# Cyclic Groups

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Definition. Let G be a group and a be any arbitrary element in G (a ∈ G). Then, the set ⟨a⟩={an | n ∈ Z} is a subgroup of G. We call it the cyclic subgroup generated by a. Futhermore, it is the smallest subgroup of G that contains a.

If we are using the addition notation, ⟨a⟩ = {na | n ∈ ℤ}.

Proof.

• a0 = e, so the identity is in ⟨a⟩.
• If g and h ∈ ⟨a⟩ ⇒ g = am, h = an for some m, n ∈ Z ⇒ gh = am+n, g-1 = a-n where m+n and -n are obviously integers, so both gh and g-1∈ ⟨a⟩.

Futhermore, any subgroup H of G containing a, must also contain all the power of a by the group closure’s property, and therefore ⟨a⟩ is the smallest subgroup of G that contains a.

More generally, if g1, g2 ….gr ∈ G, ⟨g1, g2,…, gr⟩ = ⟨{g1, g2,…, gr}⟩ is the subgroup generated by g1, g2,…, and gr.

Examples.

• 2Z = {··· -4, -2, 0, 2, 4, 6, ···} ≤ ℤ, 3Z = {··· -6, -3, 0, 3, 6, 9, ···} ≤ ℤ, and 4ℤ = {··· -8, -4, 0, 4, 8, 12, ···} ≤ ℤ. 2Z, 3Z, and 4Z are infinite cyclic groups generated by 2, 3, and 4 respectively. Futhermore, ⟨4, 2⟩ = ⟨2⟩ = 2ℤ = {2n | n ∈ ℤ} = {…, -2, 0, 2, 4, 6, ···} ≤ ℚ.

• Let (ℤ, +), n ∈ ℤ, nℤ = {r·n | r ∈ ℤ} = ⟨n⟩ = ⟨-n⟩ is a cyclic subgroup of ℤ for all integers n.

• Consider (ℚ*, ·), the multiplicative group of nonzero rational numbers, let H be the set of all powers of 2, H = {2n | n ∈ ℤ} = {··· 1/4, 1/2, 1, 2, 4, 8,···}. H is a subgroup of ℚ* determined or generated by 2, i.e., H = ⟨2⟩ ≤ ℚ*

Proof:

∀a, b ∈ H, a = 2m, b = 2n, ab-1 = 2m2-n = 2m-n ∈ H since m-n ∈ ℤ∎

Definition. Let G be a group. A cyclic group G is a group that can be generated by a single element a. More formally, G is cyclic if there exists at least an element a ∈ G: G = ⟨a⟩ = {an | n ∈ ℤ}. We call such an element a generator of G.

The order of a is the number of elements in ⟨a⟩, that is, the order of the cyclic subgroup that it generates. For a finite cyclic group G of order n we have ⟨a⟩ = {e, a, a2,…, an-1}, where e is the identity element, ai = aj whenever i≡j (mod n); in particular an = a0 = e, and a-1 = an-1. In other words, the order of a is the smallest positive integer n such that an = e, and we write it as |a| = n. If no such positive integer exists, we say that a has infinite order, |a| = ∞.

Examples.

• (ℤ, +) is cyclic. It is generated by 1 and -1, ⟨1⟩ =⟨-1⟩ = ℤ. Therefore, ℤ is an infinite cyclic group with two generators, 1 and -1 because each integer n can be written uniquely as n·1 and (-n)(-1).

• n = {[0], [1], …, [n-1]} is also cyclic. It is generated by [1]. Consider ℤn under addition. Then, mℤn = ⟨m⟩ = {km | k ∈ ℤn} is a cyclic subgroup of ℤn for all congruence classes m. 1 and n-1 are always generators, but there can be more.n = 1ℤn = ⟨1⟩ = ⟨n-1⟩.

Abusing notation, we refer m instead of [m], that is, just using the representatives of their equivalent classes.

