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Proposition 16.5.11. Let \( H = \left( {A \mid {I}_{k}}\right) \) be an \( k \times n \) canonical parity-check matrix and \( G = \left( \begin{matrix} {I}_{n - k} \\ A \end{matrix}\right) \) be the corresponding \( n \times \left( {n - k}\right) \) standard generator matrix. Then \( {HG} = \mathbf{0} \), where \( \mat... | Proof. It is possible to prove this by writing out the matrix product \( {HG} \) using summation notation (see Chapter 8). This is however somewhat long-winded. A much easier way is to multiply \( H \) and \( G \) as block matrices. \( {}^{1} \) Since the block sizes are compatible, we have\n\n\[ \n{HG} = \left( {A \mi... | Yes |
Proposition 16.5.14. Let \( {\mathbf{e}}_{i} \) be the binary \( n \) -tuple with a 1 in the \( i \) th coordinate and 0 ’s elsewhere and suppose that \( H \in {\mathbb{M}}_{m \times n}\left( {\mathbb{Z}}_{2}\right) \) . Then \( H{\mathbf{e}}_{i} \) is the \( i \) th column of the matrix \( H \) . | Proposition 16.5.14 is a well-known fact in linear algebra, so we refer the reader to a linear algebra textbook for proof. | No |
Proposition 16.6.3. Let the \( k \times n \) 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. The... | Proof. \( H\mathbf{x} = H\left( {\mathbf{c} + \mathbf{e}}\right) = H\mathbf{c} + H\mathbf{e} = \mathbf{0} + H\mathbf{e} = H\mathbf{e} \) . | Yes |
Example 16.6.5. Consider the matrix\n\n\[ H = \left( \begin{array}{llllll} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right) \]\n\nand suppose that the 6-tuples \( \mathbf{x} = \left( {111110}\right) ,\mathbf{y} = \left( {111111}\right) \), and \( \mathbf{z} = \left( {010111}\ri... | Then\n\n\[ H\mathbf{x} = \left( \begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right), H\mathbf{y} = \left( \begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right), H\mathbf{z} = \left( \begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) . \]\n\nHence, \( \mathbf{x} \) has an error in the third bit and \( \mathbf{z} \) has an error in t... | Yes |
Proposition 16.6.11. 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 |
Example 16.6.12. Table 16.7 is a decoding table for the code \( C \) given in Example 16.6.9. If \( \mathbf{x} = \left( {01111}\right) \) is received, then its syndrome can be computed to be\n\n\[ H\mathbf{x} = \left( \begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right) \] | Examining the decoding table, we determine that the coset leader is (00010). It is now easy to decode the received codeword. | Yes |
In the Symmetries chapter we also saw some examples of isomorphic groups. In particular, we saw that \( {\mathbb{Z}}_{4} \), the \( {4}^{\text{th }} \) roots of unity, and the rotations of a square act exactly the same under modular addition, modular multiplication, and function composition respectively. Let's remind o... | (1) Comparing \( {\mathbb{Z}}_{4} \) and \( \langle i\rangle \), notice that if we take the Cayley table for \( {\mathbb{Z}}_{4} \) and make the following replacements:\n\n\[ 0 \rightarrow 1\;1 \rightarrow i\;2 \rightarrow - 1\;3 \rightarrow - i, \]\n\nthen the result exactly matches the Cayley table for \( \langle i\r... | Yes |
Proposition 17.2.5. Given that \( \phi : G \rightarrow H \) is an isomorphism, then \( \phi \) takes the identity to the identity: that is, if \( e \) is the identity of \( G \), then \( \phi \left( e\right) \) is the identity of \( H \) (see Figure 17.2.1). | Exercise 17.2.6. Fill in the blanks in the following proof of Proposition 17.2.5:\n\nGiven that \( e \) is the identity of \( < 1 > \) and \( h \) is an arbitrary element of \( < 2 > \) . Since \( \phi \) is a bijection, then there exists \( g \in < 3 > \) such that \( \phi \left( { < 4 > }\right) = h \) . Then we have... | No |
Proposition 17.2.7. Given that \( \phi : G \rightarrow H \) is an isomorphism, then \( \phi \) preserves the operation of inverse: that is, for any \( g \in G \) we have\n\n\[ \phi \left( {g}^{-1}\right) = {\left( \phi \left( g\right) \right) }^{-1}. \] | Exercise 17.2.8. Fill in the blanks in the following proof of Proposition 17.2.7:\n\nLet \( e \) and \( f \) be the identities of \( G \) and \( H \), respectively. Given that \( g \in \) <1 > , we have:\n\n\[ \phi \left( g\right) \circ \phi \left( {g}^{-1}\right) = \phi \left( {g \cdot {g}^{-1}}\right) \]\n(definition... | No |
Proposition 17.2.10. Isomorphism is an equivalence relation on groups. | Exercise 17.2.11. Prove Proposition 17.2.10. (*Hint*) | No |
When elements of \( {2}^{\mathbb{Z}} \) are multiplied together, their exponents add: we know this from basic algebra. This suggests there should be an isomorphism between \( \mathbb{Z} \) and \( {2}^{\mathbb{Z}} \) . In fact, we may define the function \( \phi : \mathbb{Z} \rightarrow {2}^{\mathbb{Z}} \) by \( \phi \l... | (a) We may compute\n\n\[ \phi \left( {m + n}\right) = {2}^{m + n} = {2}^{m}{2}^{n} = \phi \left( m\right) \phi \left( n\right) . \]\n\n(b) By definition the function \( \phi \) is onto the subset \( \left\{ {{2}^{n} : n \in \mathbb{Z}}\right\} \) of \( {\mathbb{Q}}^{ * } \) . To show that the map is injective, assume t... | Yes |
Consider the groups \( {\mathbb{Z}}_{8} \) and \( {\mathbb{Z}}_{12} \). Can you tell right away that there can't be an isomorphism between them? Remember, an isomorphism is a one-to-one and onto function: but since \( \left| {\mathbb{Z}}_{12}\right| > \left| {\mathbb{Z}}_{8}\right| \) there is no onto function from \( ... | Similarly it can be shown that any two finite groups that have differing numbers of elements can't be isomorphic to each other. | Yes |
The following are the Cayley tables for \( {\mathbb{Z}}_{4} \) and \( U\left( 5\right) \). | Notice that the main diagonals (left to right) of the Cayley tables seem to have a different pattern. The main diagonal for \( {\mathbb{Z}}_{4} \) is the alternating sequence, \( 0,2,0,2 \), while the main diagonal of \( U\left( 5\right) \) is the non-alternating sequence \( 1,4,4,1 \) . It appears at first sight that ... | Yes |
