Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 4.13. Let \( E \) be a finite free module of rank \( r \) over the ring \( R \) . For each \( p = 1,\ldots, r \) there is a unique homomorphism\n\n\[ \n{d}_{p} : {\bigwedge }^{p}E \otimes {SE} \rightarrow {\bigwedge }^{p - 1}E \otimes {SE} \n\]\n\nsuch that\n\n\[ \n{d}_{i}\left( {\left( {{x}_{1} \land \cdots \l... | Proof. The above definitions are merely examples of the Koszul complex for the symmetric algebra \( {SE} \) with respect to the regular sequence consisting of some basis of \( E \) .\n\nSince \( {d}_{p} \) maps \( \mathop{\bigwedge }\limits^{p}E \otimes {S}^{q}E \) into \( \mathop{\bigwedge }\limits^{{p - 1}}E \otimes ... | Yes |
Theorem 4.15. (Hilbert Syzygy Theorem). Let \( k \) be a field and\n\n\[ A = k\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \]\n\nthe polynomial ring in \( r \) variables. Let \( M \) be a graded module over \( A \), and let\n\n\[ 0 \rightarrow K \rightarrow {L}_{r - 1} \rightarrow \cdots \rightarrow {L}_{0} \rig... | Proof. From the Koszul complex we know that \( {\operatorname{Tor}}_{i}\left( {M, k}\right) = 0 \) for \( i > r \) and all \( M \) . By dimension shifting, it follows that\n\n\[ {\operatorname{Tor}}_{i}\left( {K, k}\right) = 0\;\text{ for }\;i > 0. \]\n\nThe theorem is then a consequence of the next result. | Yes |
Theorem 4.16. Let \( F \) be a graded finite module over \( A = k\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \) . If \( {\operatorname{Tor}}_{1}\left( {F, k}\right) = 0 \) then \( F \) is free. | Proof. The method is essentially to do a Nakayama type argument in the case of the non-local ring \( A \) . First note that \[ F \otimes k = F/{IF} \] where \( I = \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) . Thus \( F \otimes k \) is naturally an \( A/I = k \) -module. Let \( {v}_{1},\ldots ,{v}_{n} \) be homogeneous ... | Yes |
Lemma 4.17. Let \( N \) be a graded module over \( A = k\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \) . Let\n\n\( I = \left( {{x}_{1},\ldots ,{x}_{r}}\right) \) . If \( N/{IN} = 0 \) then \( N = 0 \) . | Proof. This is immediate by using the grading, looking at elements of \( N \) of smallest degree if they exist, and using the fact that elements of \( I \) have degree \( > 0 \) . | No |
Theorem 4.18. Let \( R \) be a commutative local ring and let \( A = R\left\lbrack {{x}_{1},\ldots ,{x}_{r}}\right\rbrack \) be the polynomial ring in \( r \) variables. Let \( M \) be a graded finite module over \( A \) , projective over \( R \) . Let\n\n\[ 0 \rightarrow K \rightarrow {L}_{r - 1} \rightarrow \cdots \r... | Proof. Replace \( k \) by \( R \) everywhere in the proof of the Hilbert syzygy theorem. We use the fact that a finite projective module over a local ring is free. Not a word needs to be changed in the above proof with the following exception. We note that the projectivity propagates to the kernels and cokernels in the... | Yes |
Corollary 1. (Hermite-Lindemann). If \( \alpha \) is algebraic (over \( \mathbf{Q} \) ) and \( \neq 0 \) , then \( {e}^{\alpha } \) is transcendental. Hence \( \pi \) is transcendental. | Proof. Suppose that \( \alpha \) and \( {e}^{\alpha } \) are algebraic. Let \( K = \mathbf{Q}\left( {\alpha ,{e}^{\alpha }}\right) \) . The two functions \( z \) and \( {e}^{z} \) are algebraically independent over \( K \) (trivial), and the ring \( K\left\lbrack {z,{e}^{z}}\right\rbrack \) is obviously mapped into its... | No |