• In ℤ6=$\frac{ℤ}{6ℤ}$, ⟨0⟩ = {0}, ⟨1⟩ = ⟨5⟩ = $\frac{ℤ}{6ℤ}$. ⟨2⟩ = ⟨4⟩ = {0, 2, 4} and ⟨3⟩ = {0, 3} are nontrivial proper subgroups of $\frac{ℤ}{6ℤ}$. To sum up, 2ℤ6 = {0, 2, 4} = ⟨2⟩ = ⟨4⟩ ≤ ℤ6.

• [Let G be the dihedral group] Dn = ⟨r, s |rn = s2 = 1, rs = sr-1⟩, n ≥ 3, and let’s consider H the subgroup of all rotations of the n-gon, then H is cyclic, H = ⟨r⟩ = {1, r, r2, ···, rn-1}, |H| = n.

• 5 = ⟨1⟩ = ⟨2⟩ = ⟨3⟩ = ⟨4⟩. Observe that all of them are coprime to 5. Every element is a generator except the identity. We can generalize this result. The generators of ℤn are all its elements which are coprime to n.

Proof.

Let p ∈ ℤn & p coprime to n ⇒ [By the Euclidean Algorithm] ∃a, b such that 1 = ap + bn ⇒ ap ≡ 1 (mod n) ⇒ [Obviously, ap ∈ ⟨p⟩] 1 ∈ ⟨p⟩ ⇒ ℤn = ⟨1⟩ = ⟨p⟩∎

• The multiplicative group of integers modulo n, e.g., U10 = {1, 3, 7, 9} (Figure 1.a.) is a cyclic group, 3 and 7 are generators. U10 = {30, 31, 33, 32} = ⟨3⟩ = ⟨7⟩. U9 = {1, 2, 4, 5, 7, 8} is cyclic of order 6 generated by the element 2. U9 = ⟨2⟩ = {20 = 1, 21 = 2, 22 = 4, 23 = 8, 24 = 16 ≡9 7, 25 = 32 ≡9 5} and 26 = 64 ≡9 1. U14 = {1, 3, 5, 9, 11, 13} = ⟨3⟩ = ⟨5⟩. U17 = {1, 2, 3, 4, 5, . . . , 16} is cyclic of order 16 generated by the element 3.

Not every group is a cyclic group, e.g., S3, the symmetry group of an equilateral triangle and U8 are not cyclic groups. U8 = {1, 3, 5, 7}. Observe that 328 1, 528 1, 728 1, and therefore ⟨1⟩ = {1}, ⟨3⟩ = {1, 3}, ⟨5⟩ = {1, 5}, ⟨7⟩ = {1, 7}, U8 is not a cyclic group, U8 ≠ ⟨a⟩ for any a ∈ U8. In other words, there is no element of order 4.

Similarly, observe the subgroups of D3 = {id (e), rotations (ρ1 -120° around the center anticlockwise-, ρ2 -240°-), reflections or mirror images on all its vertices (μ1, μ2, and μ3)}. No single element generates the entire group (Figure 1.b.).

A subgroup lattice is a diagram or graph whose nodes are the subgroups of a group G, with an edge drawn between every pair of subgroups H, K satisfying H ≤ K and there are not intermediate subgroups, that is, $H \nleq L \nleq K$ drawn in an eye-pleasing fashion.

# Properties of Cyclic Groups

Theorem. Let G be a group, let a be an arbitrary element of G, a ∈ G. If a has infinite order, |⟨a⟩| = ∞ then ai=aj iff (if and only if) i=j. If a has finite order n, |⟨a⟩| = n < ∞, then ⟨a⟩ = {e, a, a2,..., an-1}, they are all distinct elements of ⟨a⟩, and ai=aj iff (if and only if) n|(i-j), n divides i-j.

Proof.

