Let $G$=PSU$(3,q)$ be projective special unitary group where $q$ is prime power. I would like to know why there is not any prime $r$ such that the number of Sylow $r$-subgroups of $G$ is $r+1$?
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3$\begingroup$ What makes you think this is the case? $\endgroup$– Steven LandsburgCommented Jan 30, 2013 at 17:02
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$\begingroup$ While Nick's answer looks reasonable (and better formulated than the question), the question itself isn't carefully stated and might better be closed. $\endgroup$– Jim HumphreysCommented Jan 31, 2013 at 0:10
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This can be answered on a case-by-case basis. Work in $SU(3,q)$ because it's easier, and observe that if $r>3$ then $r$ divides one of $q, q+1, q-1, q^2-q+1$. Now go through these one at a time.
E.g. if $r$ divides $q$, then $r=p$. Now either calculate the size of the normalizer of a Sylow $p$ or just observe $r+1$ is less than the minimal index of a subgroup of $G$. Similarly if $r$ divides $q^2-q+1$, then the normalizer of a Sylow has size $3(q^2-q+1)$ and its index is certainly bigger than $r+1$ for any $r$ dividing $q^2-q+1$. The other cases are all similar.
