Order of an Element

How many times must you rotate a square before it returns to its starting position? How many times must you shuffle a deck by a fixed procedure before it returns to its original arrangement? Both questions ask for the order of an element. This is the smallest number of times an operation must be repeated to reach the identity. This single integer, when it exists, controls an enormous amount of the internal structure of a group. The order of an element is one of the first genuinely structural invariants a student meets in group theory. It is more refined than merely asking whether a group is finite or infinite. Two elements in the same finite group can have wildly different orders. The set of possible orders is heavily constrained by the size of the group itself. Lagrange's Theorem will later make this precise. Even without invoking that theorem, the order of an element already determines a whole subgroup. This is the cyclic subgroup it generates. Understanding this subgroup is the gateway to understanding Cyclic Groups and Generators. The big picture is this: every element aa of a group GG generates a subgroup ⟨a⟩\langle a \rangle. It does so purely by taking its powers a0,a1,a−1,a2,a−2,…a^0, a^1, a^{-1}, a^2, a^{-2}, \dots. The order of aa is exactly the size of this subgroup. If the powers of aa never repeat, ⟨a⟩\langle a \rangle is an infinite cyclic group isomorphic to Z\mathbb{Z}. If they do repeat, there is a smallest positive exponent that brings you back to the identity. That exponent is the order of aa. Then ⟨a⟩\langle a \rangle is a finite cyclic group of exactly that size. It is isomorphic to Z/nZ\mathbb{Z}/n\mathbb{Z} under addition. This article gives the rigorous definition of the order of an element. It proves the fundamental theorem identifying ∣⟨a⟩∣|\langle a\rangle| with ord⁡(a)\operatorname{ord}(a). It works out the arithmetic of orders under powers. It also separates the phenomenon from cases where it degenerates — infinite order, or the trivial element. Building on Basic Consequences of the Group Axioms and Examples: Cyclic Groups and Modular Arithmetic, it sets up the machinery needed to classify cyclic groups completely.

Key takeaways
  • The order of a∈Ga \in G is the smallest positive integer nn with an=ea^n = e; if no such nn exists, aa has infinite order.
  • The cyclic subgroup generated by aa, written ⟨a⟩\langle a \rangle, consists of all integer powers of aa.
  • Main theorem: ∣⟨a⟩∣=ord⁡(a)|\langle a \rangle| = \operatorname{ord}(a) — the order of an element equals the size of the subgroup it generates.
  • If ord⁡(a)=n\operatorname{ord}(a) = n, then ak=ea^k = e if and only if n∣kn \mid k; this pins down all the relations among powers of aa.
  • Orders behave predictably under powers: ord⁡(ak)=n/gcd⁡(n,k)\operatorname{ord}(a^k) = n/\gcd(n,k) when ord⁡(a)=n\operatorname{ord}(a) = n.
  • In a finite group every element has finite order dividing ∣G∣|G| (a consequence proved rigorously once Lagrange's Theorem is available; here we establish the order-cyclic-subgroup correspondence that makes that later theorem meaningful).
  • Key idea: the order of an element is precisely the length of the cycle its powers trace out before returning to the identity.

Where this lives today

The order of an element is the load-bearing concept behind the classification of finite abelian groups and the structure theorem for finitely generated abelian groups, both of which describe every such group as a product of cyclic pieces determined by element orders. In number theory, the multiplicative order of an integer modulo nn — exactly the order of that integer's residue class in (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^{\times} — governs the periods of decimal expansions, the structure of pseudorandom number generators, and primality tests such as the Lucas–Lehmer test. In cryptography, the security of Diffie–Hellman key exchange and ElGamal encryption rests on choosing a generator whose order is a large prime, so that the discrete logarithm problem in the corresponding cyclic group is computationally hard; RSA's correctness likewise depends on Euler's theorem, a direct generalization of the order-divides-a-power fact proved here. In physics and crystallography, the order of a symmetry operation (a rotation, reflection, or screw axis) determines the periodicity of a crystal lattice's point group, dictating which rotational symmetries (2-fold, 3-fold, 4-fold, 6-fold) are geometrically possible. In computer science, the order of a permutation produced by a shuffling algorithm determines how many repeated applications are needed to return a deck (or any data structure) to its original arrangement, a fact used in the analysis of pseudorandom permutation ciphers.

