Let $G=p^{1+2n}{.}Q$, $n>1$, be a finite extension group of an extra-special $p$-group $N=p^{1+2n}$ by a group $Q$, where $Q$ is a linear group of dimension $2n$ over $GF(p)$. It seems that the action (by conjugation) of $G$ (or $Q$) on the $p-1$ irreducible faithful characters of $N$ is transitive. If it is the case, how does one show it? For example, let we consider the maximal subgroup $3_+^{1+4}{:}4S_6$ of $Co_3$.
12
-
3$\begingroup$ It you have a direct product, (and $p>2$) it is certainly not transitive. Also if $|Q|<p-1$. $\endgroup$– YCorCommented Sep 19, 2021 at 9:01
-
2$\begingroup$ Since you haven't given us any information about $Q$, this is impossible to answer. $Q$ could be the trivial group. $\endgroup$– Derek HoltCommented Sep 19, 2021 at 11:15
-
1$\begingroup$ That still doesn't rule out $Q$ being trivial. $\endgroup$– Derek HoltCommented Sep 19, 2021 at 13:15
-
3$\begingroup$ $Q$ 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. $\endgroup$– Will SawinCommented Sep 20, 2021 at 3:04
-
2$\begingroup$ 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. $\endgroup$– Derek HoltCommented Sep 20, 2021 at 8:01
|
Show 7 more comments