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On every Riemann surface \( M \) we can introduce a \( {C}^{\infty } \) -Riemannian metric consistent with the conformal structure. | Let \( f \) be any non-constant meromorphic function on \( M \) . Let \( \left\{ {{P}_{1},{P}_{2},\ldots }\right\} \) be the set of poles and critical values (those \( P \in M \) with \( {df}\left( P\right) = \) \( 0) \) of \( f \) . Let \( {z}_{j} \) be a local coordinate vanishing at \( {P}_{j} \) with the sets \( \l... | Yes |
Corollary 2. If \( F \) is an isomorphism (surjective), then \( {F}^{ * } \) is a homeomorphism. | Since the holomorphic mapping (11.16.2) between compact Riemann surfaces induces the homomorphism (11.16.1) between their function fields that is defined by (11.16.3), the functor under discussion establishes an equivalence between two categories. | No |
Corollary 1. There is a homomorphism\n\n\[ \n\lambda : \text{ Aut }M \rightarrow \operatorname{Perm}\left( {W\left( M\right) }\right) .\n\]\n\nFurthermore, \( \lambda \) is injective unless \( M \) is hyperelliptic, in which case Kernel \( \lambda = \) \( \langle J\rangle \), where \( J \) is the hyperelliptic involuti... | Proof. The existence of \( \lambda \) follows from the proposition. If \( M \) is not hyperelliptic, then there are more than \( {2g} + 2 \) Weierstrass points. By Proposition V.1.1, only the identity fixes all the Weierstrass points and thus \( \lambda \) is injective.\n\nIf \( M \) is hyperelliptic, then by Propositi... | Yes |
Corollary 2 (Schwarz). If \( M \) is a surface of genus \( g \geq 2 \), then Aut \( M \) is a finite group. | Proof. We have produced a homomorphism of Aut \( M \) into a finite group (the permutation group of a finite set), and the homomorphism has a finite kernel. | Yes |
Corollary 1. If ord \( T \) is prime, then\n\n\[ v\left( T\right) = 2 + \frac{{2g} - {2\gamma }\left( {\operatorname{ord}T}\right) }{\operatorname{ord}T - 1} \]\n\nwhere \( \gamma \) is the genus of \( M/\langle T\rangle \) . In this case, equality holds in (1.5.1) if and only if \( \gamma = 0 \) . | Proof. If ord \( T \) is prime, then \( v\left( T\right) = v\left( {T}^{j}\right), j = 1,\ldots \), ord \( T - 1 \) . | No |
Corollary 2. In general, \( v\left( T\right) \leq {2g} - 1 \) if \( M \) is not hyperelliptic. | Proof of Corollary. From Corollary 1, an automorphism of order 2 has precisely \( {2g} + 2 - {4\gamma } \) fixed points. Since \( M \) is not hyperelliptic, \( \gamma \geq 1 \) . Hence for such an automorphism \( v\left( T\right) \leq {2g} - 2 \) . If ord \( T \geq 3 \), then \( v\left( T\right) \leq \) \( 2 + g \) . N... | Yes |
Corollary 1. The \( \gamma \) -hyperelliptic involution on a surface of genus \( g \) is unique (if it exists) provided \( g > {4\gamma } + 1 \) . | Proof. The \( \gamma \) -hyperelliptic involution has \( {2g} + 2 - {4\gamma } \) fixed points, and \( {2g} + 2 - {4\gamma } > {4\gamma } + 4 \) if and only if \( g > {4\gamma } + 1 \) . The uniqueness now follows from (1.9.1). | Yes |
Corollary 2. If \( \gamma, r \) are non-negative integers with \( r > 0 \) and \( g > {4\gamma } + 1 + {2r} \) , then a surface of genus \( g \) cannot be both \( \gamma \) -hyperelliptic and \( \left( {\gamma + r}\right) \) -hyperelliptic. | Proof of Corollary 2. Suppose \( M \) is both \( \left( {\gamma + r}\right) \) - and \( \gamma \) -hyperelliptic, then \( {J}_{\gamma + r} \) has\n\n\[ \n{2g} + 2 - 4\left( {\gamma + r}\right) > 4\left( {\gamma + 1}\right) \n\] \n\nfixed points. Thus \( {J}_{\gamma + r} = {J}_{\gamma } \), which is only possible if \( ... | No |
Corollary 1. Let \( {e}^{\left( j\right) } \) be the \( j \) -th column of \( {I}_{g} \) . Then \[ \theta \left( {z + {e}^{\left( j\right) },\tau }\right) = \theta \left( {z,\tau }\right) \text{, all}z \in {\mathbb{C}}^{g}\text{, all}\tau \in {\mathfrak{S}}_{g}\text{.} \] | Proof. Take \( {\mu }^{\prime } = {e}^{\left( j\right) },\mu = 0 \) . | No |
Let \( {\tau }^{\left( k\right) } \) be the \( k \) -th column of \( \tau \), and \( {\tau }_{kk} \) the \( \left( {k, k}\right) \) -entry of \( \tau \). Then \[ \theta \left( {z + {\tau }^{\left( k\right) },\tau }\right) = \exp {2\pi i}\left\lbrack {-{z}_{k} - \frac{{\tau }_{kk}}{2}}\right\rbrack \theta \left( {z,\tau... | Proof. Take \( {\mu }^{\prime } = 0,\mu = {e}^{\left( k\right) } \). | No |
Example 2. The trefoil knot (Figure 278). In terms of generators \( {a}_{1},{a}_{2} \) (clockwise round the right and left holes respectively), | \[ p = {a}_{1}{a}_{2}{a}_{1}{a}_{2}^{-1}{a}_{1}^{-1}{a}_{2}^{-1} \] which gives a standard presentation of the trefoil knot group (cf. 4.2.5), \( \left\langle {{a}_{1},{a}_{2};{a}_{1}{a}_{2}{a}_{1} = {a}_{2}{a}_{1}{a}_{2}}\right\rangle \). | Yes |
Adding a one to a block of consecutive ones. | It is assumed that the tape is initially blank except for finitely many ones on consecutive squares. The machine starts on the leftmost of these squares in state \( {q}_{1} \) . \n\n\( {q}_{1}{11}\mathrm{R}{q}_{1}\; \) Move right as long as the scanned symbol is 1\n\n\( {q}_{1}▱1\mathrm{R}{q}_{2}\; \) Write 1 in the fi... | Yes |
Example 2. Doubling a block of ones. | Under the same input/output convention as in Example 1, this machine computes \( f\left( n\right) = {2n} \) . It does so by \ | No |
Example 3. Deciding whether a number is even. | This example shows how machine states play a role like mental states. The machine has a state \( {q}_{\mathrm{{YES}}} \) that \ | No |
Theorem 3.1. Let \( \mathrm{f} : \mathrm{A} \rightarrow \mathrm{B} \) be a function, with \( \mathrm{A} \) nonempty.\n\n(i) \( \mathrm{f} \) is injective if and only if there is a map \( \mathrm{g} : \mathrm{B} \rightarrow \mathrm{A} \) such that \( \mathrm{{gf}} = {1}_{\mathrm{A}} \).\n\n(ii) If \( \mathrm{A} \) is a ... | PROOF. Since every identity map is bijective, (11) and (12) prove the implications \( \left( \Leftarrow \right) \) in (i) and (ii). Conversely if \( f \) is injective, then for each \( {b\varepsilon f}\left( A\right) \) there is a unique \( {a\varepsilon A} \) with \( f\left( a\right) = b \) . Choose a fixed \( {a}_{0}... | Yes |
Theorem 4.1. If \( \mathrm{A} \) is a nonempty set, then the assignment \( \mathrm{R} \mapsto \mathrm{A}/\mathrm{R} \) defines a bijection from the set \( \mathrm{E}\left( \mathrm{A}\right) \) of all equivalence relations on \( \mathrm{A} \) onto the set \( \mathrm{Q}\left( \mathrm{A}\right) \) of all partitions of \( ... | SKETCH OF PROOF. If \( R \) is an equivalence relation on \( A \), then the set \( A/R \) of equivalence classes is a partition of \( A \) by (18),(19), and (21) so that \( R \mapsto A/R \) defines a function \( f : E\left( A\right) \rightarrow Q\left( A\right) \) . Define a function \( g : Q\left( A\right) \rightarrow... | No |
