Why is the following an implication of the First Sylow Theorem?

Question

Humphrey's A course in Group Theory states, on page 100:

A $p$-group is a group in which the order of every element is a power of $p$

Furthermore, it is stated that

The First Sylow Theorem shows that if $G$ is a finite $p$-group then $|G|$ must be a power of $p$.

Why is this the case? Let us suppose that $|G|=p^nk$ where $k$ is not a multiple of $p$.

Then we can find a Sylow-$p$ subgroup of $G$ (subgroup with $p^n$ elements) by the First Sylow Theorem. What I don't understand is how this contradicts the fact that $G$ is a finite p-group.

Can someone give me a hint?

Answer 1

Take a prime factor $q$ of $k$ and look at the Sylow-$q$ subgroup, instead. The elements of this subgroup cannot have order which is a power of $p$.

Answer 2

If $|G|$ is divisible by a prime power with a base other than $p$, say, $q^n$ for prime $q$, then the first Sylow theorem says there is a subgroup of order $q^n > 1$ which contains elements that have an order dividing $q^n$; in particular, there would be elements whose order is not a power of $p$.

Answer 3

Just to mention that Cauchy suffices. If $k>1$, then $k$ has a prime factor $q\ne p$, and hence (Cauchy) $G$ has a subgroup of order $q$, which is not a power of $p$. Contradiction. Therefore, $k=1$