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Theorem 2.9 Division Algorithm. Let \( a \) and \( b \) be integers, with \( b > 0 \) . Then there exist unique integers \( q \) and \( r \) such that\n\n\[ a = {bq} + r \]\n\nwhere \( 0 \leq r < b \) . | Proof. This is a perfect example of the existence-and-uniqueness type of proof. We must first prove that the numbers \( q \) and \( r \) actually exist. Then we must show that if \( {q}^{\prime } \) and \( {r}^{\prime } \) are two other such numbers, then \( q = {q}^{\prime } \) and \( r = {r}^{\prime } \) .\n\nExisten... | Yes |
Theorem 2.10. Let \( a \) and \( b \) be nonzero integers. Then there exist integers \( r \) and \( s \) such that\n\n\[ \gcd \left( {a, b}\right) = {ar} + {bs}. \]\n\nFurthermore, the greatest common divisor of \( a \) and \( b \) is unique. | Proof. Let\n\n\[ S = \{ {am} + {bn} : m, n \in \mathbb{Z}\text{ and }{am} + {bn} > 0\} . \]\n\nClearly, the set \( S \) is nonempty; hence, by the Well-Ordering Principle \( S \) must have a smallest member, say \( d = {ar} + {bs} \) . We claim that \( d = \gcd \left( {a, b}\right) \) . Write \( a = {dq} + {r}^{\prime ... | Yes |
Let us compute the greatest common divisor of 945 and 2415. | First observe that\n\n\[ \n{2415} = {945} \cdot 2 + {525} \n\] \n\n\[ \n{945} = {525} \cdot 1 + {420} \n\] \n\n\[ \n{525} = {420} \cdot 1 + {105} \n\] \n\n\[ \n{420} = {105} \cdot 4 + 0. \n\] \n\nReversing our steps, 105 divides 420, 105 divides 525, 105 divides 945, and 105 divides 2415. Hence, 105 divides both 945 an... | Yes |
Lemma 2.13 Euclid. Let \( a \) and \( b \) be integers and \( p \) be a prime number. If \( p \mid {ab} \), then either \( p \mid a \) or \( p \mid b \) . | Proof. Suppose that \( p \) does not divide \( a \) . We must show that \( p \mid b \) . Since \( \gcd \left( {a, p}\right) = 1 \) , there exist integers \( r \) and \( s \) such that \( {ar} + {ps} = 1 \) . So\n\n\[ b = b\left( {{ar} + {ps}}\right) = \left( {ab}\right) r + p\left( {bs}\right) . \]\n\nSince \( p \) div... | Yes |
Theorem 2.14 Euclid. There exist an infinite number of primes. | Proof. We will prove this theorem by contradiction. Suppose that there are only a finite number of primes, say \( {p}_{1},{p}_{2},\ldots ,{p}_{n} \) . Let \( P = {p}_{1}{p}_{2}\cdots {p}_{n} + 1 \) . Then \( P \) must be divisible by some \( {p}_{i} \) for \( 1 \leq i \leq n \) . In this case, \( {p}_{i} \) must divide... | Yes |
Theorem 2.15 Fundamental Theorem of Arithmetic. Let \( n \) be an integer such that \( n > 1 \) . Then\n\n\[ n = {p}_{1}{p}_{2}\cdots {p}_{k} \]\n\nwhere \( {p}_{1},\ldots ,{p}_{k} \) are primes (not necessarily distinct). Furthermore, this factorization is unique; that is, if\n\n\[ n = {q}_{1}{q}_{2}\cdots {q}_{l} \]\... | Proof. Uniqueness. To show uniqueness we will use induction on \( n \) . The theorem is certainly true for \( n = 2 \) since in this case \( n \) is prime. Now assume that the result holds for all integers \( m \) such that \( 1 \leq m < n \), and\n\n\[ n = {p}_{1}{p}_{2}\cdots {p}_{k} = {q}_{1}{q}_{2}\cdots {q}_{l} \]... | Yes |
The following examples illustrate integer arithmetic modulo \( n \) : | \[ 7 + 4 \equiv 1\;\left( {\;\operatorname{mod}\;5}\right) \] \[ 7 \cdot 3 \equiv 1\;\left( {\;\operatorname{mod}\;5}\right) \] \[ 3 + 5 \equiv 0\;\left( {\;\operatorname{mod}\;8}\right) \] \[ 3 \cdot 5 \equiv 7\;\left( {\;\operatorname{mod}\;8}\right) \] \[ 3 + 4 \equiv 7\;\left( {\;\operatorname{mod}\;{12}}\right) \]... | Yes |
Most, but not all, of the usual laws of arithmetic hold for addition and multiplication in \( {\mathbb{Z}}_{n} \) . For instance, it is not necessarily true that there is a multiplicative inverse. Consider the multiplication table for \( {\mathbb{Z}}_{8} \) in Table 3.3. Notice that \( 2,4 \), and 6 do not have multipl... | \[ \begin{matrix} \cdot & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 0 & 2 & 4 & 6 & 0 & 2 & 4 & 6 \\ 3 & 0 & 3 & 6 & 1 & 4 & 7 & 2 & 5 \\ 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 & 4 \\ 5 & 0 & 5 & 2 & 7 & 4 & 1 & 6 & 3 \\ 6 & 0 & 6 & 4 & 2 & 0 & 6 & 4 & 2 \\ 7 ... | Yes |
Proposition 3.4. Let \( {\mathbb{Z}}_{n} \) be the set of equivalence classes of the integers \( {\;\operatorname{mod}\;n} \) and \( a, b, c \in {\mathbb{Z}}_{n} \) .\n\n1. Addition and multiplication are commutative:\n\n\[ a + b \equiv b + a\;\left( {\;\operatorname{mod}\;n}\right) \]\n\n\[ {ab} \equiv {ba}\;\left( {\... | Proof. We will prove (1) and (6) and leave the remaining properties to be proven in the exercises.\n\n(1) Addition and multiplication are commutative modulo \( n \) since the remainder of \( a + b \) divided by \( n \) is the same as the remainder of \( b + a \) divided by \( n \) . | No |
The integers \( \mathbb{Z} = \{ \ldots , - 1,0,1,2,\ldots \} \) form a group under the operation of addition. | The binary operation on two integers \( m, n \in \mathbb{Z} \) is just their sum. Since the integers under addition already have a well-established notation, we will use the operator + instead of \( \circ \) ; that is, we shall write \( m + n \) instead of \( m \circ n \) . The identity is 0, and the inverse of \( n \i... | Yes |
The integers \( {\;\operatorname{mod}\;n} \) form a group under addition modulo \( n \) . Consider \( {\mathbb{Z}}_{5} \), consisting of the equivalence classes of the integers \( 0,1,2,3 \), and 4 . We define the group operation on \( {\mathbb{Z}}_{5} \) by modular addition. We write the binary operation on the group ... | By Proposition 3.4, \( {\mathbb{Z}}_{n} = \{ 0,1,\ldots, n - 1\} \) is a group under the binary operation of addition \( {\;\operatorname{mod}\;n} \) . | Yes |
Not every set with a binary operation is a group. For example, if we let modular multiplication be the binary operation on \( {\mathbb{Z}}_{n} \), then \( {\mathbb{Z}}_{n} \) fails to be a group. | The element 1 acts as a group identity since \( 1 \cdot k = k \cdot 1 = k \) for any \( k \in {\mathbb{Z}}_{n} \) ; however, a multiplicative inverse for 0 does not exist since \( 0 \cdot k = k \cdot 0 = 0 \) for every \( k \) in \( {\mathbb{Z}}_{n} \) . Even if we consider the set \( {\mathbb{Z}}_{n} \smallsetminus \{... | Yes |
