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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 2.2 Cyclic Groups

Subsection Cyclic Groups

“Madness is the exception in individuals and the rule in groups.”
―Friedrich Nietzsche
We zoom in now on the wondrous case in which a group can be generated by a single element.

Definition 2.14. Cylic Group.

If \(G\) is a generated by a single element, i.e. \(G= \langle x \rangle\) for some \(x \in G\text{,}\) then \(G\) is called a cyclic group.
Recalling Lemma 2.3, we can describe the elements of a cylic group explicitely.

Proof.

By Lemma 2.3, the group \(G\) has the following elements \(G=\{x^i \mid i\in \mathbb{Z}\}\text{.}\) We show that
  • \(|G|\geq n\) by showing the elements \(x^0, x^1, \dots, x^{n-1}\) are distinct. Indeed, if \(0\leq i<j<n\) and \(x^i=x^j\) then \(x^{j-i}=e\) and \(1\leq j-i<n\text{,}\) contradicting the minimality of the order of \(x\text{.}\)
  • \(|G|\leq n\) by showing \(G\subseteq\{x^0, x^1, \dots, x^{n-1}\}\) (this implies \(G=\{e,x,\ldots,x^{n-1}\}\)). Indeed, for any \(m\in \mathbb{Z}\) division by \(n\) yields integers \(q,r\) with \(0\leq r\leq n-1\) such that \(m=nq+r\text{.}\) Then \(x^m=x^{nq+r}=(x^n)^qx^r=x^r\in\{x^0, x^1, \dots, x^{n-1}\}\text{.}\)
One quick way of seeing if a finite group is cyclic is to find an element with the same order as the group.

Example 2.17. Cyclic Groups.

  1. \((\Z,+)\) is a cyclic group.
  2. \((\Z/n,+)\) is a cyclic group.
Answer.
  1. \((\Z,+)=\igen 1\text{,}\) for example.
  2. \((\Z/n,+)\igen{1}\text{,}\) for example.

Exercise 2.18. Not Quite Cyclic Groups.

  1. Prove that \((\mathbb Q, +)\) is not a cyclic group.
  2. Prove that \(\operatorname{GL}_2(\mathbb Z/2)\) is not cyclic.
Generators are not unique.

Exercise 2.19. Cyclic Generators are not Unique.

Let \(G=\langle x\rangle\) be a cyclic group. Then \(G=\langle x^{-1}\rangle\text{.}\)
Here is a more general criteria for determining when an element of a cyclic group is a generator.
Conveniently enough, cyclic groups are always abelian.

Example 2.22. Roots of Unity.

For a fixed \(n \geq 1\text{,}\)
\begin{equation*} \mu_n := \{z \in \mathbb{C}\mid z^n = 1\} \end{equation*}
is a subgroup of \((\mathbb{C}\setminus \{0\}, \cdot)\text{.}\) Since \(\|z\|^n = \| z^n \|\) and so if \(z \in \mu\text{,}\) then \(\|z\| = 1\) and hence \(z = e^{ri}\) for some real number \(r\text{.}\) Moreover, \(1 = z^n = e^{nri}\) implies that \(nr\) is an integer multiple of \(2 \pi\text{.}\) It follows that
\begin{equation*} \mu_n = \{1, e^{2 \pi i/n}, e^{4 \pi i/n}, \cdots , e^{(n-1) 2 \pi i/n}\} \end{equation*}
and that \(e^{2 \pi i/n}\) generates \(\mu_n\text{.}\) So, \(\mu_n\) is cyclic or order \(n\text{.}\) It is therefore isomorphic to \(\mathbb{Z}/n\text{,}\) via the map \(\overline{j} \mapsto e^{2 j \pi i/n}\text{.}\)
One of the first things one does when encountering a new group is to examine its subgroups. As it turns out, cyclic groups have some very special properties when it comes to subgroups, though proving them will be more technical than anything we have encountered thus far.

Proof.