Theorem 5.2. Let \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) be a family of sets indexed by \( \mathrm{I} \) . Then there exists a set \( \mathrm{D} \), together with a family of maps \( \left\{ {{\pi }_{\mathrm{i}} : \mathrm{D} \rightarrow {\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} ... | PROOF OF 5.2. (Existence) Let \( D = \mathop{\prod }\limits_{{i \in I}}{A}_{i} \) and let the maps \( {\pi }_{i} \) be the projections onto the \( i \) th components. Given \( C \) and the maps \( {\varphi }_{i} \), define \( \varphi : C \rightarrow \mathop{\prod }\limits_{{i \in I}}{A}_{i} \) by \( c \mapsto {f}_{c} \... | Yes |
Theorem 6.1. (Principle of Mathematical Induction) If \( \mathrm{S} \) is a subset of the set \( \mathbf{N} \) of natural numbers such that \( {0\varepsilon }\mathrm{S} \) and either\n\n(i) \( \mathrm{n}\varepsilon \mathrm{S} \Rightarrow \mathrm{n} + {1\varepsilon }\mathrm{S}\; \) for all \( \mathrm{n}\varepsilon \math... | PROOF. If \( \mathbf{N} - S \neq \varnothing \), let \( n \neq 0 \) be its least element. Then for every \( m < n \) , we must have \( m \notin \mathbf{N} - S \) and hence \( {m\varepsilon S} \) . Consequently either (i) or (ii) implies \( {n\varepsilon S} \), which is a contradiction. Therefore \( \mathbf{N} - S = \va... | Yes |
Theorem 6.2. (Recursion Theorem) If \( \mathrm{S} \) is a set, \( \mathrm{a}\varepsilon \mathrm{S} \) and for each \( \mathrm{n}\varepsilon \mathrm{N},{\mathrm{f}}_{\mathrm{n}} : \mathrm{S} \rightarrow \mathrm{S} \) is a function, then there is a unique function \( \varphi : \mathbf{N} \rightarrow \mathbf{S} \) such th... | SKETCH OF PROOF. We shall construct a relation \( R \) on \( \mathbf{N} \times S \) that is the graph of a function \( \varphi : \mathbf{N} \rightarrow S \) with the desired properties. Let \( \mathcal{G} \) be the set of all subsets \( Y \) of \( \mathbf{N} \times S \) such that\n\n\[ \left( {0, a}\right) \in Y;\text{... | Yes |
Theorem 6.3. (Division Algorithm) If \( \mathrm{a},\mathrm{b},\mathrm{e}\mathbf{Z} \) and \( \mathrm{a} \neq 0 \), then there exists unique integers \( \mathrm{q} \) and \( \mathrm{r} \) such that \( \mathrm{b} = \mathrm{{aq}} + \mathrm{r} \), and \( 0 \leq \mathrm{r} < \left| \mathrm{a}\right| \) . | SKETCH OF PROOF. Show that the set \( S = \{ b - {ax} \mid {x\varepsilon }\mathbf{Z}, b - {ax} \geq 0\} \) is a nonempty subset of \( \mathbf{N} \) and therefore contains a least element \( r = b - {aq} \) (for some \( q \in \mathbf{Z} \) ). Thus \( b = {aq} + r \) . Use the fact that \( r \) is the least element in \(... | No |
Theorem 6.5. If \( {\mathrm{a}}_{1},{\mathrm{a}}_{2},\ldots ,{\mathrm{a}}_{\mathrm{n}} \) are integers, not all 0, then \( \left( {{\mathrm{a}}_{1},{\mathrm{a}}_{2},\ldots ,{\mathrm{a}}_{\mathrm{n}}}\right) \) exists. Furthermore there are integers \( {\mathrm{k}}_{1},{\mathrm{k}}_{2},\ldots ,{\mathrm{k}}_{\mathrm{n}} ... | SKETCH OF PROOF. Use the Division Algorithm to show that the least positive element of the nonempty set \( S = \left\{ {{x}_{1}{a}_{1} + {x}_{2}{a}_{2} + \cdots + {x}_{n}{a}_{n} \mid {x}_{i} \in \mathbf{Z},\mathop{\sum }\limits_{i}{x}_{i}{a}_{i} > 0}\right\} \) is the greatest common divisor of \( {a}_{1},\ldots ,{a}_{... | No |
Theorem 6.6. If \( \mathrm{a} \) and \( \mathrm{b} \) are relatively prime integers and \( \mathrm{a} \mid \mathrm{{bc}} \), then \( \mathrm{a} \mid \mathrm{c} \) . | SKETCH OF PROOF. By Theorem 6.5 \( 1 = {ra} + {sb} \), whence \( c = {rac} + {sbc} \) . Therefore \( a \mid c \) . | No |
Theorem 6.7. (Fundamental Theorem of Arithmetic) Any positive integer \( \mathrm{n} > 1 \) may be written uniquely in the form \( \mathrm{n} = {\mathrm{p}}_{1}{}^{{\mathrm{t}}_{1}}{\mathrm{p}}_{2}{}^{{\mathrm{t}}_{2}}\cdots {\mathrm{p}}_{\mathrm{k}}{}^{{\mathrm{t}}_{\mathrm{k}}} \), where \( {\mathrm{p}}_{1} < {\mathrm... | The proof, which proceeds by induction, may be found in Shockley [51, p.17]. | No |
Theorem 6.8. Let \( \mathrm{m} > 0 \) be an integer and \( \mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\varepsilon \mathbf{Z} \) .\n\n(i) Congruence modulo \( \mathrm{m} \) is an equivalence relation on the set of integers \( \mathbf{Z} \), which has precisely \( \mathrm{m} \) equivalence classes. | PROOF. (i) The fact that congruence modulo \( m \) is an equivalence relation is an easy consequence of the appropriate definitions. Denote the equivalence class of an integer \( a \) by \( \bar{a} \) and recall property (20), which can be stated in this context as:\n\n\[ \n\bar{a} = \bar{b} \Leftrightarrow a \equiv b\... | Yes |
Theorem 7.1. (Principle of Transfinite Induction) If \( \mathbf{B} \) is a subset of a well-ordered set \( \left( {A, \leq }\right) \) such that for every \( {a\varepsilon A} \) ,\n\n\[ \{ {c\varepsilon A} \mid c < a\} \subset B \Rightarrow {a\varepsilon B}, \]\n\nthen \( \mathrm{B} = \mathrm{A} \) . | PROOF. If \( A - B \neq \varnothing \), then there is a least element \( {a\varepsilon A} - B \) . By the definitions of least element and \( A - B \) we must have \( \{ c \in A \mid c < a\} \subset B \) . By hypothesis then, \( a \in B \) so that \( a \in B \cap \left( {A - B}\right) = \varnothing \), which is a contr... | Yes |
Theorem 8.1. Equipollence is an equivalence relation on the class \( § \) of all sets. | PROOF. Exercise; note that \( \varnothing \sim \varnothing \) since \( \varnothing \subset \varnothing \times \varnothing \) is a relation that is (vacuously) a bijective function. \( {}^{3} \) | No |
Theorem 8.5. If \( \mathrm{A} \) is a set and \( \mathrm{P}\left( \mathrm{A}\right) \) its power set, then \( \left| \mathrm{A}\right| < \left| {\mathrm{P}\left( \mathrm{A}\right) }\right| \) . | SKETCH OF PROOF. The assignment \( a \mapsto \{ a\} \) defines an injective map \( A \rightarrow P\left( A\right) \) so that \( \left| A\right| \leq \left| {P\left( A\right) }\right| \) . If there were a bijective map \( f : A \rightarrow P\left( A\right) \), then for some \( {a}_{0} \in A, f\left( {a}_{0}\right) = B \... | No |
Theorem 8.6. (Schroeder-Bernstein) If \( \mathrm{A} \) and \( \mathrm{B} \) are sets such that \( \left| \mathrm{A}\right| \leq \left| \mathrm{B}\right| \) and \( \left| \mathrm{B}\right| \leq \left| \mathrm{A}\right| \), then \( \left| \mathrm{A}\right| = \left| \mathrm{B}\right| \) . | SKETCH OF PROOF. By hypothesis there are injective maps \( f : A \rightarrow B \) and \( g : B \rightarrow A \) . We shall use \( f \) and \( g \) to construct a bijection \( h : A \rightarrow B \) . This will imply that \( A \sim B \) and hence \( \left| A\right| = \left| B\right| \) . If \( {a\varepsilon A} \), then ... | No |