Ask the tutor: How does the order of an element change when you pass to a quotient group or a direct product of groups?

DefinitionOrder of an element

Let (G,⋅)(G, \cdot) be a group with identity ee, and let a∈Ga \in G. If there exists a positive integer nn such that an=ea^n = e, the order of aa, denoted ord⁡(a)\operatorname{ord}(a) or ∣a∣|a|, is the smallest such positive integer:

ord⁡(a)=min⁡{n∈Z>0:an=e}.\operatorname{ord}(a) = \min\{ n \in \mathbb{Z}_{>0} : a^n = e \}.

If no positive integer nn satisfies an=ea^n = e, we say aa has infinite order, written ord⁡(a)=∞\operatorname{ord}(a) = \infty. Here ana^n for n∈Zn \in \mathbb{Z} is defined recursively in the usual way: a0=ea^0 = e, an+1=an⋅aa^{n+1} = a^n \cdot a for n≥0n \ge 0, and a−n=(an)−1a^{-n} = (a^n)^{-1} for n>0n > 0. This makes aman=am+na^m a^n = a^{m+n} and (am)n=amn(a^m)^n = a^{mn} hold for all m,n∈Zm, n \in \mathbb{Z}, as established among the Basic Consequences of the Group Axioms. The cyclic subgroup generated by aa is

⟨a⟩={ak:k∈Z}.\langle a \rangle = \{ a^k : k \in \mathbb{Z} \}.

Intuition

Think of repeatedly applying a group element as taking steps around a track. Each application of aa moves you one step forward; a−1a^{-1} moves you one step back. Two things can happen. Either you keep visiting new points forever — the track is infinitely long, and aa has infinite order — or after some number of steps you land back exactly where you started. Because the group operation is completely determined by shifting by aa each time, once you return to the start the whole pattern repeats, and the track closes up into a cycle. The order of aa is the length of that cycle. This picture immediately explains why ak=ea^k = e exactly when n=ord⁡(a)n = \operatorname{ord}(a) divides kk: the only way to return to the starting point after kk steps is to go around the closed loop of length nn some whole number of times. It also explains why the distinct powers of aa are precisely a0,a1,…,an−1a^0, a^1, \dots, a^{n-1} — these are the nn points on the cycle — and why this set is closed under the group operation, forming the subgroup ⟨a⟩\langle a \rangle.

Figure
The cyclic subgroup generated by an element of order 6, sitting inside a larger group
⟨a⟩, order n = 6 e = a⁰ a¹ a² a³ a⁴ a⁵ G (larger group) ⟨a⟩ ≤ G

The relationship, precisely

TheoremOrder equals the size of the cyclic subgroup

Let GG be a group and a∈Ga \in G. 1. If aa has infinite order, then the map Z→⟨a⟩\mathbb{Z} \to \langle a \rangle, k↦akk \mapsto a^k, is a bijection; in particular ⟨a⟩\langle a \rangle is infinite, all powers aka^k (k∈Zk \in \mathbb{Z}) are distinct, and ak=ea^k = e only for k=0k = 0. 2. If aa has finite order nn, then

⟨a⟩={e,a,a2,…,an−1},\langle a \rangle = \{ e, a, a^2, \dots, a^{n-1} \},

these nn elements are pairwise distinct, so ∣⟨a⟩∣=n=ord⁡(a)|\langle a \rangle| = n = \operatorname{ord}(a). Moreover, for any integer kk,

ak=e  ⟺  n∣k,a^k = e \iff n \mid k,

and more generally ai=aj  ⟺  i≡j(modn)a^i = a^j \iff i \equiv j \pmod n.

Proof

Technique: direct computation with the division algorithm, plus a pigeonhole/minimality argument. All facts about exponents (am+n=amana^{m+n} = a^m a^n, (am)n=amn(a^m)^n = a^{mn}, (am)−1=a−m(a^m)^{-1} = a^{-m}) are taken from Basic Consequences of the Group Axioms and used freely.

