Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Given \( \sigma = \left( {135}\right) ,\tau = \left( {27}\right) ,\sigma ,\tau \in {S}_{7} \) ; let us compute \( {\sigma \tau } \) . | Notice right away that every number affected by \( \tau \) is unaffected by \( \sigma \) ; and vice versa. Since the two cycles always remain separate, it is appropriate to represent \( {\sigma \tau } \) as (135)(27), because the cycles don’t reduce any farther. \( \blacklozenge \) | Yes |
Example 11.3.21. Suppose \( \mu \in {S}_{7} \) and \( \mu = \left( \begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 1 & 4 & 3 & 7 & 2 & 5 \end{array}\right) \) . | Then\n\n- \( 1 \rightarrow 6,6 \rightarrow 2 \), and \( 2 \rightarrow 1 \) ; therefore we have the cycle (162).\n\n- \( 3 \rightarrow 4 \) and \( 4 \rightarrow 3 \) ; therefore we have (34).\n\n- Finally, \( 5 \rightarrow 7 \) and \( 7 \rightarrow 5 \) ; therefore we have (57).\n\nHence \( \mu = \left( {162}\right) \le... | Yes |
Proposition 11.3.27. Disjoint cycles commute: that is, given two disjoint cycles \( \sigma = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{j}}\right) \) and \( \tau = \left( {{b}_{1},{b}_{2},\ldots ,{b}_{k}}\right) \) we have\n\n\[ \n{\sigma \tau } = {\tau \sigma } = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{j}}\right) \left( {{b}_{... | Proof. We present this proof as a fill-in-the-blanks exercise:\n\nExercise 11.3.28. Fill in the blanks to complete the proof:\n\nRecall that permutations are defined as bijections on a set \( X \) . In order to show that the two permutations \( {\sigma \tau } \) and \( {\tau \sigma } \) are equal, it’s enough to show t... | No |
Given the permutations \( \mu = \left( {257}\right) \left( {134}\right) \) and \( \rho = \) (265)(137) in \( {S}_{7} \), write \( {\mu \rho } \) in cycle notation. | \[ \text{-}1 \rightarrow 3,3 \rightarrow 3,3 \rightarrow 4\text{, and}4 \rightarrow 4\text{; therefore}1 \rightarrow 4\text{.} \]\n\[ \text{-}4 \rightarrow 4,4 \rightarrow 4,4 \rightarrow 1\text{, and}1 \rightarrow 1\text{; therefore}4 \rightarrow 1\text{.} \]\nThis gives us the cycle (14). Continuing,\n\[ \text{-}2 \r... | Yes |
Example 11.3.32. Find the product (156)(2365)(123) in \( {S}_{6} \) . | \[ \text{-}1 \rightarrow 2,2 \rightarrow 3\text{, and}3 \rightarrow 3\text{; therefore}1 \rightarrow 3\text{.} \]\n\n- \( 3 \rightarrow 1,1 \rightarrow 1 \), and \( 1 \rightarrow 5 \) ; therefore \( 3 \rightarrow 5 \) .\n\n\[ \text{-}5 \rightarrow 5,5 \rightarrow 2\text{, and}2 \rightarrow 2\text{; therefore}5 \rightar... | Yes |
(a) Every permutation \( \sigma \) in \( {S}_{n} \) can be written either as the identity, a single cycle, or as the product of disjoint cycles. | Proof. Let’s begin with (a) We can assume that \( X = \{ 1,2,\ldots, n\} \) . Let \( \sigma \in {S}_{n} \), and define \( {X}_{1} = \left\{ {1,\sigma \left( 1\right) ,{\sigma }^{2}\left( 1\right) ,\ldots }\right\} \) . The set \( {X}_{1} \) is finite since \( X \) is finite. Therefore the sequence \( 1,\sigma \left( 1\... | Yes |
We know that every permutation in \( {S}_{5} \) is the product of disjoint cycles. Let us list all possible cycle lengths and number of cycles for the permutations of \( {S}_{5} \) . | - First of all, \( {S}_{5} \) contains the identity, which has no cycles.\n\n- Second, some permutations in \( {S}_{5} \) consist of a single cycle. The single cycle could have length \( 2,3,4 \), or 5 (remember, we don’t count cycles of length 1).\n\n- Third, some permutations in \( {S}_{5} \) consist of the product o... | Yes |
Consider the product (1264)(1264), which we may also write as \( {\left( {1264}\right) }^{2} \). | (1) Notice for all elements \( x \neq 1,2,6,4, x \) stays put in (1264); hence \( x \) stays put in \( {\left( {1264}\right) }^{2} \) . So the product \( {\left( {1264}\right) }^{2} \) does not involve any elements except \( 1,2,6 \) and 4 .\n\n(2) Now let’s look at what happens when \( x = 1,2,6 \), or 4 . By squaring... | Yes |
Proposition 11.4.6. The order of a cycle is always equal to the cycle's length. | Proof. To prove this, we essentially have to prove two things:\n\n(A) If \( \sigma \) is a cycle of length \( k \), then \( {\sigma }^{k} = \mathrm{{id}} \) ;\n\n(B) If \( \sigma \) is a cycle of length \( k \), then \( {\sigma }^{j} \neq \operatorname{id}\forall j : 1 \leq j < k \) .\n\nThe proof for (A) follows the s... | No |
Example 11.4.9. Here's a nice application of Proposition 11.4.6, which simply uses rules of function composition. | \[ {\left( {1264}\right) }^{6} = {\left( {1264}\right) }^{4}{\left( {1264}\right) }^{2} = \mathsf{{id}}\;\left( {16}\right) \left( {24}\right) = \left( {16}\right) \left( {24}\right) \] | Yes |
Example 11.4.13. Let \( \tau = \left( {24}\right) \left( {16}\right) \) . Notice that (24) and (16) are disjoint, so they commute (recall Proposition 11.3.27). We also know that permutations are associative under composition. So we may compute \( {\tau }^{2} \) as follows: | \[ {\tau }^{2} = \left( {\left( {24}\right) \left( {16}\right) }\right) \left( {\left( {24}\right) \left( {16}\right) }\right) \]\n\[ = \left( {24}\right) \left( {\left( {16}\right) \left( {24}\right) }\right) \left( {16}\right) \;\text{(associative)} \]\n\[ = \left( {24}\right) \left( {\left( {24}\right) \left( {16}\r... | Yes |
Proposition 11.4.17. If \( \sigma \) and \( \tau \) are disjoint cycles, then\n\n\[ \left| {\sigma \tau }\right| = \operatorname{lcm}\left( {\left| \sigma \right| ,\left| \tau \right| }\right) \]\n\nwhere 'lcm' denotes least common multiple. | Proof. Let \( j \equiv \left| \sigma \right|, k \equiv \left| \tau \right| \), and \( m \equiv \operatorname{lcm}\left( {k, j}\right) \) . Then it’s enough to prove:\n\n(i) \( {\left( \sigma \tau \right) }^{m} = \mathrm{{id}} \) ;\n\n(ii) \( {\left( \sigma \tau \right) }^{n} \neq \) id if \( n \in \mathbb{N} \) and \( ... | Yes |
