Cyclic Groups

7.3.1 Definition. Let (G, ⋅) a group, and let g ∈ G. and i ∈ ℤ. We define power gⁱ of g with exponent i the following element of G:

The usual rules for manipulation of powers hold:

On the other hand, (gh)ⁱ ≠ gⁱhⁱ in general.

These definition are applied whatever the operation * of the group. Doing the power, consists in iterating staring from x or x⁻¹ the operations of the group.

Note. When the group operation is addition, we write ka in place of a^k.

Consider a group G = {a,b,c, ...}, which may or may not be finite. Suppose that all the elements of G can be created by taking the powers of an appropriately chosen element, say a, of G. Then G is called a cyclic group, and a is called a generator of the group. For instance, consider the element i of a finite group G = {1,−1, i, −i} with ordinary multiplication as group operation. We notice that i¹ = i, i² = −1, i³ = −i, i⁴ = 1, the whole group is generated by taking the positive powers of the element i. If we take higher powers of i, elements of the group start repeating themselves. Thus G is a cyclic group and i is the generator of the group. We may also generate this group by taking the positive powers of −i. The generator of a cyclic group is therefore not necessarily unique.

Next consider an infinite group of even integers, viz.,

in which the group operation is addition. The group can be generated by 2; By this we mean that the element 2 can be combined with itself using only the group operation and inverses to produce the entire group. The element −2 is another generator of this group.

Cyclic subgroups can be constructed as follows. If G is a group and a ∈ G, let ⟨a⟩ denote the set of all powers of a:

7.3.2 Proposition. If G is a group and a ∈ G, then ⟨a⟩ = {aⁿ | n ∈ ℤ} is a subgroup of G.

Proof. The product of any two elements of ⟨a⟩ is also in ⟨a⟩ because aⁱa^j = a^i+j. The inverse of a^k, which is also in ⟨a⟩. By Theorem 7.2.2, ⟨a⟩ is a subgroup of G.  □

The group ⟨a⟩ is called the cyclic subgroup generated by a. If the subgroup ⟨a⟩ is the entire group G, we say that G is a cyclic group. Note that every cyclic group is abelian since aⁱa^j = a^i+j = a^jaⁱ.

Nota bene. If we pick any element x of G, then ⟨x⟩ will always be a cyclic subgroup of G.

7.3.3 Example. Consider the multiplicative group G = {1, 2, 3, 4, 5, 6} under modulo-7 multiplication. Taking the powers of 3, we obtain

We see that the above six powers of 3 give all the six elements of G. Hence, G is a cyclic group and 3 is a generator of G. The element of 5 is also a generator of G as shown below,

7.3.4 Definition. Let X be a subset of a group G. We define subgroup generated by X, denoted by ⟨X⟩, the smallest subgroup of G containing X. It coincides with the intersection of all the subgroups of G containing X. It is indicated with ⟨X⟩. Thus

In the case in which X = {g} ⊆ G is made by a single element, then

The subgroup ⟨g⟩ is called the cyclic subgroup of G generated by g.
So a group G is called cyclic if there exists x ∈ G such that G is generated by the singleton set {x}. Any such g is called a generator of G; we say that 'x generates G' instead of {x} generates G. One often writes G = ⟨x⟩. It turns out, however, that a finite cyclic group can have several generators, so the expression G = ⟨x⟩ is not necessarily unique.

We show now that the subgroup ⟨S⟩ is the smallest subgroup containing S. Let S be a subset of G. Consider the family 𝓔 of all subgroups of G containing S. That is

Since G is in 𝓔, 𝓔 is not empty. Let M be the intersection of all subgroups A in 𝓔. Then M is a subgroup of G and S ⊂ M. If M' is a subgroup and S ⊂ M', then M' ∈ 𝓔. Hence, M ⊂ M'. Therefore M is the smallest subgroup containing S.

More generally, a subset X of a group G 'generates' G if every element of G may be written as a product of elements of X and of inverses of elements of X (where repetitions are allowed).

7.3.4 Proposition. Let X = {x₁, x₂, ..., x_n} a subset (finite or infinite) of a group G. Then

with X⁻¹ = {x⁻¹ | x ∈ X}.

Proof. Clearly X ⊂ ⟨X⟩. For any two elements a = t₁ ⋅ t₂ ⋅⋅⋅⋅ t_r and b = s₁ ⋅ s₂ ⋅⋅⋅⋅ s_r in X,

Hence, ⟨X⟩, is a subgroup of G. Let M' be any subgroup of G containing X. Then for each x ∈ X, x ∈ M'. Hence x₁⁻¹ ∈ M'. Therefore M' contains all finite products t₁ ⋅ t₂ ⋅⋅⋅⋅ t_r such that x_i ∈ X or x⁻¹ ∈ X, i=1,...,n. Hence, M ⊂ M', proving that M is the smallest subgroup containing X and, therefore, the subgroup generated by X.  □

Let's see some examples of groups generated by a subset of G and study the difference between abelian and non-abelian groups.