Part 1 (infinite order). Suppose ord⁡(a)=∞\operatorname{ord}(a) = \infty, meaning an≠ea^n \ne e for every positive integer nn. We show the map ϕ:Z→⟨a⟩\phi: \mathbb{Z} \to \langle a \rangle, ϕ(k)=ak\phi(k) = a^k, is injective; since it is surjective onto ⟨a⟩\langle a \rangle by definition of ⟨a⟩\langle a \rangle, this makes it a bijection. Suppose ϕ(i)=ϕ(j)\phi(i) = \phi(j) for integers i,ji, j with, say, i≥ji \ge j. Then ai=aja^i = a^j. Multiplying both sides on the right by (aj)−1=a−j(a^j)^{-1} = a^{-j} gives

aia−j=aja−j=e,a^i a^{-j} = a^j a^{-j} = e,

and aia−j=ai−ja^i a^{-j} = a^{i-j}, so ai−j=ea^{i-j} = e where i−j≥0i - j \ge 0. If i−j>0i - j > 0, this contradicts the assumption that an≠ea^n \ne e for all positive nn. Hence i−j=0i - j = 0, i.e., i=ji = j. This proves injectivity, so ϕ\phi is a bijection and ⟨a⟩\langle a \rangle is in bijection with Z\mathbb{Z}, hence infinite. Distinctness of all powers and the claim "ak=ea^k = e only for k=0k=0" are immediate: ak=e=a0a^k = e = a^0 forces k=0k = 0 by the injectivity just proved.

Part 2 (finite order nn). Suppose now that aa has finite order nn, i.e., nn is the smallest positive integer with an=ea^n = e. Step A: ak=e  ⟺  n∣ka^k = e \iff n \mid k. (⇐\Leftarrow) If n∣kn \mid k, write k=nqk = nq for some integer qq. Then

ak=anq=(an)q=eq=e.a^k = a^{nq} = (a^n)^q = e^q = e.

(⇒\Rightarrow) Suppose ak=ea^k = e. By the division algorithm for integers, write k=nq+rk = nq + r with q∈Zq \in \mathbb{Z} and 0≤r<n0 \le r < n. Then

e=ak=anq+r=anq⋅ar=(an)q⋅ar=eq⋅ar=ar.e = a^k = a^{nq + r} = a^{nq} \cdot a^r = (a^n)^q \cdot a^r = e^q \cdot a^r = a^r.

So ar=ea^r = e with 0≤r<n0 \le r < n. Since nn is the smallest positive integer with an=ea^n = e, and r<nr < n, the only way ar=ea^r = e can hold is r=0r = 0 (if rr were a positive integer less than nn with ar=ea^r = e, this would contradict minimality of nn). Hence r=0r = 0, so k=nqk = nq, i.e., n∣kn \mid k. Step B: ai=aj  ⟺  i≡j(modn)a^i = a^j \iff i \equiv j \pmod n, for arbitrary integers i,ji,j. We have ai=aj  ⟺  ai−j=ea^i = a^j \iff a^{i-j} = e (multiply/divide by a−j=(aj)−1a^{-j} = (a^j)^{-1} as in Part 1)   ⟺  n∣(i−j)\iff n \mid (i - j) by Step A   ⟺  i≡j(modn)\iff i \equiv j \pmod n. This proves Step B. Step C: ⟨a⟩={e,a,…,an−1}\langle a \rangle = \{e, a, \dots, a^{n-1}\} and these nn elements are distinct. Every integer kk can be written as k=nq+rk = nq + r with 0≤r<n0 \le r < n (division algorithm), and then ak=anqar=e⋅ar=ara^k = a^{nq} a^r = e \cdot a^r = a^r. So every power aka^k equals one of a0,a1,…,an−1a^0, a^1, \dots, a^{n-1}; combined with the definition ⟨a⟩={ak:k∈Z}\langle a \rangle = \{a^k : k \in \mathbb{Z}\}, this shows

⟨a⟩⊆{e,a,…,an−1}.\langle a \rangle \subseteq \{e, a, \dots, a^{n-1}\}.

