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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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I'm reluctant to upvote the question, since it should start with a search of the available literature on Suzuki groups (which you seem to know something about). As Nick points out, the original two-part paper by Michio Suzuki (Ann. of Math. 1962, 1964) already contains lots of relevant detail. Theorem 9 occurs in Part I. These papers are available online via JSTOR if you have access. – Jim Humphreys May 23 '12 at 19:22

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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@Nick, thank you very much. – R K May 23 '12 at 13:58

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