Maximum number of Sylow subgroups 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?
 A: Possibly a partial answer - we first consider a particular prime and then we prove something slightly stronger. Let $p$ be a prime dividing the order of the group $G$. Now it is standard that $n_p:=|Syl_p(G)|=[G:N_G(P)]$. But $P \subseteq N_G(P)$ and certainly $|P| \geq p$, implying $|N_G(P)| \geq p$. Hence $n_p \leq \frac{1}{p}|G| \leq \frac{1}{2}|G|$.
Corollary Let $|G|=p^aq^b$, with $p$ and $q$ prime numbers and $p \lt q$ and $a$ and $b$ positive integers.
(1) If $p=2$ and $q \geq 7$, then the total number of Sylow subgroups of $G$ is smaller than $\frac{2}{3}|G|$. 
(2) If $p$ is odd, then then the total number of Sylow subgroups of $G$ is smaller than $\frac{2}{3}|G|$.
Proof Apparently, the total number of Sylow subgroups of $G$ equals $n_p + n_q \leq (\frac{1}{p}+\frac{1}{q})|G|$. In case (1) we have $(\frac{1}{2}+\frac{1}{q})|G| \lt \frac{2}{3}|G|$, since $q \gt 6$. In case (2) $(\frac{1}{p}+\frac{1}{q})|G| \leq (\frac{1}{3}+\frac{1}{5})|G| \lt \frac{2}{3}|G|. \square$
Note: this principle can be further generalized to more prime factors, for example if $|G|=p^aq^br^c$, with $5 \leq p \lt q \lt r$, the bound of $\frac{2}{3}|G|$ still holds.
A: Here's a strongly simplified proof when the bound $2/3$ is replaced with 1 (using the same starting point as in Matt's answer).
Write $|G|=\prod_p p^{a_p}$. Write $\mathcal{P}$ for the set of primes, $\mathcal{P}_2=\{p\in \mathcal{P}:a_p\ge 2\}$ and $\mathcal{P}_2=\{p\in \mathcal{P}:a_p=1\}$ and $q=\min(p_1)$. Then the number of (nontrivial) Sylow subgroups satisfies:
$$N/|G|\le \sum_{p\in\mathcal{P}_2}\frac{1}{p^2}+\frac{1}{q}$$
(where $q=\infty$, $1/q=0$ if $\mathcal{P}_1=\emptyset$).
In particular, it satisfies 
$$N/|G|\le \sum_{p\in\mathcal{P}}\frac{1}{p^2}+\frac{1}{(q-1)}-\frac{1}{q^2}$$
(see this post about $\sum p^{-2}$)
using that $\sum_{p\in\mathcal{P}}\frac{1}{p^2}\le 453/1000$, we see that if $q\ge 5$ then $N/|G|<453/100+1/4-1/25=663/1000<2/3$.
For $q=3$ it yields $N/|G|<46/100+1/2-1/9<85/100<1$.
For $q=2$ it only yields $N/|G|<46/100+1-1/4=121/100$, so to prove $N<|G|$ we need a little more. If $q=2$, then the signature gives a homomorphism onto the group on 2 elements, and all $p$-Sylow for odd $p$ are contained in the index 2 kernel. If $s_p$ is the set of elements of order $p$ and $N_p$ is the number of $p$-Sylow, we always have 
$$\frac{N}{|G|}\le \sum_{p\in\mathcal{P}_2}\frac{N_p}{|G|}+\sum_{p\in\mathcal{P}_1}\frac{s_p}{|G|(p-1)}$$
if $q=2$ we thus have $\frac{N_p}{|G|}\le\frac{1}{2p^2}$ for all $p>2$, also the unique subgroup of index 2 consists exactly of those elements with odd order. So $s_2\le |G|/2$ and $\sum_{p>2}s_p\le |G|/2$, so $\sum_{p>2}s_p/(p-1)\le |G|/4$. Hence 
$$\frac{N}{|G|}\le \frac12\left(\sum_{p\in\mathcal{P}}\frac{1}{p^2}-\frac{1}{4}\right)+\frac{3}{4}<86/100.$$
(Discussing on whether $a_3=1$ improves this to $<80/100$.)
To conclude, using nothing else than Sylow theorem, signature in the symmetric group, and an upper bound for $\sum p^{-2}$, we get $N<(86/100)|G|$.
A: 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.
