5.1: Introduction to Cyclic Groups

Certain groups and subgroups of groups have particularly nice structures.

Definition: Cyclic

A group is cyclic if it is isomorphic to \(\mathbb{Z}_n\) for some \(n\geq 1\text{,}\) or if it is isomorphic to \(\mathbb{Z}\text{.}\)

Example \(\PageIndex{1}\)

Examples/nonexamples of cyclic groups.

1. \(n\mathbb{Z}\) and \(\mathbb{Z}_n\) are cyclic for every \(n\in \mathbb{Z}^+\text{.}\)
2. \(\mathbb{R}\text{,}\) \(\mathbb{R}^*\text{,}\) \(\mathbb{M}_2(\mathbb{R})\text{,}\) and \(GL(2,\mathbb{R})\) are uncountable and hence can't be cyclic.
3. \(\mathbb{Z}_2^2\) is not cyclic since it would have to be isomorphic to \(\mathbb{Z}_4\) if it were (since it has order \(4\)).
4. \(\mathbb{Q}\) isn't cyclic. If it were cyclic it would have to be isomorphic to \(\mathbb{Z}\text{,}\) since \(\mathbb{Q}\) is an infinite group (so can't be isomorphic to \(\mathbb{Z}_n\) for any \(n\)). But we showed in Example \(3.3.5\) that \(\mathbb{Q} \not\simeq \mathbb{Z}\text{.}\)

Remark

Clearly, cyclicity is a group invariant.

Theorem \(\PageIndex{1}\)

If \(G\) is cyclic, then \(G\) is abelian; however, \(G\) can be abelian but not cyclic.

Proof

Since \(\mathbb{Z}\) and each \(\mathbb{Z}_n\) are abelian, every cyclic group is abelian. But, for instance, \(\mathbb{R}\) is an abelian noncyclic group (it can't be cyclic because it is uncountable).

Definition: Cyclic Subgroup of G Generated by a

Let \(a\) be an element of a group \(G\text{.}\) Then

\begin{equation*} \langle a\rangle = \{a^n:n\in \mathbb{Z}\} \end{equation*}

is called the (cyclic) subgroup of \(G\) generated by \(a\). (We will later see why the words “cyclic” and “generated” are used here.)

Remark

Note that in the above definition, we were using multiplicative notation. Using additive notation, we have

\(\langle a \rangle =\{na:n\in \mathbb{Z}\}=\{\ldots, -2a, -a, 0a, 1a, 2a, \ldots\}.\)

Theorem \(\PageIndex{2}\)

Let \(a\) be an element of a group \(G\text{.}\) Then \(\langle a\rangle \leq G\text{.}\)

Proof

Let \(x,y∈ \langle a \rangle\). Then \(x=a^i\), \(y=a^j\) for some \(i,j∈\mathbb{Z}\). So \(xy=a^ia^j=a^{i+j}∈\langle a \rangle\). Next, \(e_G=a^0∈ \langle a \rangle\). Finally, \((a^i)^{−1}=a^{−i}∈ \langle a \rangle\). So \(\langle a \rangle\) is a subgroup of \(G\).

Let \(x,y\in \langle a\rangle\text{.}\) Then \(x=a^i\text{,}\) \(y=a^j\) for some \(i,j\in \Z\text{.}\) So \(xy=a^ia^j=a^{i+j}\in \langle a\rangle\text{.}\) Next, \(e_G=a^0\in \langle a \rangle\text{.}\) Finally, \((a^i)^{-1}=a^{-i}\in \langle a \rangle\text{.}\) So \(\langle a\rangle\) is a subgroup of \(G\text{.}\)

We next prove a lemma to which we will refer when proving the theorem which succeeds it.

Lemma \(\PageIndex{1}\): Clock Lemma

Let \(G\) be a group and let \(a \in G\text{.}\) Suppose there is a positive power of \(a\) which equals the identity element \(e\) of \(G\text{.}\) Let \(n\) be the least such positive integer. Then for all \(s,t\in \mathbb{Z}\text{,}\) \(a^s=a^t\) if and only if \(s\) is congruent to \(t\) modulo \(n\text{.}\) In particular, if \(s\in \mathbb{Z}^+\) with \(a^s=e,\) then \(n\) divides \(s\text{.}\)

Proof

Let \(s,t∈\mathbb{Z}\). By the Division Algorithm, there exist \(q∈\mathbb{Z}\) and \(r∈\mathbb{Z}_n\) with \(s−t=qn+r\). Then \(a^{s−t}=a^{qn+r}=(a^n)^qa^r=e^qa^r=a^r\). Since \(0≤r<n\), be definition of nn we have that \(a^{s−t}=e\) if and only if \(r=0\); that is, \(a^s=a^t\) if and only if \(s\) and \(t\) are congruent modulo \(n\).

