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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.