On the maximum number of Sylow subgroups I have some doubts regarding the first sentences below the fourth case in the more elementary solution  by Prof. Samuel to problem 11856 of the Monthly. Here you have a screenshot of the relevant part of the solution under consideration:

It's not totally clear to me how it was that Prof. Samuel obtained the inequality $N_{2} \leq \frac{|G|}{12}$. If $p_{1}=2$, the first sentence in the aforecited paragraph only implies that $N_{2} \leq \frac{|G|}{4}$.
Would anybody here be so kind as to elaborate on this part of Prof. Samuel's solution to this gorgeous Monthly problem? Do we actually need to appeal to Burnside's transfer theorem?

As Derek says in a comment, it should be $|N_2|\le |G|/8$, which just barely is enough to still make the $2/3$ bound. The transfer theorem tells us that the Sylow subgroup is not in the center of its normalizer. If the order is $4$, that means the normalizer has order at least $12$. Since the number of Sylow subgroups is the index of the normalizer, and the normalizer has at least $8$ elements, we get the bound.
I edited that post so many times I got a friendly warning about it last year, so I think I'm going to leave it alone.
