9.2 Cyclic groups and cyclic subgroups

An important family of (sub)groups are those arising as collections of powers of a single element.

Notation:

Given a (multiplicatively-written) group G and an element x\in G, we’ll define \langle x\rangle to be the subset \{x^i:i\in\mathbb{Z}\} of G consisting of all powers of x.

It is easy to check that:

Proposition 9.5:

The set \langle x\rangle is a subgroup of G.

Proof.

We need to check the conditions of Theorem 9.3.

  • If x^i,x^j are powers of x, then x^ix^j=x^{i+j} is a power of x, so \langle x\rangle is closed.

  • We have e=x^0 is a power of x, so e\in\langle x\rangle.

  • If x^i\in\langle x\rangle, then (x^i)^{-1}=x^{-i}\in\langle x\rangle, so \langle x\rangle is closed under taking inverses.

Definition 9.6:
If x\in G, then \langle x\rangle is called the cyclic subgroup of G generated by x.

The following explicit description of \langle x\rangle follows immediately from Propositions 8.31 and 8.33.

Proposition 9.7:

If \operatorname{ord}(x)=\infty then \langle x\rangle is infinite with x^i\neq x^j unless i=j.

If \operatorname{ord}(x)=n then \langle x\rangle=\{e,x,\dots,x^{n-1}\} is finite, with n distinct elements.

Remark:
This justifies using the word “order” in two different ways. The order (as an element) of x is the same as the order (as a group) of \langle x\rangle.
Example:
If G=D_{2n} is the dihedral group of order 2n, then \langle a\rangle is the group \{e,a,\dots a^{n-1}\} of rotations, and \langle b\rangle=\{e,b\}.
Example:
Let G=(\mathbb{R}\setminus\{0\},\times). Then \langle -1\rangle=\{1,-1\} and \langle 2\rangle=\{\dots, \frac{1}{4},\frac{1}{2},1,2,4,\dots\} consists of all powers of 2 (and of \frac{1}{2}, since we include negative powers of 2).
Example:
Let G=(\mathbb{R},+). Since the operation is now addition, \langle x\rangle consists of all multiples of x. So, for example, \langle 1\rangle=\mathbb{Z} and \langle 3\rangle=3\mathbb{Z}=\{3n: n\in\mathbb{Z}\}.
Definition 9.8:
A group G is called cyclic if G=\langle x\rangle for some x\in G. Such an element x is called a generator of G.
Example:
Let G=(\mathbb{Z},+), then G is cyclic, since \mathbb{Z}=\langle 1\rangle=\langle -1\rangle. Hence 1 and -1 are generators.

This example shows that there may be more than one possible choice of generator. However, not every element is a generator, since, for example, \langle 0\rangle=\{0\} and \langle 2\rangle = 2\mathbb{Z} \neq \mathbb{Z}.

Example:
Let H be the cyclic subgroup of D_{12} generated by a. Then a and a^{-1} are generators of H, but a^2 is not, since \langle a^2\rangle=\{e,a^2,a^4\} contains only the even powers of a.
Example:
Let H be the cyclic subgroup of D_{14} generated by a. Then a^2 is a generator of H, since \langle a\rangle=\{e,a^2,a^4,a^6,a^8=a,a^{10}=a^3,a^{12}=a^5\} contains all the powers of a. In fact, all elements of H except e are generators.
Proposition 9.9:

Every cyclic group is abelian.

Proof.

Suppose G=\langle x\rangle is cyclic with a generator x. Then if g,h\in G, g=x^i and h=x^j for some integers i and j. So gh=x^ix^j=x^{i+j}=x^jx^i=hg, and so G is abelian.

Of course, this means that not every group is cyclic, since no non-abelian group is. But there are also abelian groups, even finite ones, that are not cyclic.

Proposition 9.10:

Let G be a finite group with |G|=n. Then G is cyclic if and only if it has an element of order n. An element x\in G is a generator if and only if \operatorname{ord}(x)=n.

Proof.

Suppose x\in G. Then |\langle x\rangle|=\operatorname{ord}(x), and since \langle x\rangle\leq G, \langle x\rangle=G if and only if \operatorname{ord}(x)=|\langle x\rangle|=|G|=n.

Example:
Let G=D_8 and let H=\{e,a^2,b,a^2b\}. Then H is an abelian subgroup of G (check this), but it is not cyclic, since |H|=4 but \operatorname{ord}(a^2)=\operatorname{ord}(b)=\operatorname{ord}(a^2b)=2 and \operatorname{ord}(e)=1, so H has no element of order 4.
Theorem 9.11:

Every subgroup of a cyclic group is cyclic.

Proof.

Let G=\langle x\rangle be a cyclic group with generator x, and let H\leq G be a subgroup.

If H=\{e\} is the trivial subgroup, then H=\langle e\rangle is cyclic. Otherwise, x^i\in H for some i\neq0, and since also x^{-i}\in H since H is closed under taking inverses, we can assume i>0.

Let m be the smallest positive integer such that x^m\in H. We shall show that H=\langle x^m\rangle, and so H is cyclic.

Certainly \langle x^m\rangle\subseteq H, since every power of x^m is in H. Suppose x^k\in H and write k=mq+r where q,r\in\mathbb{Z} and 0\leq r<m. Then x^r=x^{k-mq}=x^k(x^m)^q\in H, since x^k\in H and x^m\in H. But 0\leq r<m, and m is the smallest positive integer with x^m\in H, so r=0 or we have a contradiction. So x^k=(x^m)^q\in\langle x^m\rangle. Since x^k was an arbitrary element of H, H\subseteq\langle x^m\rangle.