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.

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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.

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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 \(\operatorname{ord}(x)=|\langle x\rangle|=|G|=n\).

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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 \(H\), \(H\subseteq\langle x^m\rangle\).

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