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A cyclic group is a group that can be generated by repeating a single group element $g$ and its inverse element:For an additive group, the same expression is written $\{kg:k\in \mathbb{Z}\}$: $kg$ means repeated addition, including subtraction when $k$ is negative. A finite cyclic group has a repeating cycle, while an infinite cyclic group is isomorphic to the additive integers. The notation $C_n$ refers to a finite cyclic group of order $n$.

This definition compresses a whole composition rule into one repeatable operation. Once $g$ is chosen, $g^a{g}^b=g^{a+b}$, so working with group elements becomes working with integer exponents. In the finite case those exponents are reduced using modular arithmetic. It also proves that every cyclic group is an abelian group, because $a+b=b+a$. A finite cyclic group of order $n$ has exactly $d$ solutions of $x^d=e$ for every positive integer divisor $d$ of $n$: in exponents, $n\mid dk$ has exactly $d$ solutions modulo $n$. The standard finite and infinite models are described in Group Theory (v4.01) by James S. Milne (section 1.16).

The finite cyclic group $C_4=\mathbb{Z}/4\mathbb{Z}$ and the Klein four-group $V_4=C_2\times C_2$ are both abelian groups of order four. In $C_4$, composition is addition modulo four. In $V_4$, label the elements $00,01,10,11$ and add each coordinate modulo two. For example, $01+10=11$ and $11+11=00$.

The order of a group element is preserved by a group isomorphism: an equation $g^m=e$ holds exactly when its image satisfies the same equation. In $C_4$, both $1$ and $3$ have order four. In $V_4$, each nonidentity element has order two. Hence these groups are not isomorphic, despite having the same order of a finite group. No relabelling can repair this discrepancy.

For a geometric model of $V_4$, take the symmetries of a non-square rectangle: the identity element, reflection in its horizontal axis, reflection in its vertical axis, and rotation through $180^{\circ }$. The two reflections commute, and their composite is the half-turn. Every nonidentity motion is its own inverse element. This makes the two binary switches concrete: choosing whether to apply each reflection specifies all four motions. The case of abelian groups of prime-square order generalizes this distinction from $p=2$ to any prime $p$.

An endomorphism is a morphism $f:X\to X$: its source and target are the same object, with the same specified structure. For a group it is an operation-preserving group homomorphism back into that group; for a vector space it is a linear endomorphism; for a ring it is a ring endomorphism. A self-map must satisfy the chosen compatibility conditions to qualify. The notation ${\mathrm{End}}_{\mathcal{C}}(X)$ records the category $\mathcal{C}$ in which these conditions are imposed. See Riehl, Category Theory in Context, Definition 1.1.10 and the following paragraph.

An endomorphism can lose information or miss part of its target. On the additive integers, $d(k)=2k$ preserves addition and is injective, but is not surjective: no integer maps to $1$. On the vector space ${\mathbb{R}}^2$ over the real numbers, the linear endomorphism $P(x,y)=(x,0)$ loses the second coordinate, so it is neither injective nor surjective. Neither is an automorphism. The same doubling rule becomes a group isomorphism when its target is the additive even integers, as explained under isomorphism.

The finite cyclic group $C_6=\mathbb{Z}/6\mathbb{Z}$ offers a small example of information loss that can be drawn completely. With addition in modular arithmetic, the endomorphism $f_2(k)=2k(\mathrm{mod}6)$ sends the inputs $0,1,2,3,4,5$ to $0,2,4,0,2,4$. Its image of a group homomorphism is $\{0,2,4\}$ and its kernel of a group homomorphism is $\{0,3\}$. Each reached output has two inputs; the other three outputs are missed.

For an endomorphism of a group or ring with a finite underlying set, injectivity is equivalent to surjectivity by counting; either condition makes it an automorphism. The additive integers example shows why the finiteness assumption matters. A linear endomorphism of a finite-dimensional vector space also has equivalent injectivity and surjectivity, this time by the rank-nullity theorem even when its underlying set is infinite.

