4.1: Cyclic Groups
Recall that if \(G\) is a group and \(g\in G\) , then the cyclic subgroup generated by \(g\) is given by \[\langle g\rangle =\{g^k\mid k\in\mathbb{Z}\}.\] It is important to point out that \(\langle g\rangle\) may be finite or infinite. In the finite case, the Cayley diagram with generator \(g\) gives us a good indication of where the word “cyclic" comes from (see Problem \(\PageIndex{11}\)). If there exists \(g\in G\) such that \(G=\langle g\rangle\) , then we say that \(G\) is a cyclic group .
List all of the elements in each of the following cyclic subgroups.
- \(\langle r\rangle\) , where \(r\in D_3\)
- \(\langle r\rangle\) , where \(r\in R_4\)
- \(\langle rs\rangle\) , where \(rs\in D_4\)
- \(\langle r^2\rangle\) , where \(r^2\in R_6\)
- \(\langle i\rangle\) , where \(i\in Q_8\)
- \(\langle 6\rangle\) , where \(6\in \mathbb{Z}\) and the operation is ordinary addition
Consider the group of invertible \(2\times 2\) matrices with real number entries under the operation of matrix multiplication. This group is denoted by \(\mathrm{GL}_2(\mathbb{R})\) . List the elements in the cyclic subgroups generated by each of the following matrices.
- \(\begin{bmatrix} 0 & -1\\ -1 & 0\end{bmatrix}\)
- \(\begin{bmatrix} 0 & -1\\ 1 & 0\end{bmatrix}\)
- \(\begin{bmatrix} 2 & 0\\ 0 & 1\end{bmatrix}\)
Determine whether each of the following groups is cyclic. If the group is cyclic, find at least one generator.
- \(S_2\)
- \(R_3\)
- \(R_4\)
- \(V_4\)
- \(R_5\)
- \(R_6\)
- \(D_3\)
- \(R_7\)
- \(R_8\)
- \(\text{Spin}_{1\times 2}\)
- \(D_4\)
- \(Q_8\)
Determine whether each of the following groups is cyclic. If the group is cyclic, find at least one generator. If you believe that a group is not cyclic, try to sketch an argument.
- \((\mathbb{Z},+)\)
- \((\mathbb{R},+)\)
- \((\mathbb{R}^+,\cdot)\)
- \((\{6^n\mid n\in\mathbb{Z}\},\cdot)\)
- \(\textrm{GL}_2(\mathbb{R})\) under matrix multiplication
- \(\{(\cos(\pi/4) +i\sin(\pi/4))^n\mid n\in \mathbb{Z}\}\) under multiplication of complex numbers
If \(G\) is a cyclic group, then \(G\) is abelian.
Provide an example of a finite group that is abelian but not cyclic.
Provide an example of an infinite group that is abelian but not cyclic.
If \(G\) is a group and \(g\in G\) , then \(\langle g\rangle=\langle g^{-1}\rangle\) .
If \(G\) is a cyclic group such that \(G\) has exactly one element that generates all of \(G\) , then the order of \(G\) is at most order 2.
If \(G\) is a group such that \(G\) has no proper nontrivial subgroups, then \(G\) is cyclic.
Recall that the order of a group \(G\) , denoted \(|G|\) , is the number of elements in \(G\) . We define the order of an element \(g\) , written \(|g|\) , to be the order of \(\langle g\rangle\) . That is, \(|g|=|\langle g\rangle|\) . It is clear that \(G\) is cyclic with generator \(g\) if and only if \(|G|=|g|\) .
What is the order of the identity in any group?
Find the orders of each of the elements in each of the groups in Problem \(\PageIndex{3}\).
Consider the group \((\mathbb{Z},+)\) . What is the order of 1? Are there any elements in \(\mathbb{Z}\) with finite order?
Find the order of each of the matrices in Problem \(\PageIndex{2}\).
The next result follows immediately from Theorem \(\PageIndex{2}\).
If \(G\) is a group and \(g\in G\) , then \(|g|=|g^{-1}|\) .
The next result should look familiar and will come in handy a few times in this chapter. We’ll take the result for granted and not worry about proving it.