Corollary 2. (Gelfond-Schneider). If \( \alpha \) is algebraic \( \neq 0,1 \) and if \( \beta \) is algebraic irrational, then \( {\alpha }^{\beta } = {e}^{\beta \log \alpha } \) is transcendental. | Proof. We proceed as in Corollary 1, considering the functions \( {e}^{\beta t} \) and \( {e}^{t} \) which are algebraically independent because \( \beta \) is assumed irrational. We look at the numbers \( \log \alpha ,2\log \alpha ,\ldots, m\log \alpha \) to get a contradiction as in Corollary 1. | No |
Lemma 1. Let\n\n\\[ \n{a}_{11}{x}_{1} + \cdots + {a}_{1n}{x}_{n} = 0 \n\\]\n\n\\[ \n{a}_{r1}{x}_{1} + \cdots + {a}_{rn}{x}_{n} = 0 \n\\]\n\nbe a system of linear equations with integer coefficients \\( {a}_{ij} \\), and \\( n > r \\) . Let \\( A \\) be a number such that \\( \\left| {a}_{ij}\\right| \\leqq A \\) for al... | Proof. We view our system of linear equations as a linear equation \\( L\\left( X\\right) = 0 \\), where \\( L \\) is a linear map, \\( L : {\\mathbf{Z}}^{\\left( n\\right) } \\rightarrow {\\mathbf{Z}}^{\\left( r\\right) } \\), determined by the matrix of coefficients. If \\( B \\) is a positive number, we denote by \\... | Yes |
Lemma 2. Let \( K \) be a finite extension of \( \mathbf{Q} \). Let\n\n\[ \n{\alpha }_{11}{x}_{1} + \cdots + {\alpha }_{1n}{x}_{n} = 0 \n\]\n\n\[ \n{\alpha }_{r1}{x}_{1} + \cdots + {\alpha }_{rn}{x}_{n} = 0 \n\]\n\nbe a system of linear equations with coefficients in \( {I}_{K} \), and \( n > r \). Let \( A \) be a num... | Proof. Let \( {\omega }_{1},\ldots ,{\omega }_{M} \) be a basis of \( {I}_{K} \) over \( \mathbf{Z} \). Each \( {x}_{j} \) can be written\n\n\[ \n{x}_{j} = {\xi }_{j1}{\omega }_{1} + \cdots + {\xi }_{jM}{\omega }_{M} \n\]\n\nwith unknowns \( {\xi }_{j\lambda } \). Each \( {\alpha }_{ij} \) can be written\n\n\[ \n{\alph... | Yes |
Lemma 3. Let \( K \) be of finite degree over \( \mathbf{Q} \). Let \( {f}_{1},\ldots ,{f}_{N} \) be functions, holomorphic on a neighborhood of a point \( w \in \mathbf{C} \), and assume that \( D = d/{dz} \) maps the ring \( K\left\lbrack {{f}_{1},\ldots ,{f}_{N}}\right\rbrack \) into itself. Assume that \( {f}_{i}\l... | Proof. There exist polynomials \( {P}_{i}\left( {{T}_{1},\ldots ,{T}_{N}}\right) \) with coefficients in \( K \) such that \[ D{f}_{i} = {P}_{i}\left( {{f}_{1},\ldots ,{f}_{N}}\right) \] Let \( h \) be the maximum of their degrees. There exists a unique derivation \( \bar{D} \) on \( K\left\lbrack {{T}_{1},\ldots ,{T}_... | Yes |
Proposition 1.1. Let \( D \) be an infinite subset of \( {\mathbf{Z}}^{ + } \) . Then \( D \) is denumerable, and in fact there is a unique enumeration of \( D \), say \( \left\{ {{k}_{1},{k}_{2},\ldots }\right\} \) such that \[ {k}_{1} < {k}_{2} < \cdots < {k}_{n} < {k}_{n + 1} < \cdots . \] | Proof. We let \( {k}_{1} \) be the smallest element of \( D \) . Suppose inductively that we have defined \( {k}_{1} < \cdots < {k}_{n} \), in such a way that any element \( k \) in \( D \) which is not equal to \( {k}_{1},\ldots ,{k}_{n} \) is \( > {k}_{n} \) . We define \( {k}_{n + 1} \) to be the smallest element of... | Yes |
Corollary 1.2. Let \( S \) be a denumerable set and \( D \) an infinite subset of \( S \). Then \( D \) is denumerable. | Proof. Given an enumeration of \( S \), the subset \( D \) corresponds to a subset of \( {\mathbf{Z}}^{ + } \) in this enumeration. Using Proposition 1.1, we conclude that we can enumerate \( D \) . | No |
Proposition 1.3. Every infinite set contains a denumerable subset. | Proof. Let \( S \) be an infinite set. For every non-empty subset \( T \) of \( S \), we select a definite element \( {a}_{T} \) in \( T \) . We then proceed by induction. We let \( {x}_{1} \) be the chosen element \( {a}_{S} \) . Suppose that we have chosen \( {x}_{1},\ldots ,{x}_{n} \) having the property that for ea... | Yes |
