As mentioned in Question on the unsolvability a group, a solvable finite group can't have exact 6 Sylow 5-subgroups. And I'm trying to prove it.
Suppose the opposite that $\Sigma=\{ P_1,...P_6\}$ is the set of Sylow 5-subgroups of $G$ and let $G$ acts on it by conjugacy. Then $G/H$ can be embedded into $S_6$ where $H= \bigcap_{i=1}^6 N_G(P_i)$. Since the number of Sylow 5-subgroups of $G/H$ is at most 6, and $S_6$ has 36 Sylow 5-subgroups, $G/H$ is a proper subgroup of $S_6$. Then I lost my may. Since $A_5$ has exact 6 Sylow 5-subgroups and it's simple, I guess there's something to do with $G/H$ and $A_5$. Then lead to a contradiction as $G$ is supposed to be solvable. Am I right? What to do next? Thanks in advance.