If \(n\) is a positive integer and \(m\) is any integer, then there exist unique integers \(q\) (called the quotient ) and \(r\) (called the remainder ) such that \(m=nq+r\) , where \(0\leq r<n\) .
Suppose \(G\) is a group and let \(g\in G\) . The subgroup \(\langle g\rangle\) is finite if and only if there exists \(n\in\mathbb{N}\) such that \(g^n=e\) .*
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For the forward implication, if \(\langle g\rangle\) is finite, then there exists distinct positive integers \(i\) and \(j\) such that \(g^i=g^j\) . Can you find a useful way to rewrite this equation? For the reverse implication, let \(m\in\mathbb{Z}\) and use the Division Algorithm with \(m\) and \(n\) .
If \(G\) is a finite group, then for all \(g\in G\) , there exists \(n\in\mathbb{N}\) such that \(g^n=e\) .
Suppose \(G\) is a group and let \(g\in G\) such that \(\langle g\rangle\) is a finite group. If \(n\) is the smallest positive integer such that \(g^n=e\) , then \(\langle g\rangle = \{e, g, g^2, \ldots, g^{n-1}\}\) and this set contains \(n\) distinct elements. *
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Note that Theorem \(\PageIndex{7}\) together with the Well-Ordering Principle guarantees the existence of a smallest positive integer \(n\) such that \(g^n=e\) . The claim that the set contains \(n\) distinct elements is not immediate. You need to argue that there are no repeats in the list. Choose distinct \(i,j\in\{0,1,\ldots,n-1\}\) such that \(i\neq j\) and then show that \(g^i\neq g^j\) . Consider a proof by contradiction and try to contradict the minimality of \(n\) .
The next result provides an extremely useful interpretation of the order of an element.
If \(G\) is a group and \(g\in G\) such that \(\langle g\rangle\) is a finite group, then the order of \(g\) is the smallest positive integer \(n\) such that \(g^n=e\) .
Suppose \(G\) is a finite cyclic group such that \(G=\langle g\rangle\) . Using the generating set \(\{g\}\) , what does the Cayley diagram for \(G\) look like?
Suppose \(G\) is a finite cyclic group of order \(n\) with generator \(g\) . If we write down the group table for \(G\) using \(e, g, g^2, \ldots, g^{n-1}\) as the labels for the rows and columns, are there any interesting patterns in the table?
Notice that in the definition for \(\langle g\rangle\) , we allow the exponents on \(g\) to be negative. Explain why we only need to use positive exponents when \(\langle g\rangle\) is a finite group
The Division Algorithm should come in handy when proving the next theorem.
Suppose \(G\) is a group and let \(g\in G\) such that \(|g|=n\) . Then \(g^i=g^j\) if and only if \(n\) divides \(i-j\) .
Suppose \(G\) is a group and let \(g\in G\) such that \(|g|=n\) . If \(g^k=e\) , then \(n\) divides \(k\) .
Recall that for \(n\geq3\) , \(R_n\) is the group of rotational symmetries of a regular \(n\) -gon, where the operation is composition of actions.
For all \(n\geq 3\) , \(R_n\) is cyclic.
Suppose \(G\) is a finite cyclic group of order \(n\) . Then \(G\) is isomorphic to \(R_n\) if \(n\geq 3\) , \(S_2\) if \(n=2\) , and the trivial group if \(n=1\) .
Most of the previous results have involved finite cyclic groups. What about infinite cyclic groups?
Suppose \(G\) is a group and let \(g\in G\) . The subgroup \(\langle g\rangle\) is infinite if and only if each \(g^k\) is distinct for all \(k\in\mathbb{Z}\) .*
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For the forward implication, try a proof by contradiction and suppose there exists integers \(i\) and \(j\) such that \(g^i=g^j\) .
If \(G\) is an infinite cyclic group, then \(G\) is isomorphic to \(\mathbb{Z}\) (under the operation of addition).
The upshot of Theorems \(\PageIndex{13}\) and \(\PageIndex{11}\) is that up to isomorphism, we know exactly what all of the cyclic groups are.
We now turn our attention to two new groups. Recall that two integers are relatively prime if the only positive integer that divides both of them is 1. That is, integers \(n\) and \(k\) are relatively prime if and only if \(\gcd(n,k)=1\) .
