Order of cyclic subgroup is order of generator

Dependencies:

  1. Cyclic Group
  2. Order of element in finite group is finite
  3. Integer Division Theorem

Let $G$ be a finite group and $g \in G$. Then $|\langle g \rangle| = \operatorname{order}(g)$.

Proof

Let $k = \operatorname{order}(g)$. Therefore, $k \ge 1$.

Case 1: $k = \infty$

Order of an element of a finite group is finite. Therefore, $|\langle g \rangle|$ is finite implies $k$ is finite. The contrapositive is that $k = \infty \implies |\langle g \rangle| = \infty$. Therefore, $|\langle g \rangle| = k$.

Case 2: $k$ is finite

Let $H = \{g^0, g^1, \ldots, g^{k-1}\}$. Therefore, $|H| = k$ and $H \subseteq \langle g \rangle$.

Let $i \in \mathbb{Z}$. By the integer division theorem, $\exists q \exists r$ such that $i = qk + r$ where $0 \le r < k$. \[ g^i = g^{qk+r} = (g^{k})^q * g^r = g^r \in H \]

Therefore, $\langle g \rangle \subseteq H \implies \langle g \rangle = H$. Therefore, $|\langle g \rangle| = k$.

Dependency for:

  1. Order of element divides order of group

Info:

Transitive dependencies:

  1. Group
  2. Identity of a group is unique
  3. Order of element in finite group is finite
  4. Subgroup
  5. Inverse of a group element is unique
  6. Conditions for a subset to be a subgroup
  7. Cyclic Group
  8. Integer Division Theorem