The order of a cyclic group equals the order of its generator

Given a finite-ordered group element $𝑎 \in 𝐺$ such that $| 𝑎 | = 𝑛$ , it follows that $| ⟨ 𝑎 ⟩ | = 𝑛$ . #m/thm/group

Proof

Clearly $𝑆 = {𝑒, 𝑎 1, \dots, 𝑎 𝑛 - 1} \subseteq ⟨ 𝑎 ⟩$ . Additionally, since $𝑛$ is the smallest positive integer such that $𝑎 𝑛 = 𝑒$ , each of these elements is unique: we need only show they be exhausted. Let $𝑎 𝑘 \in ⟨ 𝑎 ⟩$ . By the division algorithm $𝑘 = 𝑝 𝑛 + 𝑟$ where $0 \leq 𝑟 < 𝑛$ . Then $𝑎 𝑘 = 𝑎 𝑝 𝑛 + 𝑟 = 𝑎 𝑟 \in 𝑆$ .

#state/tidy | #lang/en | #SemBr