TITLE: If H is a subgroup of G, then H has no more Sylow subgroups than G
QUESTION [20 upvotes]: If $H$ is a subgroup of the finite group $G$, then how do I show that $n_p(H) \leq n_p(G)$? Here $n_p(X)$ is the number of Sylow $p$-subgroups in the finite group $X$.

Here is my attempt: Suppose the order of $G$ is $n$, the order of $H$ is $m$, then $m$ divides $n$ by Lagrange's theorem. By Sylow's counting theorem $n_p(H) = 1 \pmod p$ and $n_p(H)$ divides $m$, $n_p(G) =1 \pmod p$, $n_p(G)$ divides $n$. 
How do I continue after this?

REPLY [2 votes]: Bonus information: there is a paper of prof Gabriel Navarro from 2003 that implies that for a solvable group $G$, and any prime $p$ dividing $|G|$, $n_p(H) \mid n_p(G)$ for every subgroup $H$ of $G$. Here $n_p(X)=\#Syl_p(X)$ (which is understood to be $1$ if $p \nmid |X|$).