The last statement follows from the fact that \(n\) divides any positive integer to which it is congruent modulo itself.

Theorem \(\PageIndex{3}\)

Let \(G\) be a group with identity element \(e\text{,}\) and let \(a\in G\text{.}\) Then the group \(\langle a\rangle\) is cyclic.

Proof

Case 1. There is no positive integer \(k\) with \(a^k=e\). In this case, we claim \(\langle a \rangle \simeq \mathbb{Z}\). Indeed: Let \(ϕ:\mathbb{Z}→ \langle a \rangle\) be defined by \(ϕ(i)=a^i\). Clearly, \(ϕ\) is a homomorphism that is onto. So to show that \(ϕ\) is an isomorphism, it suffices to show that \(ϕ\) is one-to-one. Let \(i,j∈\mathbb{Z}\) with \(ϕ(i)=ϕ(j)\). Then \(a^i=a^j\). Without loss of generality, we may assume \(i≥j\). Then, multiplying both sides of the equation by \(a^{−j}\) (on the right or on the left) we obtain \(a^{i−j}=e\). Since \(i−j∈ \mathbb{N}\) and there is no positive integer \(k\) with \(a^k=e\), we must have \(i−j=0\), so \(i=j\). Thus, \(ϕ\) is an isomorphism from \(\mathbb{Z}\) to \(\langle a \rangle\).

Case 2. There is a positive integer \(k\) such that \(a^k=e\). In this case, let \(n\) be the least such positive integer. Define \(ϕ:\mathbb{Z}_n→ \langle a \rangle\) be defined by \(ϕ(i)=a^i\). We claim that \(ϕ\) is an isomorphism.

First, let \(i,j∈\mathbb{Z}\). We want to show that \(ϕ(i+_nj)=ϕ(i)ϕ(j)\), that is, that \(a^{i+_nj}=a^ia^j\). Since \(i+j\) is congruent to \(i+_nj\) modulo \(n\), \(a^ia^j=a^{i+j}=a^{i+_nj}\), by the Clock Lemma. So \(ϕ\) is a homomorphism. Next, we show that \(ϕ\) is one-to-one. Let \(i,j∈ \mathbb{Z}_n\) with \(ϕ(i)=ϕ(j)\), so \(a^i=a^j\). Then, by the Clock Lemma, \(i\) and \(j\) are congruent modulo \(n\). Since they're both in \(\mathbb{Z}_n\), they must be equal, so \(ϕ\) is one-to-one. Finally, we show that \(ϕ\) is onto. let \(x∈ \langle a \rangle\). Then \(x=a^i\) for some \(i∈ \mathbb{Z}\). Letting \(r\) be the remainder when we divide \(i\) by \(n\), we have \(r∈\mathbb{Z}_n\) with \(i\) congruent to \(r\) modulo \(n\); so, again using the Clock Lemma, \(x=a^i=a^r\). Since \(r∈\mathbb{Z}_n, x=ϕ(r)\). So \(ϕ\) is onto.

Thus, \(ϕ\) is an isomorphism, and so in this case we have \(\mathbb{Z}_n \simeq \langle a \rangle\).

Definition: Order

Let \(G\) be a group and let \(a\in G\text{.}\) We define the order of \(a\text{,}\) denoted \(o(a)\text{,}\) to be \(|\langle a\rangle |\text{.}\) (Note: If there exists a positive integer \(k\) such that \(a^k=e\text{,}\) then the least such integer is the order of \(a\text{;}\) otherwise, \(o(a)=\infty\text{.}\))

Remark

Do not confuse the order of a group with the order of an element of a group. These are related concepts, but they are distinct, and have distinct notations: as we've seen, the order of a group, \(G\text{,}\) is denoted by \(|G|\text{,}\) while the order of an element \(a\) of a group is denoted by \(o(a)\text{.}\)

Theorem \(\PageIndex{4}\)

Let element \(a\) in group \(G\) have \(o(a)=k \lt \infty\text{.}\) Then for \(m\in \mathbb{Z}^+\text{,}\) \(a^m=e\) if and only if \(k\) divides \(m\text{.}\)

Proof

This follows from the last statement of the Clock Lemma.

This follows from the last statement of the Clock Lemma.

We have the following handy theorem.

Theorem \(\PageIndex{5}\)

Let \(a\in G\text{.}\) Then \(o(a)=o(a^{-1})\text{.}\)

Proof

First, assume \(o(a)=n<∞\). Then

\((a^{−1})^n=(a^n)^{−1}=e^{−1}=e\),

so \(o(a^{−1})≤n=o(a)\). Using the same argument, we have \(n=o(a)≤o(a^{−1})\). Since \(o(a^{−1})≤n\) and \(n≤o(a^{−1})\), \(o(a^{−1})=n=o(a)\).