Proposition 11.4.24. Every cycle can be written as the product of transpositions: | Proof. The proof involves checking that left and right sides of the equation agree when they act on any \( {a}_{j} \) . We know that the cycle acting on \( {a}_{j} \) gives \( {a}_{j + 1} \) (or \( {a}_{1} \), if \( j = n \) ); while the product of transpositions sends \( {a}_{j} \) first to \( {a}_{1} \), then to \( {... | No |
Proposition 11.4.25. Any permutation of a finite set containing at least two elements can be written as the product of transpositions. | Proof. First write the permutation as a product of cycles: then write each cycle as a product of transpositions. | No |
Proposition 11.4.29. Suppose \( \mu \) is a cycle: \( \mu = \left( {{a}_{1}{a}_{2}\ldots {a}_{n}}\right) \) . Then \( {\mu }^{-1} = \left( {{a}_{1}{a}_{n}{a}_{n - 1}\ldots {a}_{2}}\right) . \) | Proof. By Proposition 11.4.24 we can write\n\n\[ \mu = \left( {{a}_{1}{a}_{n}}\right) \left( {{a}_{1}{a}_{n - 1}}\right) \cdots \left( {{a}_{1}{a}_{3}}\right) \left( {{a}_{1}{a}_{2}}\right) . \]\n\nNow consider first just the last two transpositions in this expression. In the Functions chapter, we proved the formula \(... | Yes |
Proposition 11.6.18. The number of even permutations in \( {S}_{n}, n \geq 2 \), is equal to the number of odd permutations; hence, \( \left| {A}_{n}\right| = n!/2 \) . | Proof. The key to the proof is showing that there is a bijection between \( {A}_{n} \) and \( {B}_{n} \) . Since a bijection is one-to-one and onto, this means that \( {A}_{n} \) and \( {B}_{n} \) must have exactly the same number of elements.\n\nTo construct a bijection, notice that \( \left( {12}\right) \in {S}_{n} \... | No |
The set \( \mathbb{R} \smallsetminus \{ 0\} \) of non-zero real numbers is written as \( {\mathbb{R}}^{ * } \) . Let’s prove that \( \left( {{\mathbb{R}}^{ * }, \cdot }\right) \) is a group. | (1) Closure:\n\nSuppose \( a, b \in {\mathbb{R}}^{ * } \) . Then to prove closure we must show \( {ab} \in {\mathbb{R}}^{ * } \) ; that is, we must show (i) \( {ab} \in \mathbb{R} \) and (ii) \( {ab} \neq 0 \) :\n\n(i): Since \( a, b \in \mathbb{R} \), and we know \( \mathbb{R} \) is closed under multiplication, then \... | No |
Let \( S = \mathbb{R} \smallsetminus \{ - 1\} \) and define a binary operation on \( S \) by \( a * b = a + b + {ab} \) . It turns out that \( \left( {S, * }\right) \) is an abelian group. We will prove closure and the commutative property; the rest of the proof will be left to you. | (1) Closure: Suppose \( a, b \in S \) . We need to show that \( a * b \in S \) ; i.e. that (i) \( a * b \in \mathbb{R} \) and (ii) \( a * b \neq - 1 \) .\n\n(i) By the closure of \( \left( {\mathbb{R}, + }\right) ,\left( {a + b}\right) \in \mathbb{R} \) . By the closure of \( \left( {\mathbb{R}, * }\right) ,\left( {ab}... | No |
Proposition 12.3.1. 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 12.3.3. If \( g \) is any element in a group \( G \), then the inverse of \( g \) is unique. | Exercise 12.3.4. Fill in the blanks to complete the following proof of Proposition 12.3.3.\n\n(a) By the definition of inverse, if \( {g}^{\prime } \) is an inverse of an element \( g \) in a group \( G \), then \( g \cdot < 1 > = {g}^{\prime } \cdot < 2 > = e \) .\n\n(b) Similarly, if \( {g}^{\prime \prime } \) is an ... | No |
Proposition 12.3.6. Let \( G \) be a group. If \( a, b \in G \), then \( {\left( ab\right) }^{-1} = {b}^{-1}{a}^{-1} \) . | Proof. By the inverse property, \( \exists {a}^{-1},{b}^{-1} \in G \) . By the closure property, \( {ab} \in G \) and \( {b}^{-1}{a}^{-1} \in G \) . So we only need to verify that \( {b}^{-1}{a}^{-1} \) satisfies the definition of inverse (from Proposition 12.3.3, we know the inverse is unique). First, we have:\n\n\[ \... | No |
Proposition 12.3.10. Let \( G \) be a group. For any \( a \in G,{\left( {a}^{-1}\right) }^{-1} = a \) . | Proof. If \( a \in G \), then since \( G \) is a group, then \( {a}^{-1} \in G \) exists. And again, since \( G \) is a group, there also exists \( {\left( {a}^{-1}\right) }^{-1} \in G \) .\n\nNow, by the definition of inverse, \( {a}^{-1}{\left( {a}^{-1}\right) }^{-1} = e \) . Consequently, multiplying both sides of t... | No |
Proposition 12.3.13. 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 \( G \). | Suppose that \( {ax} = b \) . First we must show that such an \( x \) exists. Since \( a \in G \) and \( G \) is a group, it follows that \( {a}^{-1} \) exists. Multiplying both sides of \( {ax} = b \) on the left by \( {a}^{-1} \), we have\n\n\[ \n{a}^{-1}\left( {ax}\right) = {a}^{-1}b \n\]\n\n\[ \n\left( {{a}^{-1}a}\... | Yes |
Proposition 12.3.21. If \( G \) is a group and \( a, b, c \in G \), then \( {ba} = {ca} \) implies \( b = c \) and \( {ab} = {ac} \) implies \( b = c \) . | This proposition tells us that the right and left cancellation laws are true in groups. We leave the proof as an exercise. | No |
Proposition 12.3.25. In a group, the usual laws of exponents hold; that is, for all \( g, h \in G \) ,\n\n1. \( {g}^{m}{g}^{n} = {g}^{m + n} \) for all \( m, n \in \mathbb{Z} \) ;\n\n2. \( {\left( {g}^{m}\right) }^{n} = {g}^{mn} \) for all \( m, n \in \mathbb{Z} \) ;\n\n3. \( {\left( gh\right) }^{n} = {\left( {h}^{-1}{... | Proof. We will prove part (1), and you will do the rest. We can break part (1) into four cases: (a) \( m, n \geq 0 \) ; (b) \( m, n < 0 \) ; (c) \( m \geq 0, n < 0 \) ; (d) \( m < 0, n \geq 0 \) .\n\nConsider first case (a). Using Definition 12.3.23, we have\n\n\[ \n{g}^{m}{g}^{n} = \underset{m\text{ times }}{\underbra... | No |
