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If a group $H$ has order $255$ then the Sylow theorems tell us that it must have a Sylow $p$-subgroup of order $5$ and there are either $1$ or $51$ of them, also there is either $1$ Sylow $p$-subgroup of order $3$ or $85$ Sylow p-subgroups of order $3$. How can I show that it is not the case that there are both $51$ order $5$ Sylow $p$-subgroups and $85$ order $3$ Sylow p-subgroups? In general what is the best way to deal with these cases?

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Suppose for contradiction that there were $51$ Sylow $5$-subgroups and $85$ Sylow $3$-subgroups. Any non-identity element of a Sylow $5$-subgroup must have order $5$, and and any non-identity element of a Sylow $3$-subgroup must have order $3$, so a Sylow $5$-subgroup and a Sylow $3$-subgroup will intersect only in $e$. Thus, the total number of elements in $51$ Sylow $5$-subgroups and $85$ Sylow $3$-subgroups would be $51(4)+85(2)+1=375>255$.

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  • $\begingroup$ great that is exactly what I was looking for. Tnx $\endgroup$ – tmpys May 27 '14 at 3:34

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