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 $G$.
- $\order {\map Z G} \divides \order G$
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
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51 \beta$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 51.2$ The quotient group $G / Z$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Corollary $10.22$