Consider the set of even integers \( 2\mathbb{Z} = \{ \ldots , - 2,0,2,4,\ldots \} \) . A more mathematically concise definition is:\n\n\[ 2\mathbb{Z} = \{ x \in \mathbb{Z} \mid x = {2n}\text{ for some }n \in \mathbb{Z}\} \]\n\n\( 2\mathbb{Z} \) is actually a subgroup of \( \mathbb{Z} \), under the operation of additio... | (a) and (b) can be dispatched in short order. From our work in Chapters 1 and 2, we know \( \mathbb{Z} \) is a group under addition: this takes care of (a). For item (b), we have that any element \( m \in 2\mathbb{Z} \) can be written as \( m = {2n} \), where \( n \in \mathbb{Z} \) : hence \( m \in \mathbb{Z} \) also.\... | Yes |
Proposition 12.4.15. 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. We first prove the \ | No |
Using the proposition above, let’s re-prove that \( \mathbb{T} \) is a subgroup of \( {\mathbb{C}}^{ * } \) . | Proof. Based on the proposition, there are four things we need to show:\n\n(a) \( {\mathbb{C}}^{ * } \) is a group;\n\n(b) \( \mathbb{T} \neq \varnothing \) ;\n\n(c) \( \mathbb{T} \subset {\mathbb{C}}^{ * } \) ;\n\n(d) Given \( x, y \in \mathbb{T}, x{y}^{-1} \in \mathbb{T} \) .\n\nItems (a), (b), and (c) we have shown ... | Yes |
Consider the group \( \mathbb{Z} \) . Let us try to find the smallest subgroup of \( \mathbb{Z} \) that contains the number 1 . | (1) We start with the smallest subset possible, \( P = \{ 1\} \) .\n\n(2) The subset has to be a group under addition. But so far \( P \) does not contain an additive identity. So we need to add 0 to the set, giving us \( P = \{ 0,1\} \) .\n\n(3) Zero is its own inverse under addition, but notice that our set does not ... | Yes |
Consider the group \( {\mathbb{Z}}_{6}.\langle 1\rangle \) is computed as follows: | \[ \text{-}1 \equiv 1 \] \[ \text{-}1 + 1 \equiv 2 \] \[ \text{-}1 + 1 + 1 \equiv 3 \] \[ \text{-}1 + 1 + 1 + 1 \equiv 4 \] \[ \text{-}1 + 1 + 1 + 1 + 1 \equiv 5 \] \[ \text{-}1 + 1 + 1 + 1 + 1 + 1 \equiv 0 \] - Notice that we’ve already generated all the elements in \( {\mathbb{Z}}_{6} \) . So we don't have to worry a... | Yes |
The group of units, \( U\left( 9\right) \) is a cyclic group. | As a set, \( U\left( 9\right) \) is \( \{ 1,2,4,5,7,8\} \) . Computing \( \langle 2\rangle \), we get\n\n\[ \langle 2\rangle = \left\{ {{2}^{1} \equiv 2,{2}^{2} \equiv 4,{2}^{3} \equiv 8,{2}^{4} \equiv 7,{2}^{5} \equiv 5,{2}^{6} \equiv 1}\right\} \]\n\n\[ = \{ 2,4,8,7,5,1\} \]\n\n\[ = U\left( 9\right) \]\n\nSo \( \lang... | Yes |
Proposition 12.5.16. Let \( G \) be a group and \( a \) be any element in \( G \) . Then the set\n\n\[ \n\langle a\rangle = \left\{ {{a}^{k} : k \in \mathbb{Z}}\right\} \n\]\n\nis a subgroup of \( G \) . Furthermore, any subgroup of \( G \) which contains \( a \) must also contain \( \langle a\rangle \) . | 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 |
Let us consider the orders of different elements in the infinite group \( \mathbb{Z} \). | - First, what is \( \left| 0\right| \) ? According to Definition 12.5.19, we need to find the smallest positive integer such that \( n \cdot 0 = 0 \) (remember, \( \mathbb{Z} \) is an additive group. We get \( n = 1 \), so \( \left| 0\right| = 1 \), and the cyclic subgroup generated by 0 is \( \langle 0\rangle = \{ 0\}... | Yes |
Example 12.5.22. The order of \( 2 \in {\mathbb{Z}}_{6} \) is 3, because under repeated modular addition we have\n\n\[ 2 \oplus 2 = 4;\;2 \oplus 2 \oplus 2 = 0. \] | Therefore the cyclic subgroup generated by 2 is \( \langle 2\rangle = \{ 0,2,4\} \) . | Yes |
Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle \( {D}_{3} \) (which is the same as \( {S}_{3} \) ). \( {D}_{3} \) has 6 elements: we saw the Cayley table for \( {D}_{3} \) in Chapter 10 (see Table 10.1. | You may verify by using the table that no single element generates the entire group, so \( {D}_{3} \) is not cyclic.. The cyclic subgroups of \( {S}_{3} \) are shown in Figure 12.5.1. | Yes |
Proposition 12.5.30. Given a group \( G \) and a subgroup \( H \subset G \), and suppose that \( a \in H \) . Then \( \langle a\rangle \subset H \) . | ## Exercise 12.5.31. Prove Proposition 12.5.30\n\n\( \diamond \) | No |
Example 12.5.32. We showed in Example 12.5.29 that \( {D}_{3} \) has 4 cyclic subgroups, and that every element of \( {D}_{3} \) is in at least one of these subgroups. Proposition 12.5.30 shows that, for example, any subgroup containing \( {\rho }_{1} \) must also contain \( {id} \) and \( {\rho }_{2} \), since \( \lef... | If we add any other element (which must be \( {\mu }_{k} \) for some \( k = 1,2 \) or 3), then we must also add \( {\rho }_{1}{\mu }_{k} \) and \( {\rho }_{2}{\mu }_{k} \), which means that \( H \) contains all 6 elements of \( {D}_{3} \) . It follows that \( H = {D}_{3} \) . Similarly, if we try to find a subgroup \( ... | Yes |
Proposition 12.5.34. 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 \( h = {a}^{s} \) . It follows that\n\n\[ \n{gh} = {a}^{r}{a}^{s} = {a}^{r + s} = {a}^{s + r} = {a}^{s}{a}^{r} = {hg}.\n\]\n... | Yes |
Proposition 12.5.35. The group \( \left( {\mathbb{R}, + }\right) \) is not cyclic. | Proof. We'll use our old workhorse, proof by contradiction. Suppose that \( a \) is a generator of \( \mathbb{R} \), i.e. \( \langle a\rangle = \mathbb{R} \) . Then since \( a/2 \in \mathbb{R} \), it follows that \( a/2 \in \langle a\rangle \) . This means \( a/2 = {ka} \) for some integer \( \mathrm{k} \) . Rearrangin... | Yes |
Proposition 12.5.37. Every subgroup of a cyclic group is cyclic. | Proof. The main tools used in this proof are the division algorithm, which we mentioned in Proposition 3.2.3, and the Principle of Well-Ordering, which we mentioned in Section 1.2.2.\n\nLet \( G \) be a cyclic group generated by \( a \) and suppose that \( H \) is a subgroup of \( G \) . If \( H = \{ e\} \), then trivi... | Yes |