Theorem 8.7. The class of all cardinal numbers is linearly ordered by \( \leq \) . If \( \alpha \) and \( \beta \) are cardinal numbers, then exactly one of the following is true:\n\n\[ \alpha < \beta ;\;\alpha = \beta ;\;\beta < \alpha \;\text{ (Trichotomy Law). } \] | SKETCH OF PROOF. It is easy to verify that \( \leq \) is a partial ordering. Let \( \alpha ,\beta \) be cardinals and \( A, B \) be sets such that \( \left| A\right| = \alpha ,\left| B\right| = \beta \) . We shall show that \( \leq \) is a linear ordering (that is, either \( \alpha \leq \beta \) or \( \beta \leq \alpha... | Yes |
Theorem 8.8. Every infinite set has a denumerable subset. In particular, \( {\aleph }_{0} \leq \alpha \) for every infinite cardinal number \( \alpha \) . | SKETCH OF PROOF. If \( B \) is a finite subset of the infinite set \( A \), then \( A - B \) is nonempty. For each finite subset \( B \) of \( A \), choose an element \( {x}_{B}{\varepsilon A} - B \) (Axiom of Choice). Let \( F \) be the set of all finite subsets of \( A \) and define a map \( f : F \rightarrow F \) by... | No |
Lemma 8.9. If \( \mathrm{A} \) is an infinite set and \( \mathrm{F} \) a finite set then \( \left| {\mathrm{A} \cup \mathrm{F}}\right| = \left| \mathrm{A}\right| \) . In particular, \( \alpha + \mathrm{n} = \alpha \) for every infinite cardinal number \( \alpha \) and every natural number (finite cardinal) \( \mathrm{n... | SKETCH OF PROOF. It suffices to assume \( A \cap F = \varnothing \) (replace \( F \) by \( F - A \) if necessary). If \( F = \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right\} \) and \( D = \left\{ {{x}_{i} \mid i \in {\mathbf{N}}^{ * }}\right\} \) is a denumerable subset of \( A \) (Theorem 8.8), verify that \( f : A \... | Yes |
Theorem 8.10. If \( \alpha \) and \( \beta \) are cardinal numbers such that \( \beta \leq \alpha \) and \( \alpha \) is infinite, then \( \alpha + \beta = \alpha \) . | SKETCH OF PROOF. It suffices to prove \( \alpha + \alpha = \alpha \) (simply verify that \( \alpha \leq \alpha + \beta \leq \alpha + \alpha = \alpha \) and apply the Schroeder-Bernstein Theorem to conclude \( \alpha + \beta = \alpha ) \) . Let \( A \) be a set with \( \left| A\right| = \alpha \) and let \( \mathfrak{F}... | No |
Theorem 8.11. If \( \alpha \) and \( \beta \) are cardinal numbers such that \( 0 \neq \beta \leq \alpha \) and \( \alpha \) is infinite, then \( {\alpha \beta } = \alpha \) ; in particular, \( \alpha {\aleph }_{0} = \alpha \) and if \( \beta \) is finite \( {\aleph }_{0}\beta = {\aleph }_{0} \) . | SKETCH OF PROOF. Since \( \alpha \leq {\alpha \beta } \leq {\alpha \alpha } \) it suffices (as in the proof of Theorem 8.10) to prove \( {\alpha \alpha } = \alpha \) . Let \( A \) be an infinite set with \( \left| A\right| = \alpha \) and let \( \mathcal{F} \) be the set of all bijections \( f : X \times X \rightarrow ... | No |
Theorem 8.12. Let \( \mathrm{A} \) be a set and for each integer \( \mathrm{n} \geq 1 \) let \( {\mathrm{A}}^{\mathrm{n}} = \mathrm{A} \times \mathrm{A} \times \cdots \times \mathrm{A} \) (n factors).\n\n(i) If \( \mathrm{A} \) is finite, then \( \left| {\mathrm{A}}^{\mathrm{n}}\right| = {\left| \mathrm{A}\right| }^{\m... | SKETCH OF PROOF. (i) is trivial if \( \left| A\right| \) is finite and may be proved by induction on \( n \) if \( \left| A\right| \) is infinite (the case \( n = 2 \) is given by Theorem 8.11). | No |
Corollary 8.13. If \( \mathrm{A} \) is an infinite set and \( \mathrm{F}\left( \mathrm{A}\right) \) the set of all finite subsets of \( \mathrm{A} \), then \( \left| {F\left( A\right) }\right| = \left| A\right| \) . | PROOF. The map \( A \rightarrow F\left( A\right) \) given by \( a \mapsto \{ a\} \) is injective so that \( \left| A\right| \leq \left| {F\left( A\right) }\right| \) . For each \( n \) -element subset \( S \) of \( A \), choose \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \varepsilon {A}^{n} \) such that \( S = \left\{ {... | Yes |
Theorem 1.2. If \( \mathrm{G} \) is a monoid, then the identity element \( \mathrm{e} \) is unique. | SKETCH OF PROOF. If \( {e}^{\prime } \) is also a two-sided identity, then \( e = e{e}^{\prime } = {e}^{\prime } \). | No |
Proposition 1.3. Let \( \mathrm{G} \) be a semigroup. Then \( \mathrm{G} \) is a group if and only if the following conditions hold:\n\n(i) there exists an element \( \mathrm{e} \in \mathrm{G} \) such that \( \mathrm{{ea}} = \mathrm{a} \) for all \( \mathrm{a} \in \mathrm{G} \) (left identity element);\n\n(ii) for each... | SKETCH OF PROOF OF 1.3. ( \( \Rightarrow \) ) Trivial. ( \( \Leftarrow \) ) Note that Theorem 1.2(i) is true under these hypotheses. \( G \neq \varnothing \) since \( e \in G \) . If \( a \in G \), then by (ii) \( \left( {a{a}^{-1}}\right) \left( {a{a}^{-1}}\right) \) \( = a\left( {{a}^{-1}a}\right) {a}^{-1} = a\left( ... | Yes |
Proposition 1.4. Let \( \mathrm{G} \) be a semigroup. Then \( \mathrm{G} \) is a group if and only if for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{G} \) the equations \( \mathrm{{ax}} = \mathrm{b} \) and \( \mathrm{{ya}} = \mathrm{b} \) have solutions in \( \mathrm{G} \) . | PROOF. Exercise; use Proposition 1.3. | No |
Theorem 1.5. Let \( \mathrm{R}\left( \sim \right) \) be an equivalence relation on a monoid \( \mathrm{G} \) such that \( {\mathrm{a}}_{1} \sim {\mathrm{a}}_{2} \) and \( {\mathrm{b}}_{1} \sim {\mathrm{b}}_{2} \) imply \( {\mathrm{a}}_{1}{\mathrm{\;b}}_{1} \sim {\mathrm{a}}_{2}{\mathrm{\;b}}_{2} \) for all \( {\mathrm{... | PROOF OF 1.5. If \( {\bar{a}}_{1} = {\bar{a}}_{2} \) and \( {\bar{b}}_{1} = {\bar{b}}_{2}\left( {{a}_{i},{b}_{i}{\varepsilon G}}\right) \), then \( {a}_{1} \sim {a}_{2} \) and \( {b}_{1} \sim {b}_{2} \) by (20) of Introduction, Section 4. Then by hypothesis \( {a}_{1}{b}_{1} \sim {a}_{2}{b}_{2} \) so that \( \overline{... | Yes |
Theorem 1.6. (Generalized Associative Law) If \( \mathrm{G} \) is a semigroup and \( {\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}}\varepsilon \mathrm{G} \) , then any two meaningful products of \( {\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}} \) in this order are equal. | PROOF. We use induction to show that for every \( n \) any meaningful product \( {a}_{1}\cdots {a}_{n} \) is equal to the standard \( n \) product \( \mathop{\prod }\limits_{{i = 1}}^{n}{a}_{i} \) . This is certainly true for \( n = 1,2 \) . If \( n > 2 \), then by definition \( \left( {{a}_{1}\cdots {a}_{n}}\right) = ... | Yes |
Corollary 1.7. (Generalized Commutative Law) If \( \mathrm{G} \) is a commutative semigroup and \( {\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}}\varepsilon \mathrm{G} \), then for any permutation \( {\mathrm{i}}_{1},\ldots ,{\mathrm{i}}_{\mathrm{n}} \) of \( 1,2,\ldots \mathrm{n},{\mathrm{a}}_{1}{\mathrm{a}}_{2}\c... | PROOF. Exercise. | No |