The symmetries of an equilateral triangle described in Section 3.1 form a nonabelian group. As we observed, it is not necessarily true that \( {\alpha \beta } = {\beta \alpha } \) for two symmetries \( \alpha \) and \( \beta \) . | Using Table 3.7, which is a Cayley table for this group, we can easily check that the symmetries of an equilateral triangle are indeed a group. We will denote this group by either \( {S}_{3} \) or \( {D}_{3} \), for reasons that will be explained later. | No |
We use \( {\mathbb{M}}_{2}\left( \mathbb{R}\right) \) to denote the set of all \( 2 \times 2 \) matrices. Let \( G{L}_{2}\left( \mathbb{R}\right) \) be the subset of \( {\mathbb{M}}_{2}\left( \mathbb{R}\right) \) consisting of invertible matrices; that is, a matrix\n\n\[ A = \left( \begin{array}{ll} a & b \\ c & d \end... | The set of invertible matrices forms a group called the general linear group. The identity of the group is the identity matrix\n\n\[ I = \left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \]\n\nThe inverse of \( A \in G{L}_{2}\left( \mathbb{R}\right) \) is\n\n\[ {A}^{-1} = \frac{1}{{ad} - {bc}}\left( \begin{mat... | Yes |
Let \( {\mathbb{C}}^{ * } \) be the set of nonzero complex numbers. Under the operation of multiplication \( {\mathbb{C}}^{ * } \) forms a group. The identity is 1 . If \( z = a + {bi} \) is a nonzero complex number, then | \[ {z}^{-1} = \frac{a - {bi}}{{a}^{2} + {b}^{2}} \] is the inverse of \( z \) . It is easy to see that the remaining group axioms hold. | Yes |
Proposition 3.17. The identity element in a group \( G \) is unique; that is, there exists only one element \( e \in G \) such that \( {eg} = {ge} = g \) for all \( g \in G \) . | Proof. Suppose that \( e \) and \( {e}^{\prime } \) are both identities in \( G \) . Then \( {eg} = {ge} = g \) and \( {e}^{\prime }g = g{e}^{\prime } = g \) for all \( g \in G \) . We need to show that \( e = {e}^{\prime } \) . If we think of \( e \) as the identity, then \( e{e}^{\prime } = {e}^{\prime } \) ; but if ... | Yes |
Proposition 3.20. Let \( G \) be a group. For any \( a \in G,{\left( {a}^{-1}\right) }^{-1} = a \) . | Proof. Observe that \( {a}^{-1}{\left( {a}^{-1}\right) }^{-1} = e \) . Consequently, multiplying both sides of this equation by \( a \), we have\n\n\[ \n{\left( {a}^{-1}\right) }^{-1} = e{\left( {a}^{-1}\right) }^{-1} = a{a}^{-1}{\left( {a}^{-1}\right) }^{-1} = {ae} = a.\n\] | Yes |
Proposition 3.21. Let \( G \) be a group and a and \( b \) be any two elements in \( G \) . Then the equations \( {ax} = b \) and \( {xa} = b \) have unique solutions in \( G \). | Proof. Suppose that \( {ax} = b \) . We must show that such an \( x \) exists. We can multiply both sides of \( {ax} = b \) by \( {a}^{-1} \) to find \( x = {ex} = {a}^{-1}{ax} = {a}^{-1}b \) .\n\nTo show uniqueness, suppose that \( {x}_{1} \) and \( {x}_{2} \) are both solutions of \( {ax} = b \) ; then \( a{x}_{1} = ... | Yes |
Consider the set of nonzero real numbers, \( {\mathbb{R}}^{ * } \), with the group operation of multiplication. The identity of this group is 1 and the inverse of any element \( a \in {\mathbb{R}}^{ * } \) is just \( 1/a \) . We will show that\n\n\[ \n{\mathbb{Q}}^{ * } = \{ p/q : p\text{ and }q\text{ are nonzero integ... | The identity of \( {\mathbb{R}}^{ * } \) is 1 ; however, \( 1 = 1/1 \) is the quotient of two nonzero integers. Hence, the identity of \( {\mathbb{R}}^{ * } \) is in \( {\mathbb{Q}}^{ * } \) . Given two elements in \( {\mathbb{Q}}^{ * } \), say \( p/q \) and \( r/s \) , their product \( {pr}/{qs} \) is also in \( {\mat... | Yes |
Example 3.25. Recall that \( {\mathbb{C}}^{ * } \) is the multiplicative group of nonzero complex numbers. Let \( H = \{ 1, - 1, i, - i\} \) . Then \( H \) is a subgroup of \( {\mathbb{C}}^{ * } \) . | It is quite easy to verify that \( H \) is a group under multiplication and that \( H \subset {\mathbb{C}}^{ * } \) . | No |
Let \( S{L}_{2}\left( \mathbb{R}\right) \) be the subset of \( G{L}_{2}\left( \mathbb{R}\right) \) consisting of matrices of determinant one; that is, a matrix\n\n\[ A = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \]\n\nis in \( S{L}_{2}\left( \mathbb{R}\right) \) exactly when \( {ad} - {bc} = 1 \) . To ... | The \( 2 \times 2 \) identity matrix is in \( S{L}_{2}\left( \mathbb{R}\right) \), as is the inverse of the matrix \( A \) :\n\n\[ {A}^{-1} = \left( \begin{matrix} d & - b \\ - c & a \end{matrix}\right) \]\n\nIt remains to show that multiplication is closed; that is, that the product of two matrices of determinant one ... | No |
It is important to realize that a subset \( H \) of a group \( G \) can be a group without being a subgroup of \( G \) . For \( H \) to be a subgroup of \( G \) it must inherit \( G \) ’s binary operation. The set of all \( 2 \times 2 \) matrices, \( {\mathbb{M}}_{2}\left( \mathbb{R}\right) \), forms a group under the ... | If we add two invertible matrices, we do not necessarily obtain another invertible matrix. Observe that\n\n\[ \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) + \left( \begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}\right) = \left( \begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right) \]\n\nbut the zero matrix is ... | Yes |
Proposition 3.30. A subset \( H \) of \( G \) is a subgroup if and only if it satisfies the following conditions.\n\n1. The identity \( e \) of \( G \) is in \( H \) .\n\n2. If \( {h}_{1},{h}_{2} \in H \), then \( {h}_{1}{h}_{2} \in H \) .\n\n3. If \( h \in H \), then \( {h}^{-1} \in H \) . | Proof. First suppose that \( H \) is a subgroup of \( G \) . We must show that the three conditions hold. Since \( H \) is a group, it must have an identity \( {e}_{H} \) . We must show that \( {e}_{H} = e \), where \( e \) is the identity of \( G \) . We know that \( {e}_{H}{e}_{H} = {e}_{H} \) and that \( e{e}_{H} = ... | Yes |
Proposition 3.31. Let \( H \) be a subset of a group \( G \). Then \( H \) is a subgroup of \( G \) if and only if \( H \neq \varnothing \), and whenever \( g, h \in H \) then \( g{h}^{-1} \) is in \( H \). | Proof. First assume that \( H \) is a subgroup of \( G \). We wish to show that \( g{h}^{-1} \in H \) whenever \( g \) and \( h \) are in \( H \). Since \( h \) is in \( H \), its inverse \( {h}^{-1} \) must also be in \( H \). Because of the closure of the group operation, \( g{h}^{-1} \in H \). Conversely, suppose th... | Yes |