Proposition 12.5.41. 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. Since \( G = \langle a\rangle \) it follows from Proposition 12.5.27 that \( \left| a\right| = n \) . In Exercise 12.5.26 (a) we proved that \( {a}^{k} = {a}^{\;\operatorname{mod}\;\left( {m, n}\right) } \) . Let \( r = {\;\operatorname{mod}\;\left( {m, n}\right) } \) . If \( r = 0 \) (which is the same thing as... | No |
Proposition 12.5.42. 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 12.5.41, this is the smallest integer \( m \) such that \( n \) divides \( {km} \) or, equivalently, \( n/d \) divides \( m\left( {k/d}\right) \) . (Note that \( n/d \) and \( k/d \) are both integers, since \( d... | Yes |
Example 12.5.44. 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,9 is a generator because: | \[ 1 \cdot 9 = 9\;2 \cdot 9 = 2\;3 \cdot 9 = {11} \] \[ 4 \cdot 9 = 4\;5 \cdot 9 = {13}\;6 \cdot 9 = 6 \] \[ 7 \cdot 9 = {15}\;8 \cdot 9 = 8\;9 \cdot 9 = 1 \] \[ {10} \cdot 9 = {10}\;{11} \cdot 9 = 3\;{12} \cdot 9 = {12} \] \[ {13} \cdot 9 = 5\;{14} \cdot 9 = {14}\;{15} \cdot 9 = 7. \] | No |
Example 13.1.2. Key exchange between a sender and receiver (Moses and Rachael) using the DHKE is shown in the following steps. | Step 1. Prior to sending data, Moses and Rachael agree \( p = {13} \) and \( g = 7 \) ;\n\nStep 2. Moses chooses \( n = 2 \), and sends Rachael \( {\;\operatorname{mod}\;\left( {{7}^{2},{13}}\right) } = {10} \) ;\n\nStep 3. Rachael chooses \( m = 8 \), and sends Moses \( {\;\operatorname{mod}\;\left( {{7}^{8},{13}}\rig... | Yes |
It is easy to calculate \( {\;\operatorname{mod}\;\left( {{2}^{m},{11}}\right) } \) for different values of \( m \) : for example, when \( m = 8 \) then we get \( {\;\operatorname{mod}\;\left( {{2}^{8},{11}}\right) } = {\;\operatorname{mod}\;\left( {{256},{11}}\right) } = \) 3. However, when you try to invert the proce... | \[
{\;\operatorname{mod}\;\left( {{2}^{1},{11}}\right) } = 2
\]
\[
{\;\operatorname{mod}\;\left( {{2}^{2},{11}}\right) } = 4
\]
\[
{\;\operatorname{mod}\;\left( {{2}^{3},{11}}\right) } = 8
\]
\[
{\;\operatorname{mod}\;\left( {{2}^{4},{11}}\right) } = 5
\]
\[
{\;\operatorname{mod}\;\left( {{2}^{5},{11}}\right) } = {... | Yes |
Given E: \( {y}^{2} = {x}^{3} + {ax} + b,{P}_{1} = \left( {{x}_{1},{y}_{1}}\right) ,{P}_{2} = \left( {{x}_{2},{y}_{2}}\right) \) , and \( {P}_{3} = \left( {{x}_{3},{y}_{3}}\right) \) . Find \( {P}_{3} \), where \( {P}_{3} = {P}_{1} + {P}_{2} \) . | The steps of this calculation are as follows:\n\n(a) Compute the slope of the line \( m \) through \( {P}_{1} \) and \( {P}_{2} \) as follows:\n\n\[ m = \left( {{y}_{2} - {y}_{1}}\right) \cdot {\left( {x}_{2} - {x}_{1}\right) }^{-1}\text{, for }{P}_{1} \neq {P}_{2} \]\n\n(b) Use the point-slope formula \( y - {y}_{1} =... | No |
Given E: \( {y}^{2} = {x}^{3} - {2x},{P}_{1} = \left( {0,0}\right) ,{P}_{2} = \left( {-1,1}\right) \), Find \( {P}_{3} = {P}_{1} + {P}_{2} \) | (a) Slope : \( m = \left( {1 - 0}\right) \cdot {\left( -1 - 0\right) }^{-1} = - 1 \) .\n\n(b) Equation of line: \( y - 0 = - 1\left( {x - 0}\right) \) or \( y = - x \) .\n\n(c) Use \( {x}_{3} = {m}^{2} - {x}_{1} - {x}_{2} \) to obtain: \( {x}_{3} = {\left( -1\right) }^{2} - 0 - \left( {-1}\right) = 2 \) .\n\n(d) Use eq... | Yes |
Example 13.2.6. Given E: \( {y}^{2} = {x}^{3} - x,{P}_{1} = \left( {2,\sqrt{6}}\right) ,{P}_{2} = \left( {3, - \sqrt{24}}\right) \) from Figure 13.2.5 above. Find \( {P}_{3} \), where \( {P}_{3} = {P}_{1} + {P}_{2} \) . | (a) Slope: \( m = \left( {-\sqrt{24} - \sqrt{6}}\right) \cdot {\left( 3 - 2\right) }^{-1} = - 3\sqrt{6} \)\n\n(b) Line: \( y + \sqrt{24} = - 3\sqrt{6}\left( {x - 3}\right) \), which simplifies to \( y = - 3\sqrt{6}x + 7\sqrt{6} \)\n\n(c) Use \( {x}_{3} = {m}^{2} - {x}_{1} - {x}_{2} \) to obtain: \( {x}_{3} = {\left( -3... | Yes |
Given E: \( {y}^{2} = {x}^{3} + {2x} + 2 \), and \( P = \left( {5,1}\right) \), we want to find \( {2P} \), where \( {2P} = P + P \) . | First we use the slope of the tangent line, using the same formula we did for the real case in Example 13.2.3:\n\n\[ m = \left( {3{x}_{1}^{2} + a}\right) \cdot {\left( 2{y}_{1}\right) }^{-1} \]\n\nWe may then calculate (remember we’re doing arithmetic in \( {\mathbb{Z}}_{17} \) !)\n\n\[ m \equiv \left( {\left( {3 \cdot... | Yes |
Let \( P \) be the set of all professional basketball players in the \( {\mathrm{{NBA}}}^{1} \) and let \( T \) be the set of NBA teams. \( {f}_{T} : P \rightarrow T \) as follows:\n\n\[ \n{f}_{T}\left( p\right) = \text{the team that } p \text{ plays for.} \n\]\n\nAlternatively, \( {f}_{T} \) can be represented as the ... | This is not a function, because many NBA players have played on more than one team. | No |
Let \( P \) be the set of all professional basketball players in the NBA. Consider the following subset of \( P \times P \) :\n\n\[ \left\{ {\left( {p,{p}^{\prime }}\right) \in P \times T \mid {p}^{\prime }}\right. \text{is the tallest teammate of}\left. p\right\} \text{.} \] | This is a binary relation, according to Definition! 14.1.2, and it also can be identified with the function \( {f}_{h} : P \rightarrow P \) defined by: \( {f}_{h}\left( p\right) = \) the tallest teammate of \( p \) . | No |
Consider the graphs in Figure 14.1.1, where the set \( A \times B \) is indicated at the top of each graph. Which are relations in \( A \times B \) ? Which are binary relations? Which are functions? | - All three graphs are relations because all graphs are subsets of \( A \times B \) (as specified at the top of each graph).\n\n- The first and third relations are binary relations, but the second relation isn’t because \( A \neq B \) .\n\n- The first is not a function, because e.g. both \( \left( {1,1}\right) \) and \... | Yes |