Theorem 1.9. If \( \mathrm{G} \) is a group [resp. semigroup, monoid] and \( \mathrm{a}\varepsilon \mathrm{G} \), then for all \( \mathrm{m},\mathrm{n}\varepsilon \mathbf{Z} \) [resp. \( {\mathrm{N}}^{ * },\mathrm{N} \) ]:\n\n(i) \( {\mathrm{a}}^{\mathrm{m}}{\mathrm{a}}^{\mathrm{n}} = {\mathrm{a}}^{\mathrm{m} + \mathrm... | SKETCH OF PROOF. Verify that \( {\left( {a}^{n}\right) }^{-1} = {\left( {a}^{-1}\right) }^{n} \) for all \( {n\varepsilon }\mathbf{N} \) and that \( {a}^{-n} = {\left( {a}^{-1}\right) }^{n} \) for all \( {n\varepsilon }\mathbf{Z} \) . (i) is true for \( m > 0 \) and \( n > 0 \) since the product of a standard \( n \) p... | No |
Theorem 2.3. Let \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) be a homomorphism of groups. Then\n\n(i) \( \mathrm{f} \) is a monomorphism if and only if \( \operatorname{Ker}\mathrm{f} = \{ \mathrm{e}\} \) ;\n\n(ii) \( \mathrm{f} \) is an isomorphism if and only if there is a homomorphism \( {\mathrm{f}}^{-1} :... | PROOF. (i) If \( f \) is a monomorphism and \( a \in \operatorname{Ker}f \), then \( f\left( a\right) = {e}_{H} = f\left( e\right) \) , whence \( a = e \) and \( \operatorname{Ker}f = \{ e\} \) . If \( \operatorname{Ker}f = \{ e\} \) and \( f\left( a\right) = f\left( b\right) \), then \( {e}_{H} = f\left( a\right) f{\l... | Yes |
Theorem 2.5. Let \( \mathrm{H} \) be a nonempty subset of a group \( \mathrm{G} \). Then \( \mathrm{H} \) is a subgroup of \( \mathrm{G} \) if and only if \( {\mathrm{{ab}}}^{-1}\varepsilon \mathrm{H} \) for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{H} \). | PROOF. ( \( \Leftarrow \) ) There exists \( {a\varepsilon H} \) and hence \( e = a{a}^{-1}{\varepsilon H} \). Thus for any \( {b\varepsilon H},{b}^{-1} \) \( = e{b}^{-1}{\varepsilon H} \). If \( a,{b\varepsilon H} \), then \( {b}^{-1}{\varepsilon H} \) and hence \( {ab} = a{\left( {b}^{-1}\right) }^{-1}{\varepsilon H} ... | No |
Corollary 2.6. If \( \mathrm{G} \) is a group and \( \left\{ {{\mathrm{H}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) is a nonempty family of subgroups, then \( \bigcap {\mathrm{H}}_{\mathrm{i}} \) is a subgroup of \( \mathrm{G} \) . \( {ieI} \) | PROOF. Exercise. | No |
Theorem 2.8. If \( \mathrm{G} \) is a group and \( \mathrm{X} \) is a nonempty subset of \( \mathrm{G} \), then the subgroup \( \langle \mathrm{X}\rangle \) generated by \( \mathrm{X} \) consists of all finite products \( {\mathrm{a}}_{1}{}^{{\mathrm{n}}_{1}}{\mathrm{a}}_{2}{}^{{\mathrm{n}}_{2}}\cdots {\mathrm{a}}_{\ma... | SKETCH OF PROOF. Show that the set \( H \) of all such products is a subgroup of \( G \) that contains \( X \) and is contained in every subgroup containing \( X \) . Therefore \( H < \langle X\rangle < H. \) | No |
Theorem 3.1. Every subgroup \( \mathrm{H} \) of the additive group \( \mathbf{Z} \) is cyclic. Either \( \mathrm{H} = \langle 0\rangle \) or \( \mathrm{H} = \langle \mathrm{m}\rangle \), where \( \mathrm{m} \) is the least positive integer in \( \mathrm{H} \). If \( \mathrm{H} \neq \langle 0\rangle \), then \( \mathrm{... | PROOF. Either \( H = \langle 0\rangle \) or \( H \) contains a least positive integer \( m \). Clearly \( \langle m\rangle = \{ {km} \mid {k\varepsilon }\mathbf{Z}\} \subset H \). Conversely if \( {h\varepsilon H} \), then \( h = {qm} + r \) with \( q,{r\varepsilon }\mathbf{Z} \) and \( 0 \leq r < m \) (division algori... | Yes |
Every infinite cyclic group is isomorphic to the additive group \( \mathbf{Z} \) and every finite cyclic group of order \( \mathbf{r} \) is isomorphic to the additive group \( {\mathbf{Z}}_{\mathrm{m}} \) . | If \( G = \langle a\rangle \) is a cyclic group then the map \( \alpha : \mathbf{Z} \rightarrow G \) given by \( k \mapsto {a}^{k} \) is an epimorphism by Theorems 1.9 and 2.8. If Ker \( \alpha = 0 \), then \( \mathbf{Z} \cong G \) by Theorem 2.3 (i). Otherwise Ker \( \alpha \) is a nontrivial subgroup of \( \mathbf{Z}... | No |
Theorem 3.4. Let \( \mathrm{G} \) be a group and \( \mathrm{a}\varepsilon \mathrm{G} \) . If \( \mathrm{a} \) has infinite order, then\n\n(i) \( {\mathrm{a}}^{\mathrm{k}} = \mathrm{e} \) if and only if \( \mathrm{k} = 0 \) ;\n\n(ii) the elements \( {\mathrm{a}}^{\mathrm{k}}\left( {\mathrm{k}\varepsilon \mathbf{Z}}\righ... | SKETCH OF PROOF. (i)-(vi) are immediate consequences of the proof of Theorem 3.2. (vii) \( {\left( {a}^{k}\right) }^{m/k} = {a}^{m} = e \) and \( {\left( {a}^{k}\right) }^{r} \neq e \) for all \( 0 < r < m/k \) since otherwise \( {a}^{kr} = e \) with \( {kr} < k\left( {m/k}\right) = m \) contradicting (iii). Therefore,... | No |
Theorem 3.5. Every homomorphic image and every subgroup of a cyclic group \( \mathbf{G} \) is cyclic. In particular, if \( \mathrm{H} \) is a nontrivial subgroup of \( \mathrm{G} = \langle \mathrm{a}\rangle \) and \( \mathrm{m} \) is the least positive integer such that \( {\mathrm{a}}^{\mathrm{m}}\varepsilon \mathrm{H... | SKETCH OF PROOF. If \( f : G \rightarrow K \) is a homomorphism of groups, then \( \operatorname{Im}f = \langle f\left( a\right) \rangle \) . To prove the second statement simply translate the proof of Theorem 3.1 into multiplicative notation (that is, replace every \( t \in \mathbf{Z} \) by \( {a}^{t} \) throughout). ... | No |
Theorem 3.6. Let \( \mathrm{G} = \langle \mathrm{a}\rangle \) be a cyclic group. If \( \mathrm{G} \) is infinite, then \( \mathrm{a} \) and \( {\mathrm{a}}^{-1} \) are the only generators of \( \mathrm{G} \) . If \( \mathrm{G} \) is finite of order \( \mathrm{m} \), then \( {\mathrm{a}}^{\mathrm{k}} \) is a generator o... | SKETCH OF PROOF. It suffices to assume either that \( G = \mathbf{Z} \), in which case the conclusion is easy to prove, or that \( G = {Z}_{m} \) . If \( \left( {k, m}\right) = 1 \), there are \( c, d \in \mathbf{Z} \) such that \( {ck} + {dm} = 1 \) ; use this fact to show that \( \bar{k} \) generates \( {Z}_{m} \) . ... | No |
Theorem 4.2. Let \( \mathrm{H} \) be a subgroup of a group \( \mathrm{G} \) .\n\n(i) Right [resp. left] congruence modulo \( \mathrm{H} \) is an equivalence relation on \( \mathrm{G} \) .\n\n(ii) The equivalence class of \( \mathrm{a}\varepsilon \mathrm{G} \) under right [resp. left] congruence modulo \( \mathrm{H} \) ... | PROOF OF 4.2. We write \( a \equiv b \) for \( a \equiv {}_{r}b\left( {\;\operatorname{mod}\;H}\right) \) and prove the theorem for right congruence and right cosets. Analogous arguments apply to left congruence.\n\n(i) Let \( a, b, c \in G \) . Then \( a \equiv a \) since \( a{a}^{-1} = e \in H \) ; hence \( \equiv \)... | Yes |
