TITLE: Maximum number of Sylow subgroups
QUESTION [19 upvotes]: I've been studying Sylow-$p$ subgroups, and I've come across this problem.

Let $G$ be a finite group. Show that the number of Sylow subgroups of $G$ is at most $\frac{2}{3}|G|$ . ($|G|$ is the number of elements of $G$).

I am having trouble figuring this one out, I was wondering if anyone could help?

REPLY [2 votes]: I record this in case it helps anyone to produce a simpler proof. I interpret "Sylow subgroup" to mean "Non-identity Sylow subgroup". If possible, choose a finite group of minimal order subject to having strictly more than $\frac{2|G|}{3}$ Sylow subgroups. I claim that there is no prime divisor $p$ of $|G|$ such that $G$ has a normal $p$-complement. Suppose otherwise, and let $G = PN$ where $N \lhd G$ has order prime to $p$ and $ 1 \neq P \in {\rm Syl}_{p}(G)$.
Then (using minimality) $G$ has at most $\frac{2|N|}{3}$ Sylow subgroups other than Sylow $p$-subgroups, and $G$ has at most $\frac{|G|}{|P|}$ Sylow $p$-subgroups. Hence $G$ has at most $\frac{|G|}{|P|}\left( \frac{2}{3} +1\right)$ Sylow subgroups.
Hence $\frac{5}{3} \frac{|G|}{|P|} > \frac{2|G|}{3},$ so $|P| < \frac{5}{2}$ and $|P| = 2.$ Let us examine the case $|P| = 2$ more carefully
If $N_{G}(P) = P$, then $G$ has an Abelian subgroup $N$ of odd order and index $2$. It is then clear that $G$ has at most $|N| + \log_{3}(|N|)$ Sylow subgroups. This is at most $\frac{4|N|}{3}$ by elementary calculus since $|N| \geq 3.$ This contradicts the fact that $G$ has more than $\frac{2|G|}{3}$ Sylow subgroups. Hence $N_{G}(P) \neq P$, so $|N_{G}(P)| \geq 3|P|.$
Hence $G$ has at most $\frac{2|G|}{6} + \frac{|G|}{6}$ Sylow subgroups ( using minimality again for the odd Sylow subgroups), so at most $\frac{|G|}{2}$ Sylow subgroups, a contradiction.