If \( A = \{ 1,2,3\} \) and \( B = \{ 4,5,6\} \), some examples of relations from \( A \) to \( B \) are: | \[ \{ \left( {1,4}\right) ,\left( {2,5}\right) ,\left( {3,6}\right) \} \] \[ \{ \left( {1,6}\right) ,\left( {3,4}\right) \} \] \[ \{ \left( {2,5}\right) ,\left( {3,5}\right) \} , \] \[ \varnothing \text{,} \] \[ \{ \left( {1,4}\right) ,\left( {1,5}\right) ,\left( {1,6}\right) ,\left( {2,4}\right) ,\left( {2,5}\right) ,... | Yes |
Example 14.1.11. Let \( A = \{ 1,2,3,4,5\} \) . We can define a binary relation \( {R}_{1} \) on \( A \) by letting\n\n\[ \n{R}_{1} = \left\{ {\left( {x, y}\right) \mid x \neq y\text{ and }{x}^{2} + y \leq {10}}\right\} .\n\] | This binary relation is represented by the digraph in Figure 14.1.2:\n\n\n\nFigure 14.1.2. Digraph of the binary relation of \( {R}_{1} \)\n\nNote that there’s a bidirectional arrow between 1 and 3 because \( \left( ... | Yes |
In this example we will define a binary relation from the given partition, and show that the relation has the above three properties.\n\n(a) Given the partition in Example 14.2.3(a), we can define a binary relation \( { \sim }_{R} \) on \( \mathbb{R} \) by\n\n\[ x{ \sim }_{R}y\text{ iff }\left( {x, y \in \mathbb{Q}\tex... | - First property (reflexive): \( x{ \sim }_{R}x \) ( \( x \) is always in the same set, \( \mathbb{Q} \) or \( \mathbb{I} \), as itself);\n\n- Second property (symmetric): \( x{ \sim }_{R}y \Rightarrow y{ \sim }_{R}x \) ( \( x \) is in the same set as \( y \) implies \( y \) is in the same set as \( x \) );\n\n- Third ... | Yes |
Proposition 14.2.9. Given a partition \( \mathcal{P} \) on set \( A \), define a binary relation of \( A \) as \( a \sim b \) iff there exists a subset \( C \in \mathcal{P} \) such that \( a \) and \( b \) are both elements of \( C \), then the binary relation, \( \sim \), satisfies the following 3 properties:\n\n(a) r... | Proof. Earlier in the chapter we showed that the binary relation \( { \sim }_{\mathcal{P}} \) from the partition in Figure 14.2.1 was reflexive, symmetric, and transitive. The arguments that we used are generally applicable, and can be used for any partition. | No |
Consider the following binary relations on \( \mathbb{R} \) :\n\n(a) \( = \) is reflexive, symmetric, and transitive. | - Reflexive: any real number \( x \) equals itself, so \( x = x\forall x \in \mathbb{R} \).\n\n- Symmetric: for any real numbers \( x \) and \( y \), if \( x = y \), then \( y = x \).\n\n- Transitive: for any real numbers \( x, y \), and \( z \), if \( x = y \) and \( y = z \) , then \( x = z \).\n\n- Therefore \( = \)... | Yes |
Given the set \( B = \{ 1,2,3\} \), consider the relation \( \sim \) on \( B \) defined by\n\n\[ \n{B}_{ \sim } = \{ \left( {1,1}\right) ,\left( {2,2}\right) ,\left( {3,3}\right) ,\left( {1,2}\right) ,\left( {2,1}\right) ,\left( {2,3}\right) ,\left( {3,2}\right) \} \n\] | - \( \sim \) is reflexive, because \( 1 \sim 1,2 \sim 2 \), and \( 3 \sim 3 \) (Note we had to check all elements of the set \( B \) ),\n\n- \( \sim \) is symmetric, because, for each \( \left( {a, b}\right) \in \sim \), the reversal \( \left( {b, a}\right) \) is also in \( \sim \) .\n\n- \( \sim \) is not transitive, ... | Yes |
Is the relation in A transitive? Let's consider Remember how transitivity is defined: if \( a \sim b \) and \( b \sim c \) then \( a \sim c \) . In more prosaic terms, if there’s an arrow from \( a \) to \( b \) and another arrow from \( b \) to \( c \), then there’s an arrow directly from \( a \) to \( c \) . We may c... | Therefore this relation is transitive. If \( 3 \sim 2 \) is removed from this binary relation, then the relation isn't transitive because it's still be possible to get from 3 to 2 via 1, but there's no longer a direct route. | Yes |
Define a binary relation \( \sim \) on \( \mathbb{R} \) by \( x \sim y \) iff \( {x}^{2} = {y}^{2} \) . Then \( \sim \) is an equivalence relation. | Proof. We wish to show that \( \sim \) is reflexive, symmetric, and transitive.\n\n(reflexive) Given \( x \in \mathbb{R} \), we have \( {x}^{2} = {x}^{2} \), so \( x \sim x \).\n\n(symmetric) Given \( x, y \in \mathbb{R} \), such that \( x \sim y \), we have \( {x}^{2} = {y}^{2} \). Since equality is symmetric, this im... | Yes |
Define a binary relation \( \sim \) on \( \mathbb{N} \times \mathbb{N} \) by \( \left( {{a}_{1},{b}_{1}}\right) \sim \left( {{a}_{2},{b}_{2}}\right) \) iff \( {a}_{1} + {b}_{2} = {a}_{2} + {b}_{1} \) . Then \( \sim \) is an equivalence relation. | Proof. We wish to show that \( \sim \) is reflexive, symmetric, and transitive.\n\n(reflexive) Given \( \left( {a, b}\right) \in \mathbb{N} \times \mathbb{N} \), we have \( a + b = a + b \), so \( \left( {a, b}\right) \sim \left( {a, b}\right) \).\n\n(symmetric) Given \( \left( {{a}_{1},{b}_{1}}\right) ,\left( {{a}_{2}... | Yes |
Proposition 14.3.6. Suppose \( f : A \rightarrow B \) . If we define a binary relation \( \sim \) on \( A \) by\n\n\[ \n{a}_{1} \sim {a}_{2}\; \Leftrightarrow \;f\left( {a}_{1}\right) = f\left( {a}_{2}\right) \n\]\n\nthen \( \sim \) is an equivalence relation on \( A \) . | Exercise 14.3.7. Prove Proposition 14.3.6: that is, prove that the relation defined in the proposition is (a) reflexive, (b) symmetric, and (c) transitive. (If you like, you may model your proof on the discussion prior to Exercise Example 14.2.7, where we proved the three properties for binary relations arising from pa... | No |
