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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.

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Here is one way to do it, but it involves using a few results about subgroups of $S_n$. Since all Sylow subgroups are conjugate, the image $G/H$ of the action in $S_6$ is transitive. An element of order $5$ generates a Sylow $5$-subgroup, so it normalizes that, and hence fixes one point in the action, but it acts as a $5$-cyclc on the other $5$ points. So the image acts doubly transitively.

Now if $G$ was solvable, then $G/H$ would be solvable, so it would have a nontrivial normal elementary $p$-subgroup for some prime $p$. But any nontrivial normal subgroup of a doubly transitive permutation group acts transitively, and so would have order divisible by $6$, contradiction. So $G$ cannot be solvable.

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