Self-maps can be iterated without changing the object: $f^2=f\circ f$, $f^3=f\circ f\circ f$, and so on. Composition in a category and the identity morphism make the endomorphisms of a fixed object in a locally small category into a monoid. This packages repeated compatible operations, including irreversible ones, into a single structure.

The general language has a documented historical milestone in the 1945 paper of Samuel Eilenberg and Saunders Mac Lane. They defined categories of objects and compatible maps, and distinguished maps admitting inverses within that category. Their examples included groups, vector spaces, and topological spaces. This gave a common framework for compatible self-maps and their invertible cases. See Eilenberg and Mac Lane, General Theory of Natural Equivalences, pp. 234 and 237–239.

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 finite abelian group combines a finite set of reversible operations with commutativity. All finite cyclic groups qualify, but direct products of groups supply examples that need more than one generator of a group. The Fundamental theorem of finitely generated abelian groups says that every finite abelian group is a direct product of groups of finite cyclic groups; the finite case has no infinite cyclic factors.^[ref]

These factors distinguish different ways to have the same total number of states. A single four-position dial gives $C_4$, while two independent binary switches give $C_2\times C_2$. Both have four states and commutative composition, yet only the dial permits a single operation whose repetition visits all four states. The following Cayley tables make that distinction visible without relying on the labels alone.

A finite cyclic group of order $n$ is isomorphic to the additive group $C_n=\mathbb{Z}/n\mathbb{Z}$. The group isomorphism sends the congruence class of $k$ to $g^k$, where $g$ is a chosen generator of a group. For $n=6$, the identity element is $0$, $4+5=3$ modulo six, and the inverse element of $4$ is $2$.

One finite cyclic group can have several choices of generator of a group. Adding $1$ visits all six elements before returning to zero; adding $5$ also visits all six, in the reverse order. Adding $2$ reaches only $\{0,2,4\}$ from zero, and adding $3$ reaches only $\{0,3\}$. The following permutations of the six elements show what each repeated step can reach.

This gives a precise version of a familiar scheduling question: does a fixed step visit every position, or repeatedly miss some? The generators of a finite cyclic group are exactly the steps coprime to the total number of positions. Changing the step changes the orbit through the same underlying group; the diagram does not replace its group operation with a different one.

A finite group is a group whose underlying set has finitely many elements. Its order of a finite group is the number of elements, counting the identity element. The group operation still has to satisfy all the group axioms: merely having a finite list of states does not make a group.

For example, the six congruence classes modulo six form a finite cyclic group under addition, whereas the permutations of three objects form the symmetric group $S_3$, also with six elements. The first is an abelian group; the second is a non-abelian group. Thus the order of a finite group is a useful first measurement, but the composition rule contains information that the count alone loses.

Finiteness makes group structure accessible to exact calculation. A Cayley table for a finite group of order $n$ has $n^2$ entries, and every group element eventually returns to the identity element when repeated. Lagrange's theorem strengthens this observation: the order of a group element, being the size of its cyclic subgroup, divides $n$. The cyclic group of order four and Klein four-group give a small example of how that constraint can distinguish different composition rules.

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.

A group automorphism is a bijective group homomorphism $\alpha :G\to G$. It preserves the group operation, the identity element, and inverse elements; its inverse is another group homomorphism. Thus it is a group isomorphism with identical source and target. These maps form the automorphism group $\mathrm{Aut}(G)$ under function composition.^[ref]

For the additive finite cyclic group $C_6$, take $f_5(k)=5k(\mathrm{mod}6)$. Its output list on $0,1,2,3,4,5$ is $0,5,4,3,2,1$, a permutation of all six elements. It preserves addition because $5(a+b)=5a+5b(\mathrm{mod}6)$, and applying it twice gives $25k=k(\mathrm{mod}6)$. Hence it is a group automorphism and its own inverse function. Compare the collapsing $f_2$ in the figure under endomorphism. The general criterion under endomorphism of a finite cyclic group says exactly when such a map is reversible: its multiplier must be coprime to the size of the finite cyclic group.