  1. Claim 1: For any \(\{e\}\neq H\leq G\text{,}\) setting \(k=\min\{i \mid i\in \mathbb{Z}, i>0, g^i\in H\}\) gives that \(H=\langle g^k \rangle\text{.}\)
    Since \(H\subseteq G=\langle g\rangle\) any element of \(H\) is of the form \(g^m\) for some \(m\in \mathbb{Z}\text{.}\) By the Division Algorithm \(m=kq+r\) for some \(r\in \mathbb{Z}\text{,}\) \(0\leq r<k\text{.}\) Since \(g^m\) and \(g^k\) are elements of \(H\text{,}\) \(g^r=g^m(g^{k})^{-q}\in H\text{.}\) Since \(r<k\) and \(g^r\in H\text{,}\) by the minimality of \(k\) it follows that \(r\) cannot be positive and thus \(r=0\text{.}\) Therefore \(m=kq\) and we have shown that \(H\subseteq \langle g^k \rangle\text{.}\) The opposite containment follows because \(g^k\in H\) and \(\langle g^k \rangle\) is the smallest subgroup of \(G\) containing \(g^k\text{.}\) Thus \(H=\langle g^k \rangle\text{.}\)
    Let \(\Phi:\{\text{subgroups of } G\} \to \{\text{divisors of } |G|\}\) be given by \(\Phi(H)=|H|\text{.}\)
    Claim 2: For any divisor \(d\) of \(|G|\text{,}\) we have \((\Psi\circ\Phi)(d)=d\text{.}\)
    We have \((\Psi\circ\Phi)(d)=\left| \langle g^{\frac{|G|}{d}}\rangle\right|=\left| g^{\frac{|G|}{d}}\right|=\frac{|G|}{\operatorname{gcd}(|G|/d,|G|)}=\frac{|G|}{|G|/d}=d\text{.}\)
    Claim 3: For any subgroup \(H\) of \(G\text{,}\) we have \((\Phi\circ\Psi)(H)=H\text{.}\)
    By Claim 1, any \(H\leq G\) is either \(H=\{e\}\text{,}\) for which \((\Phi\circ\Psi)(\{e\})=\Phi(1)=\langle g^{|G|} \rangle=\{e\}\) or is of the form \(H=\langle g^k \rangle\text{.}\) In the latter case, setting \(|G|=n\) we have \(|H|=|g^k|=\frac{n}{\operatorname{gcd}(n,k)}\) and \((\Phi\circ\Psi)(H)=\langle g^{\frac{n}{n/\operatorname{gcd}(n,k)}} \rangle=\langle g^{\operatorname{gcd}(n,k)}\rangle.\text{.}\) It remains to show that \(\langle g^{\operatorname{gcd}(n,k)}\rangle=\langle g^k\rangle=H\text{.}\) By Bézout’s Identity, \(\operatorname{gcd}(n,k)=an+bk\) for some integers \(a,b\text{.}\) Since \(g^{\operatorname{gcd}(n,k)}=(g^n)^a(g^k)^b=e^a(g^k)^b=(g^k)^b\in \langle g^k \rangle\) if follows that \(\langle g^{\operatorname{gcd}(n,k)}\rangle \subseteq \langle g^k \rangle\text{.}\) On the other hand \(k\) is a multiple of \(\operatorname{gcd}(n,k)\) so \(g^k\in \langle g^{\operatorname{gcd}(n,k)}\rangle\) and thus \(\langle g^k \rangle \subseteq \langle g^{\operatorname{gcd}(n,k)}\rangle\text{.}\) Finally, we conclude that \((\Phi\circ\Psi)(\langle g^k \rangle)=\langle g^{\operatorname{gcd}(n,k)}\rangle=\langle g^k \rangle\) for any \(k\in \mathbb{Z},k>0\text{.}\)
    Claims 2 and 3 establish that \(\Phi\) is a two sided inverse to \(\Phi\text{,}\) thus \(\Phi\) is a bijection.

Exercise 2.24. Cyclic Groups of Small Order.

