer mathoverflow.net/a/204173/2383)? | |
| S Apr 28, 2015 at 17:22 | history | suggested | LSpice | CC BY-SA 3.0 |
n^h -> n h
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| Apr 28, 2015 at 17:06 | comment | added | Jeremy Rouse | According to Magma, ${\rm PGL}(n,q^h)$ is isomorphic to a subgroup of ${\rm PGL}(nh,q)$ for $n = 2$, $h = 3$ and $2 \leq q \leq 5$, but ${\rm PGL}(3,4)$ is not isomorphic to a subgroup of ${\rm PGL}(6,2)$. | |
| Apr 28, 2015 at 15:51 | review | Suggested edits | |||
| S Apr 28, 2015 at 17:22 | |||||
| Apr 28, 2015 at 15:08 | answer | added | Derek Holt | timeline score: 8 | |
| Apr 28, 2015 at 14:30 | history | asked | Tom De Medts | CC BY-SA 3.0 |