A permutation need not be a group automorphism, even if it fixes the identity element. Let $\pi$ exchange $1$ and $2$ in $C_6$ and fix $0,3,4,5$. It is a bijection, and $\pi (0)=0$, butThus it fails to preserve addition and is not even a group homomorphism. Reversibility alone does not preserve the group operation.

A group automorphism carries each cyclic subgroup onto the cyclic subgroup generated by the image of its defining element, preserving the order of a group element. In particular, it carries a generator of a group to another generator of a group. For $C_6$, the existing calculation of generators of a finite cyclic group leaves only the images $1$ and $5$ for $1$, so its automorphism group consists exactly of the identity function and $f_5$. This turns the problem of compatible relabeling into a much smaller choice than enumerating arbitrary permutations.

For the trivial action of the finite cyclic group $C_m$ on $\mathbb{Z}$,for $k\ge 1$. The periodic resolution of a finite cyclic group proves this directly.

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.

For a positive integer $n$ and a congruence class $k$ in the additive finite cyclic group $\mathbb{Z}/n\mathbb{Z}$, the order of a group element isIndeed, write $d=\gcd (n,k)$, $n=d{n}^{\prime }$ and $k=d{k}^{\prime }$. Returning to zero after $m$ steps means $n\mid mk$, equivalently $n^{\prime }\mid m{k}^{\prime }$. Since $n^{\prime }$ and $k^{\prime }$ are coprime, this holds exactly when $n^{\prime }\mid m$; the least positive such $m$ is $n^{\prime }$. This argument also covers $k=0$, since $\gcd (n,0)=n$.

For $n=6$, the orders of $0,1,2,3,4,5$ are respectively $1,6,3,2,3,6$. The translation in a group $x\mapsto x+k$ has the same period from every starting element. Consequently its cycles each have length $n/d$ and there are $d$ of them. They are exactly the cosets of the cyclic subgroup generated by $k$, which explains both the lengths and the number of components in the picture.

Use the positive-exponent quantum Fourier transform convention $F_N|x⟩=N^{-1/2}{\sum }_t{e}^{2\pi ixt/N}|t⟩$. The period of a function here is the least positive translation leaving the function unchanged on the cyclic group ${\mathbb{Z}}_N$. The injective promise means $f(x)=f(x^{\prime })$ exactly when $x\equiv x^{\prime }(\mathrm{mod}r)$. In particular $r\mid N$: the subgroup of period translations is generated by $r$, and wrapping the argument by $N$ preserves its value. This is exact quantum period finding on a finite cyclic group, rather than the truncated, nondividing-period version requiring continued fractions.

Prepare a uniform superposition state in the first quantum register using $F_N|0⟩$, and perform a reversible computation of the function into a second quantum register, cleaning its workspace by uncomputation:A quantum measurement in the computational basis of the second quantum register leaves a periodic coset state in the first:For the quantum Fourier transform of this periodic coset state, the finite geometric series vanishes unless the measured label is $t=mN/r$. On those labels its normalized amplitude is $r^{-1/2}{e}^{2\pi iam/r}$. Thus $m$ has the uniform distribution on a finite set on $\{0,\dots ,r-1\}$, irrespective of $a$.

Compute the greatest common divisor by the Euclidean algorithm, and set $q=N/\gcd (t,N)$, with $\gcd (0,N)=N$. Since $t=mN/r$,The exact period recovery from a Fourier sample is therefore certified by testing $f(q\mathrm{mod}N)=f(0)$. Every candidate $q$ divides $r$ and lies between $1$ and $r$, so the injective promise makes that equality equivalent to $q=r$. This includes $r=N$, where the test uses input $0$, and $r=1$, which always succeeds. The test does not assert that an arbitrary equality of function values certifies an arbitrary period; the Fourier-derived divisor restriction is essential.

The concise result isHere $\phi$ is the Euler totient function, with $\phi (1)=1$. The totient lower bound from the Mertens product gives $\phi (r)/r=\Omega (1/\max \{1,\log \log N\})$ for $r\le N$. Each trial uses polynomial time quantum Fourier transforms, function evaluations and Euclidean algorithm arithmetic in $\log N$. Repeating $O(\max \{1,\log \log N\})$ independent trials gives constant success probability; the herald accepts only correct answers.