  1. Every group of orders \(1,2,3\) are cyclic.
  2. Every abelian group of order \(6\) is cyclic.
Finally, we end with two results that will prove invaluable later in the course. Though we won’t see them for some time, proving them now will be good practice and save us time later on.
In particular:

Subsection Uniqueness of Cyclic Groups

“There is no way to be in cyclic existence without creating the causes of suffering.”
―Jetsunma Ahkon Lhamo

Proof.

Recall that either \(G = \{e,x,x^2, \dots, x^{n-1}\}\) (with no repetitions) if \(|x| = n\) or \(G = \{ \cdots, x^{-2}, x^{-1}, e, x, x^e, \dots \}\) (with no repetitions) if \(|x| = \infty\text{.}\)
Uniqueness: We show that if \(\varphi:G\to H\) is a group homomorphism, then \(\varphi(x^i)=y^i\) for all \(i\in \mathbb{Z}\text{.}\)
  • if \(i=0\) then \(\varphi(x^0)=\varphi(e_G)=e_H=y^0\)
  • if \(i>0\) then \(\varphi(x^i)=\varphi(\underbrace{x\cdots x}_{i \text{ times}})=\underbrace{\varphi(x)\cdots \varphi(x)}_{i \text{ times}}=y^i\)
  • if \(i<0\) then \(\varphi(x^i)=\varphi\left((x^{-i})^{-1}\right)=\varphi\left((x^{-i})\right)^{-1}=(y^{-i})^{-1}=y^i\text{,}\) using the formula above for \(-i>0\)
Existence: In either case, define \(\varphi(x^i) = y^i\) for all relevant \(i\) (i.e., in the first case, for \(0 \leq i \leq n-1\) and in the second for all \(i \in\mathbb{Z}\)). We need to show this function is a well-defined group homomorphism. To see that \(\varphi\) is well defined, suppose \(x^i=x^j\) for some \(i,j\in \mathbb{Z}\text{.}\) Then, since \(x^{i-j}=e_G\text{,}\) using [provisional cross-reference: prop96] or the definition for order we have
\begin{equation*} \begin{cases} n\mid i-j & \text{ if } |x|=n\\ i-j=0 & \text{ if } |x|=\infty\\ \end{cases} \Rightarrow \begin{cases} y^ {i-j}=y^{nk} & \text{ if } |x|=n\\ y^{i-j}=y^0 & \text{ if } |x|=\infty\\ \end{cases} \Rightarrow y^ {i-j}=e_H \Rightarrow y^ i=y^j. \end{equation*}
Thus, if \(x^i=x^j\) then \(\varphi(x^i)=y^i=y^j=\varphi(x^j)\text{.}\)
The homomorphism property is immediate: \(\varphi^(x^ix^j)=\varphi(x^{i+j})=y^{i+j}=y^iy^j=\varphi(x^i)\varphi(x^j)\text{.}\)

Remark 2.28.

This is a particular case of the universal mapping property of a presentation, since a cyclic group is either presented by \(\langle x | x^n = e \rangle\) or \(\langle x \mid -\rangle\text{.}\)

Proof.

Suppose \(G = \langle x \rangle\) with \(|x| = n\) or \(|x| = \infty\) and set \(H=C_n\) in the first case and \(H=C_\infty\) in the second case. Then by Proposition 2.27, there are homomorphisms \(f: G \to H\) and \(g: G \to H\) such that \(f(x) = a\) and \(g(a) =x\text{.}\) So \(g \circ f\) is an endomorphism of \(G\) mapping \(x\) to \(x\text{.}\) But the identity map also has this property, and so the uniqueness clause gives \(g \circ f = \text{id}_G\text{.}\) Similarly, \(f \circ g = \text{id}_H\text{.}\)

Convention 2.30.

Moving forward, it is customary to denote the cyclic group of order \(n\) with the notation \(C_n\text{.}\) We similarly denote the infinite cyclic group \(C_\infty\text{.}\)
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