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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]

A cyclic group has one generator of a group: $G=⟨a⟩=\{a^j:j\in \mathbb{Z}\}$. An abelian group has $xy=yx$ for every pair of elements. Powers of one element commute because $a^r{a}^s=a^{r+s}=a^{s+r}=a^s{a}^r$, so every cyclic group is Abelian. The Klein four-group $C_2\times C_2$ is Abelian but is not a cyclic group: every nonidentity element has order two, whereas a generator of a group for a four-element cyclic group would have order four.

Fix a generator of a group $x$ of $C_n$. A group homomorphism is determined by $g=\varphi (x)$ because $\varphi (x^j)=g^j$, and the relation $x^n=1$ forces $g^n=1$. Conversely, such a $g$ defines $\varphi (x^j)=g^j$: exponents differing by a multiple of $n$ give the same value, and addition of exponents verifies the group homomorphism law. Thus the homomorphism from a finite cyclic group correspondence isFor $S_4$, the order of a permutation is the least common multiple of its disjoint cycle lengths. The condition ${\sigma }^4=1$ allows identity, transpositions, two disjoint transpositions, and four-cycles. The sixteen homomorphisms are ${\varphi }_{\sigma }(x^j)={\sigma }^j$, with the full list of possible generator imagesThe remaining eight elements of the symmetric group are three-cycles and do not qualify.