Let $G$ be a D22: Group such that

(i) | $p$ is a D571: Prime integer |

(ii) | \begin{equation} |G| = p \end{equation} |

Then $G$ is a D1303: Cyclic group.

Result R822
on D1303: Cyclic group

Group of prime order is cyclic

Formulation 0

Let $G$ be a D22: Group such that

(i) | $p$ is a D571: Prime integer |

(ii) | \begin{equation} |G| = p \end{equation} |

Then $G$ is a D1303: Cyclic group.

Proofs

Let $G$ be a D22: Group such that

(i) | $p$ is a D571: Prime integer |

(ii) | \begin{equation} |G| = p \end{equation} |

Let $g \in G$ be a non-identity element in $G$. Such an element exists because $|G| = p \geq 2$. Let $\langle g \rangle$ denote the D1301: Generated subgroup of $g$ within $G$. Now R468: Lagrange's theorem states
\begin{equation}
|\langle g \rangle| [G : \langle g \rangle] = |G| = p
\end{equation}
That is, $|\langle g \rangle|$ divides $p$. Since $p$ is prime, then $|\langle g \rangle| \in \{ 1, p \}$. Since $g$ is a non-identity element, the generated subgroup $\langle g \rangle$ must contain at least two elements. This is because $g$ is included by default and the identity element needs to be included in order for it to qualify as a group, and the two elements are distinct. Thus $|\langle g \rangle| = p$. Since $|\langle g \rangle| = G$ and since $\langle g \rangle \subseteq G$, R2745: The only subset of finite set with same cardinality is the ambient set itself implies that $\langle g \rangle = G$. $\square$