If \( H = \left\{ {{2}^{n} : n \in \mathbb{Z}}\right\} \), then \( H \) is a subgroup of the multiplicative group of nonzero rational numbers, \( {\mathbb{Q}}^{ * } \). | If \( a = {2}^{m} \) and \( b = {2}^{n} \) are in \( H \), then \( a{b}^{-1} = {2}^{m}{2}^{-n} = {2}^{m - n} \) is also in \( H \). By Proposition 3.31, \( H \) is a subgroup of \( {\mathbb{Q}}^{ * } \) determined by the element 2. | Yes |
Theorem 4.3. Let \( G \) be a group and a be any element in \( G \) . Then the set\n\n\[ \langle a\rangle = \left\{ {{a}^{k} : k \in \mathbb{Z}}\right\} \]\n\nis a subgroup of \( G \) . Furthermore, \( \langle a\rangle \) is the smallest subgroup of \( G \) that contains a. | Proof. The identity is in \( \langle a\rangle \) since \( {a}^{0} = e \) . If \( g \) and \( h \) are any two elements in \( \langle a\rangle \), then by the definition of \( \langle a\rangle \) we can write \( g = {a}^{m} \) and \( h = {a}^{n} \) for some integers \( m \) and \( n \) . So \( {gh} = {a}^{m}{a}^{n} = {a... | Yes |
Notice that a cyclic group can have more than a single generator. Both 1 and 5 generate \( {\mathbb{Z}}_{6} \) ; hence, \( {\mathbb{Z}}_{6} \) is a cyclic group. | Not every element in a cyclic group is necessarily a generator of the group. The order of \( 2 \in {\mathbb{Z}}_{6} \) is 3 . The cyclic subgroup generated by 2 is \( \langle 2\rangle = \{ 0,2,4\} \) . | No |
The group of units, \( U\left( 9\right) \), in \( {\mathbb{Z}}_{9} \) is a cyclic group. | As a set, \( U\left( 9\right) \) is \( \{ 1,2,4,5,7,8\} \) . The element 2 is a generator for \( U\left( 9\right) \) since\n\n\[ \n{2}^{1} = 2\;{2}^{2} = 4 \]\n\n\[ \n{2}^{3} = 8\;{2}^{4} = 7 \]\n\n\[ \n{2}^{5} = 5\;{2}^{6} = 1. \]\n | Yes |
Theorem 4.9. Every cyclic group is abelian. | Proof. Let \( G \) be a cyclic group and \( a \in G \) be a generator for \( G \) . If \( g \) and \( h \) are in \( G \) , then they can be written as powers of \( a \), say \( g = {a}^{r} \) and \( h = {a}^{s} \) . Since\n\n\[ \n{gh} = {a}^{r}{a}^{s} = {a}^{r + s} = {a}^{s + r} = {a}^{s}{a}^{r} = {hg}, \n\]\n\n\( G \... | Yes |
Theorem 4.10. Every subgroup of a cyclic group is cyclic. | Proof. The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let \( G \) be a cyclic group generated by \( a \) and suppose that \( H \) is a subgroup of \( G \) . If \( H = \{ e\} \), then trivially \( H \) is cyclic. Suppose that \( H \) contains some other element \( g \) d... | Yes |
Proposition 4.12. Let \( G \) be a cyclic group of order \( n \) and suppose that \( a \) is a generator for \( G \) . Then \( {a}^{k} = e \) if and only if \( n \) divides \( k \) . | Proof. First suppose that \( {a}^{k} = e \) . By the division algorithm, \( k = {nq} + r \) where \( 0 \leq r < n \) ; hence,\n\n\[ e = {a}^{k} = {a}^{{nq} + r} = {a}^{nq}{a}^{r} = e{a}^{r} = {a}^{r}. \]\n\nSince the smallest positive integer \( m \) such that \( {a}^{m} = e \) is \( n, r = 0 \) .\n\nConversely, if \( ... | Yes |
Theorem 4.13. Let \( G \) be a cyclic group of order \( n \) and suppose that \( a \in G \) is a generator of the group. If \( b = {a}^{k} \), then the order of \( b \) is \( n/d \), where \( d = \gcd \left( {k, n}\right) \) . | Proof. We wish to find the smallest integer \( m \) such that \( e = {b}^{m} = {a}^{km} \) . By Proposition 4.12, this is the smallest integer \( m \) such that \( n \) divides \( {km} \) or, equivalently, \( n/d \) divides \( m\left( {k/d}\right) \) . Since \( d \) is the greatest common divisor of \( n \) and \( k, n... | Yes |
Let us examine the group \( {\mathbb{Z}}_{16} \). The numbers \( 1,3,5,7,9,{11},{13} \), and 15 are the elements of \( {\mathbb{Z}}_{16} \) that are relatively prime to 16. Each of these elements generates \( {\mathbb{Z}}_{16} \). | For example,\n\n\( 1 \cdot 9 = 9 \)\n\( 2 \cdot 9 = 2 \)\n\( 3 \cdot 9 = {11} \)\n\( 4 \cdot 9 = 4 \)\n\( 5 \cdot 9 = {13} \)\n\( 6 \cdot 9 = 6 \)\n\( 7 \cdot 9 = {15} \)\n\( 8 \cdot 9 = 8 \)\n\( 9 \cdot 9 = 1 \)\n\( {10} \cdot 9 = {10} \)\n\( {11} \cdot 9 = 3 \)\n\( {12} \cdot 9 = {12} \)\n\( {13} \cdot 9 = 5 \)\n\( {... | Yes |
Let \( z = 2 + {3i} \) and \( w = 1 - {2i} \). Then | \[ z + w = \left( {2 + {3i}}\right) + \left( {1 - {2i}}\right) = 3 + i \] and \[ {zw} = \left( {2 + {3i}}\right) \left( {1 - {2i}}\right) = 8 - i. \] Also, \[ {z}^{-1} = \frac{2}{13} - \frac{3}{13}i \] \[ \left| z\right| = \sqrt{13} \] \[ \bar{z} = 2 - {3i} \] | Yes |
Suppose that \( z = 2\operatorname{cis}{60}^{ \circ } \) . Then | \[ a = 2\cos {60}^{ \circ } = 1 \] and \[ b = 2\sin {60}^{ \circ } = \sqrt{3}. \] Hence, the rectangular representation is \( z = 1 + \sqrt{3}i \) . | Yes |
Theorem 4.22 DeMoivre. Let \( z = r\operatorname{cis}\theta \) be a nonzero complex number. Then\n\n\[{\left\lbrack r\operatorname{cis}\theta \right\rbrack }^{n} = {r}^{n}\operatorname{cis}\left( {n\theta }\right)\]\n\nfor \( n = 1,2,\ldots \) | Proof. We will use induction on \( n \) . For \( n = 1 \) the theorem is trivial. Assume that the theorem is true for all \( k \) such that \( 1 \leq k \leq n \) . Then\n\n\[{z}^{n + 1} = {z}^{n}z\]\n\n\[= {r}^{n}\left( {\cos {n\theta } + i\sin {n\theta }}\right) r\left( {\cos \theta + i\sin \theta }\right)\]\n\n\[= {r... | Yes |
Suppose that \( z = 1 + i \) and we wish to compute \( {z}^{10} \). | Rather than computing \( {\left( 1 + i\right) }^{10} \) directly, it is much easier to switch to polar coordinates and calculate \( {z}^{10} \) using DeMoivre's Theorem:\n\n\[ \n{z}^{10} = {\left( 1 + i\right) }^{10} \n\]\n\n\[ \n= {\left( \sqrt{2}\operatorname{cis}\left( \frac{\pi }{4}\right) \right) }^{10} \n\]\n\n\[... | Yes |
Proposition 4.24. The circle group is a subgroup of \( {\mathbb{C}}^{ * } \) . | Although the circle group has infinite order, it has many interesting finite subgroups. Suppose that \( H = \{ 1, - 1, i, - i\} \) . Then \( H \) is a subgroup of the circle group. Also, \( 1, - 1, i \) , and \( - i \) are exactly those complex numbers that satisfy the equation \( {z}^{4} = 1 \) . The complex numbers s... | No |
Theorem 4.25. If \( {z}^{n} = 1 \), then the nth roots of unity are\n\n\[ z = \operatorname{cis}\left( \frac{2k\pi }{n}\right) \]\n\nwhere \( k = 0,1,\ldots, n - 1 \) . Furthermore, the nth roots of unity form a cyclic subgroup of \( \mathbb{T} \) of order \( n \) | Proof. By DeMoivre's Theorem,\n\n\[ {z}^{n} = \operatorname{cis}\left( {n\frac{2k\pi }{n}}\right) = \operatorname{cis}\left( {2k\pi }\right) = 1. \]\n\nThe \( z \) ’s are distinct since the numbers \( {2k\pi }/n \) are all distinct and are greater than or equal to 0 but less than \( {2\pi } \) . The fact that these are... | Yes |
