Cyclic group ($C_n$)

A cyclic group is generated by one element. A cyclic group of order $n$ has exactly $d$ solutions of $x^d=1$ for every positive integer divisor $d$ of $n$.

The cyclic subgroup generated by an element $g$ is $⟨g⟩=\{g^n:n\in \mathbb{Z}\}$. Its size is the order of $g$.

In a finite group, a cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function. Each element generates exactly one cyclic subgroup, so dividing the number of elements of order $m$ by $\phi (m)$ counts these subgroups. Include the identity subgroup using $m=1$ and $\phi (1)=1$. In a symmetric group, element counts come from cycle type and order of a group element is the least common multiple of cycle lengths.

A finite cyclic group of order $n$ is isomorphic to $C_n=\mathbb{Z}/n\mathbb{Z}$.

For any group $G$, evaluation at a chosen generator of a group $x$ of $C_n$ gives a bijection between group homomorphisms $C_n\to G$ and elements satisfying $g^n=1$. A group homomorphism must send $x^j$ to $g^j$. Conversely, this formula is well-defined because exponents differing by a multiple of $n$ give equal powers, and it respects multiplication. The displayed correspondence is a bijection of sets; it is not claimed to be a group isomorphism for a nonabelian target. In a symmetric group, the order of a group element is the least common multiple of its disjoint cycle lengths, making this correspondence easy to enumerate.

An infinite cyclic group is generated by an element of infinite order and is isomorphic to the additive group of integers.

Every finite subgroup of ${\mathbb{C}}^{\times }$ is cyclic. If its exponent is $n$, it is a subgroup of the cyclic group of $n$th roots of unity.

A generator of a cyclic group $G$ is an element $g$ such that every element of $G$ is an integer power of $g$.