An expert is a person who has made all the mistakes that can be made in a very narrow field, Neils Bohr.
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| = ∞.
Theorem: Every cyclic group is Abelian.
Proof: If g and h ∈⟨a⟩ ⇒ g = a^{m}, h = a^{n} for some m, n ∈ Z ⇒ gh = a^{m+n} = a^{n+m} = a^{n}a^{m} = hg∎
Theorem. Every subgroup of a cyclic group is cyclic.
Proof. Let G be cyclic, there exists at least an element a ∈ G: ⟨a⟩ = G. Suppose that H is a subgroup of G, H ≤ G. There are two possibilities, namely:
If H is the trivial subgroup, H = {e} ⇒ H = ⟨e⟩, and H is obviously cyclic.
Otherwise, if H is not the trivial subgroup, H ≠ {e} ⇒ ∃g ∈ H: g ≠ e, g = a^{n} ⇒ [Well-ordering principle. Every non-empty set of positive integers contains a least element.] Let m be the smallest natural number such that h = a^{m} ∈ H
We claim that H = ⟨h⟩ = ⟨a^{m}⟩. Let h’ be an arbitrary element of H, h’ ∈ H ⇒ h’ = a^{k}.
By the division algorithm, ∃q, r: k = mq + r where 0 ≤ r < m ⇒ a^{k} = a^{mq+r} = a^{mq}a^{r} = (a^{m})^{q}a^{r} = h^{q}a^{r} ⇒ a^{r} = a^{k}h^{-q} ∈ H [h’ = a^{k} and h^{-q}∈ H], but r < m, and h = a^{m} is the smallest element in H ⇒ r = 0 ⇒ k = mq, h’ = a^{k} = a^{mq} = h^{q}, and therefore, since h’ was an arbitrary element of H and it is generated by h = a^{m} ⇒ H = ⟨h⟩ = ⟨a^{m}⟩∎
Theorem. Let G be a finite cyclic group, G = ⟨g⟩. Then, G has exactly one subgroup of order d for each d | |G|, namely H = ⟨g^{n/d}⟩ ≤ G
Proof
Let |G| = n. By assumption d | n, let us consider h = g^{n/d}. It is obvious that h^{d} = (g^{n/d})^{d} = g^{n} = [G = ⟨g⟩, |G| = n] e.
If a ∈ ℤ, 1 ≤ a < d ⇒ [a < d ⇒ an/d < dn/d(=n)] 1 ≤ an/d < n. Thus, h^{a} = g^{na/d} ≠ e because 1 ≤ an/d < n and no power less than n can get g to the identity. Then, d is the smaller power that can get h to the identity ⇒ [By definition] |h| = d ⇒ |⟨h⟩| = d. So there is at least one such subgroup.
Let’s prove uniqueness. Let k ∈ G: |k| = d. G is cyclic so ∃m: k = g^{m}. g^{dm} = (g^{m})^{d} = k^{d} = e ⇒ [g^{a} =e ↭ |g| | a] n | dm ⇒ ∃l ∈ ℤ: nl = dm ⇒ (n/d)l = m ⇒ k = g^{m} = (g^{n/d})^{l} = h^{l} ∈ ⟨h⟩ ⇒ k = h^{l} ∈⟨h⟩ ⇒ ⟨k⟩ ⊆ ⟨h⟩, but these are two finite sets with the same order, so they are the same set ∎
Corollary. The subgroups of ℤ are exactly nℤ and the subgroups of ℤ_{n} are exactly mℤ_{n}.
Proof.
(ℤ, +) is an infinite cyclic group. nℤ = {nk | k ∈ ℤ} are all cyclic subgroups of ℤ by definition, but we have demonstrated that being ℤ cyclic (ℤ = ⟨1⟩), then all its subgroups are necessarily cyclic ⇒ These are exactly all the subgroups of ℤ, and more importantly, there are no other subgroups left.
By similar reasoning, ℤ_{n} is a cyclic finite group, all their cyclic groups are mℤ_{n} = {mk | k ∈ ℤ_{n}} by definition, and as was reasoned before, their only subgroups are cyclic subgroups, so the subgroups of ℤ_{n} are exactly mℤ_{n}.
Theorem. Let G be a finite cyclic group generated by an element a, G = ⟨a⟩, |G| = n. Then a^{k} = e iff n divides k.
Proof.
⇒) Let’s suppose that a^{k} = e. By the division algorithm, k = nq +r, where 0 ≤ r < n ⇒ a^{k} = a^{nq+r} = a^{nq}a^{r} ⇒ [a^{k} = e, a^{nq}=a^{0}=e because a^{n} = e or it could also be argued that nq≡0 (mod n)] e = a^{r} ⇒ r = 0 because r < n and n is the smallest integer m such that a^{m} = e ⇒ k = nq ⇒ n | k.
