4.4 Reading Questions
| 1. What is the order of the element \boldsymbol{3} in \boldsymbol{U(20)}? U(20)=\{1,3,5,7,9,11,13,17,19\}. 3^1=3, \ \ 3^2=9, \ \ 3^3 = 27 \equiv 7 \pmod{20}, \ \ 3^4=81 \equiv 1 \pmod{20}. \implies |3|=4. |
| 2. What is the order of the element \boldsymbol{5} in \boldsymbol{U(23)} ? 5^1=5, \ \ 5^2=25 \equiv 2 \pmod{23}, \ \ …, \ \ 5^{12} \equiv 1 \pmod{23}. \implies |5| = 12. |
| 3. Find three generators of \boldsymbol{\mathbb{Z}_8} 1, 3, and 5 all generate \mathbb{Z}_8. In general, an element a \in \mathbb{Z}_n, a \geq 1, will generate \mathbb{Z}_n if \gcd(a,n)=1 (i.e. if a and n are co-prime). |
| 4. Find three generators of the 5th roots of unity. The nth roots of unity are: z_k=cis(\frac{2k \pi}{n}), where cis(\theta)=e^{i \theta}, 0 \leq k<n. The 5th roots of unity are: z_0= e^{0}=1. z_1=e^{\frac{2 \pi}{5}}. z_2=e^{\frac{4 \pi}{5}}. z_3=e^{\frac{6 \pi}{5}}. z_4=e^{\frac{8 \pi}{5}}. In this case, any element other than \varepsilon = 1 will generate the group. This can be demonstrated by checking for each element. Alternatively, we can use the fact that the group of the nth roots of unity is isomorphic to \mathbb{Z}_n. Since 5 is prime, all non-identity elements in \mathbb{Z}_5 will generate the group (because they are all co-prime to 5). |
4.5 Exercises
| 1. Prove or disprove each of the following statements. (a) All of the generators of \mathbb{Z}_{60} are prime. This is not true. We will demonstrate this with a counterexample. Any generator of \mathbb{Z}_{60} will be co-prime to 60. To find a counterexample, we must find a nonprime number less than 60 whose prime factorization does not include numbers that divide 60. Consider 7 \times 7 = 49. 49 is a nonprime number that is coprime to 60. Thus, it generates \mathbb{Z}_{60}. Therefore, the statement is false. □ (b) U(8) is cyclic. U(8)=\{1,3,5,7\}. 1^2=3^2=5^2=7^2=1=\varepsilon. None of the elements of U(8) generate U(8). Therefore, it is not a cyclic group. □ (c) \mathbb{Q} is cyclic. Assume \mathbb{Q} is cyclic. Then \mathbb{Q} = \langle \frac{a}{b} \rangle, for some constants a,b \in \mathbb{Z}. \langle \frac{a}{b} \rangle = \{n \frac{a}{b} : n \in \mathbb{Z}\}. Consider \frac{a}{2b}. Clearly \frac{a}{2b} \notin \langle \frac{a}{b} \rangle. However, \frac{a}{2b} \in \mathbb{Q}. Contradiction. Therefore, \mathbb{Q} is not cyclic. □ |
| 2. Find the order of each of the following elements. (a) 5 \in \mathbb{Z}_{12}. \gcd(5,12)=1. \implies |5|=12. (b) \sqrt{3} \in \mathbb{R}. (\sqrt{3})^n = n \sqrt{3}. n \sqrt{3} \neq 0 for any n \geq 1. \implies | \sqrt{3} | = \infty. (c) \sqrt{3} \in \mathbb{R}^{*}. (\sqrt{3})^n = 3^{\frac{n}{2}}. 3^{\frac{n}{2}} \neq 1 for any n \geq 1. \implies | \sqrt{3} | = \infty. (d) -i \in \mathbb{C}^{*}. (-i)^1=-i, \ \ (-i)^2=-1, \ \ (-i)^3=i, \ \ (-i)^4=1. \implies |-i|=4. (e) 72 \in \mathbb{Z}_{240} . d = \gcd(72,240)=24. n=240. |72|=\frac{n}{d}=\frac{240}{24}. \implies |72|=10. (f) 312 \in \mathbb{Z}_{471}. d = \gcd(312,471)=3. n=471. |312|=\frac{n}{d}=\frac{471}{3}. \implies |312|=157. |
| 3. List all of the elements in each of the following subgroups. (a) The subgroup of \mathbb{Z} generated by 7. \langle 7 \rangle = \{ …, -7, 0, 7, 14, …\} = 7 \mathbb{Z}. (b) The subgroup of \mathbb{Z}_{24} generated by 15. \langle 7 \rangle = \{ 0, 15, 6, 21, 12, 3, 18, 9 \}. (c) All subgroups of \mathbb{Z}_{12}. \langle 0 \rangle = \{0\}. \langle 1 \rangle = \langle 5 \rangle = \langle 7 \rangle = \langle 11 \rangle = \mathbb{Z}_{12}. \langle 2 \rangle = \langle 10 \rangle = \{0, 2, 4, 6, 8, 10\}. \langle 3 \rangle = \langle 9 \rangle = \{0, 3, 6, 9\}. \langle 4 \rangle = \langle 8 \rangle = \{0, 4, 8\}. \langle 6 \rangle = \{0, 6\}. (d) All subgroups of \mathbb{Z}_{60}. \langle 0 \rangle = \{0\}. \langle 1 \rangle = \mathbb{Z}_{60}. \langle 2 \rangle = \{0,2,4,…,58\}. \langle 3 \rangle =\{0, 3, 6, …, 57\}. \langle 4 \rangle =\{0, 4, 8, …, 56\}. \langle 5 \rangle =\{0,5,10,…,55\}. \langle 6 \rangle=\{0,6,12,…,54\}. \langle 10 \rangle=\{0,10,20,…,50\}. \langle 12 \rangle=\{0, 12, 24, 36, 48\}. \langle 15 \rangle=\{0,15,30,45\}. \langle 20 \rangle=\{0,20,40\}. \langle 30 \rangle=\{0,30\}. (e) All subgroups of \mathbb{Z}_{13}. 