Proposition 14.3.12. Suppose \( f : A \rightarrow B \) . For each \( b \in \operatorname{Range}\left( f\right) \) define a subset \( {A}_{b} \subset A \) as follows:\n\n\[ \n{A}_{b} \mathrel{\text{:=}} \{ a \in A \mid f\left( a\right) = b\} .\n\]\n\nThen the collection of sets \( \mathcal{P} \mathrel{\text{:=}} \left\{... | Proof. The proof is broken up into steps in the following exercise.\n\nExercise 14.3.13.\n\n(a) Given any \( b \in \operatorname{Range}\left( f\right) \), show that \( {A}_{b} \) is nonempty\n\n(b) Given \( {b}_{1},{b}_{2} \in \operatorname{Range}\left( f\right) \) with \( {b}_{1} \neq {b}_{2} \), show that \( {A}_{{b}... | No |
Definition 14.4.2. Suppose \( \sim \) is an equivalence relation on a set \( A \) . For each \( a \in A \), the equivalence class of \( a \) is the following subset of \( A \) : | \[ \left\lbrack a\right\rbrack = \{ s \in A \mid s \sim a\} . \] | Yes |
Example 14.4.5. Suppose \( A = \{ 1,2,3,4,5\} \) and\n\n\( R = \)\n\n\[ \n\{ \left( {1,1}\right) ,\left( {1,3}\right) ,\left( {1,4}\right) ,\left( {2,2}\right) ,\left( {2,5}\right) ,\left( {3,1}\right) ,\left( {3,3}\right) ,\left( {3,4}\right) ,\left( {4,1}\right) ,\left( {4,3}\right) ,\left( {4,4}\right) ,\left( {5,2}... | One can verify that \( R \) is an equivalence relation on \( A \) . The equivalence classes are:\n\n\[ \n\left\lbrack 1\right\rbrack = \left\lbrack 3\right\rbrack = \left\lbrack 4\right\rbrack = \{ 1,3,4\} ,\;\left\lbrack 2\right\rbrack = \left\lbrack 5\right\rbrack = \{ 2,5\} .\n\] | Yes |
Proposition 14.4.7. Suppose \( \sim \) is an equivalence relation on a set \( S \) . Then:\n\n(a) For all \( a \in S \), we have \( a \in \left\lbrack a\right\rbrack \) .\n\n(b) For all \( a \in S \), we have \( \left\lbrack a\right\rbrack \neq \varnothing \) .\n\n(c) The union of the equivalence classes is all of \( S... | Exercise 14.4.8. Prove the assertions in Proposition 14.4.7. You may use the following hints:\n\n(a) Use the reflexive property of \( \sim \), together with Definition 14.4.2\n\n(b) Use part (a).\n\n(c) This can be done by showing:\n\n(i) \( \mathop{\bigcup }\limits_{{a \in S}}\left\lbrack a\right\rbrack \subset S \)\n... | No |
Proposition 14.4.9. Suppose \( \sim \) is an equivalence relation on a set \( S \) . Then any two equivalence classes are either equal or disjoint; that is, either they have exactly the same elements, or they have no elements in common. | Proof. It’s enough to show that any two equivalence classes \( \left\lbrack {a}_{1}\right\rbrack \) and \( \left\lbrack {a}_{2}\right\rbrack \) that are not disjoint must in fact be equal.\n\n(a) Since the equivalence classes are not disjoint, their intersection is nonempty: so there is some \( a \in \left\lbrack {a}_{... | No |
Proposition 14.4.12. Suppose \( \sim \) is an equivalence relation on a set \( A \) . Then\n\n\[ \{ \left\lbrack a\right\rbrack \mid a \in A\} \]\n\n is a partition of \( A \) . | Proof. From parts (b), (c), and (e) of Proposition 14.4.7, we know that the equivalence classes are nonempty, that their union is \( A \), and that they are pairwise disjoint. | Yes |
We have \( {\left\lbrack 1\right\rbrack }_{3} + {\left\lbrack 2\right\rbrack }_{3} = {\left\lbrack 1 + 2\right\rbrack }_{3} = {\left\lbrack 3\right\rbrack }_{3} \) . However, since 3 and 0 are in the same equivalence class, we have \( {\left\lbrack 3\right\rbrack }_{3} = {\left\lbrack 0\right\rbrack }_{3} \), so the ab... | Example 14.5.7 illustrates the following general rule: If \( r \) is the remainder when \( a + b \) is divided by 3, then \( \bar{a} + \bar{b} = \bar{r} \) . | No |
Proposition 14.5.12. For any \( n \in {\mathbb{N}}^{ + } \), we have\n\n\[{\mathbb{Z}}_{n} = \{ \overline{0},\overline{1},\overline{2},\ldots ,\overline{n - 1}\}\]\n\nand \( \overline{0},\overline{1},\overline{2},\ldots ,\overline{n - 1} \) are all distinct. | Exercise 14.5.13. Prove Proposition 14.5.12. It is sufficient to show (a) \( \overline{0},\overline{1},\overline{2},\ldots ,\overline{n - 1} \) are distinct; and (b) for any integer, the equivalence class \( \bar{k} \) is one of \( \bar{0},\bar{1},\bar{2},\ldots ,\overline{n - 1} \) . | No |
Let \( H \) be the subgroup of \( {S}_{3} \) defined by the permutations \( \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \). Find the left cosets of \( H \) in \( {S}_{3} \). | \[ \text{(1)}H = \left( {123}\right) H = \left( {132}\right) H = \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \text{,} \]\n\[ \left( {12}\right) H = \left( {13}\right) H = \left( {23}\right) H = \{ \left( {12}\right) ,\left( {13}\right) ,\left( {23}\right) \} .\n\]\nThere are 2 left cosets, and each ... | Yes |
Proposition 15.2.4. 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. The proof has two parts, namely (1) Cosets are disjoint; and (2) The union of cosets is all of \( G \) .\n\n(1) 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 \... | No |
Proposition 15.2.13. 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 \( L \) and \( R \) denote the set of left and right cosets of \( H \) in \( G \) , respectively. If we can define a bijection \( \phi : L \rightarrow R \), then the proposition will be proved. If \( {gH} \in L \), let \( \phi \left( {gH}\right) = H{g}^{-1} \) . By Proposition 15.2.1, the map \( \phi \) is w... | Yes |
Proposition 15.3.2. 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 a bijection; 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} ... | Yes |
Proposition 15.3.7. 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 |
Proposition 15.3.9. Suppose that \( G \) is a finite group and \( g \in G \) . Then the order of \( g \) must divide the number of elements in \( G \) . | Proof. The order of a group element \( g \), which is denoted as \( \left| g\right| \), is defined in Definition 12.5.19 in Section 12.5.2. We indicated in Exercise 12.5.26 in that same section that \( \left| g\right| \) is equal to \( \left| {\langle g\rangle }\right| \), which is the order of the cylic subgroup gener... | No |