We will compute \( {271}^{321}\left( {\;\operatorname{mod}\;{481}}\right) \) . | Notice that\n\n\[ \n{321} = {2}^{0} + {2}^{6} + {2}^{8} \n\] \n\nhence, computing \( {271}^{321}\left( {\;\operatorname{mod}\;{481}}\right) \) is the same as computing \n\n\[ \n{271}^{{2}^{0} + {2}^{6} + {2}^{8}} \equiv {271}^{{2}^{0}} \cdot {271}^{{2}^{6}} \cdot {271}^{{2}^{8}}\;\left( {\;\operatorname{mod}\;{481}}\ri... | Yes |
Theorem 5.1. The symmetric group on \( n \) letters, \( {S}_{n} \), is a group with \( n \) ! elements, where the binary operation is the composition of maps. | Proof. The identity of \( {S}_{n} \) is just the identity map that sends 1 to 1,2 to \( 2,\ldots, n \) to \( n \) . If \( f : {S}_{n} \rightarrow {S}_{n} \) is a permutation, then \( {f}^{-1} \) exists, since \( f \) is one-to-one and onto; hence, every permutation has an inverse. Composition of maps is associative, wh... | No |
Permutation multiplication is not usually commutative. Let\n\n\[ \n\sigma = \left( \begin{array}{llll} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{array}\right) \]\n\n\[ \n\tau = \left( \begin{array}{llll} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{array}\right) \]\n | Then\n\n\[ \n{\sigma \tau } = \left( \begin{array}{llll} 1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2 \end{array}\right) \]\n\nbut\n\n\[ \n{\tau \sigma } = \left( \begin{array}{llll} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4 \end{array}\right) \]\n | Yes |
The permutation\n\n\[ \sigma = \left( \begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 3 & 5 & 1 & 4 & 2 & 7 \end{array}\right) = \left( {162354}\right) \]\n\nis a cycle of length 6 , whereas\n\n\[ \tau = \left( \begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 4 & 2 & 3 & 5 & 6 \end{array}\right) = \left( {243... | Not every permutation is a cycle. Consider the permutation\n\n\[ \left( \begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 4 & 1 & 3 & 6 & 5 \end{array}\right) = \left( {1243}\right) \left( {56}\right) . \]\n\nThis permutation actually contains a cycle of length 2 and a cycle of length 4 . | Yes |
It is very easy to compute products of cycles. Suppose that\n\n\[ \sigma = \left( {1352}\right) \text{ and }\tau = \left( {256}\right) . \] | If we think of \( \sigma \) as\n\n\[ 1 \mapsto 3,\;3 \mapsto 5,\;5 \mapsto 2,\;2 \mapsto 1, \]\n\nand \( \tau \) as\n\n\[ 2 \mapsto 5,\;5 \mapsto 6,\;6 \mapsto 2, \]\n\nthen for \( {\sigma \tau } \) remembering that we apply \( \tau \) first and then \( \sigma \), it must be the case that\n\n\[ 1 \mapsto 3,\;3 \mapsto ... | Yes |
The cycles (135) and (27) are disjoint; however, the cycles (135) and (347) are not. Calculating their products, we find that | \[ \left( {135}\right) \left( {27}\right) = \left( {135}\right) \left( {27}\right) \] \[ \left( {135}\right) \left( {347}\right) = \left( {13475}\right) \text{.} \] | Yes |
Proposition 5.8. Let \( \sigma \) and \( \tau \) be two disjoint cycles in \( {S}_{X} \) . Then \( {\sigma \tau } = {\tau \sigma } \). | Proof. Let \( \sigma = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{k}}\right) \) and \( \tau = \left( {{b}_{1},{b}_{2},\ldots ,{b}_{l}}\right) \) . We must show that \( {\sigma \tau }\left( x\right) = \) \( {\tau \sigma }\left( x\right) \) for all \( x \in X \) . If \( x \) is neither in \( \left\{ {{a}_{1},{a}_{2},\ldots ,{a... | Yes |
Theorem 5.9. Every permutation in \( {S}_{n} \) can be written as the product of disjoint cycles. | Proof. We can assume that \( X = \{ 1,2,\ldots, n\} \) . If \( \sigma \in {S}_{n} \) and we define \( {X}_{1} \) to be \( \left\{ {\sigma \left( 1\right) ,{\sigma }^{2}\left( 1\right) ,\ldots }\right\} \), then the set \( {X}_{1} \) is finite since \( X \) is finite. Now let \( i \) be the first integer in \( X \) that... | Yes |
Let\n\n\\[ \n\\sigma = \\left( \\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\\\ 6 & 4 & 3 & 1 & 5 & 2 \\end{array}\\right) \n\\]\n\n\\[ \n\\tau = \\left( \\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\\\ 3 & 2 & 1 & 5 & 6 & 4 \\end{array}\\right) \n\\]\n\nUsing cycle notation, we can write\n\n\\[ \n\\sigma = \\left( {... | \\[ \n{\\sigma \\tau } = \\left( {136}\\right) \\left( {245}\\right) \n\\]\n\n\\[ \n{\\tau \\sigma } = \\left( {143}\\right) \\left( {256}\\right) \\text{.} \n\\] | Yes |
However, as it turns out, no permutation can be written as the product of both an even number of transpositions and an odd number of transpositions. | For instance, we could represent the permutation (16) by\n\n\[ \text{(23)(16)(23)} \]\n\nor by\n\n\[ \left( {35}\right) \left( {16}\right) \left( {13}\right) \left( {16}\right) \left( {13}\right) \left( {35}\right) \left( {56}\right) ,\]\n\nbut (16) will always be the product of an odd number of transpositions. | No |
Lemma 5.14. If the identity is written as the product of \( r \) transpositions,\n\n\[ \n\mathrm{{id}} = {\tau }_{1}{\tau }_{2}\cdots {\tau }_{r} \n\]\n\nthen \( r \) is an even number. | Proof. We will employ induction on \( r \) . A transposition cannot be the identity; hence, \( r > 1 \) . If \( r = 2 \), then we are done. Suppose that \( r > 2 \) . In this case the product of the last two transpositions, \( {\tau }_{r - 1}{\tau }_{r} \), must be one of the following cases:\n\n\[ \n\left( {ab}\right)... | Yes |
Theorem 5.15. If a permutation \( \sigma \) can be expressed as the product of an even number of transpositions, then any other product of transpositions equaling \( \sigma \) must also contain an even number of transpositions. Similarly, if \( \sigma \) can be expressed as the product of an odd number of transposition... | Proof. Suppose that\n\n\[ \sigma = {\sigma }_{1}{\sigma }_{2}\cdots {\sigma }_{m} = {\tau }_{1}{\tau }_{2}\cdots {\tau }_{n} \]\n\nwhere \( m \) is even. We must show that \( n \) is also an even number. The inverse of \( \sigma \) is \( {\sigma }_{m}\cdots {\sigma }_{1} \) . Since\n\n\[ \mathrm{{id}} = \sigma {\sigma ... | No |