Let \(n\in\mathbb{N}\) and define the following sets.
- \(\mathbb{Z}_n:=\{0,1,\ldots,n-1\}\)
- \(U_n:=\{k\in\mathbb{Z}_n\mid \gcd(n,k)=1\}\)
For example, \(\mathbb{Z}_{12}=\{0,1,2,3,4,5,6,7,8,9,10,11\}\) while \(U_{12}=\{1,5,7,11\}\) since 1, 5, 7, and 11 are the only elements in \(\mathbb{Z}_{12}\) that are relatively prime to 12.
For each set in Definition: \(n\in\mathbb{N}\) , the immediate goal is to determine a binary operation that will yield a group. The key is to use modular arithmetic. Let \(n\) be a positive integer. To calculate the sum (respectively, product) of two integers modulo \(n\) (we say “mod \(n\) " for short), add (respectively, multiply) the two numbers and then find the remainder after dividing the sum (respectively, product) by \(n\) . For example, \(4+9\) is \(3\) mod \(5\) since \(13\) has remainder 3 when divided by 5. Similarly, \(4\cdot 9\) is 1 mod \(5\) since 36 has remainder 1 when divided by 5. The hope is that these two operations turn \(\mathbb{Z}_n\) and \(U_n\) into groups.
We write \(i\equiv j\pmod n\) , and say “ \(i\) is equivalent to \(j\) modulo \(n\) " or “ \(i\) is equal to \(j\) modulo \(n\) ", if \(i\) and \(j\) both have the same remainder when divided by \(n\) . It is common to abbreviate “modulo" as “mod". It is also common to write \(i\equiv_n j\) , or even \(i=j\) if the context is perfectly clear.
It is well-known, and not too hard to prove, that \(\equiv_n\) is an equivalence relation on \(\mathbb{Z}\) . The corresponding equivalence classes are called congruence classes. The elements of a single congruence class are the integers that all have the same remainder when divided by \(n\) . According to the Division Algorithm, there are \(n\) congruence classes modulo \(n\) , one for each of the remainders \(0,1,\ldots, n-1\) . We can think of \(\mathbb{Z}_n\) as the set of canonical representatives of these equivalence classes.
Let \(n\) be a positive integer and let \(i,j\in\mathbb{Z}\) . Then \(i\equiv j\pmod n\) if and only if \(n\) divides \(i-j\) .
The next result follows immediately from Theorems \(\PageIndex{14}\) and \(\PageIndex{9}\).
Suppose \(G\) is a group and let \(g\in G\) such that \(|g|=n\) . Then \(g^i=g^j\) if and only if \(i\equiv j\pmod n\) .
The set \(\mathbb{Z}_n\) is a cyclic group under addition mod \(n\) .*
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There are two things to prove here. First, you need to prove that \(\mathbb{Z}_n\) is a group under addition mod \(n\) , and then you need to argue that the group is cyclic.
The set \(U_n\) is an abelian group under multiplication mod \(n\) .
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Like the previous theorem, there are two things to prove. First, prove that \(U_n\) is a group under multiplication mod \(n\) , and then argue that the group is abelian.
Consider \(\mathbb{Z}_4\) .
- Find the group table for \(\mathbb{Z}_4\) .
- Is \(\mathbb{Z}_4\) cyclic? If so, list elements of \(\mathbb{Z}_4\) that individually generate \(\mathbb{Z}_4\) . If \(\mathbb{Z}_4\) is not cyclic, explain why.
- Is \(\mathbb{Z}_4\) isomorphic to either of \(R_4\) or \(V_4\) ? Justify your answer.
- Draw the subgroup lattice for \(\mathbb{Z}_4\) .
The next two problems illustrate that \(U_n\) may or may not be cyclic.
Consider \(U_{10}=\{1,3,7,9\}\) .
- Find the group table for \(U_{10}\) .
- Is \(U_{10}\) cyclic? If so, list elements of \(U_{10}\) that individually generate \(U_{10}\) . If \(U_{10}\) is not cyclic, explain why.