Consider the group of units of \( {\mathbb{Z}}_{8} \) and the group of units of \( {\mathbb{Z}}_{12} \) ; i.e. \( U\left( 8\right) \) and \( U\left( {12}\right) \) . We’ve seen that these consist of the elements in \( {\mathbb{Z}}_{8} \) and \( {\mathbb{Z}}_{12} \), that are relatively prime to 8 and 12, respectively, ... | An isomorphism \( \phi : U\left( 8\right) \rightarrow U\left( {12}\right) \) is given by\n\n\[ 1\overset{\phi }{ \rightarrow }1 \]\n\n\[ 3\overset{\phi }{ \rightarrow }5 \]\n\n\[ 5\overset{\phi }{ \rightarrow }7 \]\n\n\[ 7\overset{\phi }{ \rightarrow }{11}\text{.} \]\n\n\( \phi \) is one-to-one and onto by observation,... | Yes |
Proposition 17.3.20. Given a Cayley table \( T \) for a finite group \( G \), and let \( g \in G \) appears on the diagonal of \( T \) . Let \( \phi : G \rightarrow H \) be an isomorphism, and let \( {T}^{\prime } \) be a Cayley table of \( H \) . Then \( \phi \left( g\right) \) appears on the diagonal of \( {T}^{\prim... | Proof. As stated above, \( g \) appears on the diagonal of \( T \) if and only if there exists \( {g}^{\prime } \in G \) such that \( {g}^{\prime } \cdot {g}^{\prime } = g \) . Since \( \phi \) is an isomorphism, this implies \( \phi \left( {g}^{\prime }\right) \cdot \phi \left( {g}^{\prime }\right) = \phi \left( g\rig... | Yes |
Even though \( {D}_{3} \) and \( {\mathbb{Z}}_{6} \) possess the same number of elements, we might suspect that they are not isomorphic, because \( {\mathbb{Z}}_{6} \) is abelian and \( {D}_{3} \) is non-abelian. | So, suppose that the two groups are isomorphic, which means there exists an isomorphism \( \phi : {\mathbb{Z}}_{6} \rightarrow {D}_{3} \) . Let \( a, b \in {D}_{3} \) be two elements such that \( a \circ b \neq b \circ a \) . Since \( \phi \) is an isomorphism, there exist elements \( m \) and \( n \) in \( {\mathbb{Z}... | Yes |
Proposition 17.3.33. \( \mathbb{Z} \) is not isomorphic to \( \mathbb{R} \) . | Proof. We will use a proof by contradiction. Suppose that there exists an isomorphism \( \phi : \mathbb{Z} \rightarrow \mathbb{R} \) . Choose any \( x \in \mathbb{R} \), and let \( m \in \mathbb{Z} \) be the pre-image of \( x \), so that \( \phi \left( m\right) = x \) . It follows that:\n\n\[ x = \phi \left( m\right) =... | Yes |
Proposition 17.3.38. Let \( \phi : G \rightarrow H \) be an isomorphism of two groups. Then the following statements are true.\n\n(1) \( \left| G\right| = \left| H\right| \) . | Proof. Assertion (1) follows from the fact that \( \phi \) is a bijection. | Yes |
Proposition 17.4.1. If \( G \) is a cyclic group of infinite order, then \( G \) is 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 |
Proposition 17.4.3. 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 left as the next exercise. | No |
The Cayley table for \( {\mathbb{Z}}_{3} \) is | The addition table of \( {\mathbb{Z}}_{3} \) suggests that it is the isomorphic to the permutation group \( \{ \) id \( ,\left( {012}\right) ,\left( {021}\right) \} \) . One possible isomorphism is\n\n\[ 0 \mapsto \left( \begin{array}{lll} 0 & 1 & 2 \\ 0 & 1 & 2 \end{array}\right) = \mathrm{{id}} \]\n\n\[ 1 \mapsto \le... | Yes |
Proposition 17.4.12. (Cayley's theorem) Every finite group is isomorphic to a group of permutations. | Proof. Let \( G \) be a group with \( \left| G\right| \) elements. We seek a group of permutations \( P \subset {S}_{\left| G\right| } \) that is isomorphic to \( G \) . For any \( g \in G \) we may define a function \( {\phi }_{g} : G \rightarrow G \) by\n\n\[ \n{\phi }_{g}\left( a\right) \mathrel{\text{:=}} {ga}.\n\]... | No |
Proposition 17.5.1. Let \( G \) and \( H \) be groups. The set \( G \times H \) is a group under the operation \( \left( {{g}_{1},{h}_{1}}\right) \left( {{g}_{2},{h}_{2}}\right) = \left( {{g}_{1}{g}_{2},{h}_{1}{h}_{2}}\right) \) where \( {g}_{1},{g}_{2} \in G \) and \( {h}_{1},{h}_{2} \in H \) . | The proof is outlined in the following exercise.\n\n## Exercise 17.5.2.\n\n(a) Show that the set \( G \times H \) is closed under the binary operation defined in Proposition 17.5.1.\n\n(b) Show that \( \left( {{e}_{G},{e}_{H}}\right) \) is the identity of \( G \times H \), where \( {e}_{G} \) and \( {e}_{H} \) are the ... | No |
Consider\n\n\[ \n{\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} = \{ \left( {0,0}\right) ,\left( {0,1}\right) ,\left( {1,0}\right) ,\left( {1,1}\right) \} .\n\]\n\nAlthough \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \) and \( {\mathbb{Z}}_{4} \) both contain four elements, they are not isomorphic. We can prove this by notin... | So we've shown that the direct product of cyclic groups is not necessarily cyclic. | Yes |
Proposition 17.5.19. Let \( {G}_{1},{G}_{2},\ldots {G}_{n} \) be arbitrary groups, and let \( \sigma \in \) \( {S}_{n} \) be any permutation on \( \{ 1,2,\ldots n\} \) . Then\n\n\[ \n{G}_{1} \times {G}_{2} \times \ldots \times {G}_{n} \cong {G}_{\sigma \left( 1\right) } \times {G}_{\sigma \left( 2\right) } \times \ldot... | The following exercise outlines the proof of Proposition 17.5.19.\n\nExercise 17.5.20.\n\n(a) What function would you define in order to prove Proposition 17.5.19?\n\n(b) Prove that the function that you defined in (a) is a bijection by showing that it has an inverse.\n\n(c) Prove that the function that you defined in ... | No |
Proposition 17.5.23. Let \( \left( {g, h}\right) \in G \times H \) . If \( g \) and \( h \) have finite orders \( r \) and \( s \) respectively, then the order of \( \left( {g, h}\right) \) in \( G \times H \) is the least common multiple of \( r \) and \( s \) . | Proof. Suppose that \( m \) is the least common multiple of \( r \) and \( s \) and let \( n = \left| \left( {g, h}\right) \right| \) . Then\n\n\[ \n{\left( g, h\right) }^{m} = \left( {{g}^{m},{h}^{m}}\right) = \left( {{e}_{G},{e}_{H}}\right) \n\] \n\n\[ \n\left( {{g}^{n},{h}^{n}}\right) = {\left( g, h\right) }^{n} = \... | Yes |