Proposition 15.3.12. (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. First, let \( r \) be the remainder when \( a \) is divided by \( n \) . We may consider \( r \) as an element of \( U\left( n\right) \) .\n\nAs noted above, the order of \( U\left( n\right) \) is \( \phi \left( n\right) \) . Lagrange’s theorem then tells us that \( \left| r\right| \) divides \( \phi \left( n\ri... | No |
Proposition 15.4.7. Let \( G \) be a group, and let \( H \) be a subgroup of \( G \) with index 2. Then \( H \) is a normal subgroup of \( G \) . | Exercise 15.4.8. Prove Proposition 15.4.7 by proving each of the following steps.\n\n(a) Prove that \( G \smallsetminus H \) is a left coset of \( H \) in \( G \) .\n\n(b) Prove that \( G \smallsetminus H \) is a right coset of \( H \) in \( G \) .\n\n(c) Prove that \( H \) is normal in \( G \) . | No |
Proposition 15.4.12. Let \( G \) be a group and \( N \) be a subgroup of \( G \) . Then the following statements are equivalent.\n\n1. The subgroup \( N \) is normal in \( G \) .\n\n2. For all \( g \in G,{gN}{g}^{-1} \subset N \) .\n\n3. For all \( g \in G,{gN}{g}^{-1} = N \) . | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) . Since \( N \) is normal in \( G,{gN} = {Ng} \) for all \( g \in G \) . Hence, for a given \( g \in G \) and \( n \in N \), there exists an \( {n}^{\prime } \) in \( N \) such that \( {gn} = {n}^{\prime }g \) . Therefore, \( {gn}{g}^{-1} = {n}^{\prime } \in N \)... | Yes |
Consider the normal subgroup \( 3\mathbb{Z} \) of \( \mathbb{Z} \) that we started exploring at the beginning of the chapter. The cosets of \( 3\mathbb{Z} \) in \( \mathbb{Z} \) were\n\n\[ 0 + 3\mathbb{Z} = \{ \ldots , - 3,0,3,6,\ldots \}\]\n\n\[ 1 + 3\mathbb{Z} = \{ \ldots , - 2,1,4,7,\ldots \}\]\n\n\[ 2 + 3\mathbb{Z}... | As we've done many times before, we may prove that these two sets are equal by showing that all elements of the left-hand set are contained in the right-hand set, and vice versa. So let's take an arbitrary element of \( \left( {0 + 3\mathbb{Z}}\right) + \left( {1 + 3\mathbb{Z}}\right) \). We may write this element as \... | Yes |
Proposition 15.4.21. Let \( N \) be a normal subgroup of a group \( G \). If \( a, b \in G \), then \( {aN} \circ {bN} = {abN} \) . | Proof. The proof parallels the argument in Example 15.4.18. Let \( x \in {aN} \) and \( y \in {bN} \) . Using Exercise 15.4.17 part (c), we may conclude that \( {xy} \in \) \( {abN} \) . This shows that \( {aN} \circ {bN} \subset {abN} \) . On the other hand, let \( z \in {abN} \) . Then \( z = {ae} \circ {bn} \) for s... | No |
Proposition 15.4.22. Let \( N \) be a normal subgroup of a group \( G \). The cosets of \( N \) in \( G \) form a group under the operation of set composition. | Proof. We have shown that the set composition operation is well-defined and closed on the set of cosets of \( N \), provided that \( N \) is normal. Associativity follows by the associativity of the group operation defined on \( G \). Using Proposition 15.4.21 we have that \( {eN} \circ {aN} = {aN} \circ {eN} = {aN} \)... | Yes |
Consider the normal subgroup of \( {S}_{3}, H = \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \) which we started exploring in Example 15.1.5. The cosets of \( H \) in \( {S}_{3} \) were \( H \) and (12) \( N \) . Using the group operation from Defintion 15.4.23 to compose these cosets together, the q... | <table><thead><tr><th></th><th>\( N \)</th><th>\( \left( \begin{matrix} {12} \\ \end{matrix}\right) N \)</th></tr></thead><tr><td>\( N \)</td><td>\( N \)</td><td>\( \left( \begin{matrix} {12} \\ \end{matrix}\right) N \)</td></tr><tr><td>\( \left( {12}\right) N \)</td><td>\( \left( \begin{matrix} {12} \\ \end{matrix}\ri... | Yes |
Example 15.4.27. Consider the dihedral group \( {D}_{n} \) that we studied in the Symmetries chapter, which was the group of symmetries (rotations and reflections) of a regular \( n \) sided polygon. We determined in the latter part of that chapter that \( {D}_{n} \) was actually generated by the two elements \( r \) a... | Now there are \( {2n} \) symmetries in \( {D}_{n} \) and \( n \) rotations in \( {R}_{n} \) ; so Lagrange’s theorem tells us the number of cosets, \( \left\lbrack {{D}_{n} : {R}_{n}}\right\rbrack = \frac{\left| {D}_{n}\right| }{\left| {R}_{n}\right| } = \frac{2n}{n} = 2 \) .\n\nSince \( {R}_{n} \), the rotations, are o... | Yes |
Proposition 15.5.4. The alternating group \( {A}_{n} \) is generated by 3-cycles for \( n \geq 3 \) . | Proof. We know that any element \( \sigma \) of \( {A}_{n} \) is an even permutation, so \( \sigma \) can be expressed as the product of an even number of transpositions. In this expression for \( \sigma \) we may pair up the transpositions two by two, and thus obtain an expression for \( \sigma \) as a product of pair... | Yes |
Proposition 15.5.6. Let \( N \) be a normal subgroup of \( {A}_{n} \), where \( n \geq 3 \) . If \( N \) contains a 3-cycle, then \( N = {A}_{n} \) . | Proof. We will first show that \( {A}_{n} \) is generated by 3-cycles of the specific form \( \left( {ijk}\right) \), where \( i \) and \( j \) are fixed in \( \{ 1,2,\ldots, n\} \) and we let \( k \) vary. Every 3-cycle is the product of 3-cycles of this form, since\n\n\[ \left( {iaj}\right) = {\left( ija\right) }^{2}... | Yes |
Proposition 15.5.8. The alternating group, \( {A}_{n} \), is simple for \( n \geq 5 \) . | Proof. Let \( N \) be a normal subgroup of \( {A}_{n} \) . By Proposition 15.5.7, \( N \) contains a 3-cycle. By Proposition 15.5.6, \( N = {A}_{n} \) ; therefore, \( {A}_{n} \) contains no proper nontrivial normal subgroups for \( n \geq 5 \) . | Yes |
