Timeline for Realization of a subgroup in a maximal subgroup of a classical group
Current License: CC BY-SA 4.0
14 events
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| Sep 7, 2022 at 20:03 | comment | added | Richard Lyons | @user488802: I think the extra 2 is not in $N_{M_0}(E)/C_{M_0}(E)$, but it can be found in $N_M(E)/C_M(E)$, inducing non-inner automorphisms on each $H_i$. | |
| Sep 7, 2022 at 9:51 | comment | added | user488802 | @RichardLyons Sorry to trouble with a further question. I computed $N_{M_0}(E)/C_{M_0}(E)$ with |$N_{M_0}(E)$| = 6144 and |$C_{M_0}(E)$| = 32. I found it was $2^6.3$. This $3$ in the end confused me as I thought it should be $Sp_{2}(2)$... Where did I drop that $2$? Thank you! | |
| Aug 29, 2022 at 23:24 | comment | added | user488802 | @Richard Lyons Thank you very much! | |
| Aug 29, 2022 at 22:30 | comment | added | Richard Lyons | Your $E$ is then $A_1\times A_2$, and you can find elements of $N_{M_0}(E)$ inducing your $*_{3\times 2}$ in the image of $Q^*$ in $M_0$. | |
| Aug 29, 2022 at 22:29 | comment | added | Richard Lyons | I'm sorry, my comment was not germane. Let $M_0$ be the $SL_2(5)^4$ subgroup of your $M$. Let $H=H_1\times H_2\times H_3\times H_4$ where each $H_i\cong SL_2(5)$. Write $Z(H_i)=<z_i>$, $1\le i\le 4$, and let $z=z_1z_2z_3z_4$. Then there is an exact sequence $1\to\langle z\rangle\to H\to M_0\to 1$. Also let $Q_i$ be a quaternion $8$-subgroup of $H_i$ and let $Q\cong Q_8$ be a "diagonal" subgroup of $Q^*:=Q_1\times Q_2\times Q_3\times Q_4$, i.e., the projection of $Q$ on each $H_i$ is $Q_i$. Then let $A_1$ and $A_2$ be the images of $Q$ and $Z(H)$ in $M_0$. (continued) | |
| Aug 28, 2022 at 23:42 | comment | added | user488802 | Sorry for the late response. I thought not many interested. Do you mean like: $N_{G}(SL_{2}(5))$ induces all of $PGL_{2}(5)$? Would you please elaborate more? Thank you. | |
| Aug 26, 2022 at 14:25 | comment | added | Richard Lyons | I think you have understated $M$. The normalizer of one of your $SL_2(5)$'s in $M$ induces all of $PGL_2(5)$ on that $SL_2(5)$. | |
| Aug 24, 2022 at 16:47 | comment | added | Derek Holt | I think you need to define the subgroup $E$ more precisely, perhaps by giving generating matrices in ${\rm GL}(8,5)$. | |
| Aug 23, 2022 at 20:35 | history | edited | user488802 | CC BY-SA 4.0 |
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| Aug 23, 2022 at 20:25 | history | edited | user488802 | CC BY-SA 4.0 |
added 46 characters in body
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| Aug 23, 2022 at 20:23 | comment | added | user488802 | Sorry. It does mean $4^{3}$. I put it this way because it comes from $(q-1)^3$. Thank you! | |
| Aug 23, 2022 at 11:36 | comment | added | LSpice | I know the notation, e.g., $4^3$, but what does $(5 - 1)^3$ mean? | |
| Aug 23, 2022 at 10:32 | history | edited | YCor | CC BY-SA 4.0 |
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| Aug 23, 2022 at 10:20 | history | asked | user488802 | CC BY-SA 4.0 |