Theorem 5.16. The set \( {A}_{n} \) is a subgroup of \( {S}_{n} \) . | Proof. Since the product of two even permutations must also be an even permutation, \( {A}_{n} \) is closed. The identity is an even permutation and therefore is in \( {A}_{n} \) . If \( \sigma \) is an even permutation, then\n\n\[ \sigma = {\sigma }_{1}{\sigma }_{2}\cdots {\sigma }_{r} \]\n\nwhere \( {\sigma }_{i} \) ... | Yes |
Proposition 5.17. The number of even permutations in \( {S}_{n}, n \geq 2 \), is equal to the number of odd permutations; hence, the order of \( {A}_{n} \) is \( n!/2 \) . | Proof. Let \( {A}_{n} \) be the set of even permutations in \( {S}_{n} \) and \( {B}_{n} \) be the set of odd permutations. If we can show that there is a bijection between these sets, they must contain the same number of elements. Fix a transposition \( \sigma \) in \( {S}_{n} \) . Since \( n \geq 2 \), such a \( \sig... | No |
Theorem 5.23. The group \( {D}_{n}, n \geq 3 \), consists of all products of the two elements \( r \) and \( s \), satisfying the relations\n\n\[ \n{r}^{n} = 1 \]\n\n\[ \n{s}^{2} = 1 \]\n\n\[ \n{srs} = {r}^{-1}\text{.} \]\n | Proof. The possible motions of a regular \( n \) -gon are either reflections or rotations (Figure 5.21). There are exactly \( n \) possible rotations:\n\n\[ \n\text{id,}\frac{{360}^{ \circ }}{n},2 \cdot \frac{{360}^{ \circ }}{n},\ldots ,\left( {n - 1}\right) \cdot \frac{{360}^{ \circ }}{n}\text{.} \]\n\nWe will denote ... | No |
Theorem 5.28. The group of rigid motions of a cube is \( {S}_{4} \) . | Proof. From Proposition 5.27, we already know that the motion group of the cube has 24 elements, the same number of elements as there are in \( {S}_{4} \) . There are exactly four diagonals in the cube. If we label these diagonals \( 1,2,3 \), and 4, we must show that the motion group of the cube will give us any permu... | Yes |
Let \( H \) be the subgroup of \( {\mathbb{Z}}_{6} \) consisting of the elements 0 and 3 . The cosets are | \[ 0 + H = 3 + H = \{ 0,3\} \] \[ 1 + H = 4 + H = \{ 1,4\} \] \[ 2 + H = 5 + H = \{ 2,5\} . \] | Yes |
Let \( H \) be the subgroup of \( {S}_{3} \) defined by the permutations \( \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \). The left cosets of \( H \) are | \[ \left( 1\right) H = \left( {123}\right) H = \left( {132}\right) H = \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \] \[ \left( {12}\right) H = \left( {13}\right) H = \left( {23}\right) H = \{ \left( {12}\right) ,\left( {13}\right) ,\left( {23}\right) \} . \] The right cosets of \( H \) are exactly ... | Yes |
Theorem 6.4. Let \( H \) be a subgroup of a group \( G \) . Then the left cosets of \( H \) in \( G \) partition \( G \) . That is, the group \( G \) is the disjoint union of the left cosets of \( H \) in \( G \) . | Proof. Let \( {g}_{1}H \) and \( {g}_{2}H \) be two cosets of \( H \) in \( G \) . We must show that either \( {g}_{1}H \cap {g}_{2}H = \) \( \varnothing \) or \( {g}_{1}H = {g}_{2}H \) . Suppose that \( {g}_{1}H \cap {g}_{2}H \neq \varnothing \) and \( a \in {g}_{1}H \cap {g}_{2}H \) . Then by the definition of a left... | Yes |
Theorem 6.8. Let \( H \) be a subgroup of a group \( G \) . The number of left cosets of \( H \) in \( G \) is the same as the number of right cosets of \( H \) in \( G \) . | Proof. Let \( {\mathcal{L}}_{H} \) and \( {\mathcal{R}}_{H} \) denote the set of left and right cosets of \( H \) in \( G \), respectively. If we can define a bijective map \( \phi : {\mathcal{L}}_{H} \rightarrow {\mathcal{R}}_{H} \), then the theorem will be proved. If \( {gH} \in {\mathcal{L}}_{H} \) , let \( \phi \l... | Yes |
Proposition 6.9. Let \( H \) be a subgroup of \( G \) with \( g \in G \) and define a map \( \phi : H \rightarrow {gH} \) by \( \phi \left( h\right) = {gh} \) . The map \( \phi \) is bijective; hence, the number of elements in \( H \) is the same as the number of elements in \( {gH} \). | Proof. We first show that the map \( \phi \) is one-to-one. Suppose that \( \phi \left( {h}_{1}\right) = \phi \left( {h}_{2}\right) \) for elements \( {h}_{1},{h}_{2} \in H \) . We must show that \( {h}_{1} = {h}_{2} \), but \( \phi \left( {h}_{1}\right) = g{h}_{1} \) and \( \phi \left( {h}_{2}\right) = g{h}_{2} \) . S... | Yes |
Theorem 6.10 Lagrange. Let \( G \) be a finite group and let \( H \) be a subgroup of \( G \) . Then \( \left| G\right| /\left| H\right| = \left\lbrack {G : H}\right\rbrack \) is the number of distinct left cosets of \( H \) in \( G \) . In particular, the number of elements in \( H \) must divide the number of element... | Proof. The group \( G \) is partitioned into \( \left\lbrack {G : H}\right\rbrack \) distinct left cosets. Each left coset has \( \left| H\right| \) elements; therefore, \( \left| G\right| = \left\lbrack {G : H}\right\rbrack \left| H\right| \) . | Yes |
Corollary 6.12. Let \( \left| G\right| = p \) with \( p \) a prime number. Then \( G \) is cyclic and any \( g \in G \) such that \( g \neq e \) is a generator. | Proof. Let \( g \) be in \( G \) such that \( g \neq e \) . Then by Corollary 6.11, the order of \( g \) must divide the order of the group. Since \( \left| {\langle g\rangle }\right| > 1 \), it must be \( p \) . Hence, \( g \) generates \( G \) . | Yes |
Corollary 6.13. Let \( H \) and \( K \) be subgroups of a finite group \( G \) such that \( G \supset H \supset K \) . Then\n\n\[ \left\lbrack {G : K}\right\rbrack = \left\lbrack {G : H}\right\rbrack \left\lbrack {H : K}\right\rbrack \] | Proof. Observe that\n\n\[ \left\lbrack {G : K}\right\rbrack = \frac{\left| G\right| }{\left| K\right| } = \frac{\left| G\right| }{\left| H\right| } \cdot \frac{\left| H\right| }{\left| K\right| } = \left\lbrack {G : H}\right\rbrack \left\lbrack {H : K}\right\rbrack . \] | Yes |
Theorem 6.16. Two cycles \( \tau \) and \( \mu \) in \( {S}_{n} \) have the same length if and only if there exists \( {a\sigma } \in {S}_{n} \) such that \( \mu = {\sigma \tau }{\sigma }^{-1} \) . | Proof. Suppose that\n\n\[ \tau = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{k}}\right) \]\n\n\[ \mu = \left( {{b}_{1},{b}_{2},\ldots ,{b}_{k}}\right) \]\n\nDefine \( \sigma \) to be the permutation\n\n\[ \sigma \left( {a}_{1}\right) = {b}_{1} \]\n\n\[ \sigma \left( {a}_{2}\right) = {b}_{2} \]\n\n\[ \sigma \left( {a}_{k}\righ... | Yes |
