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Theorem 5.13 (Mertens' theorem). We have\n\n\[ \mathop{\prod }\limits_{{p \leq x}}\left( {1 - 1/p}\right) = \Theta \left( {1/\log x}\right) \]
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Proof. Using parts (i) and (iii) of §A1, for any fixed prime \( p \), we have\n\n\[ - \frac{1}{{p}^{2}} \leq \frac{1}{p} + \log \left( {1 - 1/p}\right) \leq 0. \]\n\n(5.9)\n\nMoreover, since\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{1}{{p}^{2}} \leq \mathop{\sum }\limits_{{i \geq 2}}\frac{1}{{i}^{2}} < \infty \]\n\nsumming the inequality (5.9) over all primes \( p \leq x \) yields\n\n\[ - C \leq \mathop{\sum }\limits_{{p \leq x}}\frac{1}{p} + \log g\left( x\right) \leq 0 \]\n\nwhere \( C \) is a positive constant, and \( g\left( x\right) \mathrel{\text{:=}} \mathop{\prod }\limits_{{p \leq x}}\left( {1 - 1/p}\right) \). From this, and from Theorem 5.10, we obtain \( \log g\left( x\right) = - \log \log x + O\left( 1\right) \), which implies that \( g\left( x\right) = \Theta \left( {1/\log x}\right) \) (see Exercise 3.11). That proves the theorem.
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No
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Theorem 5.14 (Prime number theorem). We have\n\n\[ \pi \left( x\right) \sim x/\log x. \]
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Proof. Literature-see §5.6.
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No
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Theorem 5.15. Let \( \kappa \left( x\right) \mathrel{\text{:=}} {\left( \log x\right) }^{3/5}{\left( \log \log x\right) }^{-1/5} \) . Then for some \( c > 0 \), we have\n\n\[ \pi \left( x\right) = \operatorname{li}\left( x\right) + O\left( {x{e}^{-{c\kappa }\left( x\right) }}\right) . \]
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Proof. Literature-see §5.6.
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No
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Theorem 5.17 (Euler’s identity). For every real number \( s > 1 \), we have\n\n\[ \zeta \left( s\right) = \mathop{\prod }\limits_{p}{\left( 1 - {p}^{-s}\right) }^{-1}, \]\n\nwhere the product is over all primes \( p \) .
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Proof. The rigorous interpretation of the infinite product on the right-hand side of (5.13) is as a limit of finite products. Thus, if \( {p}_{i} \) denotes the \( i \) th prime, for \( i = 1,2,\ldots \), then we are really proving that\n\n\[ \zeta \left( s\right) = \mathop{\lim }\limits_{{r \rightarrow \infty }}\mathop{\prod }\limits_{{i = 1}}^{r}{\left( 1 - {p}_{i}^{-s}\right) }^{-1}. \]\n\nNow, from the identity\n\n\[ {\left( 1 - {p}_{i}^{-s}\right) }^{-1} = \mathop{\sum }\limits_{{e = 0}}^{\infty }{p}_{i}^{-{es}}, \]\n\nwe have\n\n\[ \mathop{\prod }\limits_{{i = 1}}^{r}{\left( 1 - {p}_{i}^{-s}\right) }^{-1} = \left( {1 + {p}_{1}^{-s} + {p}_{1}^{-{2s}} + \cdots }\right) \cdots \left( {1 + {p}_{r}^{-s} + {p}_{r}^{-{2s}} + \cdots }\right) \]\n\n\[ = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{h}_{r}\left( n\right) }{{n}^{s}} \]\n\nwhere\n\n\[ {h}_{r}\left( n\right) \mathrel{\text{:=}} \left\{ \begin{array}{ll} 1 & \text{ if }n\text{ is divisible only by the primes }{p}_{1},\ldots ,{p}_{r}; \\ 0 & \text{ otherwise. } \end{array}\right. \]\n\nHere, we have made use of the fact (see §A7) that we can multiply term-wise infinite series with non-negative terms.\n\nNow, for every \( \varepsilon > 0 \), there exists \( {n}_{0} \) such that \( \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{n}^{-s} < \varepsilon \) (because the series defining \( \zeta \left( s\right) \) converges). Moreover, there exists an \( {r}_{0} \) such that \( {h}_{r}\left( n\right) = 1 \) for all \( n < {n}_{0} \) and \( r \geq {r}_{0} \) . Therefore, for all \( r \geq {r}_{0} \), we have\n\n\[ \left| {\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{h}_{r}\left( n\right) }{{n}^{s}} - \zeta \left( s\right) }\right| \leq \mathop{\sum }\limits_{{n = {n}_{0}}}^{\infty }{n}^{-s} < \varepsilon . \]\n\nIt follows that\n\n\[ \mathop{\lim }\limits_{{r \rightarrow \infty }}\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{h}_{r}\left( n\right) }{{n}^{s}} = \zeta \left( s\right) \]\n\nwhich proves the theorem.
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Yes
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Theorem 5.19. We have:\n\n(i) \( \frac{x}{\log x}\left( {1 + \frac{1}{2\log x}}\right) < \pi \left( x\right) < \frac{x}{\log x}\left( {1 + \frac{3}{2\log x}}\right) \), for \( x \geq {59} \) ;\n\n(ii) \( n\left( {\log n + \log \log n - 3/2}\right) < {p}_{n} < n\left( {\log n + \log \log n - 1/2}\right) \), for \( n \geq {20} \) ;\n\n(iii) \( x\left( {1 - \frac{1}{2\log x}}\right) < \vartheta \left( x\right) < x\left( {1 + \frac{1}{2\log x}}\right) \), for \( x \geq {563} \) ;\n\n(iv) \( \log \log x + A - \frac{1}{2{\left( \log x\right) }^{2}} < \mathop{\sum }\limits_{{p \leq x}}1/p < \log \log x + A + \frac{1}{2{\left( \log x\right) }^{2}} \) ,\n\nfor \( x \geq {286} \), where \( A \approx {0.261497212847643} \) ;\n\n(v) \( \frac{{B}_{1}}{\log x}\left( {1 - \frac{1}{2{\left( \log x\right) }^{2}}}\right) < \mathop{\prod }\limits_{{p \leq x}}\left( {1 - \frac{1}{p}}\right) < \frac{{B}_{1}}{\log x}\left( {1 + \frac{1}{2{\left( \log x\right) }^{2}}}\right) \) ,\n\nfor \( x \geq {285} \), where \( {B}_{1} \approx {0.561459483566885} \) .
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Proof. Literature-see §5.6.
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No
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Theorem 5.20 (Dirichlet’s theorem). Let \( a, d \in \mathbb{Z} \) with \( d > 0 \) and \( \gcd \left( {a, d}\right) = 1 \). Then there are infinitely many primes \( p \equiv a\left( {\;\operatorname{mod}\;d}\right) \).
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Proof. Literature-see §5.6.
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No
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Theorem 5.21. Let \( a, d \in \mathbb{Z} \) with \( d > 0 \) and \( \gcd \left( {a, d}\right) = 1 \) . Then\n\n\[ \pi \left( {x;d, a}\right) \sim \frac{x}{\varphi \left( d\right) \log x}. \]
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Proof. Literature-see §5.6.
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No
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Theorem 5.23. There exists a constant \( c \) such that for all \( a, d \in \mathbb{Z} \) with \( d \geq 2 \) and \( \gcd \left( {a, d}\right) = 1 \), the least prime \( p \equiv a\left( {\;\operatorname{mod}\;d}\right) \) is at most \( c{d}^{{11}/2} \) .
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Proof. Literature-see §5.6.
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No
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Theorem 6.2. Let \( G \) be an abelian group with binary operation \( \star \) . Then we have:\n\n(i) \( G \) contains only one identity element;\n\n(ii) every element of \( G \) has only one inverse.
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Proof. Suppose \( e,{e}^{\prime } \) are both identities. Then we have\n\n\[ e = e \star {e}^{\prime } = {e}^{\prime } \]\n\nwhere we have used part (ii) of Definition 6.1, once with \( {e}^{\prime } \) as the identity, and once with \( e \) as the identity. That proves part (i) of the theorem.\n\nTo prove part (ii) of the theorem, let \( a \in G \), and suppose that \( a \) has two inverses, \( {a}^{\prime } \) and \( {a}^{\prime \prime } \) . Then using parts (i)-(iii) of Definition 6.1, we have\n\n\[ {a}^{\prime } = {a}^{\prime } \star e\text{ (by part (ii)) } \]\n\n\[ = {a}^{\prime } \star \left( {a \star {a}^{\prime \prime }}\right) \text{(by part (iii) with inverse}{a}^{\prime \prime }\text{of}a\text{)} \]\n\n\[ = \left( {{a}^{\prime } \star a}\right) \star {a}^{\prime \prime }\text{ (by part (i)) } \]\n\n\[ = e \star {a}^{\prime \prime }\text{(by part (iii) with inverse}{a}^{\prime }\text{of}a\text{)} \]\n\n\[ = {a}^{\prime \prime }\text{ (by part (ii)). } \]\n\nThese uniqueness properties justify use of the definite article in Definition 6.1 in conjunction with the terms \
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Yes
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The set \( {\mathcal{F}}^{ * } \) of all arithmetic functions \( f \), such that \( f\left( 1\right) \neq 0 \), and with the Dirichlet product as the binary operation (see §2.9) forms an abelian group.
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The special function \( I \) is the identity, and inverses are guaranteed by Exercise 2.54.
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No
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The set of all finite bit strings under concatenation does not form an abelian group.
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Although concatenation is associative and the empty string acts as an identity element, inverses do not exist (except for the empty string), nor is concatenation commutative.
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Yes
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Theorem 6.3. Let \( G \) be an abelian group. Then for all \( a, b, c \in G \), we have:\n\n(i) if \( a + b = a + c \), then \( b = c \) ;
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Proof. These statements all follow easily from Definition 6.1 and Theorem 6.2. For (i), just add \( - a \) to both sides of the equation \( a + b = a + c \) .
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Yes
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Theorem 6.4. Let \( G \) be an abelian group. Then for all \( a, b \in G \) and \( k,\ell \in \mathbb{Z} \), we have:\n\n(i) \( k\left( {\ell a}\right) = \left( {k\ell }\right) a = \ell \left( {ka}\right) \) ;\n\n(ii) \( \left( {k + \ell }\right) a = {ka} + \ell a \) ;\n\n(iii) \( k\left( {a + b}\right) = {ka} + {kb} \) .
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Proof. The proof of this is easy, but tedious. We leave the details as an exercise to the reader.
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No
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If \( {G}_{1},\ldots ,{G}_{k} \) are abelian groups, we can form the direct product \( H \mathrel{\text{:=}} {G}_{1} \times \cdots \times {G}_{k} \), which consists of all \( k \) -tuples \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) with \( {a}_{1} \in {G}_{1} \) , \( \ldots ,{a}_{k} \in {G}_{k} \). We can view \( H \) in a natural way as an abelian group if we define the group operation component-wise:
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\[ \left( {{a}_{1},\ldots ,{a}_{k}}\right) + \left( {{b}_{1},\ldots ,{b}_{k}}\right) \mathrel{\text{:=}} \left( {{a}_{1} + {b}_{1},\ldots ,{a}_{k} + {b}_{k}}\right) . \] Of course, the groups \( {G}_{1},\ldots ,{G}_{k} \) may be different, and the group operation applied in the \( i \) th component corresponds to the group operation associated with \( {G}_{i} \). We leave it to the reader to verify that \( H \) is in fact an abelian group, where \( {0}_{H} = \) \( \left( {{0}_{{G}_{1}},\ldots ,{0}_{{G}_{k}}}\right) \) and \( - \left( {{a}_{1},\ldots ,{a}_{k}}\right) = \left( {-{a}_{1},\ldots , - {a}_{k}}\right) \).
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No
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Let \( G \) be an abelian group. An element \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) of \( {G}^{\times k} \) may be identified with the function \( f : \{ 1,\ldots, k\} \rightarrow G \) given by \( f\left( i\right) = {a}_{i} \) for \( i = 1,\ldots, k \) . We can generalize this, replacing \( \{ 1,\ldots, k\} \) by an arbitrary set \( I \) . We define \( \operatorname{Map}\left( {I, G}\right) \) to be the set of all functions \( f : I \rightarrow G \), which we naturally view as a group by defining the group operation point-wise: for \( f, g \in \operatorname{Map}\left( {I, G}\right) \), we define
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\[ \left( {f + g}\right) \left( i\right) \mathrel{\text{:=}} f\left( i\right) + g\left( i\right) \text{ for all }i \in I. \] Again, we leave it to the reader to verify that \( \operatorname{Map}\left( {I, G}\right) \) is an abelian group, where the identity element is the function that maps each \( i \in I \) to \( {0}_{G} \), and for \( f \in \operatorname{Map}\left( {I, G}\right) \), we have \( \left( {-f}\right) \left( i\right) = - \left( {f\left( i\right) }\right) \) for all \( i \in I \) .
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No
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Theorem 6.6. If \( G \) is an abelian group, and \( H \) is a subgroup of \( G \), then \( H \) contains \( {0}_{G} \) ; moreover, the binary operation of \( G \), when restricted to \( H \), yields a binary operation that makes \( H \) into an abelian group whose identity is \( {0}_{G} \).
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Proof. First, to see that \( {0}_{G} \in H \), just pick any \( a \in H \), and using both properties of the definition of a subgroup, we see that \( {0}_{G} = a + \left( {-a}\right) \in H \). Next, note that by property (i) of Definition 6.5, \( H \) is closed under addition, which means that the restriction of the binary operation \
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No
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Theorem 6.7. Let \( G \) be an abelian group, and let \( m \) be an integer. Then\n\n\[ \n{mG} \mathrel{\text{:=}} \{ {ma} : a \in G\} \n\]\n\nis a subgroup of \( G \) .
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Proof. The set \( {mG} \) is non-empty, since \( {0}_{G} = m{0}_{G} \in {mG} \) . For \( {ma},{mb} \in {mG} \), we have \( {ma} + {mb} = m\left( {a + b}\right) \in {mG} \), and \( - \left( {ma}\right) = m\left( {-a}\right) \in {mG} \) .
