Cyclic group ($C_n$)

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).

Let $G$ have group order equal to a prime number $p$. Choose any $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$, and it is greater than one. It is therefore $p$, so the cyclic subgroup generated by $g$ is all of $G$. This proves that every nonidentity element generates $G$, and $G$ is isomorphic to the additive group of integers modulo $p$.

A nontrivial group has no proper nontrivial subgroups exactly when it is a cyclic group of prime group order. To prove necessity without presupposing finiteness, take $g\ne e$. Its cyclic subgroup is nontrivial and hence equals $G$. An infinite-order generator would give the proper nontrivial subgroup $⟨g^2⟩$, so its order $N$ is finite. If $N$ were composite, a proper divisor $d$ with $1<d<N$ would make $⟨g^d⟩$ another proper nontrivial subgroup. Thus $N$ is prime. Conversely Lagrange's theorem excludes intermediate subgroup orders in a prime-order group. The qualification “nontrivial” is necessary because every group with more than one element has the proper trivial subgroup $\{e\}$.

A nontrivial subgroup $H$ of $⟨g⟩$ is generated by $g^d$, where $d$ is the least positive exponent with $g^d\in H$. Divide any exponent $k$ with $g^k\in H$ by $d$; its remainder also gives an element in $H$ and must be zero. A quotient group by a normal subgroup is generated by the coset of $g$, because every coset is a power of that coset. The trivial subgroup and quotient are also cyclic.

If a nonempty subset $A$ of ${\mathbb{Z}}_p$ is invariant under translation by a nonzero $t$, then repeated translation of any member runs through all $p$ residues, so $A={\mathbb{Z}}_p$. Thus all $p$ translates of a nontrivial subset are distinct. Partitioning the $k$-element subsets into translation classes of size $p$ proves $p\mid \left(\genfrac{}{}{0ex}{}{p}{k}\right)$ for $0<k<p$.

// Target: analysis.bigb

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 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.

The cyclic group generated by $r$ with $r^4=e$ and $r^2\ne e$ has elements $e,r,r^2,r^3$. It is realized by quarter-turn rotations of a plane. The cyclic group of order four and Klein four-group have the same order but different element orders: the latter has no element of order four.

For real $b,c$, this matrix defines a group representation of $C_4=⟨a:a^4=e⟩$ exactly when $bc=\pm 1$. Indeed its lower block $B$ satisfies $B^2=bc{I}_2$, so $T(a{)}^4=I_3$ exactly when $(bc{)}^2=1$. If $bc=-1$, $T(a{)}^2=\mathrm{diag}(1,-1,-1)\ne I_3$, so its order is four and the representation is faithful. If $bc=1$, its order is exactly two and the kernel of a group homomorphism is $\{e,a^2\}$, giving an unfaithful representation. Other values do not give representations at all, including $bc=0$, when the proposed matrix is singular.

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.

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.

For $G=⟨g⟩$ infinite, every subgroup is $⟨g^m⟩$ for a unique integer $m\ge 0$. This is the classification of subgroups of the additive group of integers transported through a group isomorphism. For $m>0$ the subgroup index is $m$ and the subgroup is again an infinite cyclic group; for $m=0$ it is trivial and has infinite subgroup index.

The integers form an infinite cyclic group under addition, with identity element zero and inverse element $-n$ for $n$. Its two generators are $1$ and $-1$. Every subgroup is $m\mathbb{Z}$ for a unique nonnegative integer $m$: if it is nonzero, take its least positive element and use division with remainder to show that every element is a multiple of it. The zero subgroup corresponds to $m=0$.

These invertible matrices, for $n\in \mathbb{Z}$, form an infinite cyclic group. The displayed multiplication rule, $B_0=I$, and $B_n^{-1}=B_{-n}$ give a group isomorphism from the additive group of integers via $n\mapsto B_n$. Its generator is $B_1$. The subgroups of an infinite cyclic group become $\{B_{mk}:k\in \mathbb{Z}\}$ for integers $m\ge 0$, and their subgroup indices are $m$ when $m>0$. The case $m=0$ is the trivial subgroup and has infinite index. As transformations, these matrices send $(x,y)$ to $(x+ny,y)$.

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 group is a single element $g$ for which $G=⟨g⟩$: every group element is an integer power of $g$. A group admitting such an element is a cyclic group. In additive notation, every element must be an integer multiple of $g$.

Distinguish a single generator of a group from a generating set of a group. The Klein four-group needs two elements to generate it, although it has only four elements; no single element visits all four. A cyclic group need not have a unique generator of a group: $\mathbb{Z}/6\mathbb{Z}$ has $1$ and $5$, while the additive integers have $1$ and $-1$.

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.