⇐) Conversely, let’s suppose n divides k ⇒ k = ns for some integer s ⇒ a^{k} = a^{ns} = (a^{n})^{s} = e^{s} = e∎
Exercise: Suppose G is cyclic, a ∈ G, a is an element of order n, and a^{24} = e. What are the possibilities of n?
a^{24}=e ⇒ k = 24 ⇒ n|24 ⇒ n needs to be 1, 2, 3, 4, 6, 8, 12 or 24.
Proposition. Let G be an arbitrary group, a ∈ G, and let m, n ∈ ℤ. If a^{n} = e and a^{m} = e, then a^{gcd(m, n)} = 1.
Proof.
By the Euclidean Algorithm, there exists r, s ∈ ℤ such that d = mr + ns, d = gcd(m, n) =_{notation} (m, n) ⇒ a^{d} = a^{mr+ns} = a^{mr}a^{ns} = (a^{m})^{r}(a^{n})^{s} = [By assumption, a^{m} = e, a^{n} = e] e^{r}e^{s} = e∎
Theorem. Let a be an arbitrary element of order n in a group G, and let k be a positive integer. Then, ⟨a^{k}⟩=⟨a^{gcd(n,k)}⟩ and |⟨a^{k}⟩| = $\frac{n}{gcd(n,k)}.$
Proof.
Let b = a^{k} and d = gcd(k, n). We need to know the smallest integer m such that e = b^{m} = a^{km}. By the previous theorem, [Theorem. Let G be a finite cyclic group generated by an element a, G = ⟨a⟩, |G| = n. Then a^{k} = e iff n divides k.] this is the smallest integer m such that n divides km or n/d divides m(k/d). d = gcd(k, n), and therefore n/d and k/d are coprimes, [n/d|m(k/d), (n/d,k/d)=1]the smallest m s.t. n/d | m is m = n/d ⇒ [Let G be a finite cyclic group, G = ⟨g⟩. Then, G has exactly one subgroup of order d for each d | |G|, namely H = ⟨g^{n/d}⟩ ≤ G] ⟨a^{k}⟩ = ⟨a^{n/d}⟩ = ⟨a^{n/gcd(k,n)}⟩.
Example. Suppose |a|=24 and k=10. ⟨a^{k}⟩= ⟨a^{10}⟩= ⟨a^{gcd(n,k)}⟩ = ⟨a^{gcd(24, 10)}⟩ = ⟨a^{2}⟩ = {a^{2}, a^{4}, a^{6}, … a^{22}, a^{0}} and |⟨a^{10}⟩| = $\frac{24}{gcd(24,10)}=\frac{24}{2}=12.$
Corollaries:
As a result, the generators of ℤ_{n} are the integers k such that |⟨a^{k}⟩| =[a = 1 in ℤ_{n} = ⟨1⟩, k = k·1] |⟨k⟩| n/d = n where d = gcd(k, n) = 1. In other words, the generators of ℤ_{n} are the integers k that are relatively prime to n.
Examples:
Theorem. Let G be a cyclic group with generator a. If the order of G is infinite, then G is isomorphic to (ℤ, +). Otherwise, G is isomorphic to (ℤ_{n}, +_{n})
Proof.
If the order of G is infinite, ∀m ∈ ℤ, m ≠ 0, a^{m} ≠ e. We claim that ∀h, k ∈ ℤ, a^{h} ≠ a^{k}. Suppose a^{h} = a^{k} and h > k ⇒ a^{h}a^{-k} = a^{h-k} = e ⊥ .i.e, contrary to our assumption that G is infinite. Therefore, ∀g ∈ G, g = a^{m} for a unique power m ∈ ℤ.
The map Φ: (G,·) → (ℤ, +), Φ(g) = Φ(a^{m}) = m is well defined (We have already demonstrated that ∀g ∈ G, g = a^{m} for a unique power m ∈ ℤ), homomorphism (∀g_{1}, g_{2} ∈ G, ∃!m_{1}, m_{2} s.t. g_{1} = a^{m1}, g_{2} = a^{m2} ⇒ Φ(g_{1}·g_{2}) = Φ(a^{m1}·a^{m2}) = Φ(a^{m1+m2}) = m_{1} + m_{2} = Φ(g_{1})+Φ(g_{2})), one to one (m_{1} = m_{2} ⇒ g_{1} = a^{m1} = a^{m2} = g_{2}), and onto ℤ (∀m ∈ ℤ, m ≠ 0, a^{m}≠e, a → m, e → 0) ⇒ G ≋ ℤ.
Let’s suppose the order of G is finite. a^{m}= e for some m ∈ ℤ, m > 0. Let n be the smallest positive integer such that a^{n} = e. The map Φ: G → ℤ^{n}, given by Φ(a^{i})=i, for i=0, 1, 2,…n-1 is well-defined, homomorphism, one to one, and onto ℤ_{n} ⇒ G ≋ ℤ_{n}
We claim that a^{0} (e), a, a^{2}, a^{3},… a^{n-1} are all distinct and comprise all elements of G = ⟨a⟩, |G| = n.
