Proof that the order of any finite $p$-group is a power of $p$ What is the most concise proof that the order of any finite $p$-group is a power of $p$?
 A: Certainly the Theorem of Cauchy would do it - if $q$ is a prime other than $p$, dividing the order of $G$, then there would be an element of order $q$, a contradiction.
A: Given $p$-group $G$ , its order is not a power of $p$, then there is a prime $q$ that devides $|G|$.
Quote: Sylow’s Theorem -- Given $G$ a finite group such that $q^n$ divides $|G|$, where $q$ is prime. Then there exists a subgroup of order $q^n$.
So $G$ has a subgroup $H$ of order $q$. As $q$ is prime, $H$ is a cyclic group, and its members, which are also $G$'s members, have order $q$. This conflicts with the $p$-group definition.
A: From Lagrange's theorem we know only that |G| can be divided by the order of the elements. So, for example, |G| = n * p^k ,where n is an integer not equal to p, and p^k is the order of some element.
By Cauchy's Theorem then, there will be an element of order p.  As Hekster notes above, this implies that there cannot be a different prime factor q of |G|, because that would lead to a violation of the p-group assumption.
Also, if |G| = n * p^k, where n is not a prime or equal to 1, there is a contradiction since n can be written as the product of primes.