Proposition 17.5.24. Let \( \\left( {{g}_{1},\\ldots ,{g}_{n}}\\right) \\in \\mathop{\\prod }\\limits_{{i = 1}}^{n}{G}_{i} \) . If \( {g}_{i} \) has finite order \( {r}_{i} \) in \( {G}_{i} \), then the order of \( \\left( {{g}_{1},\\ldots ,{g}_{n}}\\right) \) is the least common multiple of \( {r}_{1},\\ldots ,{r}_{n}... | ## Exercise 17.5.25. Prove Proposition 17.5.24 using induction. | No |
Let \( \left( {8,{56}}\right) \in {\mathbb{Z}}_{12} \times {\mathbb{Z}}_{60} \). Since \( \gcd \left( {8,{12}}\right) = 4 \), the order of 8 is \( {12}/4 = 3 \) in \( {\mathbb{Z}}_{12} \). Similarly, the order of 56 in \( {\mathbb{Z}}_{60} \) is 15 . The least common multiple of 3 and 15 is 15 ; hence, \( \left( {8,{56... | The least common multiple of 3 and 15 is 15 ; hence, \( \left( {8,{56}}\right) \) has order 15 in \( {\mathbb{Z}}_{12} \times {\mathbb{Z}}_{60} \). \( \diamond \) | Yes |
The group \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \) consists of the pairs\n\n\[ \left( {0,0}\right) ,\;\left( {0,1}\right) ,\;\left( {0,2}\right) ,\;\left( {1,0}\right) ,\;\left( {1,1}\right) ,\;\left( {1,2}\right) . \]\n\nIn this case, unlike that of \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \) and \( {\mathbb{... | We need only show that \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \) is cyclic. By trial and error, we may find that \( \left( {1,1}\right) \) is a generator for \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \), so that \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} = \langle \left( {1,1}\right) \rangle \). | Yes |
Proposition 17.5.30. The group \( {\mathbb{Z}}_{m} \times {\mathbb{Z}}_{n} \) is isomorphic to \( {\mathbb{Z}}_{mn} \) if and only if \( \gcd \left( {m, n}\right) = 1 \) . | Proof. Assume first that if \( {\mathbb{Z}}_{m} \times {\mathbb{Z}}_{n} \cong {\mathbb{Z}}_{mn} \), then \( \gcd \left( {m, n}\right) = 1 \) . To show this, we will prove the contrapositive; that is, we will show that if \( \gcd \left( {m, n}\right) = d > 1 \), then \( {\mathbb{Z}}_{m} \times {\mathbb{Z}}_{n} \) cannot... | Yes |
Proposition 17.5.31. Let \( {n}_{1},\ldots ,{n}_{k} \) be positive integers. Then\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{k}{\mathbb{Z}}_{{n}_{i}} \cong {\mathbb{Z}}_{{n}_{1}\cdots {n}_{k}} \]\n\nif and only if \( \operatorname{lcm}\left( {{n}_{1},\ldots ,{n}_{k}}\right) = \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \)... | Proof. Use the argument in Proposition 17.5.30 first with \( {n}_{1} \) and \( {n}_{2} \), then with \( {n}_{1}{n}_{2} \) and \( {n}_{3} \), then with \( {n}_{1}{n}_{2}{n}_{3} \) and \( {n}_{4} \), and so on. (The best way to do this proof is using induction.) | No |
Corollary 17.5.33. If\n\n\[ m = {p}_{1}^{{e}_{1}} \\cdots {p}_{k}^{{e}_{k}} \]\n\nwhere the \( {p}_{i} \)’s are distinct primes, then\n\n\[ {\\mathbb{Z}}_{m} \\cong {\\mathbb{Z}}_{{p}_{1}^{{e}_{1}}} \\times \\cdots \\times {\\mathbb{Z}}_{{p}_{k}^{{e}_{k}}} \] | Proof. Since \( \\gcd \\left( {{p}_{i}^{{e}_{i}},{p}_{j}^{{e}_{j}}} \\right) = 1 \) for \( i \\neq j \), the proof follows from the Corollary to 17.5.31. | Yes |
All abelian groups \( G \) of order 21 are isomorphic and cyclic. | This is because \( {21} = {3}^{1} \times {7}^{1} \), so by Proposition 17.5.35 it must be the case that \( G \cong {\mathbb{Z}}_{3} \times {\mathbb{Z}}_{7} \) . In particular, \( {Z}_{21} \cong {\mathbb{Z}}_{3} \times {\mathbb{Z}}_{7} \) . So all groups of order 21 are isomorphic to the cyclic group | Yes |
Any group \( G \) of order 54 must have an element of order 3. | This is because \( {54} = {3}^{3} \cdot 2 \), and according to Proposition 17.5.35 it must be the case that either \( G \cong {\mathbb{Z}}_{27} \times {\mathbb{Z}}_{2} \), or \( G \cong {\mathbb{Z}}_{9} \times {\mathbb{Z}}_{3} \times {\mathbb{Z}}_{2} \), or \( G \cong {\mathbb{Z}}_{3} \times {\mathbb{Z}}_{3} \times {\m... | Yes |
Suppse that \( \left| G\right| = {125} \). Let us show that \( G \) has a subgroup of order 25. | From Proposition 17.5.35, we know that there are 3 possible cases for \( G \) : (i) \( G \cong {\mathbb{Z}}_{125} \) ; (ii) \( G \cong {\mathbb{Z}}_{25} \times {\mathbb{Z}}_{5} \) ; (iii) \( G \cong {\mathbb{Z}}_{5} \times {\mathbb{Z}}_{5} \times {\mathbb{Z}}_{5} \). In case (i), then the isomorphic image of (5) is a s... | Yes |
Example 17.5.47. Suppse that \( {7}^{2} \) divides \( \left| G\right| \) . Let us show that \( G \) has a subgroup of order 49. | From Proposition 17.5.35, we know that either(i) \( G \) has a factor \( {\mathbb{Z}}_{{7}^{k}} \) where \( k \geq 2 \) ; or (ii) \( G \) has at least two factors of \( {\mathbb{Z}}_{7} \) . In case (i), then \( G \) can be written as \( {\mathbb{Z}}_{{7}^{k}} \times H \), where \( k \geq 2 \) and \( H \) is a direct p... | No |
Proposition 17.6.3. \( U\left( p\right) \) is cyclic for every prime \( p \) . | Proof. First, notice that Proposition 9.6.18 says that there are at most \( m \) solutions to the equation \( {x}^{m} = 1 \) in \( {\mathbb{Z}}_{p} \) . Since 0 is not a solution, it follow that all of these solutions are also in \( U\left( p\right) \) .\n\nAlso, according to the factorization of Abelian groups (Propos... | No |
The dihedral group \( {D}_{6} \) is an internal direct product of its two subgroups\n\n\[ H = \left\{ {\mathrm{{id}},{r}^{3}}\right\} \;\text{ and }\;K = \left\{ {\mathrm{{id}},{r}^{2},{r}^{4}, s,{r}^{2}s,{r}^{4}s}\right\} . \] | It can be shown that \( K \cong {S}_{3} \) ; consequently, \( {D}_{6} \cong {\mathbb{Z}}_{2} \times {S}_{3} \) . | No |
Proposition 17.6.10. Let \( G \) be the internal direct product of subgroups \( H \) and \( K \) . Then \( G \) is isomorphic to \( H \times K \) . | Proof. Since \( G \) is an internal direct product, we can write any element \( g \in G \) as \( g = {hk} \) for some \( h \in H \) and some \( k \in K \) . Define a map \( \phi : G \rightarrow \) \( H \times K \) by \( \phi \left( g\right) = \left( {h, k}\right) \) . The first problem that we must face is to show that... | No |