- Is \(U_{10}\) isomorphic to either of \(R_4\) or \(V_4\) ? Justify your answer.
- Is \(U_{10}\) isomorphic to \(\mathbb{Z}_4\) ? Justify your answer.
- Draw the subgroup lattice for \(U_{10}\) .
Consider \(U_{12}=\{1,5,7,11\}\) .
- Find the group table for \(U_{12}\) .
- Is \(U_{12}\) cyclic? If so, list elements of \(U_{12}\) that individually generate \(U_{12}\) . If \(U_{12}\) is not cyclic, explain why.
- Is \(U_{12}\) isomorphic to either of \(R_4\) or \(V_4\) ? Justify your answer.
- Draw the subgroup lattice for \(U_{12}\) .
The upshot of the next theorem is that for \(n\geq 3\) , \(\mathbb{Z}_n\) is just the set of exponents in the set \(R_n=\{e,r,r^2,\ldots,r^{n-1}\}\) (where \(e=r^0\) ).
\mathbb{Z}_n\) Isomorphic to \(R_n\)
For \(n\geq 3\) , \(\mathbb{Z}_n\cong R_n\) . Moreover, \(\mathbb{Z}_2\cong S_2\) and \(\mathbb{Z}_1\) is isomorphic to the trivial group.
The next result can be thought of as a repackaging of Theorems \(\PageIndex{11}\) and \(\PageIndex{13}\).
Let \(G\) be a cyclic group. If the order of \(G\) is infinite, then \(G\) is isomorphic to \(\mathbb{Z}\) . If \(G\) has finite order \(n\) , then \(G\) is isomorphic to \(\mathbb{Z}_n\) .
Now that we have a complete description of the cyclic groups, let’s focus our attention on subgroups of cyclic groups.
Suppose \(G\) is a cyclic group. If \(H\leq G\) , then \(H\) is also cyclic.
It turns out that for proper subgroups, the converse of Theorem \(\PageIndex{19}\) is not true.
Provide an example of a group \(G\) such that \(G\) is not cyclic, but all proper subgroups of \(G\) are cyclic.
The next result officially settles Problem 3.1.11(d) and also provides a complete description of the subgroups of infinite cyclic groups up to isomorphism.
The subgroups of \(\mathbb{Z}\) are precisely the groups \(n\mathbb{Z}\) for \(n\in \mathbb{Z}\) .
Let’s further explore finite cyclic groups.
If \(G\) is a finite cyclic group with generator \(g\) such that \(|G|=n\) , then for all \(m\in\mathbb{Z}\) , \(\displaystyle |g^m|=\dfrac{n}{\gcd(n,m)}\) .*
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By Corollary \(\PageIndex{2}\), the order of \(g^m\) is the smallest positive exponent \(k\) such that \((g^m)^k=e\) . First, verify that \(k=\dfrac{n}{\gcd(n,m)}\) has the desired property and then verify that it is the smallest such exponent.
If \(G\) is a finite cyclic group with generator \(g\) such that \(|G|=n\) , then \(\langle g^m\rangle=\langle g^k\rangle\) if and only if \(\gcd(m,n)=\gcd(k,n)\).*
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Use Theorem \(\PageIndex{20}\) for the forward implication. For the reverse implication, first prove that for all \(m\in\mathbb{Z}\) , \(\langle g^m\rangle=\langle g^{\gcd(m,n)}\rangle\) by proving two set containments. To show \(\langle g^m\rangle\subseteq \langle g^{\gcd(m,n)}\rangle\) , use the fact that there exists an integer \(q\) such that \(m=q\cdot \gcd(m,n)\) . For the reverse containment, you may freely use a fact known as Bezout’s Lemma, which states that \(\gcd(m,n)=nx+my\) for some integers \(x\) and \(y\). *
Suppose \(G\) is a cyclic group of order 12 with generator \(g\) .
- Find the orders of each of the following elements: \(g^2\) , \(g^7\) , \(g^8\) .
- Which elements of \(G\) individually generate \(G\) ?
Suppose \(G\) is a finite cyclic group with generator \(g\) such that \(|G|=n\) . Then \(\langle g\rangle=\langle g^k\rangle\) if and only if \(n\) and \(k\) are relatively prime. That is, \(g^k\) generates \(G\) if and only if \(n\) and \(k\) are relatively prime.
