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Every endomorphism of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$, for a positive integer $n$, has the form $f_a(k)=ak(\mathrm{mod}n)$, determined by $a=f_a(1)$. This is the self-map case of homomorphism from a finite cyclic group: repeated addition forces the value at every element. Conversely, each choice of $a$ respects addition and gives a well-defined map on congruence classes.

Let $d=\gcd (n,a)$, the greatest common divisor. The image of a group homomorphism is the cyclic subgroup generated by $a$, with size $n/d$ by order of an element of a finite cyclic group. The kernel of a group homomorphism isIndeed, $ak=0(\mathrm{mod}n)$ is equivalent to $n/d$ dividing $k$, after cancelling $d$ and using coprimality. Consequently $f_a$ is a group automorphism exactly when $a$ and $n$ are coprime. In that case a modular inverse $b$ of $a$ gives $f_a^{-1}=f_b$. This connects reversible self-maps directly to the generators of a finite cyclic group.^[ref]

An element $k$ of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$ is a generator of a group precisely when $\gcd (n,k)=1$. The order of an element of a finite cyclic group proves this immediately: its cyclic subgroup has size $n/\gcd (n,k)$, and this equals $n$ exactly when the greatest common divisor is one. There are therefore $\phi (n)$ choices, where $\phi$ is the Euler totient function.

For $C_6$, the two choices are $1$ and $5$, so $\phi (6)=2$. For $C_8$, they are $1,3,5,7$. Starting from zero and repeatedly adding $3$ modulo eight gives $0,3,6,1,4,7,2,5,0$, visiting every element once before returning. By contrast, adding $2$ gives $0,2,4,6,0$ and misses the odd integers represented in the other coset.

If the cyclic group is written multiplicatively as $⟨g⟩$ with $g$ of order $n$, the same statement says that $g^k$ is a generator of a group exactly when $k$ and $n$ are coprime. This changes notation without changing the calculation.

The order of a finite group is its cardinality, written $|G|$. The identity element is included: $|\mathbb{Z}/6\mathbb{Z}|=6$, and the trivial group has order one, rather than zero.

Keep this count distinct from the order of a group element. In the additive finite cyclic group $\mathbb{Z}/6\mathbb{Z}$, the element $1$ has order six, $2$ has order three, $3$ has order two, and $0$ has order one. Each element's order is the least positive number of repetitions needed to return to the identity element $0$. The order of an element of a finite cyclic group computes these periods directly.

The count also constrains subgroups: Lagrange's theorem gives $|G|=[G:H]|H|$ for a subgroup $H$ of a finite group $G$, where the index of a subgroup counts its cosets. For $H=\{0,2,4\}$ inside $\mathbb{Z}/6\mathbb{Z}$, the two cosets are $\{0,2,4\}$ and $\{1,3,5\}$, so $6=2\cdot 3$. Divisibility is a necessary constraint; it does not imply that every divisor occurs as the order of a group element, as the Klein four-group shows.