The reverse inclusion is trivial since each ara^r (0≤r≤n−10 \le r \le n-1) is by definition a power of aa, hence in ⟨a⟩\langle a \rangle. So the two sets are equal. For distinctness: if ai=aja^i = a^j for 0≤i,j≤n−10 \le i, j \le n-1, then by Step B, i≡j(modn)i \equiv j \pmod n. But ∣i−j∣≤n−1<n|i - j| \le n - 1 < n, so the only multiple of nn that i−ji - j can equal is 00. Hence i=ji = j. Thus the listed nn elements e,a,…,an−1e, a, \dots, a^{n-1} are pairwise distinct. Combining Step C with distinctness: ⟨a⟩\langle a \rangle has exactly nn elements, i.e.,

∣⟨a⟩∣=n=ord⁡(a).|\langle a \rangle| = n = \operatorname{ord}(a).

This completes the proof of Part 2, and with Part 1 the theorem is fully established. ■\blacksquare

∎
Corollary⟨a⟩\langle a \rangle is a subgroup

For any a∈Ga \in G, the set ⟨a⟩={ak:k∈Z}\langle a \rangle = \{a^k : k \in \mathbb{Z}\} is a subgroup of GG.

Proof

Technique: direct verification of the subgroup criterion. ⟨a⟩\langle a \rangle is nonempty since e=a0∈⟨a⟩e = a^0 \in \langle a \rangle. It is closed under the operation since aiaj=ai+j∈⟨a⟩a^i a^j = a^{i+j} \in \langle a \rangle for any integers i,ji, j. It is closed under inverses since (ai)−1=a−i∈⟨a⟩(a^i)^{-1} = a^{-i} \in \langle a \rangle. A nonempty subset of a group closed under the operation and under inverses is a subgroup (this is the standard subgroup criterion, a basic consequence of the group axioms), so ⟨a⟩≤G\langle a \rangle \le G. ■\blacksquare

∎
LemmaOrder of a power

Let a∈Ga \in G with ord⁡(a)=n<∞\operatorname{ord}(a) = n < \infty, and let kk be any integer. Then

ord⁡(ak)=ngcd⁡(n,k).\operatorname{ord}(a^k) = \frac{n}{\gcd(n,k)}.
Proof

Technique: reduce to the divisibility criterion of the main theorem and use properties of gcd⁡\gcd. Write d=gcd⁡(n,k)d = \gcd(n,k), and let m=ord⁡(ak)m = \operatorname{ord}(a^k); we must show m=n/dm = n/d. First, (ak)n/d=akn/d=(an)k/d=ek/d=e(a^k)^{n/d} = a^{kn/d} = (a^n)^{k/d} = e^{k/d} = e, where k/dk/d is an integer because d∣kd \mid k. (Here we used an=ea^n = e and that k/d∈Zk/d \in \mathbb{Z} since d=gcd⁡(n,k)∣kd = \gcd(n,k) \mid k.) Since n/dn/d is a positive integer with (ak)n/d=e(a^k)^{n/d} = e, and mm is by definition the smallest such positive integer, we get

m≤n/d.m \le n/d.

Next we show m≥n/dm \ge n/d, i.e. n/d∣mn/d \mid m using the fact that n∣mn \mid m's counterpart. By the main theorem applied to aa (Part 2, Step A), (ak)m=e(a^k)^m = e means akm=ea^{km} = e, and this forces

n∣km.n \mid km.

Write n=dn′n = d n' and k=dk′k = d k' where gcd⁡(n′,k′)=1\gcd(n', k') = 1 (dividing n,kn,k by their gcd dd produces coprime integers — a standard fact about the gcd). Then n∣kmn \mid km becomes dn′∣dk′md n' \mid d k' m, i.e. n′∣k′mn' \mid k' m. Since gcd⁡(n′,k′)=1\gcd(n', k') = 1, Euclid's lemma (a coprime integer dividing a product must divide the other factor) gives

n′∣m,i.e.nd∣m.n' \mid m, \quad \text{i.e.} \quad \frac{n}{d} \mid m.