Proposition 1.4. Let \( D \) be a denumerable set, and \( f : D \rightarrow S \) a surjective mapping. Then \( S \) is denumerable or finite. | Proof. For each \( y \in S \), there exists an element \( {x}_{y} \in D \) such that \( f\left( {x}_{y}\right) = y \) because \( f \) is surjective. The association \( y \mapsto {x}_{v} \) is an injective mapping of \( S \) into \( D \), because if\n\n\[ y, z \in S\text{ and }{x}_{y} = {x}_{z} \]\n\nthen\n\n\[ y = f\le... | Yes |
Proposition 1.5. Let \( D \) be a denumerable set. Then \( D \times D \) (the set of all pairs \( \left( {x, y}\right) \) with \( x, y \in D \) ) is denumerable. | Proof. There is a bijection between \( D \times D \) and \( {\mathbf{Z}}^{ + } \times {\mathbf{Z}}^{ + } \), so it will suffice to prove that \( {\mathbf{Z}}^{ + } \times {\mathbf{Z}}^{ + } \) is denumerable. Consider the mapping of \( {\mathbf{Z}}^{ + } \times {\mathbf{Z}}^{ + } \rightarrow {\mathbf{Z}}^{ + } \) given... | Yes |
Proposition 1.6. Let \( \\left\\{ {{D}_{1},{D}_{2},\\ldots }\\right\\} \) be a sequence of denumerable sets. Let \( S \) be the union of all sets \( {D}_{i}\\left( {i = 1,2,\\ldots }\\right) \) . Then \( S \) is denumerable. | Proof. For each \( i = 1,2,\\ldots \) we enumerate the elements of \( {D}_{i} \), as indicated in the following notation:\n\n\[ \n{D}_{1} : \\left\\{ {{x}_{11},{x}_{12},{x}_{13},\\ldots }\\right\\} \n\]\n\n\[ \n{D}_{2} : \\;\\left\\{ {{x}_{21},{x}_{22},{x}_{23},\\ldots }\\right\\} \n\]\n\n...\n\n\[ \n{D}_{i} : \\;\\lef... | Yes |
Corollary 1.7. Let \( F \) be a non-empty finite set and \( D \) a denumerable set. Then \( F \times D \) is denumerable. If \( {S}_{1},{S}_{2},\ldots \) are a sequence of sets, each of which is finite or denumerable, then the union \( {S}_{1} \cup {S}_{2} \cup \cdots \) is denumerable or finite. | Proof. There is an injection of \( F \) into \( {\mathbf{Z}}^{ + } \) and a bijection of \( D \) with \( {\mathbf{Z}}^{ + } \) . Hence there is an injection of \( F \times {\mathbf{Z}}^{ + } \) into \( {\mathbf{Z}}^{ + } \times {\mathbf{Z}}^{ + } \) and we can apply Corollary 1.2 and Proposition 1.6 to prove the first ... | No |
Let \( G \) be a group. Let \( S \) be the set of subgroups. If \( H,{H}^{\prime } \) are subgroups of \( G \), we define\n\n\[ H \leqq {H}^{\prime } \]\n\nif \( H \) is a subgroup of \( {H}^{\prime } \). One verifies immediately that this relation defines an ordering on \( S \). Given two subgroups \( H,{H}^{\prime } ... | One verifies immediately that this relation defines an ordering on \( S \). | No |
Let \( S \) be strictly inductively ordered if every non-empty totally ordered subset has a least upper bound. | To prove this, let us take Example 2. Let \( T \) be a non-empty totally ordered subset of the set of subgroups of \( G \) . This means that if \( H,{H}^{\prime } \in T \), then \( H \subset {H}^{\prime } \) or \( {H}^{\prime } \subset H \) . Let \( U \) be the union of all sets in \( T \) . Then:\n\n1. \( U \) is a su... | Yes |
Theorem 2.1. (Bourbaki). Let \( A \) be a non-empty partially ordered and strictly inductively ordered set. Let \( f : A \rightarrow A \) be an increasing mapping.\n\nThen there exists an element \( {x}_{0} \in A \) such that \( f\left( {x}_{0}\right) = {x}_{0} \) . | Proof. Suppose that \( A \) were totally ordered. By assumption, it would have a least upper bound \( b \in A \), and then\n\n\[ b \leqq f\left( b\right) \leqq b, \]\n\nso that in this case, our theorem is clear. The whole problem is to reduce the theorem to that case. In other words, what we need to find is a totally ... | Yes |