In the abelian case (ℤ[i], +) the subgroup generated by the elements 2 and 3 is ⟨2,3⟩ = {2s + 3t| s,t ∈ ℤ} which is coincident with ℤ, this is indeed the same as {t₁ + t₂ + ... + t_n | t_i ∈ {2,3} ∪ {2,3}⁻¹}.

If X = {amb), the elements of ⟨X⟩ would be of the form

that can be seen as words of different length formed by the letters a, b, a⁻¹, b⁻¹. Notice that cancellation between adjacent terms is possible (such as baa⁻¹b = bb). Moreover the product of adjacent letters can be expressed as power (e.g if a is repeated k times: aaaa⋅⋅⋅⋅a = a^k). Hence we can rewrite 7.1.1 as ab⁻¹ba³b⁻¹a²b⁻¹. Depending of the particular group we're dealing with further semplificatios are possible e.g. if the group is Abelian, or if one of the generators raised to a certain power gives as result e, the identity of the group.

Let g an element of a group (G, ⋅). It can happen that for some k ∈ ℕ, g^k = e. (identity of G): this happens for certain in the case of finite G. Indeed if the subgroup ⟨g⟩ = {gⁱ | i ∈ ℤ} must be finite, there exist two integer powers s,t (s > t) such that g^s = g^t, from which letting k = s − t ∈ ℕ, we have g^k = e. It makes sense to introduce the following definition.

7.3.5 Definition. Let (G, ⋅) a group and g ∈ G, we define order of g the least positive integer n, if there exists, such that gⁿ = e. If such an integer does not exist, we sat that g has infinite order. The order of an element g is denoted by |g| or o(g).

For example in (ℤ₄, +) the class 1̅ has order 4, the class 2̅ has order 2. In (ℤ, +), every nonnull element has infinite ordeer. In (ℂ\{0},⋅), the element i has order 4. The identity of any group has order 1.

7.3.6 Definition. A group G is said to be finite (or of finite order) if it has a finite number of elements. In this case, the number of elements in G is called the order of G and is denoted |G|. A group with infinitely many elements is said to have infinite order.

Although a cyclic group consists of all powers of its generator, these powers need not be distinct. Indeed if the group is finite then they are bound to repat (by the pigeon-hole principle). It turns out that this repetition follows a regular pattern, as the following propositions shows

7.3.7 Proposition. Let G a group. If g ∈ G has infinite order and if h ≠ k, then g^h ≠ g^k, and thus ⟨g⟩ is a infinite cyclic subgroup. Conversely if g has order n, then

thus the cardinality of ⟨g⟩ is n. Moreover g^h = g^k if and only if h ≡ k (mod n).

Proof. Let g of infinite period. Then g^h = g^k (i.e. g^h−k = e) implies h = k, by definition of infinite period. Let now n be the period of g. We notice first of all that all the elements {e, g, g², ..., g^{n − 1}} are distinct; For if g^h = g^k with 0 ≤ h < k < n then g^h−k = e with h − k < n contradicting the minimality of n. To prove that ⟨g⟩ = {e, g, g², ..., gⁿ⁻¹} it suffices to show that any power g^k ∈ {e, g, g², ..., gⁿ⁻¹} ∀k ∈ ℤ. Indeed by dividing k by n: k = nq + r, 0 ≤ r < n, from which g^k = g^nq+r = (gⁿ)^q = eg^r = eg^r = g^r. Hence g^k ∈ {e, g, g², ..., gⁿ⁻¹} ∀k ∈ ℤ.

Suppose now that g^h = g^k. Then g^h−k = e. By dividing h − k by n we have

from which g^{h − k} = (gⁿ)^q = g^r = e. Thus h ≡ k (mod n).

Conversely if h ≡ k (mod n) then g^h−k = g^nq = (gⁿ)^q = e, thus g^h = g^k.  □

Thus every element of a group, generates a cyclic subgroup of G. Generally such a subgroup will be properly contained in G.

7.3.8 Definition. A group G is said cyclic if there exists an element g ∈ G such that G = ⟨g⟩. ■

Proof. Suppose G is a cyclic group generated by element g. So G = ⟨g⟩ for some element g ∈ G and the identity element of G is g⁰.

7.3.9 Examples (Subgroups generated by a subset; Cyclic groups).

• Let G = (ℤ, +).