If a has infinite order ⇒ there is no a natural number n ∈ ℕ, n > 0, such that an = e. Therefore, ai = aj ⇒ ai-j = e ⇒ i-j = 0 ⇒ i = j.

Let’s assume that |a| = n, we are going to prove that ⟨a⟩ = {e, a, a2,…, an-1}. Certainly, e, a, a2,…, an-1 are in ⟨a⟩. Suppose that ak ∈ ⟨a⟩ ⇒ ∃q, r ∈ ℤ: k = qn + r, 0 ≤ r < n ⇒ ak = aqn + r = aqnar = (an)qar = [By assumption, |a| = n, an = e] eqar = ear = ar ∈ {e, a, a2,…, an-1} since 0 ≤ r < n.

The powers {1, a, a2, ···, an-1} are all distinct. Suppose ai = aj where 0 ≤ j < i < n ⇒ ai-j = a0 = e and 0 < i -j < n ⊥ (minimality of n, an = e)

Next, let’s assume ai = aj ⇒ ai-j = e ⇒ i - j = qn + r, 0 ≤ r < n so ai-j = (an)qar = ear = ar = e, 0 ≤ r < n ⇒ Since, by definition, n is the least positive integer such that an is the identity ⇒ r = 0 ⇒ i - j = qn ⇒ n divides i-j, i.e., n | (i-j)

Conversely, if n | (i-j) ⇒∃q such that i - j = nq ⇒ ai-j = anq = (an)q = eq = e ⇒ [ai-j = e] ai = aj.

Example: ℤ20 = {0, 1, 2, 3,…, 19}. ⟨5⟩ = {5, 10, 15, 0}, |⟨5⟩|=4, so by this theorem 5i = 5j iff 4|(i-j).

0 = 50 =[54 = 20 ≡20 0] 54 =[58 = 40 ≡20 0] 58
5 = 51 =[55 = 25 ≡20 5] 55 = 59
10 = 52 = 56 = 510
15 = 53 = 57 = 511

Corollary. For any arbitrary element, say a ∈ G, its order is equal to the order of the cyclic subgroup generated by this very element, i.e., |a| = |⟨a⟩| = {e, a, a2,..., an-1}.

Theorem. Let G = ⟨a⟩ be a cyclic group generated by a. If a has finite order n, then so does G and |G| = n. Futhermore, if b ∈ G and |b| = n, then b is a generator for G.

Proof. Since a has finite order, G = ⟨a⟩ = {e, a, a2,···, an-1} and all these elements are distinct because if k, l < n, ak = al, let’s assume k > l ⇒ [G is a group, there are inverses] aka-l = e ⇒ ak-l = e ⇒ 0 < k-l < n ⊥ (it contradicts minimality of n, |a| = n). Therefore, |G| = n.

If b ∈ G, |b| = n, by similar reasoning ⟨b⟩ = {e, b, b2, …, bn-1} and they are all distinct elements. Hence, G = ⟨b⟩ = ⟨a⟩, and b is a generator∎

Proposition. Let G be an arbitrary group, let a be an element of G of order n. Then, am = e iff |a| | m. In particular, if b = am,, then the order of b divides the order of a.

Proof.

Let |a| = n. By the division algorithm, ∃q, r ∈ ℤ: m = qn + r and 0 ≤ r < n. ar = am-qn = am(an)-q = am. Then, am = e ↭ ar = e ↭ [0 ≤ r < n, |a|=n, so by the minimality of n] r = 0 ↭ m = qn

In particular, bn = [b = am] (am)n = (an)m = em = e ⇒ [By the previous statement] |b| | n = |a|∎

The idea behind these results is that multiplication in ⟨a⟩ is essentially done by addition modulo n that is, if (i+j) mod n = k, then aiaj = ak. In other words, in the finite case, it works as addition in ℤn. Otherwise, if a has infinite order, multiplication in ⟨a⟩ is essentially done by addition. It works as addition in ℤ, that is, no modular arithmetic is needed.

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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