Timeline for The action of an extension group $G=p^{1+2n}{.}Q$ on the faithful characters of its normal subgroup $p^{1+2n}$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2021 at 8:01 | comment | added | Derek Holt | In general, $Q$ is transitive on the faithful characters of $E$ if and only if it is transitive in its conjugation action on $Z(E) \setminus \{1\}$ (where $E$ is the extraspecial group), which might make it easier to determine. | |
| Sep 20, 2021 at 8:01 | comment | added | Isaac | Just want to now, why $Q$ is necessarily a subgroup of $GSp_{2g}$? | |
| Sep 20, 2021 at 7:47 | comment | added | Derek Holt | $3_+^{1+4}:4S_6$ has trivial centre, so the two faithful characters of the extraspecial group are interchanged in that example. | |
| Sep 20, 2021 at 3:04 | comment | added | Will Sawin | $Q$ is necessarily a subgroup of $GSp_{2g}$, and there is a map $GSp_{2g}(\mathbb F_p) \to \mathbb F_p^\times$ with kernel $Sp_{2g}(\mathbb F_p)$. The action is transitive if and only if the image of $G$ under this map is surjective. | |
| Sep 19, 2021 at 22:28 | comment | added | Isaac | Let us continue this discussion in chat. | |
| Sep 19, 2021 at 22:27 | history | edited | Isaac | CC BY-SA 4.0 |
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| Sep 19, 2021 at 22:27 | comment | added | Isaac | Let we consider maximal subgroup $3_+^{1+4}{:}4S_6$ of $Co_3$. | |
| Sep 19, 2021 at 15:56 | comment | added | Derek Holt | Taking $n>1$ does not prevent $Q$ being trivial. | |
| Sep 19, 2021 at 13:55 | history | edited | Isaac | CC BY-SA 4.0 |
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| Sep 19, 2021 at 13:34 | comment | added | Isaac | Lets take $n>1$. | |
| Sep 19, 2021 at 13:15 | comment | added | Derek Holt | That still doesn't rule out $Q$ being trivial. | |
| Sep 19, 2021 at 12:56 | comment | added | Isaac | The group $Q$ Is a $2n$-dimensional linear group over $GF(p)$. | |
| Sep 19, 2021 at 12:53 | history | edited | Isaac | CC BY-SA 4.0 |
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| Sep 19, 2021 at 11:35 | review | Close votes | |||
| Oct 4, 2021 at 3:02 | |||||
| Sep 19, 2021 at 11:15 | comment | added | Derek Holt | Since you haven't given us any information about $Q$, this is impossible to answer. $Q$ could be the trivial group. | |
| Sep 19, 2021 at 9:01 | comment | added | YCor | It you have a direct product, (and $p>2$) it is certainly not transitive. Also if $|Q|<p-1$. | |
| Sep 19, 2021 at 8:58 | history | asked | Isaac | CC BY-SA 4.0 |