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Yes
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Theorem 6.8. Let \( G \) be an abelian group, and let \( m \) be an integer. Then\n\n\[ G\{ m\} \mathrel{\text{:=}} \left\{ {a \in G : {ma} = {0}_{G}}\right\} \]\n\nis a subgroup of \( G \) .
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Proof. The set \( G\{ m\} \) is non-empty, since \( m{0}_{G} = {0}_{G} \), and so \( G\{ m\} \) contains \( {0}_{G} \) . If \( {ma} = {0}_{G} \) and \( {mb} = {0}_{G} \), then \( m\left( {a + b}\right) = {ma} + {mb} = {0}_{G} + {0}_{G} = {0}_{G} \) and \( m\left( {-a}\right) = - \left( {ma}\right) = - {0}_{G} = {0}_{G}. \)
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Yes
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Let \( n \) be a positive integer, let \( m \in \mathbb{Z} \), and consider the subgroup \( m{\mathbb{Z}}_{n} \) of the additive group \( {\mathbb{Z}}_{n} \). Now, for every residue class \( \left\lbrack z\right\rbrack \in {\mathbb{Z}}_{n} \), we have \( m\left\lbrack z\right\rbrack = \left\lbrack {mz}\right\rbrack \). Therefore, \( \left\lbrack b\right\rbrack \in m{\mathbb{Z}}_{n} \) if and only if there exists \( z \in \mathbb{Z} \) such that \( {mz} \equiv b\left( {\;\operatorname{mod}\;n}\right) \).
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By part (i) of Theorem 2.5, such a \( z \) exists if and only if \( d \mid b \), where \( d \mathrel{\text{:=}} \gcd \left( {m, n}\right) \). Thus, \( m{\mathbb{Z}}_{n} \) consists precisely of the \( n/d \) distinct residue classes \[ \left\lbrack {i \cdot d}\right\rbrack \left( {i = 0,\ldots, n/d - 1}\right) ,\] and in particular, \( m{\mathbb{Z}}_{n} = d{\mathbb{Z}}_{n} \).
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Yes
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Theorem 6.9. If \( G \) is a subgroup of \( \mathbb{Z} \), then there exists a unique non-negative integer \( m \) such that \( G = m\mathbb{Z} \) . Moreover, for two non-negative integers \( {m}_{1} \) and \( {m}_{2} \) , we have \( {m}_{1}\mathbb{Z} \subseteq {m}_{2}\mathbb{Z} \) if and only if \( {m}_{2} \mid {m}_{1} \) .
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Proof. Actually, we have already proven this. One only needs to observe that a subset \( G \) of \( \mathbb{Z} \) is a subgroup if and only if it is an ideal of \( \mathbb{Z} \), as defined in \( §{1.2} \) (see Exercise 1.8). The first statement of the theorem then follows from Theorem 1.6. The second statement follows easily from the definitions, as was observed in \( §{1.2} \).
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No
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Theorem 6.10. If \( G \) is a subgroup of \( {\mathbb{Z}}_{n} \), then there exists a unique positive integer \( d \) dividing \( n \) such that \( G = d{\mathbb{Z}}_{n} \) . Also, for all positive divisors \( {d}_{1},{d}_{2} \) of \( n \), we have \( {d}_{1}{\mathbb{Z}}_{n} \subseteq {d}_{2}{\mathbb{Z}}_{n} \) if and only if \( {d}_{2} \mid {d}_{1} \) .
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Proof. Note that the second statement implies the uniqueness part of the first statement, so it suffices to prove just the existence part of the first statement and the second statement.\n\nLet \( G \) be an arbitrary subgroup of \( {\mathbb{Z}}_{n} \), and let \( H \mathrel{\text{:=}} \{ z \in \mathbb{Z} : \left\lbrack z\right\rbrack \in G\} \) . We claim that \( H \) is a subgroup of \( \mathbb{Z} \) . To see this, observe that if \( a, b \in H \), then \( \left\lbrack a\right\rbrack \) and \( \left\lbrack b\right\rbrack \) belong to \( G \), and hence so do \( \left\lbrack {a + b}\right\rbrack = \left\lbrack a\right\rbrack + \left\lbrack b\right\rbrack \) and \( \left\lbrack {-a}\right\rbrack = - \left\lbrack a\right\rbrack \), and thus \( a + b \) and \( - a \) belong to \( H \) . That proves the claim, and Theorem 6.9 implies that \( H = d\mathbb{Z} \) for some non-negative integer \( d \) . It follows that\n\n\[ G = \{ \left\lbrack y\right\rbrack : y \in H\} = \{ \left\lbrack {dz}\right\rbrack : z \in \mathbb{Z}\} = d{\mathbb{Z}}_{n}. \]\n\nEvidently, \( n \in H = d\mathbb{Z} \), and hence \( d \mid n \) . That proves the existence part of the first statement of the theorem.\n\nTo prove the second statement of the theorem, observe that if \( {d}_{1} \) and \( {d}_{2} \) are arbitrary integers, then\n\n\[ {d}_{1}{\mathbb{Z}}_{n} \subseteq {d}_{2}{\mathbb{Z}}_{n} \Leftrightarrow {d}_{2}z \equiv {d}_{1}\left( {\;\operatorname{mod}\;n}\right) \text{ for some }z \in \mathbb{Z} \]\n\n\[ \Leftrightarrow \gcd \left( {{d}_{2}, n}\right) \mid {d}_{1}\text{(by part (i) of Theorem 2.5).} \]\n\nIn particular, if \( {d}_{2} \) is a positive divisor of \( n \), then \( \gcd \left( {{d}_{2}, n}\right) = {d}_{2} \), which proves the second statement.
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Yes
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Consider the group \( {\mathbb{Z}}_{15}^{ * } \). We can enumerate its elements as \[ \left\lbrack {\pm 1}\right\rbrack ,\left\lbrack {\pm 2}\right\rbrack ,\left\lbrack {\pm 4}\right\rbrack ,\left\lbrack {\pm 7}\right\rbrack \text{.} \] Therefore, the elements of \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} \) are \[ {\left\lbrack 1\right\rbrack }^{2} = \left\lbrack 1\right\rbrack ,\;{\left\lbrack 2\right\rbrack }^{2} = \left\lbrack 4\right\rbrack ,\;{\left\lbrack 4\right\rbrack }^{2} = \left\lbrack {16}\right\rbrack = \left\lbrack 1\right\rbrack ,\;{\left\lbrack 7\right\rbrack }^{2} = \left\lbrack {49}\right\rbrack = \left\lbrack 4\right\rbrack ; \] thus, \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} \) has order 2, consisting as it does of the two distinct elements [1] and [4].
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Going further, one sees that \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{4} = \{ \left\lbrack 1\right\rbrack \} \). Thus, \( {\alpha }^{4} = \left\lbrack 1\right\rbrack \) for all \( \alpha \in {\mathbb{Z}}_{15}^{ * } \). By direct calculation, one can determine that \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{3} = {\mathbb{Z}}_{15}^{ * } \); that is, cubing simply permutes \( {\mathbb{Z}}_{15}^{ * } \). For any given integer \( m \), write \( m = {4q} + r \), where \( 0 \leq r < 4 \). Then for every \( \alpha \in {\mathbb{Z}}_{15}^{ * } \), we have \( {\alpha }^{m} = {\alpha }^{{4q} + r} = {\alpha }^{4q}{\alpha }^{r} = {\alpha }^{r} \). Thus, \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{m} \) is either \( {\mathbb{Z}}_{15}^{ * },{\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} \), or \( \{ \left\lbrack 1\right\rbrack \} \).
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Yes
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Consider the group \( {\mathbb{Z}}_{5}^{ * } = \{ \left\lbrack {\pm 1}\right\rbrack ,\left\lbrack {\pm 2}\right\rbrack \} \) . The elements of \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} \) are
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\[ {\left\lbrack 1\right\rbrack }^{2} = \left\lbrack 1\right\rbrack ,{\left\lbrack 2\right\rbrack }^{2} = \left\lbrack 4\right\rbrack = \left\lbrack {-1}\right\rbrack \] thus, \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} = \{ \left\lbrack {\pm 1}\right\rbrack \} \) and has order 2 .
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Yes
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Theorem 6.11. If \( {H}_{1} \) and \( {H}_{2} \) are subgroups of an abelian group \( G \), then so is\n\n\[ \n{H}_{1} + {H}_{2} \mathrel{\text{:=}} \left\{ {{a}_{1} + {a}_{2} : {a}_{1} \in {H}_{1},{a}_{2} \in {H}_{2}}\right\} .\n\]
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Proof. It is evident that \( {H}_{1} + {H}_{2} \) is non-empty, as it contains \( {0}_{G} + {0}_{G} = {0}_{G} \) . Consider two elements in \( {H}_{1} + {H}_{2} \), which we can write as \( {a}_{1} + {a}_{2} \) and \( {b}_{1} + {b}_{2} \) , where \( {a}_{1},{b}_{1} \in {H}_{1} \) and \( {a}_{2},{b}_{2} \in {H}_{2} \) . Then by the closure properties of subgroups, \( {a}_{1} + {b}_{1} \in {H}_{1} \) and \( {a}_{2} + {b}_{2} \in {H}_{2} \), and hence \( \left( {{a}_{1} + {a}_{2}}\right) + \left( {{b}_{1} + {b}_{2}}\right) = \left( {{a}_{1} + {b}_{1}}\right) + \left( {{a}_{2} + {b}_{2}}\right) \in \) \( {H}_{1} + {H}_{2} \) . Similarly, \( - \left( {{a}_{1} + {a}_{2}}\right) = \left( {-{a}_{1}}\right) + \left( {-{a}_{2}}\right) \in {H}_{1} + {H}_{2} \) .
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Yes
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Theorem 6.12. If \( {H}_{1} \) and \( {H}_{2} \) are subgroups of an abelian group \( G \), then so is \( {H}_{1} \cap {H}_{2} \)
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Proof. It is evident that \( {H}_{1} \cap {H}_{2} \) is non-empty, as both \( {H}_{1} \) and \( {H}_{2} \) contain \( {0}_{G} \) , and hence so does their intersection. If \( a \in {H}_{1} \cap {H}_{2} \) and \( b \in {H}_{1} \cap {H}_{2} \), then since \( a, b \in {H}_{1} \), we have \( a + b \in {H}_{1} \), and since \( a, b \in {H}_{2} \), we have \( a + b \in {H}_{2} \) ; therefore, \( a + b \in {H}_{1} \cap {H}_{2} \) . Similarly, \( - a \in {H}_{1} \) and \( - a \in {H}_{2} \), and therefore, \( - a \in {H}_{1} \cap {H}_{2} \)
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Yes
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Theorem 6.13. Let \( G \) be an abelian group and \( H \) a subgroup of \( G \) . For all \( a, b, c \in G \), we have:\n\n(i) \( a \equiv a\left( {\;\operatorname{mod}\;H}\right) \) ;\n\n(ii) \( a \equiv b\left( {\;\operatorname{mod}\;H}\right) \) implies \( b \equiv a\left( {\;\operatorname{mod}\;H}\right) \) ;\n\n(iii) \( a \equiv b\left( {\;\operatorname{mod}\;H}\right) \) and \( b \equiv c\left( {\;\operatorname{mod}\;H}\right) \) implies \( a \equiv c\left( {\;\operatorname{mod}\;H}\right) \) .
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Proof. For (i), observe that \( H \) contains \( {0}_{G} = a - a \) . For (ii), observe that if \( H \) contains \( a - b \), then it also contains \( - \left( {a - b}\right) = b - a \) . For (iii), observe that if \( H \) contains \( a - b \) and \( b - c \), then it also contains \( \left( {a - b}\right) + \left( {b - c}\right) = a - c \) .
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Yes
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Theorem 6.14. Let \( G \) be an abelian group and \( H \) a subgroup of \( G \). For all \( a, b \in G \), the function\n\n\[ f : G \rightarrow G \]\n\n\[ x \mapsto b - a + x \]\n\n is a bijection, which, when restricted to the coset \( {\left\lbrack a\right\rbrack }_{H} \), yields a bijection from \( {\left\lbrack a\right\rbrack }_{H} \) to the coset \( {\left\lbrack b\right\rbrack }_{H} \). In particular, every two cosets of \( H \) in \( G \) have the same cardinality.
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Proof. First, we claim that \( f \) is a bijection. Indeed, if \( f\left( x\right) = f\left( {x}^{\prime }\right) \), then \( b - a + x = b - a + {x}^{\prime } \), and subtracting \( b \) and adding \( a \) to both sides of this equation yields \( x = {x}^{\prime } \). That proves that \( f \) is injective. To prove that \( f \) is surjective, observe that for any given \( {x}^{\prime } \in G \), we have \( f\left( {a - b + {x}^{\prime }}\right) = {x}^{\prime } \).\n\nSecond, we claim that for all \( x \in G \), we have \( x \in {\left\lbrack a\right\rbrack }_{H} \) if and only if \( f\left( x\right) \in {\left\lbrack b\right\rbrack }_{H} \). On the one hand, suppose that \( x \in {\left\lbrack a\right\rbrack }_{H} \), which means that \( x = a + h \) for some \( h \in H \). Subtracting \( a \) and adding \( b \) to both sides of this equation yields \( b - a + x = b + h \), which means \( f\left( x\right) \in {\left\lbrack b\right\rbrack }_{H} \). Conversely, suppose that \( f\left( x\right) \in {\left\lbrack b\right\rbrack }_{H} \), which means that \( b - a + x = b + h \) for some \( h \in H \). Subtracting \( b \) and adding \( a \) to both sides of this equation yields \( x = a + h \), which means that \( x \in {\left\lbrack a\right\rbrack }_{H} \).\n\nThe theorem is now immediate from these two claims.
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Yes
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Theorem 6.15 (Lagrange’s theorem). If \( G \) is a finite abelian group, and \( H \) is a subgroup of \( G \), then the order of \( H \) divides the order of \( G \) .
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Proof. This is an immediate consequence of the previous theorem, and the fact that the cosets of \( H \) in \( G \) partition \( G \) .