  • (ℤ, +) is generated by either 1 or −1 so ℤ = ⟨1⟩ or ℤ = ⟨−1⟩. As ℤ contains an infinite number of elements we say that Z is an infinite cyclic group. Remember a generating set of a group is a subset such that every element of the group can be expressed as the combination (under the group operation) of finitely many elements of the subset and their inverses.

  • X = {3}: ⟨3⟩ = 3ℤ = {multiples of 3};

  • X = {2,3}: ⟨2,3⟩ = ℤ;

  • X = {2,4}: ⟨2,4⟩ = 2ℤ;

  • X = {m,n}: ⟨m,n⟩ = dℤ; where d = GCD(m,n).

• Let G = (ℤ_n, +). In ℤ_n = ⟨1̅⟩, 1 is always a generator, thus ℤ_n is cyclic:

  1 ≡ 1 mod n
  1 + 1 ≡ 2 mod n
  1 + 1 + 1 ≡ 3 mod n
  ...
  1 + 1 + 1 + ... +1 ≡ n ≡ 0 mod n

  If a group is cyclic, then there may exist multiple generators. For example, we know ℤ₅ is a cyclic group. The element 1 is a generator for sure. And if we take a look at 2, we can find:

  2 ≡ 2 mod 5
  2+2 ≡ 4 mod 5
  2+2+2 ≡ 6 ≡ 1 mod 5
  2+2+2+2 ≡ 8 ≡ 3 mod 5
  2+2+2+2+2 ≡ 10 ≡ 0 mod 5

  So all the group elements {0,1,2,3,4} in ℤ₅ can also be generated by 2. That is to say, 2 is also a generator for the group ℤ₅. Not every element in a group is a generator. For example, the identity element in a group will never be a generator. No matter how many times you apply the group operator to the identity element, the only element you can yield is the identity element itself. For example, in ℤ_n, 0 is the identity element and 0 + 0 + ... + 0 ≡ 0 mod n in all cases.

  The group, ℤ₄ = {0,1,2,3}, under addition is cyclic with generator 1 because

  1 = 1
  1 + 1 = 2
  1 + 1 + 1 = 3
  1 + 1 + 1 + 1 = 0

  Thus every element of ℤ₄ can be written as n ⋅ 1 for some integer n. Does ℤ₄ have any other generators?

  2 = 2
  2 + 2 = 0
  2 + 2 + 2 = 2

  etc. shows that 2 is not a generator of ℤ₄.

  3 = 3
  3 + 3 = 2
  3 + 3 + 3 = 1
  3 + 3 + 3 + 3 = 0

  shows that 3 is a generator of ℤ₄. Thus ℤ₄ = ⟨1⟩ = ⟨3⟩. But ℤ has nontrivial proper subgroups. See also The ℤ/nℤ group.  ■

• Let G = (ℚ\{0},⋅). If X = {7}, then

  ⟨7⟩ = {7ⁱ| i ∈ ℤ} = {...,1/49, 1/7,7,49,...}

• Let G = (GL₂(ℝ), ⋅). Determine ⟨(0 1; −1 0)⟩. It's simply all the powers g^k there are, as k ranges over ℤ. You have to multiply that matrix by itself till you get the I. ⟨(0 1; −1 0)⟩ = {I, (0 1; −1 0), (-1 0; 0 −1), (0 −1; 1 0)}

• The nth roots of unity.

Lemma 7.3.10. Let G = ⟨g⟩ be a finite cyclic group of order n. Then g^k with 0 < k < n is a generator of G if anf only if gcd(k,n) = 1.

Proof. Suppose gcd(k,n) = 1. Then there exists s,t ∈ ℤ with ks + nt = 1. It follows that

Therefore g is a power of g^k and hence, g^k is a generator of G.
Conversely, suppose that g^k is also a generator of G. Then there exists a power s of g^k such that g = (g^k)^s = g^ks. Hence

In general in any group, if g has order n (as it does here), then g^j = e iff n | j; Thus g^{1 − ks} = e means n | (1 − ks), hence ks ≡ 1 (mod n), and so k is a unit modulo n, which implies that gcd(n,k) = 1.  □

Subgroups of cyclic groups

Generally a group has a lot of subgroups and it can be a difficult task to find all of them. Thus it is of interest that the subgroups of a cyclic group are easy to describe. The first observation is that such subgroups are themselves cyclic.