In particular n/d≤mn/d \le m (as n/dn/d and mm are both positive). Combining m≤n/dm \le n/d and n/d≤mn/d \le m gives m=n/dm = n/d, i.e., ord⁡(ak)=n/gcd⁡(n,k)\operatorname{ord}(a^k) = n/\gcd(n,k). ■\blacksquare

∎

Worked examples

Example 1: An element of order 12 inside (Z/18Z)×(\mathbb{Z}/18\mathbb{Z})^{\times}-style additive computations — using Z/nZ\mathbb{Z}/n\mathbb{Z} under addition. Let G=Z/18ZG = \mathbb{Z}/18\mathbb{Z} under addition, and let a=5a = 5 (the residue class of 55). Note the group operation is addition, so "aka^k" means ka=5k(mod18)ka = 5k \pmod{18}, and the identity is 00. We compute the order of a=5a = 5 directly: we need the smallest positive integer nn with 5n≡0(mod18)5n \equiv 0 \pmod{18}, i.e. 18∣5n18 \mid 5n. Since gcd⁡(5,18)=1\gcd(5,18) = 1, this forces 18∣n18 \mid n, so the smallest positive such nn is n=18n = 18. Thus ord⁡(5)=18\operatorname{ord}(5) = 18, and 55 generates all of GG: ⟨5⟩=G\langle 5 \rangle = G. Now consider b=8b = 8 in the same group. We want the smallest n>0n > 0 with 8n≡0(mod18)8n \equiv 0 \pmod{18}. Since gcd⁡(8,18)=2\gcd(8, 18) = 2, we have 18∣8n  ⟺  9∣4n18 \mid 8n \iff 9 \mid 4n (dividing both sides by gcd⁡=2\gcd = 2); since gcd⁡(4,9)=1\gcd(4,9) = 1, this means 9∣n9 \mid n. The smallest positive such nn is n=9n = 9. So ord⁡(8)=9\operatorname{ord}(8) = 9, and

⟨8⟩={0,8,16,6,14,4,12,2,10}⊂Z/18Z,\langle 8 \rangle = \{0, 8, 16, 6, 14, 4, 12, 2, 10\} \subset \mathbb{Z}/18\mathbb{Z},

obtained as 0,8,16,24≡6,32≡14,40≡4,48≡12,56≡2,64≡10,72≡00, 8, 16, 24 \equiv 6, 32 \equiv 14, 40 \equiv 4, 48 \equiv 12, 56 \equiv 2, 64 \equiv 10, 72 \equiv 0 — indeed 99 distinct elements before returning to 00, confirming ∣⟨8⟩∣=9=ord⁡(8)|\langle 8 \rangle| = 9 = \operatorname{ord}(8) exactly as the theorem predicts. As a check on the order-of-a-power lemma: take a=5a = 5 (order 1818) and k=8k = 8. Then gcd⁡(18,8)=2\gcd(18,8) = 2, so the lemma predicts ord⁡(5⋅8)=ord⁡(40  18)=ord⁡(4)=18/2=9\operatorname{ord}(5 \cdot 8) = \operatorname{ord}(40 \bmod 18) = \operatorname{ord}(4) = 18/2 = 9. Direct check: 4⋅1,4⋅2,…4 \cdot 1, 4\cdot2,\dots gives 4,8,12,16,2,6,10,14,04,8,12,16,2,6,10,14,0 — that's 99 steps to return to 00, confirming ord⁡(4)=9\operatorname{ord}(4) = 9. Matches.