13 is prime \implies \mathbb{Z}_{13} has no subgroups other than \{0\} and itself. (g) The subgroup generated by 3 in U(20). U(20)=\{1,3,7,9,11,13,17,19\}. \langle 3 \rangle=\{3,9,7,1\}. (h) The subgroup generated by 5 in U(18). \langle 5 \rangle=\{5,7,17,13,11,1\}=U(18). (i) The subgroup of \mathbb{R}^{*} generated by 7. \langle 7 \rangle=\{7^n : n \in \mathbb{Z} \}=\{…,\frac{1}{49}, \frac{1}{7}, 1, 7, 49, …\}. (j) The subgroup of \mathbb{C}^{*} generated by i where i^2 = −1. i^1=i, \ \ i^2=-1, \ \ i^3=-i, \ \ i^4 = 1. \implies \langle i \rangle=\{i,-1,-i,1\}. (k) The subgroup of \mathbb{C}^{*} generated by 2i. \langle 2i \rangle = \{2i, -4, -8i, 16, 32i, …\}. |
| 4. Find the subgroups of \boldsymbol{GL_2(\mathbb{R})} generated by each of the following matrices. Recall that \boldsymbol{GL_2(\mathbb{R})} is the general linear group, defined as the group of invertible \boldsymbol{2 \times 2} matrices with real entries and with the group operation being matrix multiplication. (a) \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^2=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}. \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^3=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^4=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=I_2. \implies \langle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \rangle = \{ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^1, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^2, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^3, I_2\}. (b) \begin{pmatrix} 0 & \frac{1}{3} \\ 3 & 0 \end{pmatrix}. \begin{pmatrix} 0 & \frac{1}{3} \\ 3 & 0 \end{pmatrix}^2=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=I_2. \implies \langle \begin{pmatrix} 0 & \frac{1}{3} \\ 3 & 0 \end{pmatrix} \rangle = \{ \begin{pmatrix} 0 & \frac{1}{3} \\ 3 & 0 \end{pmatrix}, I_2\}. (c) \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}. |\begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}|=6. \implies \langle \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix} \rangle = \{ \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}^1, …, \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}^5, I_2\}. (d) \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}. \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}^2=\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}. \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}^3=\begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}. …and so on. \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}^{-2}=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}. …and so on. \implies \langle \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \rangle = \{ \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}: n \in \mathbb{N}\}. |
| 5. Find the order of every element in \boldsymbol{\mathbb{Z}_{18}}. |0|=1. |1|=|5|=|7|=|11|=|13|=|17|=18. |2|=|4|=|8|=|10|=|14|=|16|=9. |3|=|6|=|9|=|12|=|15|=6. |