Let’s try to create a function from \( {\mathbb{Z}}_{8} \) to \( \langle i\rangle \) which preserves group operations. Since there are twice as many elements in \( {\mathbb{Z}}_{8} \) as in \( \langle i\rangle \), it seems natural that 2 elements from \( {\mathbb{Z}}_{8} \) should each go to one element in \( \langle i... | Let’s take the \( {\mathbb{Z}}_{8} \) table then and start transforming it according to \( g \) . First we replace all the elements of \( {\mathbb{Z}}_{8} \) with their counterparts in \( \langle i\rangle \) :\n\n<table><thead><tr><th>\( {}^{ \oplus } \)</th><th>1</th><th>\( i \)</th><th>\( - 1 \)</th><th>\( - i \)</th... | Yes |
Does cis preserve the operations between \( \mathbb{R} \) and \( \mathbb{T} \) ? | \[ \operatorname{cis}\left( {\alpha + \beta }\right) = \cos \left( {\alpha + \beta }\right) + i\sin \left( {\alpha + \beta }\right) \] \[ = \left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta }\right) + i\left( {\sin \alpha \cos \beta + \cos \alpha \sin \beta }\right) \] \[ = \left( {\cos \alpha + i\sin \alpha }\ri... | Yes |
The circle group \( \mathbb{T} \) also gives us a completely different way of constructing a homomorphism between complex and real numbers. Every complex number in \( \mathbb{T} \) has modulus 1; i.e. they lie all on a circle of radius 1 in the complex plane. If we increase radius of the circle to 2, all of those compl... | If \( {r}_{1}\operatorname{cis}{\theta }_{1} \) and \( {r}_{2}\operatorname{cis}{\theta }_{2} \) are arbitrary nonzero complex numbers, we have: \[ f\left( {\left( {{r}_{1}\operatorname{cis}{\theta }_{1}}\right) \cdot \left( {{r}_{2}\operatorname{cis}{\theta }_{2}}\right) }\right) = f\left( {{r}_{1}\operatorname{cis}{\... | Yes |
Proposition 18.3.1. Let \( f : G \rightarrow H \) be a homomorphism of groups. Then\n\n1. If \( e \) is the identity of \( G \), then \( f\left( e\right) \) is the identity of \( H \) ;\n\n2. For any element \( g \in G, f\left( {g}^{-1}\right) = {\left\lbrack f\left( g\right) \right\rbrack }^{-1} \) ;\n\n3. If \( S \) ... | Proof.\n\n(1) Suppose that \( e \) and \( {e}^{\prime } \) are the identities of \( G \) and \( H \), respectively. Then\n\n\[ \n{e}^{\prime }f\left( e\right) = f\left( e\right) = f\left( {ee}\right) = f\left( e\right) f\left( e\right) .\n\]\n\nBy cancellation, \( f\left( e\right) = {e}^{\prime } \).\n\n(2) This statem... | Yes |
Proposition 18.3.4. Let \( f : G \rightarrow H \) be a group homomorphism. Then the kernel of \( f \) is a normal subgroup of \( G \) . | Exercise 18.3.5. Prove Proposition 18.3.4. (*Hint*) \( \diamond \) | No |
Example 18.3.9. Suppose that we wish to determine all possible homomorphisms \( f \) from \( {\mathbb{Z}}_{7} \) to \( {\mathbb{Z}}_{12} \) . Since the kernel of \( f \) must be a subgroup of \( {\mathbb{Z}}_{7} \), there are only two possible kernels, \( \{ 0\} \) and all of \( {\mathbb{Z}}_{7} \) . | The image of a subgroup of \( {\mathbb{Z}}_{7} \) must be a subgroup of \( {\mathbb{Z}}_{12} \) . Hence, there is no injective homomorphism; otherwise, \( {\mathbb{Z}}_{12} \) would have a subgroup of order 7, which is impossible. Consequently, the only possible homomorphism from \( {\mathbb{Z}}_{7} \) to \( {\mathbb{Z... | Yes |
Let \( G \) be a cyclic group with generator \( g \) . Define a map \( f : \mathbb{Z} \rightarrow G \) by \( n \mapsto {g}^{n} \) . This map is a surjective homomorphism since | \[ f\left( {m + n}\right) = {g}^{m + n} = {g}^{m}{g}^{n} = f\left( m\right) f\left( n\right) . \] Clearly \( f \) is onto. If \( \left| g\right| = m \), then \( {g}^{m} = e \) . Hence, \( \ker f = m\mathbb{Z} \) and \( \mathbb{Z}/\ker f = \mathbb{Z}/m\mathbb{Z} \cong G \) . On the other hand, if the order of \( g \) is... | Yes |
To investigate this idea of compatibility let's compare \( \left\lbrack {\left( {AB}\right) \left( {ACB}\right) }\right\rbrack .C \) and \( \left( {AB}\right) .\left\lbrack {\left( {ACB}\right) .C}\right\rbrack \) where \( \left( {AB}\right) ,\left( {ACB}\right) \in {S}_{3} \) and \( B \in \) \( X = \{ A, B, C\} \) . | So, \( \left\lbrack {\left( {AB}\right) \left( {ACB}\right) }\right\rbrack \cdot C = A = \left( {AB}\right) \cdot \left\lbrack {\left( {ACB}\right) \cdot C}\right\rbrack \), therefore, we can see compatibility in this specific case. Compatibility can be easily shown to hold for the other elements of \( X \) because \( ... | Yes |
Example 19.1.4. Recall \( R \) and \( X \) from Example 19.1.2, show that \( \left( {{r}_{40} \circ }\right. \) \( \left. {r}_{30}\right) .{x}_{15} = {r}_{40}.\left( {{r}_{30}.{x}_{15}}\right) \) . | \[ \left( {{r}_{40} \circ {r}_{30}}\right) \cdot {x}_{15}\overset{?}{ = }{r}_{40} \cdot \left( {{r}_{30} \cdot {x}_{15}}\right) \] \[ {r}_{70}.{x}_{15}\overset{?}{ = }{r}_{40}.{x}_{45} \] \[ {x}_{85} = {x}_{85} \] | Yes |
Consider \( G{L}_{2}\left( \mathbb{R}\right) \) (the group of invertible \( 2 \times 2 \) matrices) and \( {\mathbb{R}}^{2} \) . Show that \( G{L}_{2}\left( \mathbb{R}\right) \) acts on \( {\mathbb{R}}^{2} \) by left multiplication on vectors which means that \( {\mathbb{R}}^{2} \) is a \( G{L}_{2}\left( \mathbb{R}\rig... | To check we must show identity and compatibility:\n\n(1) Check identity: If \( v \in {\mathbb{R}}^{2} \) and \( I \) is the identity matrix, then \( I.v = v \) .\n\n(2) Check compatibility: If \( A \) and \( B \) are \( 2 \times 2 \) invertible matrices, then \( \left( {AB}\right) .v = A.\left( {B.v}\right) \) (see Exe... | No |
Example 19.1.11. Let \( G \) be a group and \( {\mathcal{E}}_{n} \) be the set of all subsets of \( G \) with \( n \) elements where \( n \) is a positive integer and \( n \leq \left| G\right| \) . Let \( S \in {\mathcal{E}}_{n} \) , meaning \( S \) is a subset of \( G \) with \( n \) elements. Then \( G \) acts on \( ... | (1) Check the identity condition: \( e.S = \{ {es} \mid s \in S\} = S \)\n\n(2) Check the compatibility condition: Let \( g, h \in G \), then \( \left( {gh}\right) .S = \left\{ {\left( {gh}\right) s \mid s \in }\right. \) \( S\} = \{ g\left( {hs}\right) \mid s \in S\} = g.\left( {h.S}\right) \)\n\nParts (1) and (2) ver... | Yes |