Even parity is a commonly used coding scheme, which (as we shall see) can be generalized to form powerful and versatile codes. Computers use the ASCII (American Standard Code for Information Interchange) coding system to encode the letters and special characters that appear on your keyboard (these may be considered as ... | Adding a parity check bit allows the detection of all single errors because changing a single bit either increases or decreases the number of 1 's by one, and in either case the parity has been changed from even to odd, so the new word is not a codeword. (We could equally well construct an error detection scheme based ... | Yes |
Example 16.1.3. 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 original... | In Table 16.1, we present all possible words that might be received for the transmitted codewords (000) and (111). Table 16.1 also shows the number of bits by which each received 3-tuple differs from each original codeword.\n\n<table><thead><tr><th colspan=\ | No |
Example 16.2.16. Consider the \( \left( {4,3}\right) \) code in which the first three bits carry information and the fourth is an even parity check bit. (Note that now we're putting the parity bit on the right instead of on the left as we did in Example 16.1.2. This will turn out to be more useful in the development of... | Table 16.3: Distances between 4-bit codewords\n\n<table><thead><tr><th></th><th>0000</th><th>0011</th><th>0101</th><th>0110</th><th>1001</th><th>1010</th><th>1100</th><th>1111</th></tr></thead><tr><td>0000</td><td>0</td><td>2</td><td>2</td><td>2</td><td>2</td><td>2</td><td>2</td><td>4</td></tr><tr><td>0011</td><td>2</t... | Yes |
Proposition 16.2.17. 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 |
Example 16.2.18. In Table 16.4, 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 |
Suppose that we have a code that consists of the following 7-tuples:\n\n\[ \left( {0000000}\right) \;\left( {0001111}\right) \;\left( {0010101}\right) \;\left( {0011010}\right) \]\n\n\[ \left( \begin{matrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & & \end{matrix}\right) \;\left( \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & & \end... | It's possible to verify directly (for instance, by computing the Cayley table) that this code is a group code (later we will show there are much, much quicker ways to do this). To find the minimum distance, one may compute the distances between all pairs of codewords. The result is \( {d}_{\min } = 3 \), so the code ca... | No |
Proposition 16.3.12. Let \( {d}_{\min } \) be the minimum distance for a group code \( C \) . Then \( {d}_{\min } \) is the minimum weight of all 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 1 's. Notice that an arbitrary \( n \) -tuple \( \mathbf{x} = \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \) has an even n... | \[ \mathbf{1} \cdot \mathbf{x} = \mathbf{1}{\mathbf{x}}^{\mathrm{T}} = \left( \begin{array}{llll} 1 & 1 & 1 & 1 \end{array}\right) \left( \begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \\ {x}_{4} \end{array}\right) = 0. \] | 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 \( \mathbf{x} = \left( {{x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}}\right) \) to be in the null space of \( H \) it must satisfy \( H{\mathbf{x}}^{\mathrm{T}} = \mathbf{0} \... | This means that,\n\n\[ \left( \begin{array}{lllll} 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \end{array}\right) \cdot \left( \begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \\ {x}_{4} \\ {x}_{5} \end{array}\right) = \left( \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right) \]\n\nSo the following... | No |
Example 16.4.5. Suppose that\n\n\[ \nH = \left( \begin{array}{llll} 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right) \n\] \n\nFor a 4-tuple \( \mathbf{x} = \left( {{x}_{1},{x}_{2},{x}_{3},{x}_{4}}\right) \) to be in the null space of \( H \) it must satisfy \( H{\mathbf{x}}^{\mathrm{T}} = \mathbf{0}. \) | This means that,\n\n\[ \n\left( \begin{array}{llll} 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right) \cdot \left( \begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \\ {x}_{4} \end{array}\right) = \left( \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \end{array}\right) \n\] \n\nSo the following system of equations must be satisfied.\n... | 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 7-tuple \( \mathbf{x} = \left( {0,1,0,0,1,1}\right) \) is received. It is a simple matter of matrix multiplication to deter... | Since\n\n\[ H{\mathbf{x}}^{\mathrm{T}} = \left( \begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right) \]\n\nthe received word is not a codeword. We must either attempt to correct the word or request that it be transmitted again. | Yes |
Proposition 16.5.4. Let \( H \in {\mathbb{M}}_{k \times n}\left( {\mathbb{Z}}_{2}\right) \) be a canonical parity-check matrix. Then \( \operatorname{Null}\left( H\right) \) consists of all \( \mathbf{x} \in {\mathbb{Z}}_{2}^{n} \) whose first \( n - k \) bits are arbitrary but whose last \( k \) bits are determined by... | The proof of Proposition 16.5.4 simply follows the same steps as in Example 16.5.3, except that instead of 3 equations in 6 unknowns we have \( k \) equations in \( n \) unknowns. Readers who’ve had linear algebra may recognize that this is exactly the same as the method for solving linear equations using row-echelon f... | No |
Proposition 16.5.10. Suppose that \( G \) is an \( n \times m \) standard generator matrix. Then \( C = \left\{ {\mathbf{y} : G\mathbf{x} = \mathbf{y}}\right. \) for \( \left. {\mathbf{x} \in {\mathbb{Z}}_{2}^{m}}\right\} \) is an \( \left( {n, m}\right) \) -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 |
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