Some groups are sprawling and complicated. Others are secretly the work of a single element — pick one member g, keep applying the group operation to it, and that one element grinds out the entire group all by itself. Groups like that are called cyclic, and they are the simplest, cleanest family in all of group theory: everything you could want to know about them is encoded in one generator and one number.

You already have cyclic groups in your hands without naming them. A clock face is one: start at 12 and keep stepping forward by one hour and you visit every hour before looping back — the single step "add one hour" generates all twelve positions. The rotations of a pinwheel are another. Even the plain integers, marched out by repeatedly adding 1, are cyclic. This page is about that one idea: a group grown from a single seed, taught from its definition all the way to a complete description of its subgroups.

The definition: a group with a single generator

Let G be a group and g \in G one of its elements. Write g^2 = g \ast g, g^3 = g \ast g \ast g, and generally g^k for g combined with itself k times; negative powers are the inverses, g^{-k} = (g^{-1})^k, and g^0 = e is the identity. Collect all of these powers into one set:

This set \langle g \rangle is always a subgroup of G — the cyclic subgroup generated by g. When it happens to be all of G — when a single element's powers exhaust the whole group — we say G itself is cyclic, and we call g a generator, writing G = \langle g \rangle.

Order of an element = size of the group it grows

How big is \langle g \rangle? That is exactly the question answered by the order of an element. The order of g is the smallest positive k with g^{k} = e — the number of steps before the powers first loop back to the identity. And the beautiful coincidence is that this count is the size of the group it generates:

If g has order 6, then e, g, g^2, g^3, g^4, g^5 are six distinct elements and g^6 is back to e — a cyclic group of order 6. If no power of g ever returns to e, then g has infinite order and \langle g \rangle is infinite. So "order of the element" and "order of the group it generates" are two names for the same number — a fact worth pinning down, because it lets you read the size of a cyclic group straight off its generator.

The two archetypes: \mathbb{Z} and \mathbb{Z}_n

There are really only two shapes a cyclic group can take, and you meet both in the first week of any algebra course.

• The infinite cyclic group (\mathbb{Z}, +). Here the generator is 1: every integer is a power (i.e. a repeated sum) of 1, since n = 1 + 1 + \dots + 1 and -n uses -1. No sum of 1s ever returns to 0, so 1 has infinite order. (Its only other generator is -1.)
• The finite cyclic group (\mathbb{Z}_n, +) — the integers \{0, 1, 2, \dots, n-1\} added modulo n. Again 1 generates it: step by 1 and you tour every residue before wrapping back to 0 after exactly n steps. This is the clock with n hours.

Remarkably, that is the whole list. Up to relabelling, every cyclic group is either \mathbb{Z} (if infinite) or \mathbb{Z}_n (if it has n elements). One infinite pattern and one finite pattern for each size — that is all cyclic groups there are.

Every cyclic group is abelian

Cyclic groups are automatically commutative, and the reason is almost too short to call a proof. Any two elements of \langle g \rangle are powers of the same g, say g^a and g^b. Then

because ordinary integer addition a + b = b + a already commutes. The group operation just inherits the symmetry of the exponents. So cyclic \Rightarrow abelian — no exceptions. (The converse fails: plenty of abelian groups, such as \mathbb{Z}_2 \times \mathbb{Z}_2, are not cyclic, because no single element reaches all four members. Cyclic is strictly stronger than abelian.)

Which elements are generators?

In \mathbb{Z}_n, the element k is really "step forward by k each time". Whether that tour eventually hits every residue depends entirely on how k sits relative to n:

Worked example — the generators of \mathbb{Z}_6. We need the residues coprime to 6. Checking each: \gcd(1,6)=1 ✓, \gcd(2,6)=2 ✗, \gcd(3,6)=3 ✗, \gcd(4,6)=2 ✗, \gcd(5,6)=1 ✓. Only 1 and 5 survive — so \mathbb{Z}_6 has exactly \varphi(6) = 2 generators. Watch 5 do its work, stepping by 5 each time:

every residue visited — 5 genuinely generates. (Stepping by 5 is the same as stepping backwards by 1, which is why it works: 5 \equiv -1 \pmod 6.)

See it move: does g sweep the whole circle?

Here is the whole story in one picture. The n residues sit evenly around a circle, labelled 0 to n-1. Starting at 0, each arrow jumps forward by the fixed step g = k, tracing the powers of k in \mathbb{Z}_n. If the arrows visit every point before returning to 0, then k is a generator; if they close up early into a shorter loop, k only generates a smaller cyclic subgroup — of length exactly n / \gcd(n, k). Press Refresh to draw a new random n and k and see which kind you get.

Subgroups of a cyclic group are themselves cyclic

Cyclic groups are as tidy on the inside as they are on the outside. Every subgroup of a cyclic group is again cyclic — you never find some knotted, un-generated subset lurking inside. And for the finite case the subgroups line up perfectly with the divisors of n:

Worked example — the subgroup \langle 2 \rangle in \mathbb{Z}_6. Start at 0 and step by 2: 0 \to 2 \to 4 \to 0 — and we are home already. The powers of 2 close up after three steps, so

a cyclic subgroup of order 3. Notice 3 divides 6, as the theorem promises, and the order of the element 2 is 6/\gcd(2,6) = 6/2 = 3 — the formula and the picture agree. The full list of subgroups of \mathbb{Z}_6 is one per divisor of 6: sizes 1, 2, 3, 6, namely \{0\}, \{0,3\}, \{0,2,4\}, and all of \mathbb{Z}_6 — no more, no fewer.