Example 2: A non-abelian group — S4S_4, the symmetric group on 4 letters. Let σ=(1 2 3)(4)\sigma = (1\ 2\ 3)(4), i.e. the 3-cycle sending 1→2→3→11\to2\to3\to1 and fixing 44, living in G=S4G = S_4 under composition (identity ee = the identity permutation). Compute powers: σ1=(1 2 3)\sigma^1 = (1\,2\,3), σ2=(1 3 2)\sigma^2 = (1\,3\,2) (applying the cycle twice), and σ3=e\sigma^3 = e since applying a 3-cycle three times returns every element to itself. No smaller positive power works: σ1≠e\sigma^1 \ne e (it moves 11 to 22) and σ2≠e\sigma^2 \ne e (it moves 11 to 33). So ord⁡(σ)=3\operatorname{ord}(\sigma) = 3, and

⟨σ⟩={e,(1 2 3),(1 3 2)},\langle \sigma \rangle = \{e, (1\,2\,3), (1\,3\,2)\},

a cyclic subgroup of order 33, exactly matching ∣⟨σ⟩∣=ord⁡(σ)=3|\langle \sigma \rangle| = \operatorname{ord}(\sigma) = 3 as guaranteed by the theorem. Now let τ=(1 2 3 4)\tau = (1\,2\,3\,4), the 4-cycle. Its powers are τ1=(1 2 3 4)\tau^1 = (1\,2\,3\,4), τ2=(1 3)(2 4)\tau^2 = (1\,3)(2\,4), τ3=(1 4 3 2)\tau^3 = (1\,4\,3\,2), τ4=e\tau^4 = e, and none of τ,τ2,τ3\tau, \tau^2, \tau^3 equals ee (each moves some element). So ord⁡(τ)=4\operatorname{ord}(\tau) = 4. Apply the order-of-a-power lemma with k=2k=2: gcd⁡(4,2)=2\gcd(4,2) = 2, so the lemma predicts ord⁡(τ2)=4/2=2\operatorname{ord}(\tau^2) = 4/2 = 2. Direct check: τ2=(1 3)(2 4)\tau^2 = (1\,3)(2\,4), and (τ2)2=(1 3)(2 4)⋅(1 3)(2 4)=e(\tau^2)^2 = (1\,3)(2\,4)\cdot(1\,3)(2\,4) = e while τ2≠e\tau^2 \ne e, confirming ord⁡(τ2)=2\operatorname{ord}(\tau^2) = 2. This also illustrates that a power of an element of order 44 need not itself have order 44 — the exponent's gcd with 44 genuinely reduces the order, and the reduction is not merely an upper bound but exact.

CounterexampleDropping "smallest": why minimality in the definition cannot be relaxed

The definition insists nn be the smallest positive integer with an=ea^n = e — not merely some positive integer with that property. Dropping minimality breaks the entire correspondence between order and ∣⟨a⟩∣|\langle a \rangle|. To see this concretely, take G=Z/6ZG = \mathbb{Z}/6\mathbb{Z} under addition and a=2a = 2. We have a6=6⋅2=12≡0(mod6)a^6 = 6 \cdot 2 = 12 \equiv 0 \pmod 6, so 66 is a positive integer with a6=ea^6 = e. If we (incorrectly) declared "the order" to be this non-minimal witness 66, we would predict ∣⟨a⟩∣=6|\langle a \rangle| = 6. But directly: ⟨2⟩={0,2,4}\langle 2 \rangle = \{0, 2, 4\} (since 2⋅3=6≡02\cdot 3 = 6 \equiv 0, and 2,42, 4 are the nonzero multiples before returning), so ∣⟨2⟩∣=3|\langle 2 \rangle| = 3, not 66. The true (minimal) order is ord⁡(2)=3\operatorname{ord}(2) = 3, since 2⋅1=2≠02 \cdot 1 = 2 \ne 0, 2⋅2=4≠02\cdot 2 = 4 \ne 0, 2⋅3=6≡02 \cdot 3 = 6 \equiv 0: the smallest positive witness is 33, and indeed 3=∣⟨2⟩∣3 = |\langle 2\rangle|, matching the theorem. The non-minimal witness 66 gives the wrong count entirely, because Step A of the proof (that ak=e  ⟺  n∣ka^k = e \iff n \mid k) genuinely requires nn to be minimal — the division-algorithm argument in the proof only forces the remainder rr to vanish because r<nr < n and nn was chosen as small as possible. Without minimality, that step collapses and the identification of ⟨a⟩\langle a \rangle with {e,a,…,an−1}\{e, a, \dots, a^{n-1}\} as a list of distinct elements fails.