On the other hand, assume \(o(a)=∞\). Then \(a^{−1}\) must also have infinite order, since if it had finite order \(m\), \(a\) would, by the above argument, have order less than or equal to \(m\).

First, assume \(o(a)=n\lt \infty\text{.}\) Then

\begin{equation*} (a^{-1})^n=(a^n)^{-1}=e^{-1}=e, \end{equation*}

so \(o(a^{-1})\leq n=o(a)\text{.}\) Using the same argument, we have \(n=o(a)\leq o(a^{-1})\text{.}\) Since \(o(a^{-1})\leq n\) and \(n\leq o(a^{-1})\text{,}\) \(o(a^{-1})=n=o(a)\text{.}\)

On the other hand, assume \(o(a)= \infty\text{.}\) Then \(a^{-1}\) must also have infinite order, since if it had finite order \(m\text{,}\) \(a\) would, by the above argument, have order less than or equal to \(m\text{.}\)

Unfortunately, there's no formula one can simply use to compute the order of an element in an arbitrary group. However, in the special case that the group is cyclic of order \(n\text{,}\) we do have such a formula. We present the following result without proof.

Theorem \(\PageIndex{6}\)

For each \(a\in \mathbb{Z}_n\text{,}\) \(o(a)=n/\gcd(n,a)\text{.}\)

Theorem \(\PageIndex{7}\)

Let \(m,n \in \mathbb{Z}^+\) and let \((a,b)\in \mathbb{Z}_m \times \mathbb{Z}_n\text{.}\) Let \(c, d\) be the respective orders of \(a\) in \(\mathbb{Z}_m\) and \(b\) in \(\mathbb{Z}_n\text{.}\) Then \(o((a,b))= \text{lcm}(c,d)\text{.}\)

Proof

Let \(k∈\mathbb{Z}^+\). By definition of \(c\) and \(d\) and by Theorem \(5.1.4\), we have \(k(a,b)=(ka,kb)=(0,0)\) if and only if \(c\) and \(d\) divide \(k\). Since \(\gcd(c,d)=1\), the least positive integer divisble by both \(c\) and \(d\) is \(\text{lcm}(c,d)\). Thus, \(o((a,b))=\text{lcm}(c,d)\), as desired.

Corollary \(\PageIndex{1}\)

Suppose \(m,n\in \mathbb{Z}^+\) with \(\gcd(m,n)=1\text{.}\) Then \(\mathbb{Z}_m \times \mathbb{Z}_n\) is cyclic.

Proof

Since \(\gcd(m,n)=1\), \(\text{lcm}(m,n)=mn\). So

\(o((1,1))= \text{lcm}⁡(m,n)=mn=|\mathbb{Z}_m \times \mathbb{Z}_n|\),

and we see that \((1,1)\) generates \(\mathbb{Z}_m \times \mathbb{Z}_n\).

Here are some examples of cyclic subgroups of groups, and orders of group elements.

Example \(\PageIndex{2}\)

1. In \(\mathbb{Z}\text{,}\) \(\langle 2\rangle =\{\ldots,-4,-2,0,2,4,\ldots\}=2\mathbb{Z}\text{.}\) More generally, given any \(n\in \mathbb{Z}\text{,}\) in \(\mathbb{Z}\) we have \(\langle n\rangle =n\mathbb{Z}\text{.}\) For \(a\in \mathbb{Z}\text{,}\) \(o(a)=\infty\) if \(a\neq 0\text{;}\) \(o(0)=1\text{.}\)
2. In \(\mathbb{Z}_8\text{,}\) we have \(\langle 0\rangle =\{0\}\text{,}\) \(\langle 1\rangle =\langle 3\rangle =\langle 5\rangle =\langle 7\rangle =\mathbb{Z}_8,\) \(\langle 2\rangle =\langle 6\rangle =\{0,2,4,6\},\) and \(\langle 4\rangle =\{0,4\}\text{.}\)
3. In \(\mathbb{R}\text{,}\) \(\langle \pi\rangle =\pi\mathbb{Z}\text{,}\) so \(o(\pi)=\infty\text{.}\)
4. In \(\mathbb{R}^*\text{,}\) \(\langle \pi\rangle =\{\pi^n:n\in \mathbb{Z}\}\text{.}\) Again, \(o(\pi)=\infty\text{.}\)
5. In \(\mathbb{M}_2(\mathbb{R})\text{,}\)

\begin{equation*} \left\langle \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\right\rangle =\left\{ \begin{bmatrix} c & c \\ 0 & c \end{bmatrix} : c\in \mathbb{Z} \right\}= \left\{c \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} : c\in \mathbb{Z} \right\}. \end{equation*}

    The matrix therefore has infinite order.