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No
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Theorem 6.16. Suppose \( G \) is an abelian group and \( H \) is a subgroup of \( G \) . For all \( a,{a}^{\prime }, b,{b}^{\prime } \in G \), if \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \) and \( b \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \), then we have \( a + b \equiv {a}^{\prime } + {b}^{\prime }\left( {\;\operatorname{mod}\;H}\right) .
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Proof. Now, \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \) and \( b \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \) means that \( a = {a}^{\prime } + x \) and \( b = {b}^{\prime } + y \) for some \( x, y \in H \) . Therefore, \( a + b = \left( {{a}^{\prime } + x}\right) + \left( {{b}^{\prime } + y}\right) = \left( {{a}^{\prime } + {b}^{\prime }}\right) + \left( {x + y}\right) \) , and since \( x + y \in H \), this means that \( a + b \equiv {a}^{\prime } + {b}^{\prime }\left( {\;\operatorname{mod}\;H}\right) .
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Yes
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Theorem 6.17. Suppose \( G \) is a finite abelian group and \( H \) is a subgroup of \( G \) . Then \( \left\lbrack {G : H}\right\rbrack = \left| G\right| /\left| H\right| \) . Moreover, if \( K \) is a subgroup of \( H \), then\n\n\[ \left\lbrack {G : K}\right\rbrack = \left\lbrack {G : H}\right\rbrack \left\lbrack {H : K}\right\rbrack \]
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Proof. The fact that \( \left\lbrack {G : H}\right\rbrack = \left| G\right| /\left| H\right| \) follows directly from Theorem 6.14. The fact that \( \left\lbrack {G : K}\right\rbrack = \left\lbrack {G : H}\right\rbrack \left\lbrack {H : K}\right\rbrack \) follows from a simple calculation:\n\n\[ \left\lbrack {G : H}\right\rbrack = \frac{\left| G\right| }{\left| H\right| } = \frac{\left| G\right| /\left| K\right| }{\left| H\right| /\left| K\right| } = \frac{\left\lbrack G : K\right\rbrack }{\left\lbrack H : K\right\rbrack }.\]
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Yes
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Example 6.32. Continuing with Example 6.30, let \( G \mathrel{\text{:=}} {\mathbb{Z}}_{6} \) and \( H \mathrel{\text{:=}} {3G} = \) \( \{ \left\lbrack 0\right\rbrack ,\left\lbrack 3\right\rbrack \} \) . The quotient group \( G/H \) has order 3, and consists of the cosets\n\n\[ \alpha \mathrel{\text{:=}} \{ \left\lbrack 0\right\rbrack ,\left\lbrack 3\right\rbrack \} ,\beta \mathrel{\text{:=}} \{ \left\lbrack 1\right\rbrack ,\left\lbrack 4\right\rbrack \} ,\gamma \mathrel{\text{:=}} \{ \left\lbrack 2\right\rbrack ,\left\lbrack 5\right\rbrack \} . \]
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If we write out an addition table for \( G \), grouping together elements in cosets of \( H \) in \( G \), then we also get an addition table for the quotient group \( G/H \) :\n\n<table><thead><tr><th>\( + \)</th><th>[0]</th><th>[3]</th><th>[1]</th><th>[4]</th><th>[2]</th><th>[5]</th></tr></thead><tr><td>[0]</td><td>[0]</td><td>[3]</td><td>[1]</td><td>[4]</td><td>[2]</td><td>[5]</td></tr><tr><td>[3]</td><td>[3]</td><td>[0]</td><td>[4]</td><td>[1]</td><td>[5]</td><td>[2]</td></tr><tr><td>[1]</td><td>[1]</td><td>[4]</td><td>[2]</td><td>[5]</td><td>[3]</td><td>[0]</td></tr><tr><td>[4]</td><td>[4]</td><td>[1]</td><td>[5]</td><td>[2]</td><td>[0]</td><td>[3]</td></tr><tr><td>[2]</td><td>[2]</td><td>[5]</td><td>[3]</td><td>[0]</td><td>[4]</td><td>[1]</td></tr><tr><td>[5]</td><td>[5]</td><td>[2]</td><td>[0]</td><td>[3]</td><td>[1]</td><td>[4]</td></tr></table>\n\nThis table illustrates quite graphically the point of Theorem 6.16: for every two cosets, if we take any element from the first and add it to any element of the second, we always end up in the same coset.\n\nWe can also write down just the addition table for \( G/H \) :\n\n\[ \begin{matrix} + & \alpha & \beta & \gamma \\ \alpha & \alpha & \beta & \gamma \\ \beta & \beta & \gamma & \alpha \\ \gamma & \gamma & \alpha & \beta \end{matrix} \]
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Yes
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Let us return to Example 6.26. The multiplicative group \( {\mathbb{Z}}_{15}^{ * } \), as we saw, is of order 8 . The subgroup \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} \) of \( {\mathbb{Z}}_{15}^{ * } \) has order 2 . Therefore, the quotient group \( {\mathbb{Z}}_{15}^{ * }/{\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} \) has order 4 .
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Indeed, the cosets are\n\n\[ \n{\alpha }_{00} \mathrel{\text{:=}} {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} = \{ \left\lbrack 1\right\rbrack ,\left\lbrack 4\right\rbrack \} ,\;{\alpha }_{01} \mathrel{\text{:=}} \left\lbrack {-1}\right\rbrack {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} = \{ \left\lbrack {-1}\right\rbrack ,\left\lbrack {-4}\right\rbrack \} , \n\] \n\n\[ \n{\alpha }_{10} : = \left\lbrack 2\right\rbrack {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} = \left\{ {\left\lbrack 2\right\rbrack ,\left\lbrack {-7}\right\rbrack }\right\} ,\;{\alpha }_{11} : = \left\lbrack {-2}\right\rbrack {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} = \left\{ {\left\lbrack {-2}\right\rbrack ,\left\lbrack 7\right\rbrack }\right\} . \n\] \n\nWe can write down the multiplication table for the quotient group:\n\n<table><thead><tr><th>-</th><th>\( {\alpha }_{00} \)</th><th>\( {\alpha }_{01} \)</th><th>\( {\alpha }_{10} \)</th><th>\( {\alpha }_{11} \)</th></tr></thead><tr><td>\( {\alpha }_{00} \)</td><td>\( {\alpha }_{00} \)</td><td>\( {\alpha }_{01} \)</td><td>\( {\alpha }_{10} \)</td><td>\( {\alpha }_{11} \)</td></tr><tr><td>\( {\alpha }_{01} \)</td><td>\( {\alpha }_{01} \)</td><td>\( {\alpha }_{00} \)</td><td>\( {\alpha }_{11} \)</td><td>\( {\alpha }_{10} \)</td></tr><tr><td>\( {\alpha }_{10} \)</td><td>\( {\alpha }_{10} \)</td><td>\( {\alpha }_{11} \)</td><td>\( {\alpha }_{00} \)</td><td>\( {\alpha }_{01} \)</td></tr><tr><td>\( {\alpha }_{11} \)</td><td>\( {\alpha }_{11} \)</td><td>\( {\alpha }_{10} \)</td><td>\( {\alpha }_{01} \)</td><td>\( {\alpha }_{00} \)</td></tr></table>\n\nNote that this group is essentially just a \
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Yes
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Example 6.34. As we saw in Example 6.27, \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} = \{ \left\lbrack {\pm 1}\right\rbrack \} \) . Therefore, the quotient group \( {\mathbb{Z}}_{5}^{ * }/{\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} \) has order 2 . The cosets of \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} \) in \( {\mathbb{Z}}_{5}^{ * } \) are \( {\alpha }_{0} \mathrel{\text{:=}} \{ \left\lbrack {\pm 1}\right\rbrack \} \) and \( {\alpha }_{1} \mathrel{\text{:=}} \{ \left\lbrack {\pm 2}\right\rbrack \} \), and the multiplication table looks like this:
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\[ \begin{matrix} \cdot & {\alpha }_{0} & {\alpha }_{1} \\ {\alpha }_{0} & {\alpha }_{0} & {\alpha }_{1} \\ {\alpha }_{1} & {\alpha }_{1} & {\alpha }_{0} \end{matrix} \] We see that the quotient group is essentially just a \
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Yes
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Suppose \( H \) is a subgroup of an abelian group \( G \) . We define the map\n\n\[ \rho : G \rightarrow G/H \]\n\n\[ a \mapsto {\left\lbrack a\right\rbrack }_{H} \]
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It is not hard to see that this is a group homomorphism. Indeed, this follows almost immediately from the way we defined addition in the quotient group \( G/H \) :\n\n\[ \rho \left( {a + b}\right) = {\left\lbrack a + b\right\rbrack }_{H} = {\left\lbrack a\right\rbrack }_{H} + {\left\lbrack b\right\rbrack }_{H} = \rho \left( a\right) + \rho \left( b\right) .\n\nIt is clear that \( \rho \) is surjective. It is also not hard to see that \( \operatorname{Ker}\rho = H \) ; indeed, \( H \) is the identity element in \( G/H \), and \( {\left\lbrack a\right\rbrack }_{H} = H \) if and only if \( a \in H \) . The map \( \rho \) is called the natural map from \( G \) to \( G/H \) .
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Yes
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Suppose \( G \) is an abelian group and \( m \) is an integer. The map \[ \rho : \;G \rightarrow G \] \[ a \mapsto {ma} \] is a group homomorphism.
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\[ \rho \left( {a + b}\right) = m\left( {a + b}\right) = {ma} + {mb} = \rho \left( a\right) + \rho \left( b\right) .\]
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Yes
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Suppose \( G \) is an abelian group and \( a \) is an element of \( G \) . It is easy to see that the map\n\n\[ \rho : \;\mathbb{Z} \rightarrow G \]\n\n\[ z \mapsto {za} \]\nis a group homomorphism, since
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\[\rho \left( {z + {z}^{\prime }}\right) = \left( {z + {z}^{\prime }}\right) a = {za} + {z}^{\prime }a = \rho \left( z\right) + \rho \left( {z}^{\prime }\right) .\]
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Yes
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Theorem 6.19. Let \( \rho \) be a group homomorphism from \( G \) to \( {G}^{\prime } \). Then:\n\n(i) \( \rho \left( {0}_{G}\right) = {0}_{{G}^{\prime }} \);\n\n(ii) \( \rho \left( {-a}\right) = - \rho \left( a\right) \) for all \( a \in G \);\n\n(iii) \( \rho \left( {na}\right) = {n\rho }\left( a\right) \) for all \( n \in \mathbb{Z} \) and \( a \in G \);\n\n(iv) if \( H \) is a subgroup of \( G \), then \( \rho \left( H\right) \) is a subgroup of \( {G}^{\prime } \); in particular (setting \( H \mathrel{\text{:=}} G \) ), \( \operatorname{Im}\rho \) is a subgroup of \( {G}^{\prime } \);\n\n(v) if \( {H}^{\prime } \) is a subgroup of \( {G}^{\prime } \), then \( {\rho }^{-1}\left( {H}^{\prime }\right) \) is a subgroup of \( G \); in particular (setting \( {H}^{\prime } \mathrel{\text{:=}} \left\{ {0}_{{G}^{\prime }}\right\} \) ), Ker \( \rho \) is a subgroup of \( G \);\n\n(vi) for all \( a, b \in G,\rho \left( a\right) = \rho \left( b\right) \) if and only if \( a \equiv b\left( {{\;\operatorname{mod}\;\operatorname{Ker}}\rho }\right) \);\n\n(vii) \( \rho \) is injective if and only if \( \operatorname{Ker}\rho = \left\{ {0}_{G}\right} \).
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Proof. These are all straightforward calculations.\n\n(i) We have\n\n\[ \n{0}_{{G}^{\prime }} + \rho \left( {0}_{G}\right) = \rho \left( {0}_{G}\right) = \rho \left( {{0}_{G} + {0}_{G}}\right) = \rho \left( {0}_{G}\right) + \rho \left( {0}_{G}\right) .\n\] \n\nNow cancel \( \rho \left( {0}_{G}\right) \) from both sides.\n\n(ii) We have\n\n\[ \n{0}_{{G}^{\prime }} = \rho \left( {0}_{G}\right) = \rho \left( {a + \left( {-a}\right) }\right) = \rho \left( a\right) + \rho \left( {-a}\right) ,\n\] \n\nand hence \( \rho \left( {-a}\right) \) is the inverse of \( \rho \left( a\right) \).\n\n(iii) For \( n = 0 \), this follows from part (i). For \( n > 0 \), this follows from the definitions by induction on \( n \). For \( n < 0 \), this follows from the positive case and part (ii).\n\n(iv) For all \( a, b \in H \), we have \( a + b \in H \) and \( - a \in H \); hence, \( \rho \left( H\right) \) contains \( \rho \left( {a + b}\right) = \rho \left( a\right) + \rho \left( b\right) \) and \( \rho \left( {-a}\right) = - \rho \left( a\right) .\n\n(v) \( {\rho }^{-1}\left( {H}^{\prime }\right) \) is non-empty, since \( \rho \left( {0}_{G}\right) = {0}_{G}^{\prime } \in {H}^{\prime } \). If \( \rho \left( a\right) \in {H}^{\prime } \) and \( \rho \left( b\right) \in {H}^{\prime } \), then \( \rho \left( {a + b}\right) = \rho \left( a\right) + \rho \left( b\right) \in {H}^{\prime } \), and \( \rho \left( {-a}\right) = - \rho \left( a\right) \in {H}^{\prime } .\n\n(vi) We have\n\n\[ \n\rho \left( a\right) = \rho \left( b\right) \Leftrightarrow \rho \left( a\right) - \rho \left( b\right) = {0}_{{G}^{\prime }} \Leftrightarrow \rho \left( {a - b}\right) = {0}_{{G}^{\prime }}\n\] \n\n\[ \n\Leftrightarrow a - b \in \operatorname{Ker}\rho \Leftrightarrow a \equiv b\left( {{\;\operatorname{mod}\;\operatorname{Ker}}\rho }\right) .\n\] \n\n(vii) If \( \rho \) is injective, then in particular, \( {\rho }^{-1}\left( \left\{ {0}_{{G}^{\prime }}\right\} \right) \) cannot contain any other element besides \( {0}_{G} \). If \( \rho \) is not injective, then there exist two distinct elements \( a, b \in G \) with \( \rho \left( a\right) = \rho \left( b\right) \), and by part (vi), Ker \( \rho \) contains the element \( a - b \), which is non-zero.
