Cyclic Subgroups and the Order of an Element

In the setting of Groups: Standing Notation, let $G$ be a group and let $g\in G$.

(i) Cyclic subgroup. $\langle g\rangle$ is a subgroup of $G$.

(ii) Smallest subgroup containing the element. $\langle g\rangle\subseteq H$ for every subgroup $H$ of $G$ with $g\in H$.

(iii) Exponents giving the identity. If $g$ has finite order $n=\operatorname{ord}(g)$, then for every $m\in\mathbb{Z}$, with $\mid$ as in Divisibility of Integers (i),

$$g^{m}=e\iff n\mid m.$$

(iv) Listing the powers. If $g$ has finite order $n=\operatorname{ord}(g)$, then the elements $g^{k}$ with $k\in\mathbb{N}_{0}$ and $k<n$ are pairwise distinct, and

$$\langle g\rangle=\{g^{k}:k\in\mathbb{N}_{0},\ k<n\}.$$

(v) Order of the cyclic subgroup. If $g$ has finite order $n=\operatorname{ord}(g)$, then $\langle g\rangle$, a group with the operation of $G$ restricted to it by (i) and Subgroup Criterion and Basic Examples (ii), is finite of order

$$|\langle g\rangle|=n.$$

(vi) Elements of infinite order. If $g$ has infinite order, then $g^{m}\neq g^{m'}$ for all distinct $m,m'\in\mathbb{Z}$.

(vii) Finite groups. If $G$ is finite, then every $h\in G$ has finite order, $\operatorname{ord}(h)\mid|G|$, and $h^{|G|}=e$.

(viii) Groups of prime order. If $G$ is finite and $|G|$ is a prime number, then $G$ is cyclic, and every $h\in G$ with $h\neq e$ is a generator of $G$.