Theorem 6.18 Euler’s Theorem. Let \( a \) and \( n \) be integers such that \( n > 0 \) and \( \gcd \left( {a, n}\right) = 1 \). Then \( {a}^{\phi \left( n\right) } \equiv 1\left( {\;\operatorname{mod}\;n}\right) \) . | Proof. By Theorem 6.17 the order of \( U\left( n\right) \) is \( \phi \left( n\right) \) . Consequently, \( {a}^{\phi \left( n\right) } = 1 \) for all \( a \in U\left( n\right) \) ; or \( {a}^{\phi \left( n\right) } - 1 \) is divisible by \( n \) . Therefore, \( {a}^{\phi \left( n\right) } \equiv 1\left( {\;\operatorna... | Yes |
One of the first and most famous private key cryptosystems was the shift code used by Julius Caesar. We first digitize the alphabet by letting \( \\mathrm{A} = {00},\\mathrm{\\;B} = {01},\\ldots ,\\mathrm{Z} = \) 25. The encoding function will be\n\n\[ f\\left( p\\right) = p + 3{\\;\\operatorname{mod}\\;{26}} \]\n\ntha... | Suppose we receive the encoded message DOJHEUD. To decode this message, we first digitize it:\n\n\[ 3,{14},9,7,4,{20},3\\text{.} \]\n\nNext we apply the inverse transformation to get\n\n\[ 0,{11},6,4,1,{17},0, \]\n\nor ALGEBRA. | Yes |
Suppose we receive a message that we know was encrypted by using a shift transformation on single letters of the 26-letter alphabet. To find out exactly what the shift transformation was, we must compute \( b \) in the equation \( f\left( p\right) = p + b{\;\operatorname{mod}\;{26}} \) . We can do this using frequency ... | Therefore, the most likely encrypting function is\n\n\[ f\left( p\right) = p + {14}{\;\operatorname{mod}\;{26}}. \]\n\nThe corresponding decrypting function is\n\n\[ {f}^{-1}\left( p\right) = p + {12}{\;\operatorname{mod}\;{26}}. \]\n\nIt is now easy to determine whether or not our guess is correct. | Yes |
Let us consider the affine cryptosystem \( f\left( p\right) = {ap} + b{\;\operatorname{mod}\;{26}} \) . For this cryptosystem to work we must choose an \( a \in {\mathbb{Z}}_{26} \) that is invertible. This is only possible if \( \gcd \left( {a,{26}}\right) = 1 \) . Recognizing this fact, we will let \( a = 5 \) since ... | The decryption function will be \[ {f}^{-1}\left( p\right) = {21p} - {21} \cdot 3{\;\operatorname{mod}\;{26}} = {21p} + {15}{\;\operatorname{mod}\;{26}}. \] | Yes |
Now let us examine why the RSA cryptosystem works. We know that \( {DE} \equiv 1\left( {\;\operatorname{mod}\;m}\right) \) ; hence, there exists a \( k \) such that \( {DE} = {km} + 1 = {k\phi }\left( n\right) + 1. \) | There are two cases to consider. In the first case assume that \( \gcd \left( {x, n}\right) = 1 \) . Then by Theorem 6.18, \( {y}^{D} = {\left( {x}^{E}\right) }^{D} = {x}^{DE} = {x}^{{km} + 1} = {\left( {x}^{\phi \left( n\right) }\right) }^{k}x = {\left( 1\right) }^{k}x = x{\;\operatorname{mod}\;n}. \n\nSo we see that ... | Yes |
One possible coding scheme would be to send a message several times and to compare the received copies with one another. Suppose that the message to be encoded is a binary \( n \) -tuple \( \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) . The message is encoded into a binary \( {3n} \) -tuple by simply repeating th... | \[ \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \mapsto \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n},{x}_{1},{x}_{2},\ldots ,{x}_{n},{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) . \] To decode the message, we choose as the \( i \) th digit the one that appears in the \( i \) th place in at least two of the three transmissions... | Yes |
What can or should be done with the extra bit in the ASCII coding system? | The bit can be used for error checking on the other seven bits. It is set to either 0 or 1 so that the total number of 1 bits in the representation of a character is even. Using even parity, the codes for \( \mathrm{A},\mathrm{B} \), and \( \mathrm{C} \) now become\n\n\[ \mathrm{A} = {01000001}_{2} \]\n\n\[ \mathrm{B} ... | Yes |
Suppose that our original message is either a 0 or a 1, and that 0 encodes to (000) and 1 encodes to (111). If only a single error occurs during transmission, we can detect and correct the error. For example, if a (101) is received, then the second bit must have been changed from a 1 to a 0 . The originally transmitted... | In Table 8.4, we present all possible words that might be received for the transmitted codewords (000) and (111). Table 8.4 also shows the number of bits by which each received 3-tuple differs from each original codeword. | No |
If a binary n-tuple \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is transmitted across a binary symmetric channel with probability \( p \) that no error will occur in each coordinate, then the probability that there are errors in exactly \( k \) coordinates is\n\n\[ \left( \begin{array}{l} n \\ k \end{array}\right) {q... | Proof. Fix \( k \) different coordinates. We first compute the probability that an error has occurred in this fixed set of coordinates. The probability of an error occurring in a particular one of these \( k \) coordinates is \( q \) ; the probability that an error will not occur in any of the remaining \( n - k \) coo... | Yes |
Suppose that \( p = {0.995} \) and a 500-bit message is sent. The probability that the message was sent error-free is | \[ {p}^{n} = {\left( {0.995}\right) }^{500} \approx {0.082}. \]\n\nThe probability of exactly one error occurring is\n\n\[ \left( \begin{array}{l} n \\ 1 \end{array}\right) q{p}^{n - 1} = {500}\left( {0.005}\right) {\left( {0.995}\right) }^{499} \approx {0.204}. \]\n\nThe probability of exactly two errors is\n\n\[ \lef... | Yes |
Theorem 8.13. Let \( C \) be a code with \( {d}_{\min } = {2n} + 1 \) . Then \( C \) can correct any \( n \) or fewer errors. Furthermore, any \( {2n} \) or fewer errors can be detected in \( C \) . | Proof. Suppose that a codeword \( \mathbf{x} \) is sent and the word \( \mathbf{y} \) is received with at most \( n \) errors. Then \( d\left( {\mathbf{x},\mathbf{y}}\right) \leq n \) . If \( \mathbf{z} \) is any codeword other than \( \mathbf{x} \), then\n\n\[ \n{2n} + 1 \leq d\left( {\mathbf{x},\mathbf{z}}\right) \le... | Yes |