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Yes
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Theorem 6.20. If \( \rho : G \rightarrow {G}^{\prime } \) and \( {\rho }^{\prime } : {G}^{\prime } \rightarrow {G}^{\prime \prime } \) are group homomorphisms, then so is their composition \( {\rho }^{\prime } \circ \rho : G \rightarrow {G}^{\prime \prime } \) .
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Proof. For all \( a, b \in G \), we have\n\n\[ \n{\rho }^{\prime }\left( {\rho \left( {a + b}\right) }\right) = {\rho }^{\prime }\left( {\rho \left( a\right) + \rho \left( b\right) }\right) = {\rho }^{\prime }\left( {\rho \left( a\right) }\right) + {\rho }^{\prime }\left( {\rho \left( b\right) }\right) .\n\]
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Yes
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Theorem 6.21. Let \( {\rho }_{i} : G \rightarrow {G}_{i}^{\prime } \), for \( i = 1,\ldots, k \), be group homomorphisms. Then the map\n\n\[ \rho : \;G \rightarrow {G}_{1}^{\prime } \times \cdots \times {G}_{k}^{\prime } \]\n\n\[ a \mapsto \left( {{\rho }_{1}\left( a\right) ,\ldots ,{\rho }_{k}\left( a\right) }\right) \]\n\nis a group homomorphism.
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Proof. For all \( a, b \in G \), we have\n\n\[ \rho \left( {a + b}\right) = \left( {{\rho }_{1}\left( {a + b}\right) ,\ldots ,{\rho }_{k}\left( {a + b}\right) }\right) = \left( {{\rho }_{1}\left( a\right) + {\rho }_{1}\left( b\right) ,\ldots ,{\rho }_{k}\left( a\right) + {\rho }_{k}\left( b\right) }\right) \]\n\n\[ = \rho \left( a\right) + \rho \left( b\right) \text{.} \]
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Yes
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Theorem 6.22. If \( \rho \) is a group isomorphism of \( G \) with \( {G}^{\prime } \), then the inverse function \( {\rho }^{-1} \) is a group isomorphism of \( {G}^{\prime } \) with \( G \) .
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Proof. For all \( {a}^{\prime },{b}^{\prime } \in {G}^{\prime } \), we have\n\n\[ \rho \left( {{\rho }^{-1}\left( {a}^{\prime }\right) + {\rho }^{-1}\left( {b}^{\prime }\right) }\right) = \rho \left( {{\rho }^{-1}\left( {a}^{\prime }\right) }\right) + \rho \left( {{\rho }^{-1}\left( {b}^{\prime }\right) }\right) = {a}^{\prime } + {b}^{\prime }, \]\n\nand hence \( {\rho }^{-1}\left( {a}^{\prime }\right) + {\rho }^{-1}\left( {b}^{\prime }\right) = {\rho }^{-1}\left( {{a}^{\prime } + {b}^{\prime }}\right) \) .
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Yes
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Theorem 6.23 (First isomorphism theorem). Let \( \rho : G \rightarrow {G}^{\prime } \) be a group homomorphism with kernel \( K \) and image \( {H}^{\prime } \). Then we have a group isomorphism\n\n\[ G/K \cong {H}^{\prime } \]\n\nSpecifically, the map\n\n\[ \bar{\rho } : \;G/K \rightarrow {G}^{\prime } \]\n\n\[ {\left\lbrack a\right\rbrack }_{K} \mapsto \rho \left( a\right) \]\n\nis an injective group homomorphism whose image is \( {H}^{\prime } \).
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Proof. Using part (vi) of Theorem 6.19, we see that for all \( a, b \in G \), we have\n\n\[ {\left\lbrack a\right\rbrack }_{K} = {\left\lbrack b\right\rbrack }_{K} \Leftrightarrow a \equiv b\left( {\;\operatorname{mod}\;K}\right) \Leftrightarrow \rho \left( a\right) = \rho \left( b\right) .\n\nThis immediately implies that the definition of \( \bar{\rho } \) is unambiguous \( \left( {\left\lbrack a\right\rbrack }_{K}\right. = {\left\lbrack b\right\rbrack }_{K} \) implies \( \rho \left( a\right) = \rho \left( b\right) \) ), and that \( \bar{\rho } \) is injective \( \left( {\rho \left( a\right) = \rho \left( b\right) }\right. \) implies \( \left. {{\left\lbrack a\right\rbrack }_{K} = {\left\lbrack b\right\rbrack }_{K}}\right) \). It is clear that \( \bar{\rho } \) maps onto \( {H}^{\prime } \), since every element of \( {H}^{\prime } \) is of the form \( \rho \left( a\right) \) for some \( a \in G \), and the map \( \bar{\rho } \) sends \( {\left\lbrack a\right\rbrack }_{K} \) to \( \rho \left( a\right) \). Finally, to see that \( \bar{\rho } \) is a group homomorphism, note that\n\n\[ \bar{\rho }\left( {{\left\lbrack a\right\rbrack }_{K} + {\left\lbrack b\right\rbrack }_{K}}\right) = \bar{\rho }\left( {\left\lbrack a + b\right\rbrack }_{K}\right) = \rho \left( {a + b}\right) = \rho \left( a\right) + \rho \left( b\right) = \bar{\rho }\left( {\left\lbrack a\right\rbrack }_{K}\right) + \bar{\rho }\left( {\left\lbrack b\right\rbrack }_{K}\right) .\n
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Yes
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Theorem 6.24. Let \( \rho : G \rightarrow {G}^{\prime } \) be a group homomorphism. Then for every subgroup \( H \) of \( G \) with \( H \subseteq \operatorname{Ker}\rho \), we may define a group homomorphism\n\n\[ \bar{\rho } : \;G/H \rightarrow {G}^{\prime } \]\n\n\[ {\left\lbrack a\right\rbrack }_{H} \mapsto \rho \left( a\right) . \]\n\nMoreover, \( \operatorname{Im}\bar{\rho } = \operatorname{Im}\rho \), and \( \bar{\rho } \) is injective if and only if \( H = \operatorname{Ker}\rho \) .
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Proof. Using the assumption that \( H \subseteq \operatorname{Ker}\rho \), we see that \( \bar{\rho } \) is unambiguously defined, since for all \( a, b \in G \), we have\n\n\[ {\left\lbrack a\right\rbrack }_{H} = {\left\lbrack b\right\rbrack }_{H} \Rightarrow a \equiv b\left( {\;\operatorname{mod}\;H}\right) \Rightarrow a \equiv b\left( {{\;\operatorname{mod}\;\operatorname{Ker}}\rho }\right) \Rightarrow \rho \left( a\right) = \rho \left( b\right) . \]\n\nThat \( \bar{\rho } \) is a group homomorphism, with \( \operatorname{Im}\bar{\rho } = \operatorname{Im}\rho \), follows as in the proof of Theorem 6.23. If \( H = \operatorname{Ker}\rho \), then by Theorem 6.23, \( \bar{\rho } \) is injective, and if \( H \varsubsetneq \operatorname{Ker}\rho \) , then \( \bar{\rho } \) is not injective, since if we choose \( a \in \operatorname{Ker}\rho \smallsetminus H \), we see that \( \bar{\rho }\left( {\left\lbrack a\right\rbrack }_{H}\right) = {0}_{{G}^{\prime }} \) , and hence \( \operatorname{Ker}\bar{\rho } \) is non-trivial.
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Yes
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Theorem 6.25 (Internal direct product). Let \( G \) be an abelian group with subgroups \( {H}_{1},{H}_{2} \), where \( {H}_{1} \cap {H}_{2} = \left\{ {0}_{G}\right\} \) . Then we have a group isomorphism\n\n\[ \n{H}_{1} \times {H}_{2} \cong {H}_{1} + {H}_{2} \n\]\n\ngiven by the map\n\n\[ \n\rho : \;{H}_{1} \times {H}_{2} \rightarrow {H}_{1} + {H}_{2} \n\]\n\n\[ \n\left( {{a}_{1},{a}_{2}}\right) \mapsto {a}_{1} + {a}_{2} \n\]
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Proof. We already saw that \( \rho \) is a surjective group homomorphism in Example 6.45. To see that \( \rho \) is injective, it suffices to show that \( \operatorname{Ker}\rho \) is trivial; that is, it suffices to show that for all \( {a}_{1} \in {H}_{1} \) and \( {a}_{2} \in {H}_{2} \), if \( {a}_{1} + {a}_{2} = {0}_{G} \), then \( {a}_{1} = {a}_{2} = {0}_{G} \) . But \( {a}_{1} + {a}_{2} = {0}_{G} \) implies \( {a}_{1} = - {a}_{2} \in {H}_{2} \), and hence \( {a}_{1} \in {H}_{1} \cap {H}_{2} = \left\{ {0}_{G}\right\} \), and so \( {a}_{1} = {0}_{G} \) . Similarly, one shows that \( {a}_{2} = {0}_{G} \), and that finishes the proof.
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Yes
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We can use the general theory developed so far to get a quick-and-dirty proof of the Chinese remainder theorem (Theorem 2.6). Let \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) be a pairwise relatively prime family of positive integers, and let \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \) . Consider the map\n\n\[ \rho : \;\mathbb{Z} \rightarrow {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}} \]\n\n\[ a \mapsto \left( {{\left\lbrack a\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack a\right\rbrack }_{{n}_{k}}}\right) . \]
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It is easy to see that this map is a group homomorphism; indeed, it is the map constructed in Theorem 6.21 applied with the natural maps \( {\rho }_{i} : \mathbb{Z} \rightarrow {\mathbb{Z}}_{{n}_{i}} \), for \( i = 1,\ldots, k \) . Evidently, \( a \in \operatorname{Ker}\rho \) if and only if \( {n}_{i} \mid a \) for \( i = 1,\ldots, k \), and since \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) is pairwise relatively prime, it follows that \( a \in \operatorname{Ker}\rho \) if and only if \( n \mid a \) ; that is, \( \operatorname{Ker}\rho = n\mathbb{Z} \) . Theorem 6.23 then gives us an injective group homomorphism\n\n\[ \bar{\rho } : \;{\mathbb{Z}}_{n} \rightarrow {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}} \]\n\n\[ {\left\lbrack a\right\rbrack }_{n} \mapsto \left( {{\left\lbrack a\right\rbrack }_{{n}_{1}},\ldots ,{\left\lbrack a\right\rbrack }_{{n}_{k}}}\right) . \]\n\nBut since the sets \( {\mathbb{Z}}_{n} \) and \( {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}} \) have the same size, injectivity implies surjectivity. From this, Theorem 2.6 is immediate.\n\nThe map \( \bar{\rho } \) is a group isomorphism\n\n\[ {\mathbb{Z}}_{n} \cong {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{k}} \]
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Yes
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Let \( {n}_{1},{n}_{2} \) be positive integers with \( {n}_{1} \mid {n}_{2} \) . Consider the natural map \( \rho : \mathbb{Z} \rightarrow {\mathbb{Z}}_{{n}_{1}} \) . This is a surjective group homomorphism with \( \operatorname{Ker}\rho = {n}_{1}\mathbb{Z} \) . Since \( H \mathrel{\text{:=}} {n}_{2}\mathbb{Z} \subseteq {n}_{1}\mathbb{Z} \), we may apply Theorem 6.24 with the subgroup \( H \) , obtaining the surjective group homomorphism
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\[ \bar{\rho } : \;{\mathbb{Z}}_{{n}_{2}} \rightarrow {\mathbb{Z}}_{{n}_{1}} \] \[ {\left\lbrack a\right\rbrack }_{{n}_{2}} \mapsto {\left\lbrack a\right\rbrack }_{{n}_{1}}. \]
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Yes
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Let \( n \) be a positive integer, and let \( m \) be any integer. Let \( {\rho }_{1} : \mathbb{Z} \rightarrow {\mathbb{Z}}_{n} \) be the natural map, and let \( {\rho }_{2} : {\mathbb{Z}}_{n} \rightarrow {\mathbb{Z}}_{n} \) be the \( m \) -multiplication map. The composed map \( \rho \mathrel{\text{:=}} {\rho }_{2} \circ {\rho }_{1} \) from \( \mathbb{Z} \to {\mathbb{Z}}_{n} \) is also a group homomorphism. For each \( z \in \mathbb{Z} \), we have \( \rho \left( z\right) = m{\left\lbrack z\right\rbrack }_{n} = {\left\lbrack mz\right\rbrack }_{n} \). The kernel of \( \rho \) consists of those integers \( z \) such that \( {mz} \equiv 0\left( {\;\operatorname{mod}\;n}\right) \), and so part (ii) of Theorem 2.5 implies that \( \operatorname{Ker}\rho = \left( {n/d}\right) \mathbb{Z} \), where \( d \mathrel{\text{:=}} \gcd \left( {m, n}\right) \). The image of \( \rho \) is \( m{\mathbb{Z}}_{n} \) .
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Theorem 6.23 therefore implies that the map\n\n\[ \bar{\rho } : \;{\mathbb{Z}}_{n/d} \rightarrow m{\mathbb{Z}}_{n} \]\n\n\[ {\left\lbrack z\right\rbrack }_{n/d} \mapsto m{\left\lbrack z\right\rbrack }_{n} \] \n\nis a group isomorphism.