| 23. Let \boldsymbol{a, b \in G}. Prove the following statements. (a) The order of \boldsymbol{a} is the same as the order of \boldsymbol{a^{-1}}. (b) For all \boldsymbol{g \in G}, \boldsymbol{|a| = |g^{-1}ag}|. (c) The order of \boldsymbol{ab} is the same as the order of \boldsymbol{ba}. (a) The order of a is the same as the order of a^{-1}. Consider arbitrary a \in G with |a|=n. |a|=n \implies \overbrace{aa \cdot \cdot \cdot a}^{n-times} = \varepsilon. (aa \cdot \cdot \cdot a) (a^{-1}a^{-1} \cdot \cdot \cdot a^{-1}) = \varepsilon (\overbrace{a^{-1}a^{-1} \cdot \cdot \cdot a^{-1}}^{n-times}). \varepsilon = a^{-1}a^{-1} \cdot \cdot \cdot a^{-1} \implies |a^{-1}|=n. □ (b) For all g \in G, |a| = |g^{-1}ag|. Consider arbitrary a,g \in G and let |a|=n. Now, consider (g^{-1}ag)^n. (g^{-1}ag)^n=(g^{-1}ag)(g^{-1}ag) \cdot \cdot \cdot (g^{-1}ag). (g^{-1}ag)^n=(g^{-1} \overbrace{aa \cdot \cdot \cdot a}^{n-times} g). (g^{-1}ag)^n=(g^{-1}g). (g^{-1}ag)^n = \varepsilon \implies |g^{-1}ag|=n. □ (c) The order of ab is the same as the order of ba. Consider arbitrary a,b \in G and let |ab|=n. Then (ab)^n=(ab)(ab) \cdot \cdot \cdot (ab) = \varepsilon. (ab)(ab) \cdot \cdot \cdot (ab)aa^{-1}= \varepsilon aa^{-1}. a(ba) \cdot \cdot \cdot (ba)(ba)a^{-1}= \varepsilon. a(ba)^n a^{-1}= \varepsilon. (ba)^n = \varepsilon \implies |ba|=n=|ab|. □ |
| 24. Let \boldsymbol{p} and \boldsymbol{q} be distinct primes. How many generators does \boldsymbol{\mathbb{Z}_{pq}} have? Let’s consider the elements in \mathbb{Z}_{pq}=\{0,1,2,…,(pq-1)\} that do NOT generate \mathbb{Z}_{pq}. They are exactly all the positive multiples of p and q, and 0. There are (q-1) multiples of p between 0 and (pq-1). Similarly, there are (p-1) positive multiples of q between 0 and (pq-1). Hence, the number of elements that do not generate \mathbb{Z}_{pq} is given by: (q-1) + (p-1) + 1 = q + p – 1. The number of elements that generate \mathbb{Z}_{pq} is thus given by the total number of elements in \mathbb{Z}_{pq} minus the number of elements that do NOT generate the group: pq – (q+p-1) = pq – p – q +1 = (p-1)(q-1). □ |
| 25. Let \boldsymbol{p} be prime and \boldsymbol{r} be a positive integer. How many generators does \boldsymbol{\mathbb{Z}_{p^r}} have? Again, let’s consider all of the elements that do NOT generate \mathbb{Z}_{p^r}=\{0,1,2,…,p^r-1\}. They are exactly all the positive multiples of p and 0. There are (p^{r-1}-1) multiples of p between 0 and (p^r-1) (this follows from the fact that p^r=pp^{r-1} ). Hence, the number of elements that do not generate \mathbb{Z}_{p^r} is given by: (p^{r-1}-1)+1=p^{r-1}. The number of elements that generate \mathbb{Z}_{p^r} is thus given by the total number of elements in \mathbb{Z}_{p^r} minus the number of elements that do NOT generate the group: p^r-p^{r-1}. □ |
| 26. Prove that \boldsymbol{\mathbb{Z}_p} has no nontrivial subgroups if \boldsymbol{p} is prime. The generators of \mathbb{Z}_p=\{ 0, 1, …, (p-1)\} are the integers r such that: 1 \leq r < p and \gcd(r,p)=1. (Corollary 4.14) Since p is prime, it follows that: \gcd(r,p)=1 for all r such that 1 \leq r < p. Thus, each of the elements 1,2,…,(p-1) generate \mathbb{Z}_p. Now, we need only consider 0, which generates the trivial subgroup \{0\}. Therefore, every element in \mathbb{Z}_p generates a trivial subgroup if p is prime. □ |
| 29. Prove that \boldsymbol{\mathbb{Z}_n} has an even number of generators for \boldsymbol{n > 2}. If x generates \mathbb{Z}_n, then so does x^{-1}. If x=x^{-1}, then x^2=\varepsilon \implies |G| \leq 2. If x \neq x^{-1}, then we can divide the generators into pairs. □ |
| 43. If \boldsymbol{z=r(\cos(\theta) + i\sin(\theta)} and \boldsymbol{w = s(\cos(\phi) + i\sin(\phi)} are two nonzero complex numbers, show that: \boldsymbol{zw = rs[\cos(\theta+ \phi) + i\sin(\theta+ \phi)]}. z=r(\cos(\theta) + i\sin(\theta) \implies z=re^{i\theta}. w=s(\cos(\phi) + i\sin(\phi) \implies w=se^{i\phi}. zw=re^{i\theta}se^{i\phi}. zw=rse^{i\theta + i\phi}. zw=rse^{i(\theta + \phi)}. zw=rs[\cos(\theta+ \phi) + i\sin(\theta+ \phi)]. □ |