In Table 8.15, the codewords \( {\mathbf{c}}_{1} = \left( {00000}\right) ,{\mathbf{c}}_{2} = \left( {00111}\right) ,{\mathbf{c}}_{3} = \left( {11100}\right) \) , and \( {\mathbf{c}}_{4} = \left( {11011}\right) \) determine a single error-correcting code. | <table><thead><tr><th></th><th>00000</th><th>00111</th><th>11100</th><th>11011</th></tr></thead><tr><td>00000</td><td>0</td><td>3</td><td>3</td><td>4</td></tr><tr><td>00111</td><td>3</td><td>0</td><td>4</td><td>3</td></tr><tr><td>11100</td><td>3</td><td>4</td><td>0</td><td>3</td></tr><tr><td>11011</td><td>4</td><td>3</... | Yes |
Example 8.16. Suppose that we have a code that consists of the following 7-tuples: | It is a straightforward though tedious task to verify that this code is also a subgroup of \( {\mathbb{Z}}_{2}^{7} \) and, therefore, a group code. This code is a single error-detecting and single error-correcting code, but it is a long and tedious process to compute all of the distances between pairs of codewords to d... | No |
Lemma 8.17. Let \( \mathbf{x} \) and \( \mathbf{y} \) be binary n-tuples. Then \( w\left( {\mathbf{x} + \mathbf{y}}\right) = d\left( {\mathbf{x},\mathbf{y}}\right) \) . | Proof. Suppose that \( \mathbf{x} \) and \( \mathbf{y} \) are binary \( n \) -tuples. Then the distance between \( \mathbf{x} \) and \( \mathbf{y} \) is exactly the number of places in which \( \mathbf{x} \) and \( \mathbf{y} \) differ. But \( \mathbf{x} \) and \( \mathbf{y} \) differ in a particular coordinate exactly... | Yes |
Theorem 8.18. Let \( {d}_{\min } \) be the minimum distance for a group code \( C \) . Then \( {d}_{\min } \) is the minimum of all the nonzero weights of the nonzero codewords in \( C \) . That is,\n\n\[ \n{d}_{\min } = \min \{ w\left( \mathbf{x}\right) : \mathbf{x} \neq \mathbf{0}\} .\n\] | Proof. Observe that\n\n\[ \n{d}_{\min } = \min \{ d\left( {\mathbf{x},\mathbf{y}}\right) : \mathbf{x} \neq \mathbf{y}\} \n\]\n\n\[ \n= \min \{ d\left( {\mathbf{x},\mathbf{y}}\right) : \mathbf{x} + \mathbf{y} \neq \mathbf{0}\} \n\]\n\n\[ \n= \min \{ w\left( {\mathbf{x} + \mathbf{y}}\right) : \mathbf{x} + \mathbf{y} \neq... | Yes |
Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. | Notice that an arbitrary \( n \) -tuple \( \mathbf{x} = {\left( {x}_{1},{x}_{2},\ldots ,{x}_{n}\right) }^{\mathrm{t}} \) has an even number of 1s exactly when \( {x}_{1} + {x}_{2} + \cdots + {x}_{n} = 0 \) ; hence, a 4-tuple \( \mathbf{x} = {\left( {x}_{1},{x}_{2},{x}_{3},{x}_{4}\right) }^{\mathrm{t}} \) has an even nu... | Yes |
Suppose that\n\n\[ H = \left( \begin{array}{lllll} 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \end{array}\right) \]\n\nFor a 5-tuple \( \mathbf{x} = {\left( {x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}\right) }^{\mathrm{t}} \) to be in the null space of \( H, H\mathbf{x} = \mathbf{0} \). Equivalently, the f... | The set of binary 5-tuples satisfying these equations is\n\n\[ \text{(00000) (11110) (10101) (01011).} \]\n\nThis code is easily determined to be a group code. | Yes |
Let \( C \) be the code given by the matrix\n\n\[ H = \left( \begin{array}{llllll} 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 & 1 \end{array}\right) \]\n\nSuppose that the 6-tuple \( \mathbf{x} = {\left( {010011}\right) }^{\mathrm{t}} \) is received. It is a simple matter of matrix multiplicati... | Since\n\n\[ H\mathbf{x} = \left( \begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right) \]\n\nthe received word is not a codeword. We must either attempt to correct the word or request that it be transmitted again. | Yes |
Theorem 8.26. Suppose that \( G \) is an \( n \times k \) standard generator matrix. Then \( C = \) \( \left\{ {\mathbf{y} : G\mathbf{x} = \mathbf{y}\text{for}\mathbf{x} \in {\mathbb{Z}}_{2}^{k}}\right\} \) is an \( \left( {n, k}\right) \) -block code. More specifically, \( C \) is a group code. | Proof. Let \( G{\mathbf{x}}_{1} = {\mathbf{y}}_{1} \) and \( G{\mathbf{x}}_{2} = {\mathbf{y}}_{2} \) be two codewords. Then \( {\mathbf{y}}_{1} + {\mathbf{y}}_{2} \) is in \( C \) since\n\n\[ G\left( {{\mathbf{x}}_{1} + {\mathbf{x}}_{2}}\right) = G{\mathbf{x}}_{1} + G{\mathbf{x}}_{2} = {\mathbf{y}}_{1} + {\mathbf{y}}_{... | Yes |
Lemma 8.27. Let \( H = \left( {A \mid {I}_{m}}\right) \) be an \( m \times n \) canonical parity-check matrix and \( G = \left( \frac{{I}_{n - m}}{A}\right) \) be the corresponding \( n \times \left( {n - m}\right) \) standard generator matrix. Then \( {HG} = \mathbf{0} \) . | Proof. Let \( C = {HG} \) . The \( {ij} \) th entry in \( C \) is\n\n\[ \n{c}_{ij} = \mathop{\sum }\limits_{{k = 1}}^{n}{h}_{ik}{g}_{kj} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{k = 1}}^{{n - m}}{h}_{ik}{g}_{kj} + \mathop{\sum }\limits_{{k = n - m + 1}}^{n}{h}_{ik}{g}_{kj} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{k = 1}}^{... | Yes |
Proposition 8.36. Let the \( m \times n \) binary matrix \( H \) determine a linear code and let \( \mathbf{x} \) be the received \( n \) -tuple. Write \( \mathbf{x} \) as \( \mathbf{x} = \mathbf{c} + \mathbf{e} \), where \( \mathbf{c} \) is the transmitted codeword and \( \mathbf{e} \) is the transmission error. Then ... | Proof. The proof follows from the fact that\n\n\[ H\mathbf{x} = H\left( {\mathbf{c} + \mathbf{e}}\right) = H\mathbf{c} + H\mathbf{e} = \mathbf{0} + H\mathbf{e} = H\mathbf{e}. \] | Yes |
Consider the matrix\n\n\[ \nH = \left( \begin{array}{llllll} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right) \n\]\n\nand suppose that the 6-tuples \( \mathbf{x} = {\left( {111110}\right) }^{\mathrm{t}},\mathbf{y} = {\left( {111111}\right) }^{\mathrm{t}} \), and \( \mathbf{z} =... | Hence, \( \mathbf{x} \) has an error in the third bit and \( \mathbf{z} \) has an error in the fourth bit. The transmitted codewords for \( \mathbf{x} \) and \( \mathbf{z} \) must have been (110110) and (010011), respectively. The syndrome of \( \mathbf{y} \) does not occur in any of the columns of the matrix \( H \), ... | Yes |
Let \( C \) be the \( \left( {5,3}\right) \) -linear code given by the parity-check matrix\n\n\[ H = \left( \begin{array}{lllll} 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 \end{array}\right) \]\n\nThe code consists of the codewords\n\n\[ \left( {00000}\right) \;\left( {01101}\right) \;\left( {10011}\ri... | There are \( {2}^{5 - 2} = {2}^{3} \) cosets of \( C \) in \( {\mathbb{Z}}_{2}^{5} \), each with order \( {2}^{2} = 4 \) . These cosets are listed in Table 8.40.\n\n<table><thead><tr><th>Coset Representative</th><th>Coset</th></tr></thead><tr><td>\( C \)</td><td>\( \left( {00000}\right) \left( {01101}\right) \left( {10... | Yes |