Lemma 2.2. We have \( {M}_{c} = M \) for every extreme point \( c \) of \( M \) . | Proof. It will suffice to prove that \( {M}_{c} \) is an admissible subset. Let \( x \in {M}_{c} \) . If \( x < c \) then \( f\left( x\right) \leqq c \) so \( f\left( x\right) \in {M}_{c} \) . If \( x = c \) then \( f\left( x\right) = f\left( c\right) \) is again in \( {M}_{c} \) . If \( f\left( c\right) \leqq x \), th... | Yes |
Corollary 2.4. Let \( A \) be a non-empty strictly inductively ordered set. Then \( A \) has a maximal element. | Proof. Suppose that \( A \) does not have a maximal element. Then for each \( x \in A \) there exists an element \( {y}_{x} \in A \) such that \( x < {y}_{x} \) . Let \( f : A \rightarrow A \) be the map such that \( f\left( x\right) = {y}_{x} \) for all \( x \in A \) . Then \( A, f \) satisfy the hypotheses of Theorem... | Yes |
Corollary 2.5. (Zorn's lemma). Let \( S \) be a non-empty inductively ordered set. Then \( S \) has a maximal element. | Proof. Let \( A \) be the set of non-empty totally ordered subsets of \( S \) . Then \( A \) is not empty since any subset of \( S \) with one element belongs to \( A \) . If \( X, Y \in A \) , we define \( X \leqq Y \) to mean \( X \subset Y \) . Then \( A \) is partially ordered, and is in fact strictly inductively o... | Yes |
Theorem 3.1. (Schroeder-Bernstein). Let \( A, B \) be sets, and suppose that \( \operatorname{card}\left( A\right) \leqq \operatorname{card}\left( B\right) \), and \( \operatorname{card}\left( B\right) \leqq \operatorname{card}\left( A\right) \) . Then\n\n\[\n\operatorname{card}\left( A\right) = \operatorname{card}\lef... | Proof. Let\n\n\[\nf : A \rightarrow B\text{ and }g : B \rightarrow A\n\]\n\nbe injections. We separate \( A \) into two disjoint sets \( {A}_{1} \) and \( {A}_{2} \) . We let \( {A}_{1} \) consist of all \( x \in A \) such that, when we lift back \( x \) by a succession of inverse maps,\n\n\[\nx,{g}^{-1}\left( x\right)... | Yes |
Lemma 3.2. Let \( A \) be an infinite set. Then there exists a disjoint covering of \( A \) by denumerable sets. | Proof. Let \( S \) be the set whose elements are pairs \( \left( {B,\Gamma }\right) \) consisting of a subset \( B \) of \( A \), and a disjoint covering of \( B \) by denumerable sets. Then \( S \) is not empty. Indeed, since \( A \) is infinite, \( A \) contains a denumerable set \( D \), and the pair \( \left( {D,\{... | Yes |
Theorem 3.3. Let \( A \) be an infinite set, and let \( D \) be a denumerable set. Then\n\n\[ \operatorname{card}\left( {A \times D}\right) = \operatorname{card}\left( A\right) . \] | Proof. By the lemma, we can write\n\n\[ A = \mathop{\bigcup }\limits_{{i \in I}}{D}_{i} \]\n\n as a disjoint union of denumerable sets. Then\n\n\[ A \times D = \mathop{\bigcup }\limits_{{i \in I}}\left( {{D}_{i} \times D}\right) \]\n\nFor each \( i \in I \), there is a bijection of \( {D}_{i} \times D \) on \( {D}_{i} ... | Yes |
Corollary 3.4. If \( F \) is a finite non-empty set, then\n\n\[ \operatorname{card}\left( {A \times F}\right) = \operatorname{card}\left( A\right) \] | Proof. We have\n\n\[ \operatorname{card}\left( A\right) \leqq \operatorname{card}\left( {A \times F}\right) \leqq \operatorname{card}\left( {A \times D}\right) = \operatorname{card}\left( A\right) . \]\n\nWe can then use Theorem 3.1 to get what we want. | No |
Corollary 3.5. Let \( A, B \) be non-empty sets, \( A \) infinite, and suppose\n\n\[ \operatorname{card}\left( B\right) \leqq \operatorname{card}\left( A\right) \]\n\nThen\n\n\[ \operatorname{card}\left( {A \cup B}\right) = \operatorname{card}\left( A\right) . \] | Proof. We can write \( A \cup B = A \cup C \) for some subset \( C \) of \( B \), such that \( C \) and \( A \) are disjoint. (We let \( C \) be the set of all elements of \( B \) which are not elements of \( A \) .) Then \( \operatorname{card}\left( C\right) \leqq \operatorname{card}\left( A\right) \) . We can then co... | Yes |
