Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2
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Theorem
Let $p$ be a prime number.
Let $G$ be a finite abelian group whose identity is $e$.
Let $G$ have at least $p$ elements of order $p$.
Then:
- $p^2 \divides \order G$
where:
- $\divides$ denotes divisibility
- $\order G$ denotes the order of $G$.
Proof
Let $x \in G$ be of order $p$.
Consider $\gen x$, the subgroup generated by $x$.
By Group of Prime Order p has p-1 Elements of Order p, the elements of $\gen x$ are all of order $p$ except $e$.
Thus, by hypothesis, there must exist another $y \in G$ of order $p$.
Consider the subset of $G$:
- $S := \set {x^i y^j: 0 \le i, j < p}$
By the Finite Subgroup Test, $S$ is a subgroup of $G$ which has $p^2$ elements.
- $\order S \divides \order G$
But $\order S = p^2$ and so $p^2 \divides \order G$.
$\blacksquare$
Examples
Order $3$
Let $G$ be a finite abelian group whose identity is $e$.
Let $G$ have more than $2$ elements of order $3$.
Then:
- $9 \divides \order G$
where:
- $\divides$ denotes divisibility
- $\order G$ denotes the order of $G$.
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $20$