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0
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It known the Suzuki group $Sz(q)$, where $q=2^{2n+1}$ is of order $q^2(q^{2}+1)(q-1)$. By $2^{2}$ $=-1$ mod $5$, then $2^{2n}$ $=(-1)^{n}$ mod $5$. So $q=2^{2n+1}$ $=2(-1)^{n}$ mod $5$ and so $q^{2}+1=0$ mod $5$. Therefore $5$ always divide order of $Sz(q)$. My question is: If $P$ is a Sylow $5$-subgroup of the group $Sz(q)$, then order of normalizer of $P$ in $Sz(q)$? Thanks. |
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9
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Go to Suzuki's original paper On a class of doubly transitive groups. Theorem 9 of that paper lists all of the subgroups of $G=Sz(q)$. In particular, setting $q=2^{2a+1}$ and $r=2^{a+1}$, there is a maximal subgroup $M$ of order $4(q-r+1)$. Now $M$ has a cyclic normal subgroup $C$ of order $q-r+1$ which contains $P$, a Sylow $5$-subgroups of $G$. Since a subgroup of a cyclic group is characteristic, this implies that $M$ normalizes $P$. Since $M$ is maximal we conclude that $M=N_G(P)$. |
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