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Yes
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Consider the group \( {\mathbb{Z}}_{p}^{ * } \) where \( p \) is an odd prime, and let \( \rho : {\mathbb{Z}}_{p}^{ * } \rightarrow {\mathbb{Z}}_{p}^{ * } \) be the squaring map. By definition, \( \operatorname{Im}\rho = {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \), and we proved in Theorem 2.18 that \( \operatorname{Ker}\rho = \{ \left\lbrack {\pm 1}\right\rbrack \} \) . Theorem 2.19 says that for all \( \gamma ,\beta \in {\mathbb{Z}}_{p}^{ * },{\gamma }^{2} = {\beta }^{2} \) if and only if \( \gamma = \pm \beta \) . This fact can also be seen to be a special case of part (vi) of Theorem 6.19. Theorem 6.23 says that \( {\mathbb{Z}}_{p}^{ * }/\operatorname{Ker}\rho \cong \operatorname{Im}\rho \), and since \( \left| {{\mathbb{Z}}_{p}^{ * }/\operatorname{Ker}\rho }\right| = \left| {\mathbb{Z}}_{p}^{ * }\right| /\left| {\operatorname{Ker}\rho }\right| = \left( {p - 1}\right) /2 \), we see that Theorem 2.20, which says that \( \left| {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2}\right| = \left( {p - 1}\right) /2 \), follows from this.
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Let \( H \mathrel{\text{:=}} {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \), and consider the quotient group \( {\mathbb{Z}}_{p}^{ * }/H \) . Since \( \left| H\right| = \left( {p - 1}\right) /2 \) , we know that \( \left| {{\mathbb{Z}}_{p}^{ * }/H}\right| = \left| {\mathbb{Z}}_{p}^{ * }\right| /\left| H\right| = 2 \), and hence \( {\mathbb{Z}}_{p}^{ * }/H \) consists of the two cosets \( H \) and \( \bar{H} \mathrel{\text{:=}} {\mathbb{Z}}_{p}^{ * } \smallsetminus H \) .\n\nLet \( \alpha \) be an arbitrary, fixed element of \( \bar{H} \), and consider the map\n\n\[ \tau : \;\mathbb{Z} \rightarrow {\mathbb{Z}}_{p}^{ * }/H \]\n\n\[ z \mapsto {\left\lbrack {\alpha }^{z}\right\rbrack }_{H} \]\n\nIt is easy to see that \( \tau \) is a group homomorphism; indeed, it is the composition of the homomorphism discussed in Example 6.43 and the natural map from \( {\mathbb{Z}}_{p}^{ * } \) to \( {\mathbb{Z}}_{p}^{ * }/H \) . Moreover, it is easy to see (for example, as a special case of Theorem 2.17) that\n\n\[ {\alpha }^{z} \in H \Leftrightarrow z\text{is even.} \]\n\nFrom this, it follows that \( \operatorname{Ker}\tau = 2\mathbb{Z} \) ; also, since \( {\mathbb{Z}}_{p}^{ * }/H \) consists of just the two cosets \( H \) and \( \bar{H} \), it follows that \( \tau \) is surjective. Therefore, Theorem 6.23 says that the map\n\n\[ \bar{\tau } : \;{\mathbb{Z}}_{2} \rightarrow {\mathbb{Z}}_{p}^{ * }/H \]\n\n\[ {\left\lbrack z\right\rbrack }_{2} \mapsto {\left\lbrack {\alpha }^{z}\right\rbrack }_{H} \]\n\n is a group isomorphism, under which \( {\left\lbrack 0\right\rbrack }_{2} \) corresponds to \( H \), and \( {\left\lbrack 1\right\rbrack }_{2} \) corresponds to \( \bar{H} \) .\n\nThis isomorphism gives another way to derive Theorem 2.23, which says that in \( {\mathbb{Z}}_{p}^{ * } \), the product of two non-squares is a square; indeed, the statement \
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Yes
|
Theorem 6.26. Let \( G \) and \( {G}^{\prime } \) be abelian groups, and consider the group of functions \( \operatorname{Map}\left( {G,{G}^{\prime }}\right) \) . Then\n\n\[ \operatorname{Hom}\left( {G,{G}^{\prime }}\right) \mathrel{\text{:=}} \left\{ {\sigma \in \operatorname{Map}\left( {G,{G}^{\prime }}\right) : \sigma \text{ is a group homomorphism }}\right\} \]\n\n is a subgroup of \( \operatorname{Map}\left( {G,{G}^{\prime }}\right) \) .
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Proof. First, observe that \( \operatorname{Hom}\left( {G,{G}^{\prime }}\right) \) is non-empty, as it contains the map that sends everything in \( G \) to \( {0}_{{G}^{\prime }} \) (this is the identity element of \( \operatorname{Map}\left( {G,{G}^{\prime }}\right) \) ).\n\nNext, we have to show that if \( \sigma \) and \( \tau \) are homomorphisms from \( G \) to \( {G}^{\prime } \), then so are \( \sigma + \tau \) and \( - \sigma \) . But \( \sigma + \tau = {\rho }_{2} \circ {\rho }_{1} \), where \( {\rho }_{1} : G \rightarrow {G}^{\prime } \times {G}^{\prime } \) is the map constructed in Theorem 6.21, applied with \( \sigma \) and \( \tau \), and \( {\rho }_{2} : {G}^{\prime } \times {G}^{\prime } \rightarrow {G}^{\prime } \) is as in Example 6.45. Also, \( - \sigma = {\rho }_{-1} \circ \sigma \), where \( {\rho }_{-1} \) is the \( \left( {-1}\right) \) -multiplication map.
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Yes
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Consider the additive group \( \mathbb{Z} \). This is a cyclic group, with 1 being a generator:
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\[ \langle 1\rangle = \{ z \cdot 1 : z \in \mathbb{Z}\} = \{ z : z \in \mathbb{Z}\} = \mathbb{Z}. \] For every \( m \in \mathbb{Z} \), we have \[ \langle m\rangle = \{ {zm} : z \in \mathbb{Z}\} = \{ {mz} : z \in \mathbb{Z}\} = m\mathbb{Z}. \] It follows that the only elements of \( \mathbb{Z} \) that generate \( \mathbb{Z} \) are 1 and -1: every other element generates a subgroup that is strictly contained in \( \mathbb{Z} \).
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Yes
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For \( n > 0 \), consider the additive group \( {\mathbb{Z}}_{n} \). This is a cyclic group, with [1] being a generator:
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\[ \langle \left\lbrack 1\right\rbrack \rangle = \{ z\left\lbrack 1\right\rbrack : z \in \mathbb{Z}\} = \{ \left\lbrack z\right\rbrack : z \in \mathbb{Z}\} = {\mathbb{Z}}_{n}. \] For every \( m \in \mathbb{Z} \), we have \[ \langle \left\lbrack m\right\rbrack \rangle = \{ z\left\lbrack m\right\rbrack : z \in \mathbb{Z}\} = \{ \left\lbrack {zm}\right\rbrack : z \in \mathbb{Z}\} = \{ m\left\lbrack z\right\rbrack : z \in \mathbb{Z}\} = m{\mathbb{Z}}_{n}. \] By Example 6.23, the subgroup \( m{\mathbb{Z}}_{n} \) has order \( n/\gcd \left( {m, n}\right) \). Thus, \( \left\lbrack m\right\rbrack \) has order \( n/\gcd \left( {m, n}\right) \); in particular, \( \left\lbrack m\right\rbrack \) generates \( {\mathbb{Z}}_{n} \) if and only if \( m \) is relatively prime to \( n \), and hence, the number of generators of \( {\mathbb{Z}}_{n} \) is \( \varphi \left( n\right) \).
|
Yes
|
Theorem 6.27. Let \( G \) be a cyclic group generated by \( a \) . Then for every \( m \in \mathbb{Z} \) , we have\n\n\[ \langle {ma}\rangle = {mG}\text{.} \]
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Proof. We have\n\n\[ \langle {ma}\rangle = \{ z\left( {ma}\right) : z \in \mathbb{Z}\} = \{ m\left( {za}\right) : z \in \mathbb{Z}\} = m\langle a\rangle = {mG}. \]
|
Yes
|
Consider the additive group \( G \mathrel{\text{:=}} \mathbb{Z} \times \mathbb{Z} \). Set\n\n\[ {\alpha }_{1} \mathrel{\text{:=}} \left( {1,0}\right) \in G\text{ and }{\alpha }_{2} \mathrel{\text{:=}} \left( {0,1}\right) \in G. \]\n\nIt is not hard to see that \( G = \left\langle {{\alpha }_{1},{\alpha }_{2}}\right\rangle \), since for all \( {z}_{1},{z}_{2} \in \mathbb{Z} \), we have\n\n\[ {z}_{1}{\alpha }_{1} + {z}_{2}{\alpha }_{2} = \left( {{z}_{1},0}\right) + \left( {0,{z}_{2}}\right) = \left( {{z}_{1},{z}_{2}}\right) . \]\n\nHowever, \( G \) is not cyclic.
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To see this, let \( \beta = \left( {{b}_{1},{b}_{2}}\right) \) be an arbitrary element of \( G \). We claim that one of \( {\alpha }_{1} \) or \( {\alpha }_{2} \) does not belong to \( \langle \beta \rangle \). Suppose to the contrary that both \( {\alpha }_{1} \) and \( {\alpha }_{2} \) belong to \( \langle \beta \rangle \). This would imply that there exist integers \( z \) and \( {z}^{\prime } \) such that\n\n\[ z{b}_{1} = 1 \]\n\n\[ z{b}_{2} = 0, \]\n\n\[ {z}^{\prime }{b}_{1} = 0 \]\n\n\[ {z}^{\prime }{b}_{2} = 1.\text{ \}\]\n\nMultiplying the upper left equality by the lower right, and the upper right by the lower left, we obtain\n\n\[ 1 = z{z}^{\prime }{b}_{1}{b}_{2} = 0, \]\n\nwhich is impossible.
|
Yes
|
Consider the additive group \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times {\mathbb{Z}}_{{n}_{2}} \). Set \[ {\alpha }_{1} \mathrel{\text{:=}} \left( {{\left\lbrack 1\right\rbrack }_{{n}_{1}},{\left\lbrack 0\right\rbrack }_{{n}_{2}}}\right) \in G\text{ and }{\alpha }_{2} \mathrel{\text{:=}} \left( {{\left\lbrack 0\right\rbrack }_{{n}_{1}},{\left\lbrack 1\right\rbrack }_{{n}_{2}}}\right) \in G. \] It is not hard to see that \( G = \left\langle {{\alpha }_{1},{\alpha }_{2}}\right\rangle \), since for all \( {z}_{1},{z}_{2} \in \mathbb{Z} \), we have \[ {z}_{1}{\alpha }_{1} + {z}_{2}{\alpha }_{2} = \left( {{\left\lbrack {z}_{1}\right\rbrack }_{{n}_{1}},{\left\lbrack 0\right\rbrack }_{{n}_{2}}}\right) + \left( {{\left\lbrack 0\right\rbrack }_{{n}_{1}},{\left\lbrack {z}_{2}\right\rbrack }_{{n}_{2}}}\right) = \left( {{\left\lbrack {z}_{1}\right\rbrack }_{{n}_{1}},{\left\lbrack {z}_{2}\right\rbrack }_{{n}_{2}}}\right) . \] However, \( G \) may or may not be cyclic: it depends on \( d \mathrel{\text{:=}} \gcd \left( {{n}_{1},{n}_{2}}\right) \).
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If \( d = 1 \), then \( G \) is cyclic, with \( \alpha \mathrel{\text{:=}} \left( {{\left\lbrack 1\right\rbrack }_{{n}_{1}},{\left\lbrack 1\right\rbrack }_{{n}_{2}}}\right) \) being a generator. One can see this easily using the Chinese remainder theorem: for all \( {z}_{1},{z}_{2} \in \mathbb{Z} \), there exists \( z \in \mathbb{Z} \) such that \[ z \equiv {z}_{1}\left( {\;\operatorname{mod}\;{n}_{1}}\right) \text{ and }z \equiv {z}_{2}\left( {\;\operatorname{mod}\;{n}_{2}}\right) , \] which implies \[ {z\alpha } = \left( {{\left\lbrack z\right\rbrack }_{{n}_{1}},{\left\lbrack z\right\rbrack }_{{n}_{2}}}\right) = \left( {{\left\lbrack {z}_{1}\right\rbrack }_{{n}_{1}},{\left\lbrack {z}_{2}\right\rbrack }_{{n}_{2}}}\right) . \] If \( d > 1 \), then \( G \) is not cyclic. To see this, let \( \beta = \left( {{\left\lbrack {b}_{1}\right\rbrack }_{{n}_{1}},{\left\lbrack {b}_{2}\right\rbrack }_{{n}_{2}}}\right) \) be an arbitrary element of \( G \). We claim that one of \( {\alpha }_{1} \) or \( {\alpha }_{2} \) does not belong to \( \langle \beta \rangle \). Suppose to the contrary that both \( {\alpha }_{1} \) and \( {\alpha }_{2} \) belong to \( \langle \beta \rangle \). This would imply that there exist integers \( z \) and \( {z}^{\prime } \) such that \[ z{b}_{1} \equiv 1\left( {\;\operatorname{mod}\;{n}_{1}}\right) ,\;z{b}_{2} \equiv 0\left( {\;\operatorname{mod}\;{n}_{2}}\right) , \] \[ {z}^{\prime }{b}_{1} \equiv 0\left( {\;\operatorname{mod}\;{n}_{1}}\right) ,\;{z}^{\prime }{b}_{2} \equiv 1\left( {\;\operatorname{mod}\;{n}_{2}}\right) . \] All of these congruences hold modulo \( d \) as well, and multiplying the upper left congruence by the lower right, and the upper right by the lower left, we obtain \[ 1 \equiv z{z}^{\prime }{b}_{1}{b}_{2} \equiv 0\left( {\;\operatorname{mod}\;d}\right) \] which is impossible.
|
Yes
|
Theorem 6.28. Let \( \rho : G \rightarrow {G}^{\prime } \) be a group isomorphism.\n\n(i) For all \( a \in G \), we have \( \rho \left( {\langle a\rangle }\right) = \langle \rho \left( a\right) \rangle \) .
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Proof. For all \( a \in G \), we have\n\n\[ \rho \left( {\langle a\rangle }\right) = \{ \rho \left( {za}\right) : z \in \mathbb{Z}\} = \{ {z\rho }\left( a\right) : z \in \mathbb{Z}\} = \langle \rho \left( a\right) \rangle . \]\n\nThat proves (i).
