I am scared because we have cavemen instincts and emotions, medieval institutions, modern global oligarchy, narratives and fake news spreading like wildfire, tech giant censorship and bias, and the technology of gods, and it is terrifying, Anawim, #justtothepoint.
The saddest aspect of life right now is that science gathers knowledge faster than society gathers wisdom, Isaac Asimov
Definition. Let G be a group and a be any arbitrary element in G (a ∈ G). Then, the set ⟨a⟩={a^{n} | 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.
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 g_{1}, g_{2} ….g_{r} ∈ G, ⟨g_{1}, g_{2},…, g_{r}⟩ = ⟨{g_{1}, g_{2},…, g_{r}}⟩ is the subgroup generated by g_{1}, g_{2},…, 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 a^{n}.
Consider (ℚ^{*}, ·), the multiplicative group of nonzero rational numbers, let H be the set of all powers of 2, H = {2^{n} | 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 = 2^{n}, ab^{-1} = 2^{m}2^{-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⟩ = {a^{n} | 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, a^{2},…, a^{n-1}}, where e is the identity element, a^{i} = a^{j} whenever i≡j (mod n); in particular a^{n} = a^{0} = e, and a^{-1} = a^{n-1}. In other words, the order of a is the smallest positive integer n such that a^{n} = 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 ℤ_{6}=$\frac{ℤ}{6ℤ}$, ⟨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, 2ℤ_{6} = {0, 2, 4} = ⟨2⟩ = ⟨4⟩ ≤ ℤ_{6}.
[Let G be the dihedral group] D_{n} = ⟨r, s |r^{n} = s^{2} = 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, r^{2}, ···, r^{n-1}}, |H| = n.
ℤ_{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 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⟩∎
Not every group is a cyclic group, e.g., S_{3}, the symmetry group of an equilateral triangle and U_{8} are not cyclic groups. U_{8} = {1, 3, 5, 7}. Observe that 3^{2} ≡_{8} 1, 5^{2} ≡_{8} 1, 7^{2} ≡_{8} 1, and therefore ⟨1⟩ = {1}, ⟨3⟩ = {1, 3}, ⟨5⟩ = {1, 5}, ⟨7⟩ = {1, 7}, U_{8} is not a cyclic group, U_{8} ≠ ⟨a⟩ for any a ∈ U_{8}. In other words, there is no element of order 4.
Similarly, observe the subgroups of D_{3} = {id (e), rotations (ρ_{1} -120° around the center anticlockwise-, ρ_{2} -240°-), reflections or mirror images on all its vertices (μ_{1}, μ_{2}, and μ_{3})}. 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.
Theorem. Let G be a group, let a be an arbitrary element of G, a ∈ G. If a has infinite order, |⟨a⟩| = ∞ then a^{i}=a^{j} iff (if and only if) i=j. If a has finite order n, |⟨a⟩| = n < ∞, then ⟨a⟩ = {e, a, a^{2},..., a^{n-1}}, they are all distinct elements of ⟨a⟩, and a^{i}=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 a^{n} = e. Therefore, a^{i} = 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, a^{2},…, a^{n-1}}. Certainly, e, a, a^{2},…, 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^{qn}a^{r} = (a^{n})^{q}a^{r} = [By assumption, |a| = n, a^{n} = e] e^{q}a^{r} = ea^{r} = a^{r} ∈ {e, a, a^{2},…, a^{n-1}} since 0 ≤ r < n.
The powers {1, a, a^{2}, ···, a^{n-1}} are all distinct. Suppose a^{i} = a^{j} where 0 ≤ j < i < n ⇒ a^{i-j} = a^{0} = e and 0 < i -j < n ⊥ (minimality of n, a^{n} = e)
Next, let’s assume a^{i} = a^{j} ⇒ a^{i-j} = e ⇒ i - j = qn + r, 0 ≤ r < n so a^{i-j} = (a^{n})^{q}a^{r} = ea^{r} = a^{r} = e, 0 ≤ r < n ⇒ Since, by definition, n is the least positive integer such that a^{n} 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} = (a^{n})^{q} = e^{q} = e ⇒ [a^{i-j} = e] a^{i} = a^{j}.
Example: ℤ_{20} = {0, 1, 2, 3,…, 19}. ⟨5⟩ = {5, 10, 15, 0}, |⟨5⟩|=4, so by this theorem 5^{i} = 5^{j} iff 4|(i-j).
0 = 5^{0} =[5^{4} = 20 ≡_{20} 0] 5^{4} =[5^{8} = 40 ≡_{20} 0] 5^{8}…
5 = 5^{1} =[5^{5} = 25 ≡_{20} 5] 5^{5} = 5^{9}…
10 = 5^{2} = 5^{6} = 5^{10}…
15 = 5^{3} = 5^{7} = 5^{11}…
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, a^{2},..., 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, a^{2},···, 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^{k}a^{-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, b^{2}, …, 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}(a^{n})^{-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, b^{n} = [b = a^{m}] (a^{m})^{n} = (a^{n})^{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 a^{i}a^{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.