What are the orders of maximal abelian subgroups of the simple groups $F_4(q)$ and $C_4(q)$, where $F_4(q)$ is an exceptional group and $C_4(q)$ is a symplectic group?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
For a finite group $G$, denote the maximum of the orders of its Abelian subgroups by $a(G)$. Then we have, for $G=F_4(q)$ and $q$ even, $$ q^{11}\le a(G)\le q^{17}, $$ and for $G=F_4(q)$, $q$ odd, $$ q^{9}\le a(G)\le q^{14}. $$ For $G=C_n(q)$ we have, for $n\neq 2$, $$ a(G)=q^{\frac{n(n+1)}{2}}. $$ For a reference, see the paper Maximal Orders of Abelian Subgroups in Finite Chevalley Groups by E. P. Vdovin, published in $2001$; in particular table $2$.
answered Dec 10, 2014 at 14:57
-
$\begingroup$ The paper itself is freely available online in Russian (but the English translation may require a library subscription): mathnet.ru/php/archive.phtml? wshow=paper&jrnid=mzm&paperid=521&option_lang=eng But in any case, it seems tricky in some cases to get a precise value for $a(G)$, or a description of such subgroups in terms of the Lie theory data. $\endgroup$ Commented Dec 10, 2014 at 16:26
-
2$\begingroup$ I see. It is translated in the Mathematical Notes (which might require a library subscription), see here. $\endgroup$ Commented Dec 10, 2014 at 16:28