Scope, edge cases & common traps

SituationWhat happensWhy
a=ea = eord⁡(e)=1\operatorname{ord}(e) = 1e1=ee^1 = e, and 11 is the smallest positive integer, period.
GG infinite, a≠ea \ne eorder may still be finitee.g. −1∈Z×-1 \in \mathbb{Z}^{\times} (multiplicatively) has order 22 even though Z×\mathbb{Z}^\times has only two elements; more strikingly, a finite-order element can sit inside an infinite group like GL2(R)GL_2(\mathbb{R}).
GG finiteevery element has finite orderthe powers a,a2,a3,…a, a^2, a^3, \dots can only take finitely many values, so by pigeonhole some ai=aja^i = a^j with i<ji<j, forcing aj−i=ea^{j-i}=e; a rigorous bound on which orders occur needs Lagrange's Theorem.
ak=ea^k = e for some kk not a multiple of ord⁡(a)\operatorname{ord}(a)impossiblethis is precisely ruled out by Step A of the theorem: ak=e  ⟺  ord⁡(a)∣ka^k = e \iff \operatorname{ord}(a) \mid k.
Confusing ord⁡(a)\operatorname{ord}(a) with ∣G∣|G|these can differ wildlye.g. in S4S_4 (∣G∣=24|G|=24) the element (1 2)(1\,2) has order 22, far smaller than 2424.

The mathematician behind it

Origins · 1 minJoseph-Louis Lagrange and how this was discovered
JL
Joseph-Louis Lagrange
1736–1813 · Turin and Paris, foundational work on permutations and equations

Lagrange's investigations into the solvability of polynomial equations by radicals led him to study the permutations of the roots and how repeatedly applying a substitution cycles through a finite set of arrangements before returning to the start. Although the modern axiomatic notion of a group and the abstract definition of "order of an element" postdate him — crystallizing later in the work of Cauchy, Galois, and Cayley — the phenomenon that a permutation's powers eventually cycle back to the identity, and that this cycle length divides the size of the ambient symmetric group, traces directly to Lagrange's 18th-century analysis of resolvent equations.

History
- 1770 — Lagrange studies permutations of roots of polynomials and functions invariant under them, planting the seed of what becomes permutation group theory.
- 1824 — Niels Henrik Abel uses properties of roots of unity (elements of finite multiplicative order) in proving the unsolvability of the general quintic by radicals.
- 1830s — Évariste Galois develops the theory of substitution groups, formalizing the cyclic behavior of repeated permutations underlying element order.
- 1854 — Arthur Cayley gives the first abstract definition of a group, separating the concept of "order of an element" from any specific representation as permutations or numbers.
- 1870 — Camille Jordan's treatise systematizes group-theoretic language, cementing "order" (of both groups and elements) as standard vocabulary.
What is the difference between the order of a group and the order of an element?

The order of a group GG, written ∣G∣|G|, is simply its cardinality — the number of elements it contains, possibly infinite. The order of an element a∈Ga \in G is the smallest positive integer nn with an=ea^n = e (or ∞\infty if no such nn exists). These are related but distinct: the theorem proved here shows ord⁡(a)\operatorname{ord}(a) equals the size of the cyclic subgroup ⟨a⟩\langle a \rangle, and once Lagrange's Theorem is available, one learns further that ord⁡(a)\operatorname{ord}(a) must divide ∣G∣|G| when GG is finite. But ord⁡(a)\operatorname{ord}(a) itself is generally much smaller than ∣G∣|G|, except in the special case where aa generates the entire group.

Can an element have order 0?

No. By definition, the order of an element is either a positive integer (the least n>0n>0 with an=ea^n=e) or the symbol ∞\infty signifying that no such positive integer exists. The value 00 is never assigned as an order, though a0=ea^0 = e always holds trivially for every element — that is a statement about the exponent 00, not about the order of aa. Confusing "a0=ea^0 = e trivially" with "ord⁡(a)=0\operatorname{ord}(a)=0" is a common notational slip to avoid.