Theorem 4.5. If \( \mathrm{K},\mathrm{H},\mathrm{G} \) are groups with \( \mathrm{K} < \mathrm{H} < \mathrm{G} \), then \( \left\lbrack {\mathrm{G} : \mathrm{K}}\right\rbrack = \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \left\lbrack {\mathrm{H} : \mathrm{K}}\right\rbrack \) . If any two of these indices are fi... | PROOF. By Corollary \( {4.3G} = \mathop{\bigcup }\limits_{{i\varepsilon I}}H{a}_{i} \) with \( {a}_{i}{\varepsilon G},\left| I\right| = \left\lbrack {G : H}\right\rbrack \) and the cosets \( H{a}_{i} \) mutually disjoint (that is, \( H{a}_{i} = H{a}_{j} \Leftrightarrow i = j \) ). Similarly \( H = \mathop{\bigcup }\lim... | Yes |
Corollary 4.6. (Lagrange). If \( \mathrm{H} \) is a subgroup of a group \( \mathrm{G} \), then \( \left| \mathrm{G}\right| = \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \left| \mathrm{H}\right| \) . In particular if \( \mathrm{G} \) is finite, the order \( \left| \mathrm{a}\right| \) of \( \mathrm{a}\varepsilon... | PROOF. Apply the theorem with \( K = \langle e\rangle \) for the first statement. The second is a special case of the first with \( H = \langle a\rangle \) . | No |
Theorem 4.7. Let \( \mathrm{H} \) and \( \mathrm{K} \) be finite subgroups of a group \( \mathrm{G} \) . Then \( \left| \mathrm{{HK}}\right| = \) \( \left| \mathrm{H}\right| \left| \mathrm{K}\right| /\left| {\mathrm{H} \cap \mathrm{K}}\right| \) . | SKETCH OF PROOF. \( C = H \cap K \) is a subgroup of \( K \) of index \( n = \) \( \left| K\right| /\left| {H \cap K}\right| \) and \( K \) is the disjoint union of right cosets \( C{k}_{1} \cup C{k}_{2} \cup \cdots \cup C{k}_{n} \) for some \( {k}_{i} \in K \) . Since \( {HC} = H \), this implies that \( {HK} \) is th... | No |
Proposition 4.8. If \( \mathrm{H} \) and \( \mathrm{K} \) are subgroups of a group \( \mathrm{G} \), then \( \left\lbrack {\mathrm{H} : \mathrm{H} \cap \mathrm{K}}\right\rbrack \leq \) \( \left\lbrack {\mathrm{G} : \mathrm{K}}\right\rbrack \) . If \( \left\lbrack {\mathrm{G} : \mathrm{K}}\right\rbrack \) is finite, the... | SKETCH OF PROOF. Let \( A \) be the set of all right cosets of \( H \cap K \) in \( H \) and \( B \) the set of all right cosets of \( K \) in \( G \) . The map \( \varphi : A \rightarrow B \) given by \( \left( {H \cap K}\right) h \mapsto {Kh} \) \( \left( {h \in H}\right) \) is well defined since \( \left( {H \cap K}... | No |
Proposition 4.9. Let \( \\mathrm{H} \) and \( \\mathrm{K} \) be subgroups of finite index of a group \( \\mathrm{G} \). Then \( \\left\\lbrack {\\mathrm{G} : \\mathrm{H} \\cap \\mathrm{K}}\\right\\rbrack \) is finite and \( \\left\\lbrack {\\mathrm{G} : \\mathrm{H} \\cap \\mathrm{K}}\\right\\rbrack \\leq \\left\\lbrack... | PROOF. Exercise; use Theorem 4.5 and Proposition 4.8. | No |
Theorem 5.1. If \( \mathrm{N} \) is a subgroup of a group \( \mathrm{G} \), then the following conditions are equivalent.\n\n(i) Left and right congruence modulo \( \mathbf{N} \) coincide (that is, define the same equivalence relation on \( \mathrm{G} \) );\n\n(ii) every left coset of \( \mathbf{N} \) in \( \mathbf{G} ... | PROOF. (i) \( \Leftrightarrow \) (iii) Two equivalence relations \( R \) and \( S \) are identical if and only if the equivalence class of each element under \( R \) is equal to its equivalence class under \( S \) . In this case the equivalence classes are the left and right cosets respectively of \( N \) . (ii) \( \Ri... | Yes |
Theorem 5.3. Let \( \mathrm{K} \) and \( \mathrm{N} \) be subgroups of a group \( \mathrm{G} \) with \( \mathrm{N} \) normal in \( \mathrm{G} \). Then\n\n(i) \( \mathrm{N} \cap \mathrm{K} \) is a normal subgroup of \( \mathrm{K} \);\n\n(ii) \( \mathrm{N} \) is a normal subgroup of \( \mathrm{N} \vee \mathrm{K} \);\n\n(... | PROOF. (i) If \( n \in N \cap K \) and \( a \in K \), then \( {an}{a}^{-1} \in N \) since \( N \vartriangleleft G \) and \( {an}{a}^{-1} \in K \) since \( K < G \). Thus \( a\left( {N \cap K}\right) {a}^{-1} \subset N \cap K \) and \( N \cap K \vartriangleleft K \). (ii) is trivial since \( N < N \vee K \). (iii) Clear... | Yes |
Theorem 5.4. If \( \mathrm{N} \) is a normal subgroup of a group \( \mathrm{G} \) and \( \mathrm{G}/\mathrm{N} \) is the set of all (left) cosets of \( \mathrm{N} \) in \( \mathrm{G} \), then \( \mathrm{G}/\mathrm{N} \) is a group of order \( \left\lbrack {\mathrm{G} : \mathrm{N}}\right\rbrack \) under the binary opera... | PROOF. Since the coset \( {aN} \) [resp. \( {bN},{abN} \) ] is simply the equivalence class of \( a \in G \) [resp. \( b \in G,{ab} \in G \) ] under the equivalence relation of congruence modulo \( N \), it suffices by Theorem 1.5 to show that congruence modulo \( N \) is a congruence relation, that is, that \( {a}_{1}... | Yes |
Theorem 5.5. If \( f : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups, then the kernel of \( \mathrm{f} \) is a normal subgroup of \( \mathrm{G} \) . Conversely, if \( \mathrm{N} \) is a normal subgroup of \( \mathrm{G} \), then the map \( \pi : \mathrm{G} \rightarrow \mathrm{G}/\mathrm{N} \) given by... | PROOF. If \( x \in \operatorname{Ker}f \) and \( a \in G \), then\n\n\[ f\left( {{ax}{a}^{-1}}\right) = f\left( a\right) f\left( x\right) f\left( {a}^{-1}\right) = f\left( a\right) {ef}{\left( a\right) }^{-1} = e \]\n\nand \( {ax}{a}^{-1}\varepsilon \operatorname{Ker}f \) . Therefore \( a\left( {\operatorname{Ker}f}\ri... | Yes |
Theorem 5.6. If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups and \( \mathrm{N} \) is a normal subgroup of \( \mathrm{G} \) contained in the kernel of \( \mathrm{f} \), then there is a unique homomorphism \( \bar{\mathrm{f}} : \mathrm{G}/\mathrm{N} \rightarrow \mathrm{H} \) such that ... | PROOF OF 5.6. If \( {b\varepsilon aN} \), then \( b = {an},{n\varepsilon N} \), and \( f\left( b\right) = f\left( {an}\right) = f\left( a\right) f\left( n\right) \) \( = f\left( a\right) e = f\left( a\right) \), since \( N < \operatorname{Ker}f \) . Therefore, \( f \) has the same effect on every element of the coset \... | Yes |
Corollary 5.7. (First Isomorphism Theorem) If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups, then \( \mathrm{f} \) induces an isomorphism \( \mathrm{G}/\operatorname{Ker}\mathrm{f} \cong \operatorname{Im}\mathrm{f} \) . | PROOF. \( f : G \rightarrow \operatorname{Im}f \) is an epimorphism. Apply Theorem 5.6 with \( N = \operatorname{Ker}f \) . \( \blacksquare \) | No |
Corollary 5.8. If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups, \( \mathrm{N} \vartriangleleft \mathrm{G},\mathrm{M} \vartriangleleft \mathrm{H} \), and \( \mathrm{f}\left( N\right) < \mathrm{M} \), then \( \mathrm{f} \) induces a homomorphism \( \bar{\mathrm{f}} : \mathrm{G}/\mathrm... | SKETCH OF PROOF. Consider the composition \( G\overset{f}{ \rightarrow }H\overset{\pi }{ \rightarrow }H/M \) and verify that \( N \subset {f}^{-1}\left( M\right) = \operatorname{Ker}{\pi f} \) . By Theorem 5.6 (applied to \( {\pi f} \) ) the map \( G/N \rightarrow H/M \) given by \( {aN} \mapsto \left( {\pi f}\right) \... | No |