Proposition 19.2.5. Let \( X \) be a \( G \) -set. Then \( G \) -equivalence is an equivalence relation on \( X \) . | Proof. The relation \( \sim \) is reflexive since \( {ex} = x \) . Suppose that \( x \sim y \) for \( x, y \in X \) . Then there exists a \( g \) such that \( g.x = y \) . In this case \( {g}^{-1}.y = x \) ; hence, \( y \sim x \) . To show that the relation is transitive, suppose that \( x \sim y \) and \( y \sim z \) ... | No |
Let \( G \) be the permutation group defined by\n\n\[ G = \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) ,\left( {45}\right) ,\left( {123}\right) \left( {45}\right) ,\left( {132}\right) \left( {45}\right) \} \]\n\nand \( X = \{ 1,2,3,4,5\} \) . Then \( X \) is a \( G \) -set. There are permutations in \( ... | So the orbits are \( \{ 1,2,3\} \) and \( \{ 4,5\} \). | Yes |
Let \( G \) be the rotational symmetries of a cube and \( X = \{ \) faces \( \} \cup \{ \) edges \( \} \cup \{ \) vertices \( \} \) . The fixed point set of id is: | \[ {X}_{\mathrm{{id}}} = \{ \text{ faces }\} \cup \{ \text{ edges }\} \cup \{ \text{ vertices }\} ,\] since the identity rotation leaves the entire cube unchanged. | Yes |
Let's consider the stabilizer subgroups for the faces of a cube. These contain the elements of the group \( G \) of rotations of the cube that leave each face unchanged. | \[ {G}_{{x}_{ + }} = {G}_{{x}_{ - }} = \left\{ {\mathrm{{id}},{\mathrm{r}}_{\mathrm{x}},{\mathrm{r}}_{\mathrm{x}}^{2},{\mathrm{r}}_{\mathrm{x}}^{3}}\right\} \] \[ {G}_{{y}_{ + }} = {G}_{{y}_{ - }} = \{ \mathrm{{id}},{\mathrm{r}}_{\mathrm{y}},{\mathrm{r}}_{\mathrm{y}}^{2},{\mathrm{r}}_{\mathrm{y}}^{3}\} \] \[ \text{....... | No |
Consider our old friend the group of rotational symmetries of a cube acting on \( X = \{ \) faces \( \} \cup \{ \) edges \( \} \cup \{ \) vertices \( \} \) . We’ve seen that \( {G}_{{x}_{ + }} = \left\{ {\mathrm{{id}},{\mathrm{r}}_{\mathrm{x}},{\mathrm{r}}_{\mathrm{x}}^{2},{\mathrm{r}}_{\mathrm{x}}^{3}}\right\} \) is t... | \[ \left| G\right| = \left| {G}_{{x}_{ + }}\right| \cdot \left| {\mathcal{O}}_{{x}_{ + }}\right| \] | Yes |
Proposition 19.2.20. (Counting formula): Let \( G \) be a group and \( X \) a \( G \) -set. If \( x \in X \), then \( \left| {\mathcal{O}}_{x}\right| = \left\lbrack {G : {G}_{x}}\right\rbrack \) . | Proof. In general, a good way to show that two sets are the same size is to show that there is a bijection (1-1 and onto map) between the two sets. We will define a map \( \phi \) between the orbit \( {\mathcal{O}}_{x} \) and the set of left cosets of \( {G}_{x} \) in \( G \) as follows. Let \( y \in {\mathcal{O}}_{x} ... | Yes |
Consider the group \( {\mathbb{Z}}_{5} \) which as we know is a group under addition. We will show that \( {\mathbb{Z}}_{5} \) is a \( {\mathbb{Z}}_{5} \) -set where the group action is \( g.x \mathrel{\text{:=}} g + x \). | To do this we must show the identity and compatibility conditions. First, \( 0.x = 0 + x = x \) for all \( x \in {\mathbb{Z}}_{5} \), so the identity condition is met. Secondly, we need to show compatibility: by the associative property of addition in \( {\mathbb{Z}}_{5} \), we have \( \left( {{g}_{1} + {g}_{2}}\right)... | Yes |
Proposition 19.3.6. If \( G \) is a group, and \( H \) is a subgroup of \( G \), then \( G \) is an \( H \) -set using the definition \( h.g \mathrel{\text{:=}} {hg} \) . | Exercise 19.3.7. Prove the above proposition. | No |
Example 19.3.9. Let \( L \) be the set of left cosets of \( {2\pi }\mathbb{Z} \) in the group \( \left( {\mathbb{R}, + }\right) \) . Recall from Definition 15.1.4 in Chapter 15 that the set of left cosets \( L \) is defined as \( x + {2\pi }\mathbb{Z} = \{ x + h : h \in {2\pi }\mathbb{Z}\} \) . For example, the left co... | (a) For the identity condition note that \( e \in G = 0 \) . Then, \( 0 + x + {2\pi }\mathbb{Z} = \) \( x + {2\pi }\mathbb{Z} \) for any \( x + {2\pi }\mathbb{Z} \in L \) . So the identity condition is true.\n\n(b) For the compatibility condition consider two real numbers \( a, b \) . Then, by associativity of real num... | Yes |
Exercise 19.3.12. In the following you will show that there is a bijection between cosets of the form \( a + H \), where \( a \in {\mathbb{R}}^{2} \) and points of the unit square. | (a) Let \( \left( {m, n}\right) \) be the lower left point of the lattice square which contains the point \( a \) . Using the floor function, give expressions for \( m \) and \( n \) .\n\n(b) Show that \( a + \left( {-m, - n}\right) \) is inside the unit square. This implies that \( a + H \) contains at least one point... | No |
Example 19.3.13. Continuing from Example 19.3.11: let \( b = \left( {{0.8},{0.3}}\right) \) , where \( b \in G \) . Find the point \( h = \left( {m, n}\right) \in H \) such that \( b + a + h \) is inside the unit square (recall \( a = \left( {{0.7},{0.5}}\right) \) ). | The element \( b \) acts on the coset \( a + H \) as follows:\n\n\[ b + a + H = \{ \left( {{0.8} + \left( {{0.7} + m}\right) ,{0.3} + \left( {{0.5} + n}\right) }\right) : m, n \in \mathbb{Z}\} \]\n\n\[ = \{ \left( {{0.8} + {0.7} + m,{0.3} + {0.5} + n}\right) : m, n \in \mathbb{Z}\} \]\n\n\[ = \{ \left( {{1.5} + m,{0.8}... | Yes |
Consider group elements \( a = \left( {-{0.6}, - {0.4}}\right) \) and \( b = \left( {{0.9},{1.6}}\right) \) where \( a, b \in {\mathbb{R}}^{2} \). Let’s first find the point \( h = \left( {m, n}\right) \in H \) such that \( b + a + H \) is inside the unit square. | From Exercise 19.3.14, \( m = - \lfloor - {0.6} + {0.9}\rfloor = - \lfloor {0.3}\rfloor = 0 \) and \( n = - \lfloor - {0.4} + {1.6}\rfloor = - \lfloor {1.2}\rfloor = - 1 \), so \( h = \left( {0, - 1}\right) \n\nNext, let’s find the point \( b + a + h \) in the unit square, let’s call this point c.\n\n\[ c = \left( {{0.... | Yes |
