Cyclic group

There are exactly two models. The group \((\mathbb{Z},+)\) is infinite cyclic, generated by 1 (and also by \(-1\)). The group \((\mathbb{Z}_n,+)\) is cyclic of order \(n\); in multiplicative notation it is written \(\{e,g,\dots,g^{n-1}\}\) with \(g^n=e\). Every cyclic group is isomorphic to one of the two, so up to the names of its elements there is only one cyclic group of each order. Other examples come in disguise, like \((\mathbb{Z}_7^*,\cdot)\), which is generated by 3.

A cyclic group is always abelian, because powers of \(g\) commute. The converse is false: the Klein group, of order 4, is abelian and none of its elements has order 4. Moreover, every subgroup of a cyclic group is cyclic: if \(H\ne\{e\}\), it is generated by the smallest positive power of \(g\) it contains.

Let \(G=\langle g\rangle\) have order \(n\). Then \(g^k\) has order \(n/\gcd(n,k)\), so \(g^k\) is another generator exactly when \(\gcd(n,k)=1\), and there are \(\varphi(n)\) generators. Also, for each divisor \(d\) of \(n\) there is a unique subgroup of order \(d\), the one generated by \(g^{n/d}\), and there are no other subgroups. This turns the subgroups of \(\mathbb{Z}_n\) into the list of divisors of \(n\).

A cyclic group of prime order \(p\) has no proper subgroups and any element other than the identity generates it; this is the starting point for the theorem of Lagrange and, later, for the classification of finite abelian groups, which break into cyclic pieces.