Corollary 5.9. (Second Isomorphism Theorem) If \( \mathrm{K} \) and \( \mathrm{N} \) are subgroups of a group: \( \mathrm{G} \), with \( \mathrm{N} \) normal in \( \mathrm{G} \), then \( \mathrm{K}/\left( {\mathrm{N} \cap \mathrm{K}}\right) \cong \mathrm{{NK}}/\mathrm{N} \). | PROOF. \( N \vartriangleleft {NK} = N \vee K \) by Theorem 5.3. The composition \( K\overset{ \subset }{ \rightarrow }{NK}\overset{\pi }{ \rightarrow } \) \( {NK}/N \) is a homomorphism \( f \) with kernel \( K \cap N \), whence \( \bar{f} : K/K \cap N \cong \operatorname{Im}f \) by Corollary 5.7. Every element in \( {... | Yes |
Corollary 5.10. (Third Isomorphism Theorem). If \( \mathrm{H} \) and \( \mathrm{K} \) are normal subgroups of a group \( \mathrm{G} \) such that \( \mathrm{K} < \mathrm{H} \), then \( \mathrm{H}/\mathrm{K} \) is a normal subgroup of \( \mathrm{G}/\mathrm{K} \) and \( \left( {\mathrm{G}/\mathrm{K}}\right) /\left( {\math... | PROOF. The identity map \( {1}_{G} : G \rightarrow G \) has \( {1}_{G}\left( K\right) < H \) and therefore induces an epimorphism \( I : G/K \rightarrow G/H \), with \( I\left( {aK}\right) = {aH} \) . Since \( H = I\left( {aK}\right) \) if and only if \( {a\varepsilon H} \), Ker \( I = \{ {aK} \mid {a\varepsilon H}\} =... | Yes |
Theorem 5.11. If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is an epimorphism of groups, then the assignment \( \mathrm{K} \mapsto \mathrm{f}\left( \mathrm{K}\right) \) defines a one-to-one correspondence between the set \( {\mathrm{S}}_{\mathrm{f}}\left( \mathrm{G}\right) \) of all subgroups \( \mathrm{K} \)... | SKETCH OF PROOF. By Exercise 2.9 the assignment \( K \mapsto f\left( K\right) \) defines a function \( \varphi : {S}_{f}\left( G\right) \rightarrow S\left( H\right) \) and \( {f}^{-1}\left( J\right) \) is a subgroup of \( G \) for every subgroup \( J \) of \( H \) . Since \( J < H \) implies \( \operatorname{Ker}f < {f... | No |
Corollary 5.12. If \( \mathrm{N} \) is a normal subgroup of a group \( \mathrm{G} \), then every subgroup of \( \mathrm{G}/\mathrm{N} \) is of the form \( \mathrm{K}/\mathrm{N} \), where \( \mathrm{K} \) is a subgroup of \( \mathrm{G} \) that contains \( \mathrm{N} \) . Furthermore, \( \mathrm{K}/\mathrm{N} \) is norma... | PROOF. Apply Theorem 5.11 to the canonical epimorphism \( \pi : G \rightarrow G/N \) . If \( N < K < G \), then \( \pi \left( K\right) = K/N \) . | Yes |
Corollary 6.4. The order of a permutation \( \sigma \in {\mathrm{S}}_{\mathrm{n}} \) is the least common multiple of the orders of its disjoint cycles. | PROOF. Let \( \sigma = {\sigma }_{1}\cdots {\sigma }_{r} \), with \( \left\{ {\sigma }_{i}\right\} \) disjoint cycles. Since disjoint cycles commute, \( {\sigma }^{m} = {\sigma }_{1}{}^{m}\cdots {\sigma }_{r}{}^{m} \) for all \( m \in \mathbf{Z} \) and \( {\sigma }^{m} = \left( 1\right) \) if and only if \( {\sigma }_{... | Yes |
Corollary 6.5. Every permutation in \( {\mathrm{S}}_{\mathrm{n}} \) can be written as a product of (not necessarily disjoint) transpositions. | PROOF. It suffices by Theorem 6.3 to show that every cycle is a product of transpositions. This is easy: \( \left( {x}_{1}\right) = \left( {{x}_{1}{x}_{2}}\right) \left( {{x}_{1}{x}_{2}}\right) \) and for \( r > 1,\left( {{x}_{1}{x}_{2}{x}_{3}\cdots {x}_{r}}\right) \) \( = \left( {{x}_{1}{x}_{r}}\right) \left( {{x}_{1}... | Yes |
Theorem 6.8. For each \( \mathrm{n} \geq 2 \), let \( {\mathrm{A}}_{\mathrm{n}} \) be the set of all even permutations of \( {\mathrm{S}}_{\mathrm{n}} \) . Then \( {\mathrm{A}}_{\mathrm{n}} \) is a normal subgroup of \( {\mathrm{S}}_{\mathrm{n}} \) of index 2 and order \( \left| {\mathrm{S}}_{\mathrm{n}}\right| /2 = \m... | SKETCH OF PROOF OF 6.8. Let \( C \) be the multiplicative subgroup \( \{ 1, - 1\} \) of the integers. Define a map \( f : {S}_{n} \rightarrow C \) by \( \sigma \mapsto \operatorname{sgn}\sigma \) and verify that \( f \) is an epimorphism of groups. Since the kernel of \( f \) is clearly \( {A}_{n},{A}_{n} \) is normal ... | No |
Lemma 6.11. Let \( \mathrm{r},\mathrm{s} \) be distinct elements of \( \{ 1,2,\ldots ,\mathrm{n}\} \) . Then \( {\mathrm{A}}_{\mathrm{n}}\left( {\mathrm{n} \geq 3}\right) \) is generated by the 3-cycles \( \{ \left( \mathrm{{rsk}}\right) \mid 1 \leq \mathrm{k} \leq \mathrm{n},\mathrm{k} \neq \mathrm{r},\mathrm{s}\} \) ... | PROOF. Assume \( n > 3 \) (the case \( n = 3 \) is trivial). Every element of \( {A}_{n} \) is a product of terms of the form \( \left( {ab}\right) \left( {cd}\right) \) or \( \left( {ab}\right) \left( {ac}\right) \), where \( a, b, c, d \) are distinct elements of \( \{ 1,2,\ldots, n\} \) . Since \( \left( {ab}\right)... | Yes |
Lemma 6.12. If \( \mathrm{N} \) is a normal subgroup of \( {\mathrm{A}}_{\mathrm{n}}\left( {\mathrm{n} \geq 3}\right) \) and \( \mathrm{N} \) contains a 3-cycle, then \( \mathrm{N} = {\mathrm{A}}_{\mathrm{n}} \) . | PROOF. If \( \left( {rsc}\right) {\varepsilon N} \), then for any \( k \neq r, s, c,\left( {rsk}\right) = \left( {rs}\right) \left( {ck}\right) {\left( rsc\right) }^{2}\left( {ck}\right) \left( {rs}\right) \) \( = \left\lbrack {\left( {rs}\right) \left( {ck}\right) }\right\rbrack {\left( rsc\right) }^{2}{\left\lbrack \... | Yes |
Theorem 6.13. For each \( \mathrm{n} \geq 3 \) the dihedral group \( {\mathrm{D}}_{\mathrm{n}} \) is a group of order \( 2\mathrm{n} \) whose generators \( \mathrm{a} \) and \( \mathrm{b} \) satisfy:\n\n(i) \( {a}^{n} = \left( 1\right) ;{b}^{2} = \left( 1\right) ;{a}^{k} \neq \left( 1\right) \) if \( 0 < k < n \) ;\n\n... | SKETCH OF PROOF. Verify that \( a, b \in {D}_{n} \) as defined above satisfy (i) and (ii), whence \( {D}_{n} = \langle a, b\rangle = \left\{ {{a}^{i}{b}^{j} \mid 0 \leq i < n;j = 0,1}\right\} \) (see Theorem 2.8). Then verify that the \( {2n} \) elements \( {a}^{i}{b}^{j}\left( {0 \leq i < n;j = 0,1}\right) \) are all ... | Yes |
Theorem 7.3. If \( \left( {\mathrm{P},\left\{ {\pi }_{\mathrm{i}}\right\} }\right) \) and \( \left( {\mathrm{Q},\left\{ {\psi }_{\mathrm{i}}\right\} }\right) \) are both products of the family \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathbf{I}}\right\} \) of objects of a category \( \mathrm{C} \), the... | PROOF. Since \( P \) and \( Q \) are both products, there exist morphisms \( f : P \rightarrow Q \) and \( g : Q \rightarrow P \) such that the following diagrams are commutative for each \( {i\varepsilon I} \) :\n\n\n... | Yes |