Corollary 3.7. If \( A \) is an infinite set, and \( {A}^{\left( n\right) } = A \times \cdots \times A \) is the product taken \( n \) times, then | Proof. Induction. | No |
Corollary 3.8. If \( {A}_{1},\ldots ,{A}_{n} \) are non-empty sets with \( {A}_{n} \) infinite, and\n\n\[ \n\operatorname{card}\left( {A}_{i}\right) \leqq \operatorname{card}\left( {A}_{n}\right) \n\]\n\nfor \( i = 1,\ldots, n \), then\n\n\[ \n\operatorname{card}\left( {{A}_{1} \times \cdots \times {A}_{n}}\right) = \o... | Proof. We have\n\n\[ \n\operatorname{card}\left( {A}_{n}\right) \leqq \operatorname{card}\left( {{A}_{1} \times \cdots \times {A}_{n}}\right) \leqq \operatorname{card}\left( {{A}_{n} \times \cdots \times {A}_{n}}\right) \n\]\n\nand we use Corollary 3.7 and the Schroeder-Bernstein theorem to conclude the proof. | No |
Corollary 3.9. Let \( A \) be an infinite set, and let \( \Phi \) be the set of finite subsets of \( A \) . Then \[ \operatorname{card}\left( \Phi \right) = \operatorname{card}\left( A\right) \] | Proof. Let \( {\Phi }_{n} \) be the set of subsets of \( A \) having exactly \( n \) elements, for each integer \( n = 1,2,\ldots \) . We first show that \( \operatorname{card}\left( {\Phi }_{n}\right) \leqq \operatorname{card}\left( A\right) \) . If \( F \) is an element of \( {\Phi }_{n} \), we order the elements of ... | No |
Theorem 3.10. Let \( A \) be an infinite set, and \( T \) the set consisting of two elements \( \{ 0,1\} \) . Let \( M \) be the set of all maps of \( A \) into \( T \) . Then | Proof. For each \( x \in A \) we let\n\n\[ \n{f}_{x} : A \rightarrow \{ 0,1\} \n\]\n\nbe the map such that \( {f}_{x}\left( x\right) = 1 \) and \( {f}_{x}\left( y\right) = 0 \) if \( y \neq x \) . Then \( x \mapsto {f}_{x} \) is obviously an injection of \( A \) into \( M \), so that \( \operatorname{card}\left( A\righ... | Yes |
Corollary 3.11. Let \( A \) be an infinite set, and let \( S \) be the set of all subsets of \( A \) . Then \( \operatorname{card}\left( A\right) \leqq \operatorname{card}\left( S\right) \) and \( \operatorname{card}\left( A\right) \neq \operatorname{card}\left( S\right) \) . | Proof. We leave it as an exercise. [Hint: If \( B \) is a non-empty subset of \( A \) , use the characteristic function \( {\varphi }_{B} \) such that \[ {\varphi }_{B}\left( x\right) = 1\;\text{ if }\;x \in B, \] \[ {\varphi }_{B}\left( x\right) = 0\;\text{ if }\;x \notin B. \] What can you say about the association \... | No |
The set of positive integers \( {\mathbf{Z}}^{ + } \) is well-ordered. | Any finite set can be well-ordered, and a denumerable set \( D \) can be well-ordered: Any bijection of \( D \) with \( {\mathbf{Z}}^{ + } \) will give rise to a well-ordering of \( D \) . | No |
Example 2. Let \( S \) be a well-ordered set and let \( b \) be an element of some set, \( b \notin S \) . Let \( A = S \cup \{ b\} \) . We define \( x \leqq b \) for all \( x \in S \) . Then \( A \) is totally ordered, and is in fact well-ordered. | Proof. Let \( B \) be a non-empty subset of \( A \) . If \( B \) consists of \( b \) alone, then \( b \) is a least element of \( B \) . Otherwise, \( B \) contains some element \( a \in A \) . Then \( B \cap A \) is not empty, and hence has a least element, which is obviously also a least element for B. | No |
Example 1.3. Take \( S = \mathbb{Z} \), and let \( \sim \) be the relation defined by\n\n\[ a \sim b \Leftrightarrow a - b\\text{ is even. } \]\n\nThen \( \mathbb{Z}/ \sim \) consists of two equivalence classes: | Indeed, every integer \( b \) is either even (and hence \( b - 0 \) is even, so \( b \sim 0 \), and \( b \in {\\left\\lbrack 0\\right\\rbrack }_{ \\sim } \) ) or odd (and hence \( b - 1 \) is even, so \( b \sim 1 \), and \( b \in {\\left\\lbrack 1\\right\\rbrack }_{ \\sim } \) ). | Yes |
