Prime Group is Cyclic

Theorem

Let $p$ be a prime number.

Let $G$ be a group whose order is $p$.


Then $G$ is cyclic.


Proof

Let $a \in G: a \ne e$ where $e$ is the identity of $G$.

From Group of Prime Order p has p-1 Elements of Order p, $a$ has order $p$.

Hence by definition, $a$ generates $G$.

Hence also by definition, $G$ is cyclic.

$\blacksquare$


Sources

  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.4$. Lagrange's theorem: Example $118$
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.7$ Corollary
  • 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Theorem $13$
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 43$
  • 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts: Exercise $1$
  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 44.2$ Some consequences of Lagrange's Theorem
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.19$
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