Theorem 7.8. If \( \mathcal{C} \) is a concrete category, \( \mathbf{F} \) and \( {\mathbf{F}}^{\prime } \) are objects of \( \mathcal{C} \) such that \( \mathbf{F} \) is free on the set \( \mathrm{X} \) and \( {\mathrm{F}}^{\prime } \) is free on the set \( {\mathrm{X}}^{\prime } \) and \( \left| \mathrm{X}\right| = \... | PROOF OF 7.8. Since \( F,{F}^{\prime } \) are free and \( \left| X\right| = \left| {X}^{\prime }\right| \), there is a bijection \( f : X \rightarrow {X}^{\prime } \) and maps \( i : X \rightarrow F \) and \( j : {X}^{\prime } \rightarrow {F}^{\prime } \) . Consider the map \( {jf} : X \rightarrow {F}^{\prime } \) . Si... | Yes |
Theorem 7.10. Any two universal [resp. couniversal] objects in a category \( \mathcal{C} \) are equivalent. | PROOF. Let \( I \) and \( J \) be universal objects in \( \mathcal{C} \) . Since \( I \) is universal, there is a unique morphism \( f : I \rightarrow J \) . Similarly, since \( J \) is universal, there is a unique morphism \( g : J \rightarrow I \) . The composition \( g \circ f : I \rightarrow I \) is a morphism of \... | Yes |
Theorem 8.1. If \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) is a family of groups, then\n\n(i) the direct product \( \mathop{\prod }\limits_{{i \in I}}{\mathrm{G}}_{\mathrm{i}} \) is a group;\n\n(ii) for each \( \mathrm{k} \in \mathrm{I} \), the map \( {\pi }_{\mathrm{k}} : \... | PROOF. Exercise. | No |
Theorem 8.2. Let \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) be a family of groups and \( \left\{ {{\varphi }_{\mathrm{i}} : \mathrm{H} \rightarrow {\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) a family of group homomorphisms. Then there is a unique homomorphi... | PROOF. By Introduction, Theorem 5.2, the map of sets \( \varphi : H \rightarrow \mathop{\prod }\limits_{{i \in I}}{G}_{i} \) given by \( \varphi \left( a\right) = {\left\{ {\varphi }_{i}\left( a\right) \right\} }_{i\epsilon I} \in \mathop{\prod }\limits_{{i\epsilon I}}{G}_{i} \) is the unique function such that \( {\pi... | Yes |
Theorem 8.4. If \( \\left\\{ {{G}_{i} \\mid {i\\varepsilon }\\mathbf{I}}\\right\\} \) is a family of groups, then\n\n(i) \( \\mathop{\\prod }\\limits_{{i, I}}{}^{\\mathrm{w}}{\\mathrm{G}}_{\\mathrm{i}} \) is a normal subgroup of \( \\mathop{\\prod }\\limits_{{i, I}}{\\mathrm{G}}_{\\mathrm{i}} \) ;\n\n(ii) for each \( \... | PROOF. Exercise. | No |
Theorem 8.5. Let \( \\left\\{ {{\\mathrm{A}}_{\\mathrm{i}} \\mid \\mathrm{i}\\varepsilon \\mathrm{I}}\\right\\} \) be a family of abelian groups (written additively). If \( \\mathrm{B} \) is an abelian group and \( \\left\\{ {{\\psi }_{\\mathrm{i}} : {\\mathrm{A}}_{\\mathrm{i}} \\rightarrow \\mathrm{B} \\mid \\mathrm{i... | PROOF OF 8.5. Throughout this proof all groups will be written additively. If \( 0 \\neq \\left\\{ {a}_{i}\\right\\} \\varepsilon \\sum {A}_{i} \), then only finitely many of the \( {a}_{i} \) are nonzero, say \( {a}_{{i}_{1}},{a}_{{i}_{2}},\\ldots ,{a}_{{i}_{r}} \) . Define \( \\psi : \\sum {A}_{i} \\rightarrow B \) b... | Yes |
Theorem 8.9. Let \( \\left\\{ {{\\mathbf{N}}_{\\mathrm{i}} \\mid \\mathrm{i} \\in \\mathbf{I}}\\right\\} \) be a family of normal subgroups of a group \( \\mathbf{G} \) . \( \\mathbf{G} \) is the internal weak direct product of the family \( \\left\\{ {{\\mathbf{N}}_{\\mathbf{i}} \\mid \\mathbf{i} \\in \\mathbf{I}}\\ri... | ## PROOF. Exercise. | No |
Theorem 8.10. Let \( \left\{ {{\mathrm{f}}_{\mathrm{i}} : {\mathrm{G}}_{\mathrm{i}} \rightarrow {\mathrm{H}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) be a family of homomorphisms of groups and let \( \mathrm{f} = \prod {\mathrm{f}}_{\mathrm{i}} \) be the map \( \mathop{\prod }\limits_{{\mathrm{i}{\vareps... | PROOF. Exercise. | No |
Corollary 8.11. Let \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) and \( \left\{ {{\mathrm{N}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) be families of groups such that \( {\mathrm{N}}_{\mathrm{i}} \) is a normal subgroup of \( {\mathbf{G}}_{\mathrm{i}} \) ... | PROOF. (i) For each \( i \), let \( {\pi }_{i} : {G}_{i} \rightarrow {G}_{i}/{N}_{i} \) be the canonical epimorphism. By Theorem 8.10, the map \( \prod {\pi }_{i} : \mathop{\prod }\limits_{{i \in I}}{G}_{i} \rightarrow \mathop{\prod }\limits_{{i \in I}}{G}_{i}/{N}_{i} \) is an epimorphism with kernel \( \mathop{\prod }... | Yes |
Theorem 9.1. If \( \mathrm{X} \) is a nonempty set and \( \mathrm{F} = \mathrm{F}\left( \mathrm{X}\right) \) is the set of all reduced words on \( \mathrm{X} \), then \( \mathrm{F} \) is a group under the binary operation defined above and \( \mathrm{F} = \langle \mathrm{X}\rangle \) . | SKETCH OF PROOF OF 9.1. Since 1 is an identity element and \( {x}_{1}{}^{{\delta }_{1}}\cdots {x}_{n}{}^{{\delta }_{n}} \) has inverse \( {x}_{n}^{-{\delta }_{n}} \cdot \cdot \cdot {x}_{1}^{-{\delta }_{1}} \), we need only verify associativity. This may be done by induction and a tedious examination of cases or by the ... | Yes |
Theorem 9.2. Let \( \mathrm{F} \) be the free group on a set \( \mathrm{X} \) and \( \iota : \mathrm{X} \rightarrow \mathrm{F} \) the inclusion map. If \( \mathrm{G} \) is a group and \( \mathrm{f} : \mathrm{X} \rightarrow \mathrm{G} \) a map of sets, then there exists a unique homomorphism of groups \( \widetilde{\mat... | SKETCH OF PROOF OF 9.2. Define \( \bar{f}\left( 1\right) = e \) and if \( {x}_{1}{}^{{\delta }_{1}}\cdots {x}_{n}{}^{{\delta }_{n}} \) is a nonempty reduced word on \( X \), define \( \bar{f}\left( {{x}_{1}{}^{{\delta }_{1}}\cdots {x}_{n}{}^{{\delta }_{n}}}\right) = f{\left( {x}_{1}\right) }^{{\delta }_{1}}f{\left( {x}... | No |
Every group \( \mathbf{G} \) is the homomorphic image of a free group. | Let \( X \) be a set of generators of \( G \) and let \( F \) be the free group on the set \( X \) . By Theorem 9.2 the inclusion map \( X \rightarrow G \) induces a homomorphism \( \bar{f} : F \rightarrow G \) such that \( x \mapsto x \in G \) . Since \( G = \langle X\rangle \), the proof of Theorem 9.2 shows that \( ... | Yes |
Theorem 9.5. (Van Dyck) Let \( \mathrm{X} \) be a set, \( \mathrm{Y} \) a set of (reduced) words on \( \mathrm{X} \) and \( \mathrm{G} \) the group defined by the generators \( \mathrm{x}\varepsilon \mathrm{X} \) and relations \( \mathrm{w} = \mathrm{e}\left( {\mathrm{w}\varepsilon \mathrm{Y}}\right) \) . If \( \mathrm... | PROOF OF 9.5. If \( F \) is the free group on \( X \) then the inclusion map \( X \rightarrow H \) induces an epimorphism \( \varphi : F \rightarrow H \) by Corollary 9.3. Since \( H \) satisfies the relations \( w = e\left( {w \in Y}\right), Y \subset \operatorname{Ker}\varphi \) . Consequently, the normal subgroup \(... | Yes |