Hint: s = nq + r, 0 ≤ r < n, a^{s} = a^{nq}a^{r} = (a^{n})^{q}a^{r} =e^{q}a^{r} = a^{r}, a^{r-s} = e. If 0 < s < r < n, a^{r-s} = e and 0 < r-s < n so it contradicts our choice of n, r = 0 and s = nq. Besides, a^{0} (e), a, a^{2}, a^{3},… a^{n-1} are all distinct.
is Φ homomorphism? Φ(g·g’) = Φ(g)+_{n}Φ(g’)? ∀g, g’ ∈ H, ∃i, j such that g = a^{i}, g’ = a^{j}
Let k = i +_{n} j, a^{n} = a^{0} = e ⇒ [a^{k-(i+j)} = a^{l·n} = e] a^{k} = a^{i+j} ⇒ a^{i}a^{j} = a^{k} ⇒ Φ(a^{i}a^{j}) = k = i +_{n} j = Φ(a^{i}) + Φ(a^{j}).
Corollary. Any two cyclic groups of the same order are isomorphic.
Proof
Let ⟨a⟩ and ⟨b⟩ be cyclic groups of order n, then the map Φ: ⟨a⟩ → ⟨b⟩ defined by, Φ(a^{k}) = b^{k} is well-defined and is an isomorphism (you may want to check this).
Is it well-defined? If a^{r} = a^{s} ⇒ Φ(a^{r}) = Φ(a^{s}) ↭ b^{r} = b^{s}
a^{r} = a^{s} ⇒ a^{r-s} = e ⇒ n | (r -s), say r -s = tn ⇒ r = tn +s ⇒ Φ(a^{r}) = Φ(a^{tn+s}) = b^{tn+s} = b^{tn}b^{s} = (b^{n})^{t}b^{s} = b^{s}.
Fundamental Theorem of Cyclic Groups. Every subgroup of a cyclic group is cyclic. If ⟨a⟩=n, then the order of any subgroup of ⟨a⟩ is a divisor of n. For each positive divisor k of n, the group ⟨a⟩ has exactly one subgroup of order k, namely, ⟨a^{n/k}⟩
Example. Suppose G = ⟨a⟩ and |G|=20. List the subgroups G and their orders.
Orders of subgroups of G: 1, 2, 4, 5, 10, and 20.
Corollary. For each positive divisor k of n, the set ⟨n/k⟩ is the unique subgroup of ℤ_{n} of order k.
Let φ(1) = 1 and for any n ∈ ℤ, n > 1, let φ(n) denote the number of positive integers less than n and relative prime to n, φ is the Euler phi function. If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is φ(d).
Proof.
By a previous result, Let G be a finite cyclic group, G = ⟨g⟩. Then, G has exactly one subgroup of order d for each d | |G|, namely H = ⟨g^{n/d}⟩ ≤ G. Hence, for each d | |G|, G = ⟨g⟩ has exactly one subgroup of order d, say ⟨a⟩, and we have also demonstrated that a^{k} generates ⟨a⟩ ↭ gcd(k, d) = 1 and this is by definition precisely φ(d)∎
ℤ_{12}. Order of subgroups are: 1, 2, 3, 4, 6, and 12. The number of elements of order 2, 3, 4, 6, 12, are φ(1)= 1 (0), φ(2)= 1 (6, |⟨6⟩| = 2), φ(3)= 2 (4, 8), φ(4)= 2 (3, 9), φ(6)= 2 (2, 10), φ(12)= 4 (1, 5, 7, 11) respectively (Figure 1.a.)
ℤ_{24}. Order of subgroups are: 1, 2, 3, 4, 6, 8, 12, and 24. Let’s list all generators for the subgroup of order 8.
⟨n/k⟩ is the unique subgroup of ℤ_{n} of order k. ⟨24/8⟩=⟨3⟩ is the unique subgroup of ℤ_{24} of order 8. 3 is obviously a generator, but it does not need to be the only one. ⟨3⟩=⟨3j⟩, 3j is a generator of a group of order 8 of ℤ_{24} where 24/gcd(8, j) = 24, that is, gcd(8, j) = 1, so 3 (j=1), 9(j=3), 15(j=5), 21(j=7). ⟨3⟩=⟨9⟩=⟨15⟩=⟨21⟩ = {3, 6, 9, 12, 15, 18, 21, 1}. There are four generators for the unique subgroup of ℤ_{24} of order 8, φ(8) = 4.
To sum up, in ℤ_{24}, the number of elements of order 8 is φ(8) = 4. They are ⟨3⟩=⟨9⟩=⟨15⟩=⟨21⟩ = {3, 6, 9, 12, 15, 18, 21, 1}.