6. In \(\mathbb{M}_2(\mathbb{Z}_2)\text{,}\) if \(A= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \text{,}\) then \(\langle A\rangle =\left\{A, \begin{bmatrix}0& 0 \\ 0& 0\end{bmatrix}\right\}\text{,}\) so \(A\) has order \(2\).
7. In \(GL(2,\mathbb{Z}_2)\text{,}\) \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \) has order \(2\). (Why?)
8. In \(GL(2,\mathbb{R})\text{,}\) \(\begin{bmatrix} 0 & -1 \\ 1 & \phantom{-}0 \end{bmatrix} \) has order \(4\). (Why?)
9. In \(GL(2,\mathbb{R})\text{,}\) \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \) has infinite order. (Why?)
10. In \(GL(4,\mathbb{Z}_2)\text{,}\)\(A=\ \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0& 0 & 1 \\ 1 & 0 & 0 &0 \\ 0 &1 & 0 & 0 \end{bmatrix} \) has order \(2\). (Why?)

Theorem \(\PageIndex{8}\)

Every cyclic group \(G\) is of the form \(\langle a\rangle\) for some \(a\in G\text{.}\)

Proof

Let \(G\) be cyclic. Suppose \(|G|=∞\). Then there is an isomorphism \(ϕ:\mathbb{Z}→G\). Note that \(\mathbb{Z}= \langle 1 \rangle\). So

\(\begin{array} &G &=ϕ(\mathbb{Z})\\ &={ϕ(a):a∈ \mathbb{Z}} \\&={ϕ(a(1)):a∈ \mathbb{Z}} \\&={aϕ(1):a∈ \mathbb{Z}} \\&= \langle ϕ(1) \rangle. \end{array}\)

A similar argument shows that there exists \(a∈G\) such that \(G=\langle a \rangle\)⟩ when \(|G|<∞\).

Definition: Generator of G

Let \(G\) be a group. An element \(a\in G\) is a generator of \(G\) (equivalently, \(a\) generates \(G\)) if \(\langle a\rangle =G\text{.}\)

Remark

1. Note that if \(G\) has a generator, then it is necessarily a cyclic group.
2. Note that an element \(a\) of a group \(G\) generates \(G\) if and only if every element of \(G\) is of the form \(a^n\) for some \(n\in \mathbb{Z} \text{.}\)
3. Generators of groups need not be unique. For instance, we saw in Example \(5.1.2\) that each of the elements \(1,3,5\) and \(7\) of \(\mathbb{Z}_8\) is a generator for \(\mathbb{Z}_8\text{.}\)​

Example \(\PageIndex{3}\)

1. \(\mathbb{Z}\) has generator \(1\) and generator \(-1\text{.}\)
2. Given \(n\in \mathbb{Z}\text{,}\) \(nZ\) has generator \(n\) and generator \(-n\text{.}\)
3. Given \(n\geq 2\) in \(\mathbb{Z}\text{,}\) the generators of \(\mathbb{Z}_n\) are exactly the elements \(a\in \mathbb{Z}_n\) such that \(\gcd(n,a)=1\text{.}\) (This follows from Theorem \(5.1.6\).)

Remark

Order of elements provides another group invariant.

Theorem \(\PageIndex{9}\)

Let \(\phi:G\to G'\) be a group isomorphism and let \(a\in G\text{.}\) Then \(o(\phi(a))=o(a)\text{.}\)

Proof

This follows from the fact that \(\phi(\langle a \rangle)= \langle \phi(a) \rangle\).

Corollary \(\PageIndex{2}\)

If groups \(G\) and \(G'\) are isomorphic then for any \(n\in \mathbb{Z}^+\text{,}\) the number of elements of \(G\) of order \(n\) is the same as the number of elements of \(G'\) of order \(n\text{.}\)

Example \(\PageIndex{4}\)

Since \(1\) in \(\mathbb{Z}_4\) has order \(4\) but every element in \(\mathbb{Z}_2 \times \mathbb{Z}_2\) has order less than or equal to \(2\), these groups cannot be isomorphic.

Note

Note, however, that just because the orders of elements of two groups “match up” the groups need not be isomorphic. For example, every element of \(\mathbb{Z}\) has infinite order, except for its identity element, which has order \(1\); the same is true for the group \(\mathbb{Q}\text{.}\) However, we have previously proven that these groups are not isomorphic.