Does every element of a finite group have finite order?

Yes, and this is not hard to see directly, without invoking Lagrange's theorem. Since GG is finite, the infinite list of powers a1,a2,a3,…a^1, a^2, a^3, \dots cannot consist of pairwise distinct elements, so by the pigeonhole principle ai=aja^i = a^j for some integers 1≤i<j1\le i<j. Multiplying both sides by (ai)−1(a^i)^{-1} gives aj−i=ea^{j-i}=e with j−i>0j - i > 0, so the set of positive integers nn with an=ea^n=e is nonempty, and hence has a least element by the well-ordering principle. That least element is ord⁡(a)\operatorname{ord}(a), which is therefore finite.

How is the multiplicative order used in modular arithmetic related to this definition?

Exactly the same way. If gcd⁡(a,n)=1\gcd(a,n)=1, the residue class of aa lies in the group (Z/nZ)×(\mathbb{Z}/n\mathbb{Z})^{\times} under multiplication, and its "multiplicative order mod nn" is precisely ord⁡(a)\operatorname{ord}(a) in that group: the smallest positive kk with ak≡1(modn)a^k \equiv 1 \pmod n. Euler's theorem states aφ(n)≡1(modn)a^{\varphi(n)} \equiv 1 \pmod n, which by the theorem proved here is equivalent to saying ord⁡(a)∣φ(n)\operatorname{ord}(a) \mid \varphi(n) — a special case of the later fact (via Lagrange's theorem) that element order divides group order.

If aa and bb commute and have finite orders mm and nn, does abab have order lcm⁡(m,n)\operatorname{lcm}(m,n)?

Not necessarily — this common guess is false in general, even for commuting elements, unless ⟨a⟩∩⟨b⟩={e}\langle a \rangle \cap \langle b \rangle = \{e\}. A simple counterexample: in Z/4Z\mathbb{Z}/4\mathbb{Z} (abelian, so everything commutes), take a=1a = 1 (order 4) and b=3=−1b = 3 = -1 (also order 4); then a+b=0a+b = 0 has order 11, not lcm⁡(4,4)=4\operatorname{lcm}(4,4)=4. The correct general statement — that ord⁡(ab)\operatorname{ord}(ab) divides lcm⁡(m,n)\operatorname{lcm}(m,n), with equality guaranteed under the extra hypothesis that ⟨a⟩∩⟨b⟩\langle a\rangle \cap \langle b\rangle is trivial — is a more delicate fact than the basic order-of-a-power lemma proved in this article.

Why do we care about ⟨a⟩\langle a \rangle instead of just the number ord⁡(a)\operatorname{ord}(a)?

Because ⟨a⟩\langle a \rangle is an actual subgroup of GG — a structured object with its own operation table — while ord⁡(a)\operatorname{ord}(a) is just a number describing its size. Knowing ∣⟨a⟩∣=ord⁡(a)=n|\langle a \rangle| = \operatorname{ord}(a) = n tells you ⟨a⟩\langle a\rangle is a cyclic group of order nn, isomorphic to Z/nZ\mathbb{Z}/n\mathbb{Z} via ak↔k  na^k \leftrightarrow k \bmod n. This isomorphism is exactly the bridge that lets you transport arithmetic facts about Z/nZ\mathbb{Z}/n\mathbb{Z} (like the classification of subgroups, or which elements generate the whole group) into facts about ⟨a⟩\langle a \rangle inside any group GG, which is the substance of Cyclic Groups and Generators.

Is the order of an element always well defined, i.e., does the minimum in the definition always make sense?

Yes, whenever the underlying set of positive witnesses {n>0:an=e}\{n>0 : a^n=e\} is nonempty. The well-ordering principle guarantees every nonempty set of positive integers has a least element, so the minimum in the definition genuinely exists whenever some positive power of aa equals ee. When no positive power equals ee, the set is empty and there is no minimum to take — this is exactly the infinite-order case, handled separately and explicitly in the definition rather than left ambiguous.

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