Let \( H \) be a subgroup of \( G \) and \( {L}_{H} \) the set of left cosets of \( H \) . The set \( {L}_{H} \) is a \( G \) -set under the action \( g.\left( {xH}\right) = \left( {gx}\right) H \) (note that \( {gx} \in G \) so \( \left( {gx}\right) H \) is a coset of \( H \) ). | Again, it is easy to see that the identity condition is true. Since \( \left( {g{g}^{\prime }}\right) \cdot \left( {xH}\right) = \left( {g{g}^{\prime }x}\right) H = g \cdot \left( {\left( {{g}^{\prime }x}\right) H}\right) = g \cdot \left( {{g}^{\prime } \cdot \left( {xH}\right) }\right) \) , the compatibility condition... | No |
Let \( \sigma = \left( {153}\right) \left( {276}\right) \) and \( \tau = \left( {427}\right) \left( {165}\right) \) . Then | \[ \tau = \left( \begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 7 & 3 & 2 & 1 & 5 & 4 \end{array}\right) \] Relabeling \( \sigma \) according to \( \tau \) gives the conjugate (136)(457). You can check that computing \( {\tau \sigma }{\tau }^{-1} \) will give the same result as the relabeling method. \( \black... | Yes |
Let \( \sigma = \left( {12}\right) \left( {3456}\right) \left( {789}\right) ,\mu = \left( {149}\right) \left( {2658}\right) \left( {37}\right) \). Notice that \( \sigma \) becomes \( \mu \) if we use the following relabeling: | \[ 1 \rightarrow 3;\;2 \rightarrow 7;\;3 \rightarrow 2;\;4 \rightarrow 6;\;5 \rightarrow 5;\;6 \rightarrow 8;\;7 \rightarrow 1;\;8 \rightarrow 4;\;9 \rightarrow 9. \] We can use this information to write \( \tau \) in tableau notation: \[ \tau = \left( \begin{array}{lllllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 3 & 7... | Yes |
Proposition 19.4.6. Given a permutation group \( G \), and two permutations \( \sigma ,\mu \in G \). Then \( \sigma \) and \( \mu \) are conjugate if and only if they have exactly the same cycle structure. | Proof. The \ | No |
Let \( G \) be the dihedral group \( {D}_{4} \). Recall that \( {D}_{4} \) consists of four rotations and four reflections. In fact we can write \( {D}_{4} = \left\{ {e, r,{r}^{2},{r}^{3}, s, s \circ r, s \circ {r}^{2}, s \circ {r}^{3}}\right\} \), where \( r \) is counterclockwise rotation by \( {90}^{ \circ } \), and... | For example, consider the case \( h = s \) and \( g = r \). Then \( \left( {s, r}\right) \rightarrow s \circ r \circ {s}^{-1} \). We can simplify this, since \( s \) is a reflection, so \( {s}^{-1} = s \). furthermore, by part c of Proposition 10.4.15 in Section 10.4, we can show \( r \circ s = s \circ {r}^{3} \). This... | Yes |
Proposition 19.4.10. If \( H \) is a subgroup of \( G \), then \( G \) is an \( H \) -set under conjugation. That is, we can define an action \( H \times G \rightarrow G \), by \( h.g = {hg}{h}^{-1} \) for \( h \in H \) and \( g \in G \) . | The proof is contained in the following exercise.\n\nExercise 19.4.11. Fill in the blanks to prove the proposition:\n\nFirst, we have that \( < 1 > \) is in \( H \) and \( e.g = < 2 > g < 3 > = g \) . So the identity condition for a group action holds.\n\nAlso, observing that\n\n\[ \left( {{h}_{1}{h}_{2}}\right) \cdot ... | No |
Let \( g = {r}_{x}^{2} \) then consider \( {r}_{x}^{2} \circ {r}_{z} \circ {r}_{x}^{-2} \). What will this rotation do? | First \( {r}_{x}^{-2} \) will take the top face to the bottom face and vice versa. Then \( {r}_{z} \) will rotate the face \( {z}_{ - } \) (which is now on top) \( {90}^{ \circ } \) counterclockwise and \( {z}_{ + } \) (which is now on the bottom) \( {90}^{ \circ } \) clockwise. Then \( {r}_{x}^{2} \) will rotate \( {z... | Yes |
Example 19.4.15. \( {r}_{y} \circ {r}_{z} \) is a 120 degree stabilizer of vertex \( + + + \) . Consider the conjugation of \( {r}_{y} \circ {r}_{z} \) by the group element \( {r}_{y} \), that is, \( {r}_{y} \circ \left( {{r}_{y} \circ }\right. \) \( \left. {r}_{z}\right) \circ {r}_{y}^{-1} \) . | First, \( {r}_{y}^{-1} \) takes \( + + - \) to \( + + + \) . Then \( {r}_{y} \circ {r}_{z} \) rotates \( + + - \) 120 degrees counterclockwise. Then \( {r}_{y} \) rotates \( + + - \) back to its original place. The net result is a 120 degree counterclockwise rotation of the vertex \( + + - \) . | No |
The rotation \( {r}_{z}^{2} \circ {r}_{y}^{-1} \) stabilizes the edge \( \overline{{x}_{ - }{z}_{ - }} \) . It’s a 180 degree rotation about an axis through this edges \( \overline{{x}_{ - }{z}_{ - }} \) and \( \overline{{x}_{ + }{z}_{ + }} \) . Consider the conjugate \( {r}_{z}^{2} \circ \left( {{r}_{z}^{2} \circ {r}_... | First, \( {r}_{z}^{-2} \) takes \( \overline{{x}_{ + }{z}_{ - }} \) to \( \overline{{x}_{ - }{z}_{ - }} \) . Then \( \left( {{r}_{z}^{2} \circ {r}_{y}^{-1}}\right) \) rotates about the axis through \( \overline{{x}_{ + }{z}_{ - }}{180} \) degrees, switching the two faces. Then \( {r}_{z}^{2} \) rotates \( \overline{{x}... | Yes |
Proposition 19.4.19. Let \( G \) be a group, \( g \in G \), and \( \widetilde{g} \) is conjugate to \( g \) . Then \( \left| g\right| = \left| \widetilde{g}\right| \) : that is, \( g \) has the same order as \( \widetilde{g} \) . | Proof. The proof is outlined in the following exercise. | No |
We can verify that the class equation correctly calculates the order of the group of rotational symmetries of a cube. | \n\\( \\left| G\\right| = \\mid \\) conjugacy class of 90 degree stabilizers of faces \\( \\mid \\)\n\n+ |conjugacy class of 180 degree stabilizers of faces|\n\n+ |conjugacy class of stabilizers of vertices|\n\n+ |conjugacy class of stabilizers of edges|\n\n+ |conjugacy class of identity|\n\n\\( = 6 + 3 + 8 + 6 + 1 \\)... | Yes |
Example 19.4.25. We can also create a conjugacy table for using permutation notation. Here is the conjugacy table for \( {S}_{3} \) using permutations. | <table><thead><tr><th>\( g \)</th><th>(1)</th><th>(123)</th><th>(132)</th><th>\( \left( {23}\right) \)</th><th>(13)</th><th>(12)</th></tr></thead><tr><td>\( g \circ \left( 1\right) \circ {g}^{-1} \)</td><td>(1)</td><td>(1)</td><td>(1)</td><td>(1)</td><td>(1)</td><td>(1)</td></tr><tr><td>\( g \circ \left( {123}\right) \... | Yes |
