Proving group is $p$-group by contradiction

Question

http://www.proofwiki.org/wiki/Group_is_P-Group_iff_All_Elements_have_Order_Power_of_P

Is $k$ a prime or a prime power? Sorry for this stupid question but I can't tell what $k$ is in this context :'(

Thanks guys/gals :D

Answer 1

$k$ is simply the quotient $\frac{|G|}{p^n}$, so it is some positive integer. This proof is, however, not very clearly written. Here's another take:

Suppose $|G| = k p^n$, where $k$ is not a multiple of $p$. If $k>1$, choose some prime $q|k$. Then by the First Sylow Theorem, there is a subgroup $H$ of $G$ of order $q^r$ for some $r\ge 1$, so that there is an element $h\in H$ with $|h|$ a power of $q$. But then $|h|$ as an element of $G$ is also a power of $q$; this is impossible since $q\ne p$. Therefore $k=1$ and $|G| = p^n$ as desired.