Theorem 9.6. Let \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) be a family of groups and \( \mathop{\prod }\limits_{{i \in I}}{}^{ * }{\mathrm{G}}_{\mathrm{i}} \) their free product. If \( \left\{ {{\psi }_{\mathrm{i}} : {\mathrm{G}}_{\mathrm{i}} \rightarrow \mathrm{H} \mid \mathrm{i}... | SKETCH OF PROOF. If \( {a}_{1}{a}_{2}\cdots {a}_{n} \) is a reduced word in \( \mathop{\prod }\limits_{{i \in I}}{}^{ * }{G}_{i} \) with \( {a}_{k} \in {G}_{{i}_{k}} \) , define \( \psi \left( {{a}_{1}\cdots {a}_{n}}\right) \) to be \( {\psi }_{{i}_{1}}\left( {a}_{1}\right) {\psi }_{{i}_{2}}\left( {a}_{2}\right) \cdots... | No |
Theorem 1.1. The following conditions on an abelian group \( \mathrm{F} \) are equivalent.\n\n(i) \( \mathrm{F} \) has a nonempty basis.\n\n(ii) \( \mathrm{F} \) is the (internal) direct sum of a family of infinite cyclic subgroups.\n\n(iii) \( \mathrm{F} \) is (isomorphic to) a direct sum of copies of the additive gro... | SKETCH OF PROOF OF 1.1. (i) \( \Rightarrow \) (ii) If \( X \) is a basis of \( F \), then for each \( {x\varepsilon X},{nx} = 0 \) if and only if \( n = 0 \) . Hence each subgroup \( \langle x\rangle \left( {x\varepsilon X}\right) \) is infinite cyclic (and normal since \( F \) is abelian). Since \( F = \langle X\rangl... | Yes |
Proposition 1.3. Let \( {\mathrm{F}}_{1} \) be the free abelian group on the set \( {\mathrm{X}}_{1} \) and \( {\mathrm{F}}_{2} \) the free abelian group on the set \( {\mathrm{X}}_{2} \) . Then \( {\mathrm{F}}_{1} \cong {\mathrm{F}}_{2} \) if and only if \( {\mathrm{F}}_{1} \) and \( {\mathrm{F}}_{2} \) have the same ... | SKETCH OF PROOF OF 1.3. If \( \alpha : {F}_{1} \cong {F}_{2} \), then \( \alpha \left( {X}_{1}\right) \) is a basis of \( {F}_{2} \) , whence \( \left| {X}_{1}\right| = \left| {\alpha \left( {X}_{1}\right) }\right| = \left| {X}_{2}\right| \) by Theorem 1.2. The converse is Theorem I.7.8. | Yes |
Theorem 1.4. Every abelian group \( \mathrm{G} \) is the homomorphic image of a free abelian group of rank \( \left| \mathrm{X}\right| \), where \( \mathrm{X} \) is a set of generators of \( \mathbf{G} \) . | PROOF. Let \( F \) be the free abelian group on the set \( X \) . Then \( F = \mathop{\sum }\limits_{{x \in X}}\mathbf{Z}x \) and rank \( F = \left| X\right| \) . By Theorem 1.1 the inclusion map \( X \rightarrow G \) induces a homomorphism \( \bar{f} : F \rightarrow G \) such that \( {1x} \mapsto {x\varepsilon G} \), ... | Yes |
Lemma 1.5. If \( \\left\\{ {{\\mathrm{x}}_{1},\\ldots ,{\\mathrm{x}}_{\\mathrm{n}}}\\right\\} \) is a basis of a free abelian group \( \\mathrm{F} \) and \( \\mathrm{a}\\varepsilon \\mathbf{Z} \), then for all \( \\mathrm{i} \\neq \\mathrm{j}\\left\\{ {{\\mathrm{x}}_{1},\\ldots ,{\\mathrm{x}}_{\\mathrm{j} - 1},{\\mathr... | PROOF. Since \( {x}_{j} = - a{x}_{i} + \\left( {{x}_{j} + a{x}_{i}}\\right) \), it follows that \( F = \\left\\langle {{x}_{1},\\ldots ,{x}_{j - 1},{x}_{j} + }\\right. \) \( \\left. {a{x}_{i},{x}_{j + 1},\\ldots ,{x}_{n}}\\right\\rangle \) . If \( {k}_{1}{x}_{1} + \\cdots + {k}_{j}\\left( {{x}_{j} + a{x}_{i}}\\right) +... | Yes |
Corollary 1.7. If \( \mathrm{G} \) is a finitely generated abelian group generated by \( \mathrm{n} \) elements, then every subgroup \( \mathrm{H} \) of \( \mathrm{G} \) may be generated by \( \mathrm{m} \) elements with \( \mathrm{m} \leq \mathrm{n} \) . | PROOF OF 1.7. By Theorem 1.4 there is a free abelian group \( F \) of rank \( n \) and an epimorphism \( \pi : F \rightarrow G.{\pi }^{-1}\left( H\right) \) is a subgroup of \( F \), and therefore, free of rank \( m \leq n \) by Theorem 1.6. The image under \( \pi \) of any basis of \( {\pi }^{-1}\left( H\right) \) is ... | Yes |
Theorem 2.1. Every finitely generated abelian group \( \mathrm{G} \) is (isomorphic to) a finite direct sum of cyclic groups in which the finite cyclic summands (if any) are of orders \( {\mathrm{m}}_{1},\ldots ,{\mathrm{m}}_{\mathrm{t}} \), where \( {\mathrm{m}}_{1} > 1 \) and \( {\mathrm{m}}_{1}\left| {\mathrm{\;m}}_... | PROOF. If \( G \neq 0 \) and \( G \) is generated by \( n \) elements, then there is a free abelian group \( F \) of rank \( n \) and an epimorphism \( \pi : F \rightarrow G \) by Theorem 1.4. If \( \pi \) is an isomorphism, then \( G \cong F \cong \mathbf{Z} \oplus \cdots \oplus \mathbf{Z} \) ( \( n \) summands). If n... | Yes |
Theorem 2.2. Every finitely generated abelian group \( \mathrm{G} \) is (isomorphic to) a finite direct sum of cyclic groups, each of which is either infinite or of order a power of a prime. | SKETCH OF PROOF. The theorem is an immediate consequence of Theorem 2.1 and the following lemma. Another proof is sketched in Exercise 4. | No |
Lemma 2.3. If \( \mathrm{m} \) is a positive integer and \( \mathrm{m} = {\mathrm{p}}_{1}{}^{{\mathrm{n}}_{1}}{\mathrm{p}}_{2}{}^{{\mathrm{n}}_{2}}\cdots {\mathrm{p}}_{\mathrm{t}}{}^{{\mathrm{n}}_{\mathrm{t}}}\left( {{\mathrm{p}}_{1},\ldots ,{\mathrm{p}}_{\mathrm{t}}}\right. \) distinct primes and each \( {\mathrm{n}}_... | SKETCH OF PROOF. Use induction on the number \( t \) of primes in the prime decomposition of \( m \) and the fact that\n\n\[ \n{Z}_{rn} \cong {Z}_{r} \oplus {Z}_{n}\;\text{ whenever }\;\left( {r, n}\right) = 1, \]\n\nwhich we now prove. The element \( n = {n1\varepsilon }{Z}_{rn} \) has order \( r \) (Theorem I.3.4 (vi... | No |
Corollary 2.4. If \( \mathrm{G} \) is a finite abelian group of order \( \mathrm{n} \), then \( \mathrm{G} \) has a subgroup of order \( \mathrm{m} \) for every positive integer \( \mathrm{m} \) that divides \( \mathrm{n} \) . | SKETCH OF PROOF. Use Theorem 2.2 and observe that \( G \cong \mathop{\sum }\limits_{{i = 1}}^{k}{G}_{i} \) implies that \( \left| G\right| = \left| {G}_{1}\right| \left| {G}_{2}\right| \cdots \left| {G}_{k}\right| \) and for \( i \leq r,{p}^{r - i}{Z}_{{p}^{r}} \cong {Z}_{{p}^{i}} \) by Lemma 2.5 (v) below. | No |
Lemma 2.5. Let \( \mathrm{G} \) be an abelian group, \( \mathrm{m} \) an integer and \( \mathrm{p} \) a prime integer. Then each of the following is a subgroup of \( \mathrm{G} \) :\n\n(i) \( \mathrm{{mG}} = \{ \mathrm{{mu}} \mid \mathrm{u}\varepsilon \mathrm{G}\} \) ;\n\n(ii) \( \mathrm{G}\left\lbrack \mathrm{m}\right... | SKETCH OF PROOF. (i)-(iv) are exercises; the hypothesis that \( G \) is abelian is essential \( \left( {S}_{3}\right. \) provides counterexamples for (i)-(iii) and Exercise I.3.5 for (iv)). | No |
Corollary 2.7. Two finitely generated abelian groups \( \mathrm{G} \) and \( \mathrm{H} \) are isomorphic if and only if \( \mathrm{G}/{\mathrm{G}}_{\mathrm{t}} \) and \( \mathrm{H}/{\mathrm{H}}_{\mathrm{t}} \) have the same rank and \( \mathrm{G} \) and \( \mathrm{H} \) have the same invariant factors [resp. elementar... | PROOF. Exercise. | No |
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