Proposition 20.1.19. The multiplicative identity of a ring, \( R \), is unique. | Proof. We need to show that if \( x \) is a multiplicative identity, then \( x = 1 \). \n\n\( \mathrm{x} \) is a multiplicative identity Given \n\n\[ \n x \cdot 1 = 1 \cdot x = 1 \n\] \nDefinition of Multiplicative Identity \n\n1 is a multiplicative identity Given \n\n\[ \n 1 \cdot x = x \cdot 1 = x \n\] \nDefinition o... | Yes |
Proposition 20.1.22. Given a ring, \( R \), for any \( x \in R \) we have \( x \cdot 0 = 0 \cdot x = 0 . | Proof. We will use properties of rings in our proof.\n\n\[ 0 = 0 + 0 \]\nDefiniton of Additive Identity\n\n\[ x \cdot 0 = x \cdot \left( {0 + 0}\right) \]\nSubstitution\n\n\[ x \cdot 0 = x \cdot 0 + x \cdot 0 \]\nDistributive Property\n\n\[ x \cdot 0 + - \left( {x \cdot 0}\right) = \left( {x \cdot 0 + x \cdot 0}\right)... | No |
Given two rings \( {R}_{1} \) and \( {R}_{2} \) which share the same + and - operations, show that \( {R}_{1} \cap {R}_{2} \) is a subring of both \( {R}_{1} \) and \( {R}_{2} \). | Proof. Let’s begin by showing that \( {R}_{1} \cap {R}_{2} \subset {R}_{1} \) and \( {R}_{2} \). By definition of \( \cap, a \in {R}_{1} \cap {R}_{2} \Rightarrow a \in {R}_{1} \) and \( a \in {R}_{2} \). So \( {R}_{1} \cap {R}_{2} \subset {R}_{1} \) and \( {R}_{2} \). Next, we will show the additive inverse property. L... | Yes |
Example 20.2.8. \( \{ 0,2,4\} \) is a subring of \( {\mathbb{Z}}_{6} \) without unity. | Proof. The Cayley tables for modular addition and modular multiplication of \( {\mathbb{Z}}_{6} \) are:\n\n<table><tr><td>\( \oplus \)</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>\( \odot \)</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr><tr><td>0</td><td>0</td><td>1</td><t... | Yes |
Proposition 20.3.2. \( p\\left( x\\right) = 1 + x \) has no multiplicative inverse in \( \\mathbb{Q}\\left\\lbrack x\\right\\rbrack \) . | Proof. We will show by contradiction that \( 1 + x \) has no multiplicative inverse in \( \\mathbb{Q}\\left\\lbrack x\\right\\rbrack \) . Suppose on the contrary that \( 1 + x \) has an inverse that can be written as \( q\\left( x\\right) = \\mathop{\\sum }\\limits_{{n = 0}}^{N}{a}_{n}{x}^{n} \), where \( {a}_{N} \) is... | Yes |
Find the multiplicative inverse of \( \\left( {1 + x}\\right) \) in \( \\widehat{\\mathbb{Z}}\\left\\lbrack x\\right\\rbrack \) | 1. First Method: Taylor series\n\nThe Taylor series expansion for the function \( f\\left( x\\right) \) about the point \( a = 0 \) is given by:\n\n\[ f\\left( x\\right) = f\\left( 0\\right) + {f}^{\\prime }\\left( 0\\right) x + \\frac{{f}^{\\prime \\prime }\\left( 0\\right) {x}^{2}}{2!} + \\frac{{f}^{\\prime \\prime \... | Yes |
Consider the two rings \( {R}_{1} = \mathbb{Z}\left\lbrack x\right\rbrack \) and \( {R}_{2} = \mathbb{Z}\left\lbrack y\right\rbrack \) . Obviously, these two rings are basically the same except we replace \( x \) with \( y \) . We can make a formal correspondence between the two rings by defining a function \( \phi : {... | Note that \( \phi \) is a bijection because it has an inverse (see Proposition 6.7.11). Additionally, \( \phi \) preserves the operations of addition and multiplication: \( \phi \left( {x + {}_{1}y}\right) = \phi \left( x\right) + {}_{2}\phi \left( y\right) \) and \( \phi \left( {x \cdot {}_{1}y}\right) = \phi \left( x... | No |
Consider the function \( \phi : {R}_{1} \rightarrow {R}_{2} \), where \( {R}_{1} = \{ z = a + {bi} \in \mathbb{C}\} ,{R}_{2} = \left\{ {A = \left\lbrack \begin{matrix} a & b \\ - b & a \end{matrix}\right\rbrack \in {\mathbb{M}}_{2}}\right\} \) and \( \phi \) is defined as: \( \phi \left( {a + {bi}}\right) = \left\lbrac... | To further explore the structure of the function \( \phi \), let’s consider two arbitrary elements of \( {R}_{1} \), say \( z = a + {bi} \) and \( w = c + {di} \) . Using addition of complex numbers, then applying the \( \phi \) function, we get \[ \phi \left( {z + w}\right) = \phi \left( {a + {bi} + c + {di}}\right) \... | Yes |
Show that \( f : \mathbb{C} \rightarrow \mathbb{C} \) is an isomorphism, where \( f \) maps every element of \( \mathbb{C} \) to its complex conjugate in \( \mathbb{C} \) . In other words, if \( z = a + {bi} \in \mathbb{C} \), where \( a, b \in \mathbb{R} \), then \( f\left( z\right) = a - {bi} \in \mathbb{C} \) . | Proof. In order for \( f \) to be an isomorphism, we must first show that the sets are indeed rings and that the function \( f \) is a bijection. Then we must show that the two equations above are true for all \( z, w \in \mathbb{C} \) . At the beginning of this chapter, we concluded that \( \mathbb{C} \) satisfies all... | Yes |
Show that \( f\left( {a + {bi}}\right) = a - {bi} \) is a ring automorphism from \( \mathbb{C} \) to \( \mathbb{C} \) . | We showed in Example 20.5.8 that this function is a ring isomorphism. It should be clear that the domain and range of \( f \) are the same, so \( f \) is also a ring automorphism. | No |
Define the function \( f : \mathbb{Z} \rightarrow {\mathbb{Z}}_{n} \) defined by \( f\left( x\right) = \) \( {\;\operatorname{mod}\;\left( {x, n}\right) } \), where \( n \) is a fixed integer \( > 1 \). | In Proposition 3.4.4 we showed that for any \( \ell, m \in \mathbb{Z} \) we have:\n\n(a) \( {\;\operatorname{mod}\;\left( {\ell + m, n}\right) } = {\;\operatorname{mod}\;\left( {\ell, n}\right) } \oplus {\;\operatorname{mod}\;\left( {m, n}\right) } \), and\n\n(b) \( {\;\operatorname{mod}\;\left( {\ell \cdot m, n}\right... | Yes |
Prove or disprove that \( f : \mathbb{Q} \rightarrow \mathbb{Q} \) defined by : \( f\left( x\right) = {2x} \) is a ring homomorphism. | Proof. \( f \) is not a homomorphism since it does not follow Equation 20.6.4: For example, \( f\left( {1 \cdot 1}\right) = 2\left( {1 \cdot 1}\right) = 2 \) but \( f\left( 1\right) + f\left( 1\right) = 2\left( 1\right) \cdot 2\left( 1\right) = 4 \) . Many other counterexamples can be found. | Yes |