7.3.11 Proposition. Every subgroup of a cyclic group G is a cyclic group.

Proof. Let H ≤ G = ⟨g⟩. Let H = {e}, then H is certainly cyclic. Let now g^t ∈ H, g^t ≠ e. Then also g^−t ∈ H, and thus H contains an element of the form g^h, with n ∈ ℕ. Among the elements belonging to H, let m the least positive integer such that g^m ∈ H. We prove that H = ⟨g^m⟩, and a very typical way to show set equality is to show H ⊆ ⟨g^m⟩ and ⟨g^m⟩ ⊆ H. Certainly ⟨g^m⟩ ⊆ H. Conversely to show H ⊆ ⟨g^m⟩, we must show that all elements of H are also in ⟨g^m⟩. If g^k ∈ H. We divide k by m and we get k = mq + r, 0 ≤ r < m, from which

It must result r = 0, that is k = mq, i.e. g^k = (g^m)^q ∈ ⟨g^m⟩. □

The next result tells us how to construct the subgroups of a given cyclic group.

Example 1 The set of even integers is a proper subgroup of the group of all integers under addition. The set of all multiples of 3 is also a proper subgroup of the integers under addition. In fact, if k is any fixed integer, then

is a subgroup of the integers under addition. Note that if k = 0, then we obtain the trivial subgroup {0}; whereas if k = ±1, then we obtain the entire group ℤ. Thus, unless k = ± 1, kℤ is a proper subgroup of ℤ. The subgroup 2ℤ = ⟨2⟩, with generator 2 because any even integer can be written in the form k ⋅ 2, is infinite cyclic just like its 'parent' ℤ, the same is true for any ⟨k⟩ with k ≠ 0. It is also true that 2ℤ = ⟨−2⟩, since −2 is also a genrator.

7.3.12 Proposition. Let G = ⟨g | gⁿ = e⟩, a cyclic group of order n. Then

1. The order of any subgroup is a divisor of n.

2. For each divisor k of n there exists one and only one subgroup with order k

Proof. Let H = ⟨g^m⟩ a subgroup of G. We have

We have that the period of g^m divides n see exercise. Since the order of H is period of g^m, we have proved (a).

Let now k a divisor of n. The subgroup ⟨g^n/k⟩ has order k. We'll show now this is the only subgroup of G with such order. Let H a subgroup of G with order k. This subgroup is of the form H = ⟨g^m⟩, with m the least positive integer such that g^m ∈ H (see the previous proposition to understand why). Furthermore m|n, as it can be seen dividing n by m: we get n = mk + r, 0 ≤ r < m, from which

Proof. Let H = ⟨g^k⟩. If d = (k,n) then d|k and d|n, write them as k = ds and n = dt, and gcd(s,t) = 1 [gcd(n/d,k/d) = gcd(n,k)/d = 1]. Then (g^k)^s = g^tms = g^tn = (gⁿ)^t = e. In other words, the order of g^k divides s and thus is less than or equal to s (The order, x of g^k must be a multiple of n, i.e. x = ny, since (g^k)^x = (g^k)^ny = e). So H contains the elements g^k, g^2k, g^3k, g^sk. We need to show these s elements are distinct to prove the theorem. Assume that g^ik = g^jk for some i and j such that 1 ≤ j < i ≤ s. So g^{(i − j)k} = g⁰. This means that n|(i − j)k, which means that sd| (i − j)td, which means that s | (i − j)t, which since s and t are relatively prime, means that s |i − j, which is impossible since 1 ≤ j < i ≤ s. Therefore, H has order s = n/d.  □

Since any subgroup of a cyclic group is cyclic, the theorem has some immediate corollaries. The first follows from the special ease d = 1 in the theorem. The second from the special case k = d = n/m so that m = n/d.

Proposition 7.3.15. Let G = ⟨g⟩ is a cyclic group of order n, then an element g^k is a generator of G iff gcd(k, n) = 1. Consequentely there are φ(n) generators of G.

Proof. Let g^k a generator of G; then n = ord g^k = n/ (k,n) which shows that gcd(k,n) = 1 (see Proposition 5.7.5). Conversely, suppose that k and n are coprime. Then there are integers s,t ∈ ℤ with sk + tn = 1. This implies that

Hence G = ⟨g⟩ ⊆ ⟨g^k⟩. Thus g^k is a generator of G.  □

Corollary 7.3.16. If G is a cyclic group of order n with generator g and d|n, then ⟨g^n/d⟩ has order d.

7.3.18 Example. If G = ⟨g^k⟩ is a cyclic group of order 12, then the generators of G are the powers g^k where gcd(k,12) = 1, that is g, g⁵, g⁷, and g¹¹. In the particular case of the additive cyclic group, ℤ₁₂, the generators are the integers 1, 5, 7, 11 (mod 12).   ■

Theorem 7.3.17 (Cyclic Implies Abelian) If [G, *] is cyclic, then it is abelian.

Proof. If G is cyclic, and a is a generator of G, then every element is of the form, a^r, so we have