The Number of Sylow Subgroups We've been studying Sylow $p$-subgroups, and I've come across this problem.

Let $H$ be a subgroup of $G$, and suppose $G$ is finite. Then, $n_p (H) \leq n_p (G)$, where $n_p$ denotes the number of Sylow $p$-subgroups of a group. 

I am having trouble figuring this one out, and I was wondering if anyone could help?
Thank you.
 A: This is pretty obvious if $H$ contains a Sylow $p$-subgroup of $G$, since then the Sylow $p$-subgroups of $H$ are just the Sylow $p$-subgroups of $G$ contained in $H$, so $n_p(H) \leq n_p(G)$ since $\operatorname{Syl}_p(H) \subseteq \operatorname{Syl}_p(G)$.
If $H$ is smaller though, then something strange seems possible: a Sylow $p$-subgroup of $G$ can have a ton of smaller $p$-subgroups, perhaps $H$ could conspire to have all those smaller $p$-subgroups as Sylow subgroups.  For instance, the alternating group of order 12 has only one Sylow 2-subgroup, but it has 3 subgroups of order 2. Why can't a subgroup have those 3 smaller subgroups as its Sylows?
The answer is Lagrange's theorem. If $Q_1, Q_2 \leq P$ are two $p$-subgroups of a Sylow $p$-subgroup $P$ and $Q_1, Q_2 \leq H$ are both contained in $H$, then of course the subgroup $Q=\langle Q_1 ,Q_2 \rangle$ generated by these two $p$-subgroups is contained in both $P$ and $H$. Since $Q \leq P$, it is a $p$-subgroup, and since $Q \leq H$, it is a $p$-subgroup contained in $H$. If $Q_1 \neq Q_2$, then $|Q| > \max(|Q_1|,|Q_2|)$, and so neither $Q_1$ nor $Q_2$ can be a Sylow $p$-subgroup of $H$. In particular, a Sylow $p$-subgroup $P$ of $G$ can only contribute at most one Sylow $p$-subgroup of $H$, namely $H \cap P$.
Not all $H\cap P$ are actually Sylow $p$-subgroups of $H$, but every Sylow $p$-subgroup of $H$ is contained in one of these by Sylow's containment/extension theorem: every $p$-subgroup of $G$ (such as a Sylow $p$-subgroup of $H$) is contained in a Sylow $p$-subgroup of $G$.