Prove or disprove that \( f : {\mathbb{Z}}_{6} \rightarrow {\mathbb{Z}}_{3} \) defined by: \( f\left( x\right) = \) \( {\;\operatorname{mod}\;\left( {x,3}\right) } \) is a ring homomorphsm. | Proof. \( f \) is a homomorphism since,\n\n\[ f\left( {x{ + }_{6}y}\right) = {\;\operatorname{mod}\;\left( {x{ + }_{6}y,3}\right) } = {\;\operatorname{mod}\;\left( {x,3}\right) } + {}_{3}{\;\operatorname{mod}\;\left( {y,3}\right) },\]\n\nwhere \( { + }_{n} \) is addition in \( {\mathbb{Z}}_{n} \), and\n\n\[ f\left( {x{... | Yes |
Prove or disprove that \( f : \mathbb{R}\left\lbrack x\right\rbrack \rightarrow \mathbb{R} \), defined by \( f\left( {p\left( x\right) }\right) = \) \( p\left( 0\right) \), is a homomorphism. | Proof. We will divide this proof into two parts, one for each property of ring homomorphisms:\n\n(a) Let \( p\left( x\right), q\left( x\right) \) be arbitrary elements of \( \mathbb{R}\left\lbrack x\right\rbrack \), where \( p\left( x\right) \) and \( q\left( x\right) \) have constant terms of \( {a}_{0},{b}_{0} \in \m... | Yes |
Example 20.6.12. Find the kernel of \( f : \mathbb{R}\left\lbrack x\right\rbrack \rightarrow \mathbb{R} \), given by \( f\left( {p\left( x\right) }\right) = p\left( 0\right) \) . | We are looking for the set of all \( p\left( x\right) \) such that \( p\left( 0\right) = 0 \) . We know from the polynomials chapter that \( p\left( 0\right) = 0 \) implies that \( x \) divides \( p\left( x\right) \) . (So, there is no constant term!) In summary, \( \operatorname{Ker}\left( f\right) = \{ {xp}\left( x\r... | Yes |
Proposition 20.6.17. The kernel of a homomorphism \( f : {R}_{1} \rightarrow {R}_{2} \) satisfies the following properties:\n\n1. If \( a, b \in {f}^{-1}\left( 0\right) \), then \( a + b \in {f}^{-1}\left( 0\right) \).\n\n2. If \( a \in {f}^{-1}\left( 0\right) \) and \( b \in {R}_{1} \) then \( {ab} \in {f}^{-1}\left( ... | Proof.\n\n1.\n\n\[ a, b \in {f}^{-1}\left( 0\right) \]\ngiven\n\n\[ f\left( a\right) = 0, f\left( b\right) = 0 \]\ndef. of inverse\n\n\[ f\left( {a + b}\right) = f\left( a\right) + f\left( b\right) \]\ndef. of homomorphism\n\n\[ f\left( {a + b}\right) = 0 + 0 = 0 \]\nsubstitution & zero property\n\n\[ a + b \in {f}^{-1... | Yes |
Example 20.6.22. Let \( f : \mathbb{Z} \rightarrow {\mathbb{Z}}_{7} \) be defined by : \( f\left( x\right) = {\;\operatorname{mod}\;\left( {x,7}\right) } \)\n\n(a) Prove or disprove \( f \) is a ring homomorphism. | (a) To determine whether \( f \) is a ring homomorphism, we must verify Equations (20.6.3) and (20.6.4):\n\n\[ f\left( {x + y}\right) = {\;\operatorname{mod}\;\left( {x + y,7}\right) } = {\;\operatorname{mod}\;\left( {x,7}\right) } + {{}_{7}\operatorname{mod}\;\left( {y,7}\right) }\;\text{definition of} + {}_{7} \]\n\n... | Yes |
Given the ring \( \mathbb{Z} \) and \( J = \{ 0,7,{14},{21},\cdots \} \) . Prove or disprove \( J \) is an ideal. | We can see that \( J \subset \mathbb{Z} \), and \( J \) is closed under addition however \( J \) fails properties (b) and (c). | No |
Proposition 20.7.8. For every \( a \in R \), the set \( {Ra} \) is an ideal. | Exercise 20.7.9. Prove Proposition 20.7.8 by showing that \( {Ra} \) satisfies all properties of an ideal. | No |
Consider the ring of integers \( \mathbb{Z} \). Then \( {2Z} = \{ 0, \pm 2, \pm 4,\cdots \} \) is a principal ideal and is generated by 2. | In fact, for any integer \( k \), the set \( {kZ} = \{ 0, \pm k, \pm {2k},\cdots \} \) is a principal ideal. | Yes |
Show that \( J \) is an ideal, but not a principal ideal. | Proof.\n\n\( 2 \in J \) and \( x \in J \) definition of \( J \)\n\nSuppose \( J = a\mathbb{Z}\left\lbrack x\right\rbrack \) for some \( a \in \mathbb{Z}\left\lbrack x\right\rbrack \; \) supposition\n\nSince \( 2 \in J, a = 1 \) or \( a = 2\; \) only elements to divide 2\n\nIf \( a = 1 \), then \( 1\mathbb{Z}\left\lbrac... | Yes |
Show that \( f \) is a ring homomorphism. | \[ f\left( {a + b}\right) = {\;\operatorname{mod}\;\left( {a + b,5}\right) } = {\;\operatorname{mod}\;\left( {a,5}\right) } + {\;\operatorname{mod}\;\left( {b,5}\right) } = f\left( a\right) + f\left( b\right) \] \[ \text{and}f\left( {ab}\right) = {\;\operatorname{mod}\;\left( {{ab},5}\right) } = {\;\operatorname{mod}\;... | Yes |
Show that \( {\mathbb{Z}}_{p} \) is an integral domain if \( p \) is prime. | Proof. We have shown in Example 20.1.3 that \( {\mathbb{Z}}_{p} \) is a commutative ring for all \( p \in \mathbb{Z} \) . It remains to show that \( {\mathbb{Z}}_{p} \) has no zero divisors. We will show this by contradiction.\n\nSuppose \( {\mathbb{Z}}_{p} \) has a zero divisor \( a \in {\mathbb{Z}}_{p} \) such that \... | Yes |
Prove that \( \mathbb{Q}\left\lbrack \sqrt[3]{2}\right\rbrack \) is an integral domain. | Proof. In Example 20.1.4 we showed that \( \mathbb{Q}\left\lbrack \sqrt[3]{2}\right\rbrack \) is a ring. It remains to show that \( \mathbb{Q}\left\lbrack \sqrt[3]{2}\right\rbrack \) is commutative and contains no zero divisors. We know that elements of \( \mathbb{Q}\left\lbrack \sqrt[3]{2}\right\rbrack \) are real num... | Yes |
Proposition 20.9.13. Given integral domain \( D \) and \( a, b, c \in D \) . If \( {ab} = {ac} \) and \( a \neq 0 \), then \( b = c \) . | Proof.\n\n\[ \n{ab} = {ac} \n\]\nGiven\n\n\[ \n{ab} - {ac} = 0 \n\]\nSubstitution\n\n\[ \na\left( {b - c}\right) = 0 \n\]\nDistributive Law\n\nSince \( D \) is an integral domain, then \( D \) has no zero divisors. This means that \( a = 0 \) or \( b - c = 0 \) . But we know that \( a \neq 0 \) . So \( b - c = 0 \) and... | Yes |
Suppose we are given the set \( J \subset \mathbb{Z} : J = \{ \cdots , - {12}, - 6,0,6,{12},\cdots \} \) . Prove or disprove that \( p, q \in \mathbb{Z} \) and \( p \cdot q \in J \) implies \( p \in J \) or \( q \in J \) . | Proof. We can disprove this by counterexample. Consider, for example, \( 3,4 \in \mathbb{Z} \) . It is true that \( 3 \cdot 4 = {12} \in J \), but neither 3 nor 4 is in \( J \) . (Many other counterexamples can be found.) | Yes |
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