The printed asymptotic bound has the wrong direction for this argument. In Big O notation, the stated upper bound on the number of coprime integers supplies no success lower bound and is itself false uniformly: for prime $N$, $\phi (N)=N-1$, not $O(N/\log \log N)$. Likewise the single-trial success need not be $O(1/\log \log N)$, since it tends to one for $r=N$ prime. The meaningful corrected guarantee is the lower bound above. If an upper-bounded success rate is literally required, one may deliberately retain an accepted result with an independent probability proportional to $1/\max \{1,\log \log N\}$; this discards useful successes and is unnecessary for efficient recovery.

First consider a $K_{t,t}$-free bipartite graph with parts $X,Y$, each of size at most $n$, and $m$ edges. Count pairs $(S,y)$ in which $S$ is a $t$-element subset of the graph neighbourhood of $y\in Y$. Any $t$ vertices of $X$ have at most $t-1$ common neighbours, since $t$ common neighbours would form the complete bipartite graph $K_{t,t}$. ConsequentlyIf the average degree of a vertex is bounded in terms of $t$, then $m=O(n)$ and the required result is immediate. Otherwise, convexity of $x\mapsto \left(\genfrac{}{}{0ex}{}{x}{t}\right)$ and the preceding inequality giveand hence $m=O(n^{2-1/t})$. Every graph has a bipartite spanning subgraph containing at least half its edges, obtained by choosing a random bipartition. Applying the bipartite estimate to that subgraph proves the Kővári–Sós–Turán theorem in the required form:

The stated "nice" subsets of the finite cyclic group ${\mathbb{Z}}_n$ are Sidon sets. Ifthen $a+d=c+b$. Niceness forces $(a,b)=(c,d)$; the alternative pairing would force $a=b$. Thus the $|A|(|A|-1)$ ordered nonzero differences are distinct elements of the $n-1$ nonzero residue classes. Thereforeso

Fix a real angle fraction $\alpha$, and repeatedly multiply by $g=e^{2\pi i\alpha }$ in the circle group. The generated cyclic subgroup is $H_{\alpha }=\{g^k:k\in \mathbb{Z}\}$. Equivalently, the angle fractions are the fractional parts of $k\alpha$, with one full turn identified with zero.

If $\alpha =p/q$ is a rational number in lowest terms, with $q>0$ and $p,q$ coprime, then $g^q=1$ and no smaller positive power is $1$. Hence $H_{\alpha }$ is a finite cyclic group of order $q$, consisting of equally spaced points. For $\alpha =2/7$, the successive angle fractions, starting at $k=0$, areThe subgroup is finite, so it is a closed subgroup of the circle group; it is also a discrete subgroup. It is not an open subgroup, since every ambient neighbourhood of any of its points contains an arc, rather than only these seven points.

If $\alpha$ is an irrational number, no nonzero integer $k$ satisfies $k\alpha \in \mathbb{Z}$, so the powers never repeat. The stronger statement is that $H_{\alpha }$ is a dense set in the circle group: every nonempty open arc contains a power $g^k$. Thus its closure is all of $\mathbb{T}$, while $H_{\alpha }$ itself is a countable proper subgroup and is not a closed subgroup or a discrete subgroup. See irrational rotation and everywhere interval frequency under an irrational rotation for the corresponding motion and its long-term frequencies.

For $\alpha =\sqrt{2}-1$, direct computation of the fractional parts of $k\alpha$ gives, to six decimal places,For example, $\{3\alpha \}=3\sqrt{2}-4\approx 0.242641$ and $\{5\alpha \}=5\sqrt{2}-7\approx 0.071068$. These calculations specify exactly which points the diagram displays.

This example shows why algebraic generation and closure must be distinguished: one irrational rotation generates a countable cyclic subgroup, yet its closure is the whole uncountable compact group.