Order of Finite Abelian Group with p+ Order p Elements is Divisible by p^2

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.

By Lagrange's theorem:

$\order S \divides \order G$

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 $20$