Proposition 8.43. Let \( C \) be an \( \left( {n, k}\right) \) -linear code given by the matrix \( H \) and suppose that \( \mathbf{x} \) and \( \mathbf{y} \) are in \( {\mathbb{Z}}_{2}^{n} \) . Then \( \mathbf{x} \) and \( \mathbf{y} \) are in the same coset of \( C \) if and only if \( H\mathbf{x} = H\mathbf{y} \) . ... | Proof. Two \( n \) -tuples \( \mathbf{x} \) and \( \mathbf{y} \) are in the same coset of \( C \) exactly when \( \mathbf{x} - \mathbf{y} \in C \) ; however, this is equivalent to \( H\left( {\mathbf{x} - \mathbf{y}}\right) = 0 \) or \( H\mathbf{x} = H\mathbf{y} \) . | Yes |
To show that \( {\mathbb{Z}}_{4} \cong \langle i\rangle \), define a map \( \phi : {\mathbb{Z}}_{4} \rightarrow \langle i\rangle \) by \( \phi \left( n\right) = {i}^{n} \) . We must show that \( \phi \) is bijective and preserves the group operation. | The map \( \phi \) is one-to-one and onto because\n\n\[ \phi \left( 0\right) = 1 \]\n\n\[ \phi \left( 1\right) = i \]\n\n\[ \phi \left( 2\right) = - 1 \]\n\n\[ \phi \left( 3\right) = - i \]\n\nSince\n\n\[ \phi \left( {m + n}\right) = {i}^{m + n} = {i}^{m}{i}^{n} = \phi \left( m\right) \phi \left( n\right) ,\]\n\nthe gr... | Yes |
We can define an isomorphism \( \phi \) from the additive group of real numbers \( \left( {\mathbb{R}, + }\right) \) to the multiplicative group of positive real numbers \( \left( {{\mathbb{R}}^{ + }, \cdot }\right) \) with the exponential map; | \[ \phi \left( {x + y}\right) = {e}^{x + y} = {e}^{x}{e}^{y} = \phi \left( x\right) \phi \left( y\right) . \] Of course, we must still show that \( \phi \) is one-to-one and onto, but this can be determined using calculus. | No |
The integers are isomorphic to the subgroup of \( {\mathbb{Q}}^{ * } \) consisting of elements of the form \( {2}^{n} \) . Define a map \( \phi : \mathbb{Z} \rightarrow {\mathbb{Q}}^{ * } \) by \( \phi \left( n\right) = {2}^{n} \) . | Then\n\n\[ \phi \left( {m + n}\right) = {2}^{m + n} = {2}^{m}{2}^{n} = \phi \left( m\right) \phi \left( n\right) . \]\n\nBy definition the map \( \phi \) is onto the subset \( \left\{ {{2}^{n} : n \in \mathbb{Z}}\right\} \) of \( {\mathbb{Q}}^{ * } \) . To show that the map is injective, assume that \( m \neq n \) . If... | Yes |
The groups \( U\left( 8\right) \) and \( U\left( {12}\right) \) are isomorphic. | An isomorphism \( \phi : U\left( 8\right) \rightarrow U\left( {12}\right) \) is then given by\n\n\[ 1 \mapsto 1 \]\n\n\[ 3 \mapsto 5 \]\n\n\[ 5 \mapsto 7 \]\n\n\[ 7 \mapsto {11}\text{.} \] | Yes |
Even though \( {S}_{3} \) and \( {\mathbb{Z}}_{6} \) possess the same number of elements, we would suspect that they are not isomorphic, because \( {\mathbb{Z}}_{6} \) is abelian and \( {S}_{3} \) is nonabelian. To demonstrate that this is indeed the case, suppose that \( \phi : {\mathbb{Z}}_{6} \rightarrow {S}_{3} \) ... | Let \( a, b \in {S}_{3} \) be two elements such that \( {ab} \neq {ba} \) . Since \( \phi \) is an isomorphism, there exist elements \( m \) and \( n \) in \( {\mathbb{Z}}_{6} \) such that\n\n\[ \phi \left( m\right) = a\;\text{ and }\;\phi \left( n\right) = b. \]\n\nHowever,\n\n\[ {ab} = \phi \left( m\right) \phi \left... | Yes |
Theorem 9.6. Let \( \phi : G \rightarrow H \) be an isomorphism of two groups. Then the following statements are true.\n\n1. \( {\phi }^{-1} : H \rightarrow G \) is an isomorphism.\n\n2. \( \left| G\right| = \left| H\right| \) .\n\n3. If \( G \) is abelian, then \( H \) is abelian.\n\n4. If \( G \) is cyclic, then \( H... | Proof. Assertions (1) and (2) follow from the fact that \( \phi \) is a bijection. We will prove (3) here and leave the remainder of the theorem to be proved in the exercises.\n\n(3) Suppose that \( {h}_{1} \) and \( {h}_{2} \) are elements of \( H \) . Since \( \phi \) is onto, there exist elements \( {g}_{1},{g}_{2} ... | No |
Theorem 9.7. All cyclic groups of infinite order are isomorphic to \( \mathbb{Z} \) . | Proof. Let \( G \) be a cyclic group with infinite order and suppose that \( a \) is a generator of \( G \) . Define a map \( \phi : \mathbb{Z} \rightarrow G \) by \( \phi : n \mapsto {a}^{n} \) . Then\n\n\[ \phi \left( {m + n}\right) = {a}^{m + n} = {a}^{m}{a}^{n} = \phi \left( m\right) \phi \left( n\right) . \]\n\nTo... | Yes |
Theorem 9.8. If \( G \) is a cyclic group of order \( n \), then \( G \) is isomorphic to \( {\mathbb{Z}}_{n} \) . | Proof. Let \( G \) be a cyclic group of order \( n \) generated by \( a \) and define a map \( \phi : {\mathbb{Z}}_{n} \rightarrow G \) by \( \phi : k \mapsto {a}^{k} \), where \( 0 \leq k < n \) . The proof that \( \phi \) is an isomorphism is one of the end-of-chapter exercises. | No |
Corollary 9.9. If \( G \) is a group of order \( p \), where \( p \) is a prime number, then \( G \) is isomorphic to \( {\mathbb{Z}}_{p} \) . | Proof. The proof is a direct result of Corollary 6.12. | No |
Theorem 9.10. The isomorphism of groups determines an equivalence relation on the class of all groups. | Hence, we can modify our goal of classifying all groups to classifying all groups up to \( \mathbf{isomorphism} \) ; that is, we will consider two groups to be the same if they are isomorphic. | No |
Consider the group \( {\mathbb{Z}}_{3} \) . The Cayley table for \( {\mathbb{Z}}_{3} \) is as follows.\n\n<table><thead><tr><th>\( + \)</th><th>0</th><th>1</th><th>2</th></tr></thead><tr><td>0</td><td>0</td><td>1</td><td>2</td></tr><tr><td>1</td><td>1</td><td>2</td><td>0</td></tr><tr><td>2</td><td>2</td><td>0</td><td>1... | The isomorphism here is\n\n\[ 0 \mapsto \left( \begin{array}{lll} 0 & 1 & 2 \\ 0 & 1 & 2 \end{array}\right) = \left( 0\right) \]\n\n\[ 1 \mapsto \left( \begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 0 \end{array}\right) = \left( {012}\right) \]\n\n\[ 2 \mapsto \left( \begin{array}{lll} 0 & 1 & 2 \\ 2 & 0 & 1 \end{array}\right... | Yes |
Theorem 9.12 Cayley. Every group is isomorphic to a group of permutations. | Proof. Let \( G \) be a group. We must find a group of permutations \( \bar{G} \) that is isomorphic to \( G \) . For any \( g \in G \), define a function \( {\lambda }_{g} : G \rightarrow G \) by \( {\lambda }_{g}\left( a\right) = {ga} \) . We claim that \( {\lambda }_{g} \) is a permutation of \( G \) . To show that ... | Yes |
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