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Yes
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Consider again the additive group \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times {\mathbb{Z}}_{{n}_{2}} \), discussed in Example 6.56. If \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \), then one can also see that \( G \) is cyclic as follows:
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by the discussion in Example 6.48, we know that \( G \) is isomorphic to \( {\mathbb{Z}}_{{n}_{1}{n}_{2}} \), and since \( {\mathbb{Z}}_{{n}_{1}{n}_{2}} \) is cyclic, so is \( \mathbf{G} \).
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Yes
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Consider again the subgroup \( m{\mathbb{Z}}_{n} \) of \( {\mathbb{Z}}_{n} \), discussed in Example 6.54. One can also see that this is cyclic of order \( n/d \), where \( d \mathrel{\text{:=}} \gcd \left( {m, n}\right) \).
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in Example 6.50, we constructed an isomorphism between \( {\mathbb{Z}}_{n/d} \) and \( m{\mathbb{Z}}_{n} \) , and this implies \( m{\mathbb{Z}}_{n} \) is cyclic of order \( n/d \) .
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Yes
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Theorem 6.30. Let \( G \) be a finite abelian group and let \( a \in G \) . Then \( \left| G\right| a = {0}_{G} \) and the order of a divides \( \left| G\right| \) .
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Proof. Since \( \langle a\rangle \) is a subgroup of \( G \), by Lagrange’s theorem (Theorem 6.15), the order of \( a \) divides \( \left| G\right| \) . It then follows by Theorem 6.29 that \( \left| G\right| a = {0}_{G} \) .
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Yes
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The group \( {\mathbb{Z}}_{5}^{ * } \) is cyclic, with [2] being a generator:
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\[ {\left\lbrack 2\right\rbrack }^{2} = \left\lbrack 4\right\rbrack = \left\lbrack {-1}\right\rbrack ,\;{\left\lbrack 2\right\rbrack }^{3} = \left\lbrack {-2}\right\rbrack ,\;{\left\lbrack 2\right\rbrack }^{4} = \left\lbrack 1\right\rbrack . \]
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Yes
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Consider again the additive group \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times {\mathbb{Z}}_{{n}_{2}} \), discussed in Example 6.56. If \( d \mathrel{\text{:=}} \gcd \left( {{n}_{1},{n}_{2}}\right) > 1 \), then one can also see that \( G \) is not cyclic as follows:
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for every \( \beta \in G \), we have \( \left( {{n}_{1}{n}_{2}/d}\right) \beta = {0}_{G} \), and hence by Theorem 6.29, the order of \( \beta \) divides \( {n}_{1}{n}_{2}/d \) .
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Yes
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Theorem 6.31. Let \( G \) be a cyclic group of infinite order.\n\n(i) \( G \) is isomorphic to \( \mathbb{Z} \) .\n\n(ii) There is a one-to-one correspondence between the non-negative integers and the subgroups of \( G \), where each such integer \( m \) corresponds to the cyclic group \( {mG} \) .\n\n(iii) For every two non-negative integers \( m,{m}^{\prime } \), we have \( {mG} \subseteq {m}^{\prime }G \) if or only if \( {m}^{\prime } \mid m \) .
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Proof. That \( G \cong \mathbb{Z} \) was established in our classification of cyclic groups, and so it suffices to prove the other statements of the theorem for \( G = \mathbb{Z} \) . As we saw in Example 6.53, for every integer \( m \), the subgroup \( m\mathbb{Z} \) is cyclic, as it is generated by \( m \) . This fact, together with Theorem 6.9, establishes all the other statements.
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No
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Theorem 6.32. Let \( G \) be a cyclic group of finite order \( n \) . (i) \( G \) is isomorphic to \( {\mathbb{Z}}_{n} \) .
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Proof. That \( G \cong {\mathbb{Z}}_{n} \) was established in our classification of cyclic groups, and so it suffices to prove the other statements of the theorem for \( G = {\mathbb{Z}}_{n} \) .
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No
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Theorem 6.33. If \( G \) is an abelian group of prime order, then \( G \) is cyclic.
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Proof. Let \( \left| G\right| = p \), which, by hypothesis, is prime. Let \( a \in G \) with \( a \neq {0}_{G} \), and let \( k \) be the order of \( a \) . As the order of an element divides the order of the group, we have \( k \mid p \), and so \( k = 1 \) or \( k = p \) . Since \( a \neq {0}_{G} \), we must have \( k \neq 1 \), and so \( k = p \), which implies that \( a \) generates \( G \) .
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Yes
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Theorem 6.34. If \( {G}_{1} \) and \( {G}_{2} \) are finite cyclic groups of relatively prime order, then \( {G}_{1} \times {G}_{2} \) is also cyclic. In particular, if \( {G}_{1} \) is generated by \( {a}_{1} \) and \( {G}_{2} \) is generated by \( {a}_{2} \), then \( {G}_{1} \times {G}_{2} \) is generated by \( \left( {{a}_{1},{a}_{2}}\right) \) .
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Proof. We give a direct proof, based on Theorem 6.29. Let \( {n}_{1} \mathrel{\text{:=}} \left| {G}_{1}\right| \) and \( {n}_{2} \mathrel{\text{:=}} \left| {G}_{2}\right| \), where \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \) . Also, let \( {a}_{1} \in {G}_{1} \) have order \( {n}_{1} \) and \( {a}_{2} \in {G}_{2} \) have order \( {n}_{2} \) . We want to show that \( \left( {{a}_{1},{a}_{2}}\right) \) has order \( {n}_{1}{n}_{2} \) . Applying Theorem 6.29 to \( \left( {{a}_{1},{a}_{2}}\right) \), we see that the order of \( \left( {{a}_{1},{a}_{2}}\right) \) is the smallest positive integer \( k \) such that \( k\left( {{a}_{1},{a}_{2}}\right) = \left( {{0}_{{G}_{1}},{0}_{{G}_{2}}}\right) \) . Now, for every integer \( k \), we have \( k\left( {{a}_{1},{a}_{2}}\right) = \left( {k{a}_{1}, k{a}_{2}}\right) \), and\n\n\[ \left( {k{a}_{1}, k{a}_{2}}\right) = \left( {{0}_{{G}_{1}},{0}_{{G}_{2}}}\right) \Leftrightarrow {n}_{1}\left| {k\text{ and }{n}_{2}}\right| k \]\n\n\[ \text{(applying Theorem 6.29 to}{a}_{1}\text{and}{a}_{2}\text{)} \]\n\n\[ \Leftrightarrow {n}_{1}{n}_{2} \mid k\text{ (since }\gcd \left( {{n}_{1},{n}_{2}}\right) = 1\text{ ). } \]
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Yes
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Theorem 6.35. Let \( G \) be a cyclic group. Then for every subgroup \( H \) of \( G \), both \( H \) and \( G/H \) are cyclic.
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Proof. The fact that \( H \) is cyclic follows from part (ii) of Theorem 6.31 in the case where \( G \) is infinite, and part (ii) of Theorem 6.32 in the case where \( G \) is finite. If \( G \) is generated by \( a \), then it is easy to see that \( G/H \) is generated by \( {\left\lbrack a\right\rbrack }_{H} \) .
|
No
|
Theorem 6.36. Let \( G \) be an abelian group, let \( a \in G \) be of finite order \( n \), and let \( m \) be an arbitrary integer. Then the order of \( {ma} \) is \( n/\gcd \left( {m, n}\right) \) .
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Proof. Let \( H \mathrel{\text{:=}} \langle a\rangle \), and \( d \mathrel{\text{:=}} \gcd \left( {m, n}\right) \) . By Theorem 6.27, we have \( \langle {ma}\rangle = {mH} \) , and by Theorem 6.32, we have \( {mH} = {dH} \), which has order \( n/d \) .\n\nThat proves the theorem. Alternatively, we can give a direct proof, based on Theorem 6.29. Applying Theorem 6.29 to \( {ma} \), we see that the order of \( {ma} \) is the smallest positive integer \( k \) such that \( k\left( {ma}\right) = {0}_{G} \) . Now, for every integer \( k \), we have \( k\left( {ma}\right) = \left( {km}\right) a \), and\n\n\[ \left( {km}\right) a = {0}_{G} \Leftrightarrow {km} \equiv 0\left( {\;\operatorname{mod}\;n}\right) \text{ (applying Theorem 6.29 to }a\text{ ) } \]\n\n\[ \Leftrightarrow k \equiv 0\left( {{\;\operatorname{mod}\;n}/\gcd \left( {m, n}\right) }\right) \text{(by part (ii) of Theorem 2.5).} \]
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Yes
|
Theorem 6.37. Suppose that \( a \) is an element of an abelian group, and for some prime \( p \) and integer \( e \geq 1 \), we have \( {p}^{e}a = {0}_{G} \) and \( {p}^{e - 1}a \neq {0}_{G} \) . Then \( a \) has order \( {p}^{e} \) .
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Proof. If \( m \) is the order of \( a \), then since \( {p}^{e}a = {0}_{G} \), we have \( m \mid {p}^{e} \) . So \( m = {p}^{f} \) for some \( f = 0,\ldots, e \) . If \( f < e \), then \( {p}^{e - 1}a = {0}_{G} \), contradicting the assumption that \( {p}^{e - 1}a \neq {0}_{G} \).
|
Yes
|
Theorem 6.38. Suppose \( G \) is an abelian group with \( {a}_{1},{a}_{2} \in G \) such that \( {a}_{1} \) is of finite order \( {n}_{1},{a}_{2} \) is of finite order \( {n}_{2} \), and \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \) . Then the order of \( {a}_{1} + {a}_{2} \) is \( {n}_{1}{n}_{2} \) .
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Proof. Let \( {H}_{1} \mathrel{\text{:=}} \left\langle {a}_{1}\right\rangle \) and \( {H}_{2} \mathrel{\text{:=}} \left\langle {a}_{2}\right\rangle \) so that \( \left| {H}_{1}\right| = {n}_{1} \) and \( \left| {H}_{2}\right| = {n}_{2} \). First, we claim that \( {H}_{1} \cap {H}_{2} = \left\{ {0}_{G}\right\} \) . To see this, observe that \( {H}_{1} \cap {H}_{2} \) is a subgroup of \( {H}_{1} \), and so \( \left| {{H}_{1} \cap {H}_{2}}\right| \) divides \( {n}_{1} \) ; similarly, \( \left| {{H}_{1} \cap {H}_{2}}\right| \) divides \( {n}_{2} \) . Since \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \), we must have \( \left| {{H}_{1} \cap {H}_{2}}\right| = 1 \), and that proves the claim. Using the claim, we can apply Theorem 6.25, obtaining a group isomorphism between \( {H}_{1} + {H}_{2} \) and \( {H}_{1} \times {H}_{2} \) . Under this isomorphism, the group element \( {a}_{1} + {a}_{2} \in {H}_{1} + {H}_{2} \) corresponds to \( \left( {{a}_{1},{a}_{2}}\right) \in {H}_{1} \times {H}_{2} \), which by Theorem 6.34 (again using the fact that \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \) ) has order \( {n}_{1}{n}_{2} \) .
|
Yes
|
Theorem 6.39. Let \( G \) be an abelian group of exponent \( m \). (i) For every integer \( k, k \) kills \( G \) if and only if \( m \mid k \).
|
Proof. Exercise.
|
No
|
Theorem 6.40. If \( {G}_{1} \) and \( {G}_{2} \) are abelian groups of exponents \( {m}_{1} \) and \( {m}_{2} \), then the exponent of \( {G}_{1} \times {G}_{2} \) is \( \operatorname{lcm}\left( {{m}_{1},{m}_{2}}\right) \) .
|
Proof. Exercise.
|
No
|
If an abelian group \( G \) has non-zero exponent \( m \), then \( G \) contains an element of order \( m \).
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The second statement follows immediately from the first. For the first statement, let \( m = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}^{{e}_{i}} \) be the prime factorization of \( m \) .\n\nFirst, we claim that for each \( i = 1,\ldots, r \), there exists \( {a}_{i} \in G \) such that \( \left( {m/{p}_{i}}\right) {a}_{i} \neq \) \( {0}_{G} \) . Suppose the claim were false: then for some \( i,\left( {m/{p}_{i}}\right) a = {0}_{G} \) for all \( a \in G \) ; however, this contradicts the minimality property in the definition of the exponent \( m \) . That proves the claim.\n\nLet \( {a}_{1},\ldots ,{a}_{r} \) be as in the above claim. Then by Theorem 6.37, \( \left( {m/{p}_{i}^{{e}_{i}}}\right) {a}_{i} \) has order \( {p}_{i}^{{e}_{i}} \) for each \( i = 1,\ldots, r \) . Finally, by Theorem 6.38, the group element\n\n\[ \left( {m/{p}_{1}^{{e}_{1}}}\right) {a}_{1} + \cdots + \left( {m/{p}_{r}^{{e}_{r}}}\right) {a}_{r} \]\n\nhas order \( m \) .
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Yes
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Theorem 6.42. Let \( G \) be a finite abelian group of order \( n \) . If \( p \) is a prime dividing \( n \), then \( G \) contains an element of order \( p \) .
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Proof. We can prove this by induction on \( n \) .\n\nIf \( n = 1 \), then the theorem is vacuously true.\n\nNow assume \( n > 1 \) and that the theorem holds for all groups of order strictly less than \( n \) . Let \( a \) be any non-zero element of \( G \), and let \( m \) be the order of \( a \) . Since \( a \) is non-zero, we must have \( m > 1 \) . If \( p \mid m \), then \( \left( {m/p}\right) a \) is an element of order \( p \), and we are done. So assume that \( p \nmid m \) and consider the quotient group \( G/H \), where \( H \) is the subgroup of \( G \) generated by \( a \) . Since \( H \) has order \( m, G/H \) has order \( n/m \) , which is strictly less than \( n \), and since \( p \nmid m \), we must have \( p \mid \left( {n/m}\right) \) . So we can apply the induction hypothesis to the group \( G/H \) and the prime \( p \), which says that there is an element \( b \in G \) such that the coset \( {\left\lbrack b\right\rbrack }_{H} \in G/H \) has order \( p \) . If \( \ell \) is the order of \( b \), then \( \ell b = {0}_{G} \), and so \( \ell b \equiv {0}_{G}\left( {\;\operatorname{mod}\;H}\right) \), which implies that the order of \( {\left\lbrack b\right\rbrack }_{H} \) divides \( \ell \) . Thus, \( p \mid \ell \), and so \( \left( {\ell /p}\right) b \) is an element of \( G \) of order \( p \) .