Theorem \(\PageIndex{20}\), Theorem \(\PageIndex{21}\), and Corollary \(\PageIndex{6}\) are written using multiplicative notation. Rewrite both of these results using additive notation.
Consider \(\mathbb{Z}_{18}\) .
- Find all of the elements of \(\mathbb{Z}_{18}\) that individually generate all of \(\mathbb{Z}_{18}\) .
- Draw the subgroup lattice for \(\mathbb{Z}_{18}\) . For each subgroup, list the elements of the corresponding set. Moreover, circle the elements in each subgroup that individually generate that subgroup. For example, \(\langle 2\rangle=\{0,2,4,6,8,10,12,14,16\}\) . In this case, we should circle 2, 4, 8, 10, 14, and 16 since each of these elements individually generate \(\langle 2\rangle\) and none of the remaining elements do. I’ll leave it to you to figure out why this is true.
Repeat the above exercise, but this time use \(\mathbb{Z}_{12}\) instead of \(\mathbb{Z}_{18}\) .
If \(G\) is a finite cyclic group such that \(|G|=p\) , where \(p\) is prime, then \(G\) has no proper nontrivial subgroups.
If there is exactly one group up to isomorphism of order \(n\) , then to what group are all the groups of order \(n\) isomorphic?
Suppose \(G\) is a group and \(x,y\in G\) such that \(|x|=m\) and \(|y|=n\) . Is it true that \(|xy|=mn\) ? If this is true, provide a proof. If this is not true, then provide a counterexample.
The punchline of the next two theorems is Theorem \(\PageIndex{22}\).
Suppose \(G\) is a finite abelian group and let \(x,y\in G\) such that \(|x|=m\) and \(|y|=n\) . If \(\gcd(m,n)=1\) , then \(|xy|=mn\) .*
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Hint: First, verify that \((xy)^{mn}=e\) . Now, suppose \(|xy|=k\) . What do you immediately know about the relationship between \(k\) and \(mn\) . Next, consider \((xy)^{kn}\) . Argue that \(m\) divides \(kn\) and then argue that \(m\) divides \(k\) . Similarly, \(n\) divides \(k\) . Ultimately, conclude that \(mn=k\) .
Suppose \(G\) is a finite abelian group. If \(n\) is the maximal order among all elements in \(G\) , then the order of every element in \(G\) divides \(n\).*
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Hint: Suppose \(g\in G\) such that \(|g|=n\) . Let \(h\) be an arbitrary element in \(G\) such that \(|h|=m\) . You need to show that \(m\) divides \(n\) . For sake of a contradiction, assume otherwise. Then there exists a prime \(p\) whose multiplicity as a factor of \(m\) exceeds that of \(n\) . Let \(p^a\) be the highest power of \(p\) in \(m\) and \(p^b\) be the highest power of \(p\) in \(n\) , so \(a>b\) . Consider the elements \(g^{p^a}\) and \(h^{m/p^b}\) .
Recall that every cyclic group is abelian (see Theorem \(\PageIndex{1}\)). However, we know that not every abelian group is cyclic (see Problem \(\PageIndex{5}\)). The next theorem tells us that abelian groups with some additional properties are cyclic.
If \(G\) is a finite abelian group with at most one subgroup of any order, then \(G\) is cyclic.
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Hint: Let \(n\) be the maximal order among the elements of \(G\) and let \(g\in G\) be an element with order \(n\) . Prove that \(G=\langle g\rangle\) .
Is the converse of Theorem \(\PageIndex{22}\) true for finite groups? That is, if \(G\) is a finite cyclic group, does that imply that \(G\) contains at most one subgroup of each order? If the answer is yes, then prove it. Otherwise, provide a counterexample.
We conclude this section with a couple interesting counting problems involving the number of generators of certain cyclic groups.
Let \(p\) and \(q\) be distinct primes. Find the number of generators of \(\mathbb{Z}_{pq}\) .
Let \(p\) be a prime. Find the number of generators of \(\mathbb{Z}_{p^r}\) , where \(r\) is an integer greater than or equal to 1.