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 = a^m, h = aⁿ for some m, n ∈ Z ⇒ gh = a^m+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 ….g_r ∈ G, ⟨g₁, g₂,…, g_r⟩ = ⟨{g₁, g₂,…, g_r}⟩ is the subgroup generated by g₁, g₂,…, and g_r.

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.

  Additive notation: a+b is used instead of a·b, 0 instead of e, -a instead of a^-1, and na instead of aⁿ.

• 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 = 2^m, b = 2ⁿ, ab^-1 = 2^m2^-n = 2^m-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,…, a^n-1}, where e is the identity element, aⁱ = a^j whenever i≡j (mod n); in particular aⁿ = a⁰ = e, and a^-1 = a^n-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).

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

• ℤ₅ = ⟨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 ℤ_n are all its elements which are coprime to n.

Proof.

Let p ∈ ℤ_n & p coprime to n ⇒ [By the Euclidean Algorithm] ∃a, b such that 1 = ap + bn ⇒ ap ≡ 1 (mod n) ⇒ [Obviously, ap ∈ ⟨p⟩] 1 ∈ ⟨p⟩ ⇒ ℤ_n = ⟨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 3² ≡₈ 1, 5² ≡₈ 1, 7² ≡₈ 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=a^j iff (if and only if) i=j. If a has finite order n, |⟨a⟩| = n < ∞, then ⟨a⟩ = {e, a, a²,..., a^n-1}, they are all distinct elements of ⟨a⟩, and aⁱ=a^j 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 = a^j ⇒ a^i-j = e ⇒ i-j = 0 ⇒ i = j.

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

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

Next, let’s assume aⁱ = a^j ⇒ a^i-j = e ⇒ i - j = qn + r, 0 ≤ r < n so a^i-j = (aⁿ)^qa^r = ea^r = a^r = 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 ⇒ a^i-j = a^nq = (aⁿ)^q = e^q = e ⇒ [a^i-j = e] aⁱ = a^j.

Example: ℤ₂₀ = {0, 1, 2, 3,…, 19}. ⟨5⟩ = {5, 10, 15, 0}, |⟨5⟩|=4, so by this theorem 5ⁱ = 5^j 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,..., a^n-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,···, a^n-1} and all these elements are distinct because if k, l < n, a^k = a^l, let’s assume k > l ⇒ [G is a group, there are inverses] a^ka^-l = e ⇒ a^k-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, …, b^n-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, a^m = e iff |a| | m. In particular, if b = a^m,, 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. a^r = a^m-qn = a^m(aⁿ)^-q = a^m. Then, a^m = e ↭ a^r = e ↭ [0 ≤ r < n, |a|=n, so by the minimality of n] r = 0 ↭ m = qn

In particular, bⁿ = [b = a^m] (a^m)ⁿ = (aⁿ)^m = e^m = 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 aⁱa^j = a^k. In other words, in the finite case, it works as addition in ℤ_n. 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.  

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