Finite Group is p-Group iff Order is Power of p
Jump to navigation
Jump to search
Theorem
Let $p$ be a prime number.
Let $G$ be a finite group.
Then $G$ is a $p$-group if and only if the order of $G$ is a power of $p$.
Proof
Necessary Condition
Let $G$ be a finite group whose order is $p^n$ for some $n \in \Z_{>0}$.
Let $g \in G$.
From Order of Element Divides Order of Finite Group, the order of $g$ is a divisor of $p^n$.
That is, $x$ is a $p$-element by definition.
As $x$ is arbitrary, it follows that all elements of $G$ are $p$-elements.
Thus $G$ is a $p$-group.
$\Box$
Sufficient Condition
Let $G$ be a finite $p$-group.
By definition, every element of $G$ is a $p$-element.
From Order of Finite $p$-Group is Power of $p$:
- $G$ is a finite group whose order is $p^n$ for some $n \in \Z_{\ge 0}$.
Hence the result.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Definition $11.8$