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Yes
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Theorem 6.43. Let \( G \) be a finite abelian group. Then the primes dividing the exponent of \( G \) are the same as the primes dividing its order.
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Proof. Since the exponent divides the order, every prime dividing the exponent must divide the order. Conversely, if a prime \( p \) divides the order, then since there is an element of order \( p \) in the group, the exponent must be divisible by \( p \) .
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Yes
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Theorem 6.44 (Fundamental theorem of finite abelian groups). A finite abelian group (with more than one element) is isomorphic to a direct product of cyclic groups
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\[ {\mathbb{Z}}_{{p}_{1}^{{e}_{1}}} \times \cdots \times {\mathbb{Z}}_{{p}_{r}^{{e}_{r}}} \] where the \( {p}_{i} \) ’s are primes (not necessarily distinct) and the \( {e}_{i} \) ’s are positive integers. This direct product of cyclic groups is unique up to the order of the factors.
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Yes
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Lemma 6.47. Suppose that \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{m}_{1}} \times \cdots \times {\mathbb{Z}}_{{m}_{t}} \) and \( H \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{t}} \) are isomorphic, where the \( {m}_{i} \) ’s and \( {n}_{i} \) ’s are positive integers (possibly 1) such that \( {m}_{i} \mid {m}_{i + 1} \) and \( {n}_{i} \mid {n}_{i + 1} \) for \( i = 1,\ldots, t - 1 \) . Then \( {m}_{i} = {n}_{i} \) for \( i = 1,\ldots, t \) .
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Proof. Clearly, \( \mathop{\prod }\limits_{i}{m}_{i} = \left| G\right| = \left| H\right| = \mathop{\prod }\limits_{i}{n}_{i} \) . We prove the lemma by induction on the order of the group. If the group order is 1, then clearly all the \( {m}_{i} \) ’s and \( {n}_{i} \) ’s must be 1, and we are done. Otherwise, let \( p \) be a prime dividing the group order. Now, suppose that \( p \) divides \( {m}_{r},\ldots ,{m}_{t} \) but not \( {m}_{1},\ldots ,{m}_{r - 1} \), and that \( p \) divides \( {n}_{s},\ldots ,{n}_{t} \) but not \( {n}_{1},\ldots ,{n}_{s - 1} \), where \( r \leq t \) and \( s \leq t \) . Evidently, the groups \( {pG} \) and \( {pH} \) are isomorphic. Moreover,\n\n\[ \n{pG} \cong {\mathbb{Z}}_{{m}_{1}} \times \cdots \times {\mathbb{Z}}_{{m}_{r - 1}} \times {\mathbb{Z}}_{{m}_{r}/p} \times \cdots \times {\mathbb{Z}}_{{m}_{t}/p}, \n\]\n\nand\n\n\[ \n{pH} \cong {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{s - 1}} \times {\mathbb{Z}}_{{n}_{s}/p} \times \cdots \times {\mathbb{Z}}_{{n}_{t}/p}. \n\]\n\nThus, we see that \( \left| {pG}\right| = \left| G\right| /{p}^{t - r + 1} \) and \( \left| {pH}\right| = \left| H\right| /{p}^{t - s + 1} \), from which it follows that \( r = s \), and the lemma then follows by induction.
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Yes
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Theorem 7.2. Let \( R \) be a ring. Then:\n\n(i) the multiplicative identity \( {1}_{R} \) is unique;
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Proof. Part (i) may be proved using the same argument as was used to prove part (i) of Theorem 6.2.
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No
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The set \( \mathbb{C} \) of complex numbers under the usual rules of multiplication and addition forms a ring.
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Every \( \alpha \in \mathbb{C} \) can be written (uniquely) as \( \alpha = a + {bi} \) , where \( a, b \in \mathbb{R} \) and \( i = \sqrt{-1} \) . If \( {\alpha }^{\prime } = {a}^{\prime } + {b}^{\prime }i \) is another complex number, with \( {a}^{\prime },{b}^{\prime } \in \mathbb{R} \), then\n\n\[ \alpha + {\alpha }^{\prime } = \left( {a + {a}^{\prime }}\right) + \left( {b + {b}^{\prime }}\right) i\text{ and }\alpha {\alpha }^{\prime } = \left( {a{a}^{\prime } - b{b}^{\prime }}\right) + \left( {a{b}^{\prime } + {a}^{\prime }b}\right) i. \]\n\nThe fact that \( \mathbb{C} \) is a ring can be verified by direct calculation; however, we shall see later that this follows easily from more general considerations.
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No
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Consider the set \( \mathcal{F} \) of all arithmetic functions, that is, functions mapping positive integers to reals. Let us define addition of arithmetic functions point-wise (i.e., \( \left( {f + g}\right) \left( n\right) = f\left( n\right) + g\left( n\right) \) for all positive integers \( n \) ) and multiplication using the Dirichlet product, introduced in \( §{2.9} \).
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The reader should verify that with addition and multiplication so defined, \( \mathcal{F} \) forms a ring, where the all-zero function is the additive identity, and the special function \( I \) defined in \( §{2.9} \) is the multiplicative identity.
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No
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Generalizing Example 6.19, if \( I \) is an arbitrary set and \( R \) is a ring, then \( \operatorname{Map}\left( {I, R}\right) \), which is the set of all functions \( f : I \rightarrow R \), may be naturally viewed as a ring, with addition and multiplication defined point-wise: for \( f, g \in \operatorname{Map}\left( {I, R}\right) \), we define\n\n\[ \left( {f + g}\right) \left( i\right) \mathrel{\text{:=}} f\left( i\right) + g\left( i\right) \text{and}\left( {f \cdot g}\right) \left( i\right) \mathrel{\text{:=}} f\left( i\right) \cdot g\left( i\right) \text{for all}i \in I\text{.} \]
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We leave it to the reader to verify that \( \operatorname{Map}\left( {I, R}\right) \) is indeed a ring, where the additive identity is the all-zero function, and the multiplicative identity is the all-one function.
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No
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For non-zero \( \alpha = a + {bi} \in \mathbb{C} \), with \( a, b \in \mathbb{R} \), we have \( c \mathrel{\text{:=}} N\left( \alpha \right) = \) \( {a}^{2} + {b}^{2} > 0 \). It follows that the complex number \( \bar{\alpha }{c}^{-1} = \left( {a{c}^{-1}}\right) + \left( {-b{c}^{-1}}\right) i \) is the multiplicative inverse of \( \alpha \)
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since \( \alpha \cdot \bar{\alpha }{c}^{-1} = \left( {\alpha \bar{\alpha }}\right) {c}^{-1} = 1 \) . Hence, every non-zero element of \( \mathbb{C} \) is a unit, and so \( \mathbb{C} \) is a field.
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Yes
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For rings \( {R}_{1},\ldots ,{R}_{k} \), it is easy to see that the multiplicative group of units of the direct product \( {R}_{1} \times \cdots \times {R}_{k} \) is equal to \( {R}_{1}^{ * } \times \cdots \times {R}_{k}^{ * } \).
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Indeed, by definition, \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) has a multiplicative inverse if and only if each individual \( {a}_{i} \) does.
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Yes
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Consider the ring \( \mathcal{F} \) of arithmetic functions defined in Example 7.6.
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By the result of Exercise 2.54, \( {\mathcal{F}}^{ * } = \{ f \in \mathcal{F} : f\left( 1\right) \neq 0\} \)
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No
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For \( n > 1,{\mathbb{Z}}_{n} \) is an integral domain if and only if \( n \) is prime.
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In particular, if \( n \) is composite, so \( n = {ab} \) with \( 1 < a < n \) and \( 1 < b < n \), then \( {\left\lbrack a\right\rbrack }_{n} \) and \( {\left\lbrack b\right\rbrack }_{n} \) are zero divisors: \( {\left\lbrack a\right\rbrack }_{n}{\left\lbrack b\right\rbrack }_{n} = {\left\lbrack 0\right\rbrack }_{n} \), but \( {\left\lbrack a\right\rbrack }_{n} \neq {\left\lbrack 0\right\rbrack }_{n} \) and \( {\left\lbrack b\right\rbrack }_{n} \neq {\left\lbrack 0\right\rbrack }_{n} \).
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No
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Theorem 7.3. If \( R \) is a ring, and \( a, b, c \in R \) such that \( a \neq 0 \) and \( a \) is not a zero divisor, then \( {ab} = {ac} \) implies \( b = c \) .
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Proof. \( {ab} = {bc} \) implies \( a\left( {b - c}\right) = 0 \) . The fact that \( a \neq 0 \) and \( a \) is not a zero divisor implies that we must have \( b - c = 0 \), and so \( b = c \) .
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Yes
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Theorem 7.4. Let \( R \) be a ring.\n\n(i) Suppose \( a, b \in R \), and that either \( a \) or \( b \) is not a zero divisor. Then \( a \mid b \) and \( b \mid a \) if and only if \( {ar} = b \) for some \( r \in {R}^{ * } \).
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Proof. For the first statement, if \( {ar} = b \) for some \( r \in {R}^{ * } \), then we also have \( b{r}^{-1} = a \) ; thus, \( a\left| {b\text{and}b}\right| a \) . For the converse, suppose that \( a\left| {b\text{and}b}\right| a \) . We may assume that \( b \) is not a zero divisor (otherwise, exchange the roles of \( a \) and \( b \) ). We may also assume that \( b \) is non-zero (otherwise, \( b \mid a \) implies \( a = 0 \), and so the conclusion holds with any \( r \) ). Now, \( a \mid b \) implies \( {ar} = b \) for some \( r \in R \), and \( b \mid a \) implies \( b{r}^{\prime } = a \) for some \( {r}^{\prime } \in R \), and hence \( b = {ar} = b{r}^{\prime }r \) . Canceling \( b \) from both sides of the equation \( b = b{r}^{\prime }r \), we obtain \( 1 = {r}^{\prime }r \), and so \( r \) is a unit.
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Yes
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Theorem 7.5. The characteristic of an integral domain is either zero or a prime.
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Proof. By way of contradiction, suppose that \( D \) is an integral domain with characteristic \( m \) that is neither zero nor prime. Since, by definition, \( D \) is not a trivial ring, we cannot have \( m = 1 \), and so \( m \) must be composite. Say \( m = {st} \), where \( 1 < s < m \) and \( 1 < t < m \) . Since \( m \) is the additive order of \( {1}_{D} \), it follows that \( \left( {s \cdot {1}_{D}}\right) \neq {0}_{D} \) and \( \left( {t \cdot {1}_{D}}\right) \neq {0}_{D} \) ; moreover, since \( D \) is an integral domain, it follows that \( \left( {s \cdot {1}_{D}}\right) \left( {t \cdot {1}_{D}}\right) \neq {0}_{D} \) . So we have\n\n\[ \n{0}_{D} = m \cdot {1}_{D} = \left( {st}\right) \cdot {1}_{D} = \left( {s \cdot {1}_{D}}\right) \left( {t \cdot {1}_{D}}\right) \neq {0}_{D}, \n\]\n\na contradiction.
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Yes
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Theorem 7.6. Every finite integral domain is a field.
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Proof. Let \( D \) be a finite integral domain, and let \( a \) be any non-zero element of \( D \) . Consider the \( a \) -multiplication map that sends \( b \in D \) to \( {ab} \), which is a group homomorphism on the additive group of \( D \) . Since \( a \) is not a zero-divisor, it follows that the kernel of the \( a \) -multiplication map is \( \left\{ {0}_{D}\right\} \), hence the map is injective, and by finiteness, it must be surjective as well. In particular, there must be an element \( b \in D \) such that \( {ab} = {1}_{D} \).
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Yes
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Theorem 7.7. Every finite field \( F \) must be of cardinality \( {p}^{w} \), where \( p \) is prime, \( w \) is a positive integer, and \( p \) is the characteristic of \( F \) .
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Proof. By Theorem 7.5, the characteristic of \( F \) is either zero or a prime, and since \( F \) is finite, it must be prime. Let \( p \) denote the characteristic. By definition, \( p \) is the exponent of the additive group of \( F \), and by Theorem 6.43, the primes dividing the exponent are the same as the primes dividing the order, and hence \( F \) must have cardinality \( {p}^{w} \) for some positive integer \( w \) .
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Yes
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Example 7.24. \( \mathbb{R} \) is a subring of \( \mathbb{C} \) .
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Note that for all \( \alpha \mathrel{\text{:=}} a + {bi} \in \mathbb{C} \), with \( a, b \in \mathbb{R} \) , we have \( \bar{\alpha } = \alpha \Leftrightarrow a + {bi} = a - {bi} \Leftrightarrow b = 0 \) . That is, \( \bar{\alpha } = \alpha \Leftrightarrow \alpha \in \mathbb{R} \) .
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No
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Let us determine the units of \( \mathbb{Z}\left\lbrack i\right\rbrack \) . Suppose \( \alpha \in \mathbb{Z}\left\lbrack i\right\rbrack \) is a unit, so that there exists \( {\alpha }^{\prime } \in \mathbb{Z}\left\lbrack i\right\rbrack \) such that \( \alpha {\alpha }^{\prime } = 1 \) .
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Taking norms, we obtain\n\n\[ 1 = N\left( 1\right) = N\left( {\alpha {\alpha }^{\prime }}\right) = N\left( \alpha \right) N\left( {\alpha }^{\prime }\right) . \]\n\nSince the norm of any Gaussian integer is itself a non-negative integer, and since \( N\left( \alpha \right) N\left( {\alpha }^{\prime }\right) = 1 \), we must have \( N\left( \alpha \right) = 1 \) . Now, if \( \alpha = a + {bi} \), with \( a, b \in \mathbb{Z} \), then \( 1 = N\left( \alpha \right) = {a}^{2} + {b}^{2} \), which implies that \( \alpha = \pm 1 \) or \( \alpha = \pm i \) . Conversely, it is easy to see that \( \pm 1 \) and \( \pm i \) are indeed units, and so these are the only units in \( \mathbb{Z}\left\lbrack i\right\rbrack \) .
