Understanding the third Sylow theorem I am trying to understand the following theorem:
Let $G$ be a finite group, $p$ a prime number, and let's suppose $|G|=p^ns$ s.t. $p$ doesn't divide $s$
Let $n_p$ be the number p-sylow subgroups of  G then we have

$$\begin{cases}n_p |s\\n_p  \equiv 1 \mod p\end{cases}$$

Now my problem is with $n_p$ I don't understand why $n_p$ can be anything different from 1. a p-sylow subgroup is a p-subgroup which is maximal, but I don't understand why there can be more than one maximal subgroup of a group G

Consider the set of all subgroups of a given group $G$, with the relation 'being a subgroup' it is a partial order (a lattice actually) and 'directly below' G you could have several subgroups. Now consider the same for finite dimensional vector spaces, take for example $\mathbb{R}^n$, how many different subspaces does it have with dimension $(n-1)$?. In the set of all subspaces, those subspaces would be the maximal ones, then your $p$-Sylow subgroups would be those subgroups 'directly below' $G$ when considering the set of all $p$-subgroups.
