Timeline for The finite extensions of $SL_2(q)$ [closed]
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2018 at 15:22 | comment | added | Derek Holt | For odd $q$, ${\rm PGL}_2(q)$ has Schur multiplier of order $2$, and there are two isomorphism classes of covering groups. It is difficult to come up with more concrete descriptions. The situation is similar for the symmetric groups $S_n$. | |
| Jul 3, 2018 at 15:05 | history | closed |
Derek Holt user6976 YCor Stefan Kohl♦ Neil Strickland |
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| Jul 3, 2018 at 14:33 | comment | added | sara | What are these two isomorphism classes? | |
| Jul 3, 2018 at 12:20 | comment | added | Derek Holt | For odd $q$ there are two isomorphism classes of groups $G$ that contain ${\rm SL}_2(q)$ as a subgroup of index $2$, and for which $G/Z(G) \cong {\rm PGL}_2(q)$. | |
| Jul 3, 2018 at 12:13 | history | edited | sara | CC BY-SA 4.0 |
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| Jul 3, 2018 at 12:13 | comment | added | sara | My main goal is to find the structure of such $G$ | |
| Jul 3, 2018 at 12:03 | comment | added | YCor | This changes nothing, your assumptions force equality of the cardinals. Derek means that the inclusion $Z(G)\le SL_2(q)$ fails for $GL_2(q)$. Anyway, this post is not research level and should be posted in a more appropriate forum. | |
| Jul 3, 2018 at 12:01 | review | Close votes | |||
| Jul 3, 2018 at 15:05 | |||||
| Jul 3, 2018 at 11:55 | history | edited | sara | CC BY-SA 4.0 |
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| Jul 3, 2018 at 11:55 | comment | added | sara | Yes.. I should say that a subgroup of $GL_2(q)$. I am editting it Thank you | |
| Jul 3, 2018 at 11:45 | comment | added | Derek Holt | ${\rm GL}_2(q)$ does not satisfy the condition $Z(G) \le {\rm SL}_2(q)$ when $q>3$. | |
| Jul 3, 2018 at 11:36 | review | First posts | |||
| Jul 3, 2018 at 11:46 | |||||
| Jul 3, 2018 at 11:35 | history | asked | sara | CC BY-SA 4.0 |