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Yes
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Let \( m \) be a positive integer, and let \( {\mathbb{Q}}^{\left( m\right) } \) be the set of rational numbers which can be written as \( a/b \), where \( a \) and \( b \) are integers, and \( b \) is relatively prime to \( m \). Then \( {\mathbb{Q}}^{\left( m\right) } \) is a subring of \( \mathbb{Q} \).
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since for all \( a, b, c, d \in \mathbb{Z} \) with \( \gcd \left( {b, m}\right) = 1 \) and \( \gcd \left( {d, m}\right) = 1 \), we have\n\n\[ \frac{a}{b} + \frac{c}{d} = \frac{{ad} + {bc}}{bd}\text{ and }\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \]\n\nand since \( \gcd \left( {{bd}, m}\right) = 1 \), it follows that the sum and product of any two elements of \( {\mathbb{Q}}^{\left( m\right) } \) are again in \( {\mathbb{Q}}^{\left( m\right) } \). Clearly, \( {\mathbb{Q}}^{\left( m\right) } \) contains -1, and so it follows that \( {\mathbb{Q}}^{\left( m\right) } \) is a subring of \( \mathbb{Q} \).
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Yes
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Let us define a few polynomials over the ring \( \mathbb{Z} \) :\n\n\[ a \mathrel{\text{:=}} 3 + {X}^{2}, b \mathrel{\text{:=}} 1 + {2X} - {X}^{3}, c \mathrel{\text{:=}} 5, d \mathrel{\text{:=}} 1 + X, e \mathrel{\text{:=}} X, f \mathrel{\text{:=}} 4{X}^{3}. \]
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We have\n\n\[ a + b = 4 + {2X} + {X}^{2} - {X}^{3}, a \cdot b = 3 + {6X} + {X}^{2} - {X}^{3} - {X}^{5},{cd} + {ef} = 5 + {5X} + 4{X}^{4}. \]
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Yes
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Theorem 7.9. Let \( D \) be an integral domain. Then:\n\n(i) for all \( g, h \in D\left\lbrack X\right\rbrack \), we have \( \deg \left( {gh}\right) = \deg \left( g\right) + \deg \left( h\right) \) ;\n\n(ii) \( D\left\lbrack X\right\rbrack \) is an integral domain;\n\n(iii) \( {\left( D\left\lbrack X\right\rbrack \right) }^{ * } = {D}^{ * } \) .
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Proof. Exercise.
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No
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Theorem 7.10 (Division with remainder property). Let \( R \) be a ring. For all \( g, h \in R\left\lbrack X\right\rbrack \) with \( h \neq 0 \) and \( \operatorname{lc}\left( h\right) \in {R}^{ * } \), there exist unique \( q, r \in R\left\lbrack X\right\rbrack \) such that \( g = {hq} + r \) and \( \deg \left( r\right) < \deg \left( h\right) \) .
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Proof. Consider the set \( S \mathrel{\text{:=}} \{ g - {ht} : t \in R\left\lbrack X\right\rbrack \} \) . Let \( r = g - {hq} \) be an element of \( S \) of minimum degree. We must have \( \deg \left( r\right) < \deg \left( h\right) \), since otherwise, we could subtract an appropriate multiple of \( h \) from \( r \) so as to eliminate the leading coefficient of \( r \), obtaining\n\n\[ \n{r}^{\prime } \mathrel{\text{:=}} r - h \cdot \left( {\operatorname{lc}\left( r\right) \operatorname{lc}{\left( h\right) }^{-1}{X}^{\deg \left( r\right) - \deg \left( h\right) }}\right) \in S, \n\]\n\nwhere \( \deg \left( {r}^{\prime }\right) < \deg \left( r\right) \), contradicting the minimality of \( \deg \left( r\right) \) .\n\nThat proves the existence of \( r \) and \( q \) . For uniqueness, suppose that \( g = {hq} + r \) and \( g = h{q}^{\prime } + {r}^{\prime } \), where \( \deg \left( r\right) < \deg \left( h\right) \) and \( \deg \left( {r}^{\prime }\right) < \deg \left( h\right) \) . This implies \( {r}^{\prime } - r = h \cdot \left( {q - {q}^{\prime }}\right) \) . However, if \( q \neq {q}^{\prime } \), then\n\n\[ \n\deg \left( h\right) > \deg \left( {{r}^{\prime } - r}\right) = \deg \left( {h \cdot \left( {q - {q}^{\prime }}\right) }\right) = \deg \left( h\right) + \deg \left( {q - {q}^{\prime }}\right) \geq \deg \left( h\right) , \n\]\n\nwhich is impossible. Therefore, we must have \( q = {q}^{\prime } \), and hence \( r = {r}^{\prime } \) .
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Yes
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Let \( R \) be a subring of a ring \( E \) . For all \( g, h \in R\left\lbrack X\right\rbrack \) and \( \alpha \in E \), if \( s \mathrel{\text{:=}} g + h \in R\left\lbrack X\right\rbrack \) and \( p \mathrel{\text{:=}} {gh} \in R\left\lbrack X\right\rbrack \), then we have\n\n\[ s\left( \alpha \right) = g\left( \alpha \right) + h\left( \alpha \right) \text{ and }p\left( \alpha \right) = g\left( \alpha \right) h\left( \alpha \right) . \]\n\nAlso, if \( c \in R \) is a constant polynomial, then \( c\left( \alpha \right) = c \) for all \( \alpha \in E \) .
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The statement about evaluating a constant polynomial is clear from the definitions. The proof of the statements about evaluating the sum or product of polynomials is really just symbol pushing. Indeed, suppose \( g = \mathop{\sum }\limits_{i}{a}_{i}{X}^{i} \) and \( h = \mathop{\sum }\limits_{i}{b}_{i}{X}^{i} \) . Then \( s = \mathop{\sum }\limits_{i}\left( {{a}_{i} + {b}_{i}}\right) {X}^{i} \), and so\n\n\[ s\left( \alpha \right) = \mathop{\sum }\limits_{i}\left( {{a}_{i} + {b}_{i}}\right) {\alpha }^{i} = \mathop{\sum }\limits_{i}{a}_{i}{\alpha }^{i} + \mathop{\sum }\limits_{i}{b}_{i}{\alpha }^{i} = g\left( \alpha \right) + h\left( \alpha \right) . \]\n\nAlso, we have\n\n\[ p = \left( {\mathop{\sum }\limits_{i}{a}_{i}{X}^{i}}\right) \left( {\mathop{\sum }\limits_{j}{b}_{j}{X}^{j}}\right) = \mathop{\sum }\limits_{{i, j}}{a}_{i}{b}_{j}{X}^{i + j}, \]\n\nand employing the result for evaluating sums of polynomials, we have\n\n\[ p\left( \alpha \right) = \mathop{\sum }\limits_{{i, j}}{a}_{i}{b}_{j}{\alpha }^{i + j} = \left( {\mathop{\sum }\limits_{i}{a}_{i}{\alpha }^{i}}\right) \left( {\mathop{\sum }\limits_{j}{b}_{j}{\alpha }^{j}}\right) = g\left( \alpha \right) h\left( \alpha \right) . \]
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Yes
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If \( E = R\left\lbrack X\right\rbrack \), then evaluating a polynomial \( g \in R\left\lbrack X\right\rbrack \) at a point \( \alpha \in E \) amounts to polynomial composition. For example, if \( g \mathrel{\text{:=}} {X}^{2} + X \) and \( \alpha \mathrel{\text{:=}} X + 1 \), then
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\[ g\left( \alpha \right) = g\left( {X + 1}\right) = {\left( X + 1\right) }^{2} + \left( {X + 1}\right) = {X}^{2} + {3X} + 2. \]
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Yes
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Theorem 7.12. Let \( R \) be a ring, \( g \in R\left\lbrack X\right\rbrack \), and \( x \in R \) . Then there exists a unique polynomial \( q \in R\left\lbrack X\right\rbrack \) such that \( g = \left( {X - x}\right) q + g\left( x\right) \) . In particular, \( x \) is a root of \( g \) if and only if \( \left( {X - x}\right) \) divides \( g \) .
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Proof. If \( R \) is the trivial ring, there is nothing to prove, so assume that \( R \) is nontrivial. Using the division with remainder property for polynomials, there exist unique \( q, r \in R\left\lbrack X\right\rbrack \) such that \( g = \left( {X - x}\right) q + r \), with \( q, r \in R\left\lbrack X\right\rbrack \) and \( \deg \left( r\right) < 1 \) , which means that \( r \in R \) . Evaluating at \( x \), we see that \( g\left( x\right) = \left( {x - x}\right) q\left( x\right) + r = r \) . That proves the first statement. The second follows immediately from the first.
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Yes
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Theorem 7.13. Let \( D \) be an integral domain, and let \( {x}_{1},\ldots ,{x}_{k} \) be distinct elements of \( D \) . Then for every polynomial \( g \in D\left\lbrack X\right\rbrack \), the elements \( {x}_{1},\ldots ,{x}_{k} \) are roots of \( g \) if and only if the polynomial \( \mathop{\prod }\limits_{{i = 1}}^{k}\left( {X - {x}_{i}}\right) \) divides \( g \) .
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Proof. One direction is trivial: if \( \mathop{\prod }\limits_{{i = 1}}^{k}\left( {X - {x}_{i}}\right) \) divides \( g \), then it is clear that each \( {x}_{i} \) is a root of \( g \) . We prove the converse by induction on \( k \) . The base case \( k = 1 \) is just Theorem 7.12. So assume \( k > 1 \), and that the statement holds for \( k - 1 \) . Let \( g \in D\left\lbrack X\right\rbrack \) and let \( {x}_{1},\ldots ,{x}_{k} \) be distinct roots of \( g \) . Since \( {x}_{k} \) is a root of \( g \), then by Theorem 7.12, there exists \( q \in D\left\lbrack X\right\rbrack \) such that \( g = \left( {X - {x}_{k}}\right) q \) . Moreover, for each \( i = 1,\ldots, k - 1 \), we have\n\n\[ 0 = g\left( {x}_{i}\right) = \left( {{x}_{i} - {x}_{k}}\right) q\left( {x}_{i}\right) \]\n\nand since \( {x}_{i} - {x}_{k} \neq 0 \) and \( D \) is an integral domain, we must have \( q\left( {x}_{i}\right) = 0 \) . Thus, \( q \) has roots \( {x}_{1},\ldots ,{x}_{k - 1} \), and by induction \( \mathop{\prod }\limits_{{i = 1}}^{{k - 1}}\left( {X - {x}_{i}}\right) \) divides \( q \), from which it then follows that \( \mathop{\prod }\limits_{{i = 1}}^{k}\left( {X - {x}_{i}}\right) \) divides \( g \) .
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Yes
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Theorem 7.14. Let \( D \) be an integral domain, and suppose that \( g \in D\left\lbrack X\right\rbrack \), with \( \deg \left( g\right) = k \geq 0 \) . Then \( g \) has at most \( k \) distinct roots.
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Proof. If \( g \) had \( k + 1 \) distinct roots \( {x}_{1},\ldots ,{x}_{k + 1} \), then by the previous theorem, the polynomial \( \mathop{\prod }\limits_{{i = 1}}^{{k + 1}}\left( {X - {x}_{i}}\right) \), which has degree \( k + 1 \), would divide \( g \), which has degree \( k \) -an impossibility.
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Yes
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Theorem 7.15 (Lagrange interpolation). Let \( F \) be a field, let \( {x}_{1},\ldots ,{x}_{k} \) be distinct elements of \( F \), and let \( {y}_{1},\ldots ,{y}_{k} \) be arbitrary elements of \( F \) . Then there exists a unique polynomial \( g \in F\left\lbrack X\right\rbrack \) with \( \deg \left( g\right) < k \) such that \( g\left( {x}_{i}\right) = {y}_{i} \) for \( i = 1,\ldots, k \), namely\n\n\[ g \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{k}{y}_{i}\frac{\mathop{\prod }\limits_{{j \neq i}}\left( {X - {x}_{j}}\right) }{\mathop{\prod }\limits_{{j \neq i}}\left( {{x}_{i} - {x}_{j}}\right) }.\]
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Proof. For the existence part of the theorem, one just has to verify that \( g\left( {x}_{i}\right) = {y}_{i} \) for the given \( g \), which clearly has degree less than \( k \) . This is easy to see: for \( i = 1,\ldots, k \), evaluating the \( i \) th term in the sum defining \( g \) at \( {x}_{i} \) yields \( {y}_{i} \), while evaluating any other term at \( {x}_{i} \) yields 0 . The uniqueness part of the theorem follows almost immediately from Theorem 7.14: if \( g \) and \( h \) are polynomials of degree less than \( k \) such that \( g\left( {x}_{i}\right) = {y}_{i} = h\left( {x}_{i}\right) \) for \( i = 1,\ldots, k \), then \( g - h \) is a polynomial of degree less than \( k \) with \( k \) distinct roots, which, by the previous theorem, is impossible.
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Yes
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Let \( R \) be a ring and \( x \in R \) . Consider the set\n\n\[ I \mathrel{\text{:=}} \{ g \in R\left\lbrack X\right\rbrack : g\left( x\right) = 0\} . \]
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It is not hard to see that \( I \) is an ideal of \( R\left\lbrack X\right\rbrack \) . Indeed, for all \( g, h \in I \) and \( q \in R\left\lbrack X\right\rbrack \) , we have\n\n\[ \left( {g + h}\right) \left( x\right) = g\left( x\right) + h\left( x\right) = 0 + 0 = 0\text{ and }\left( {gq}\right) \left( x\right) = g\left( x\right) q\left( x\right) = 0 \cdot q\left( x\right) = 0. \]
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Yes
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