Group of Order Prime Squared is Abelian

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Theorem

A group whose order is the square of a prime is abelian.


Proof

Let $G$ be a group of order $p^2$, where $p$ is prime.

Let $\map Z G$ be the center of $G$.

By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.

By Lagrange's Theorem:

$\order {\map Z G} \divides \order G$

It follows that $\order {\map Z G} = 1, p$ or $p^2$.

By Center of Group of Prime Power Order is Non-Trivial:

$\order {\map Z G} \ne 1$


Suppose $\order {\map Z G} = p$.

Then:

\(\ds \order {\map Z G}\) \(=\) \(\ds \index G {\map Z G}\) Definition of Index of Subgroup
\(\ds \) \(=\) \(\ds \order G / \order {\map Z G}\) Lagrange's Theorem
\(\ds \) \(=\) \(\ds p^2 / p\)
\(\ds \) \(=\) \(\ds p\)

So $G / \map Z G$ is non-trivial, and of prime order.

From Prime Group is Cyclic, $G / \map Z G$ is a cyclic group.

But by Quotient of Group by Center Cyclic implies Abelian, that cannot be the case.

Therefore $\order {\map Z G} = p^2$ and therefore $\map Z G = G$.

Therefore $G$ is abelian.

$\blacksquare$


Sources

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