Order of Finite p-Group is Power of p

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Theorem

Let $G$ be a finite group.

Let $p$ be a prime number.

Let all elements of $G$ have order a power of $p$.


Then $G$ is a $p$-group.


Proof 1

Aiming for a contradiction, suppose:

$\order G = k p^n: p \nmid k$

where $\order G$ denotes the order of $G$.

By Divisors of Power of Prime:

$k \nmid p^n$


From the First Sylow Theorem:

$\exists H \le G: \order H = k$

where $H \le G$ denotes that $H$ is a subgroup of $G$.


Thus:

$\exists h \in H: \order h \divides k \implies \order h \nmid p$

where $\divides$ denotes divisibility.

Thus:

$\exists h \in G: \order h \ne p^n: n \in \Z$

Thus by Proof by Contradiction, $\order G$ must be a power of $p$.

$\blacksquare$


Proof 2

Let every element of $G$ be a $p$-element.

Let $q$ be a prime number which is a divisor of the order $\order G$ of $G$.

By Cauchy's Lemma, there exists an element of $G$ whose order is a divisor of $q$.

But as the order of all elements of $G$ divide $p^n$ it follows that $q = p$.

Thus $G$ is a group whose order is $p^n$ for some $n \in \Z_{>0}$.

$\blacksquare$


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