Proposition 2.1. Assume \( A \neq \varnothing \), and let \( f : A \rightarrow B \) be a function. Then\n\n(1) \( f \) has a left-inverse if and only if it is injective. | Proof. Let’s prove (1).\n\n\( \left( \Rightarrow \right) \) If \( f : A \rightarrow B \) has a left-inverse, then there exists a \( g : B \rightarrow A \) such that \( g \circ f = {\operatorname{id}}_{A} \) . Now assume that \( {a}^{\prime } \neq {a}^{\prime \prime } \) are arbitrary different elements in \( A \) ; the... | Yes |
Proposition 2.3. A function is injective if and only if it is a monomorphism. | Proof. ( \( \Rightarrow \) ) By Proposition 2.1, if a function \( f : A \rightarrow B \) is injective, then it has a left-inverse \( g : B \rightarrow A \) . Now assume that \( {\alpha }^{\prime },{\alpha }^{\prime \prime } \) are arbitrary functions from another set \( Z \) to \( A \) and that\n\n\[ f \circ {\alpha }^... | Yes |
Let \( A, B \) be sets. Then there are natural projections \( {\pi }_{A},{\pi }_{B} \) defined by \[ {\pi }_{A}\left( \left( {a, b}\right) \right) \mathrel{\text{:=}} a,\;{\pi }_{B}\left( \left( {a, b}\right) \right) \mathrel{\text{:=}} b \] for all \( \left( {a, b}\right) \in A \times B \) . | Both of these maps are (clearly) surjective. | No |
Similarly, there are natural injections from \( A \) and \( B \) to the disjoint union: | obtained by sending \( a \in A \) (resp., \( b \in B \) ) to the corresponding element in the isomorphic copy \( {A}^{\prime } \) of \( A \) (resp., \( {B}^{\prime } \) of \( B \) ) in \( A \coprod B \) . | No |
If \( \sim \) is an equivalence relation on a set \( A \), there is a (clearly surjective) canonical projection | \[ A \rightarrow A/ \sim \] obtained by sending every \( a \in A \) to its equivalence class \( {\left\lbrack a\right\rbrack }_{ \sim } \) | Yes |
Example 3.2. It is hopefully crystal clear by now that sets (as objects), together with set-functions (as morphisms), form a category; if not, the reader must stop here and go no further until this assertion sheds any residual mystery \( {}^{17} \) . | - \( \operatorname{Obj}\left( \operatorname{Set}\right) = \) the class of all sets;\n\n- for \( A, B \) in \( \operatorname{Obj}\left( \operatorname{Set}\right) \) (that is, for \( A, B \) sets) \( {\operatorname{Hom}}_{\operatorname{Set}}\left( {A, B}\right) = {B}^{A} \) . | No |
Suppose \( S \) is a set and \( \sim \) is a relation on \( S \) satisfying the reflexive and transitive properties. Then we can encode this data into a category: | - objects: the elements of \( S \) ;\n\n- morphisms: if \( a, b \) are objects (that is, if \( a, b \in S \) ), then let \( \operatorname{Hom}\left( {a, b}\right) \) be the set consisting of the element \( \left( {a, b}\right) \in S \times S \) if \( a \sim b \), and let \( \operatorname{Hom}\left( {a, b}\right) = \var... | No |
Let \( S \) again be a set. Define a category \( \widehat{\mathrm{S}} \) by setting\n\n- \( \operatorname{Obj}\left( \widehat{\mathrm{S}}\right) = \mathcal{P}\left( S\right) \), the power set \( S \) (cf. \( §{1.2} \) and Exercise 2.11);\n\n- for \( A, B \) objects of \( \widehat{\mathrm{S}} \) (that is, \( A \subseteq... | The identity \( {1}_{A} \) consists of the pair \( \left( {A, A}\right) \) (which is one, and in fact the only one, morphism from \( A \) to \( A \), since \( A \subseteq A \) ). Composition is obtained by stringing inclusions: if there are morphisms\n\n\[ A \rightarrow B,\;B \rightarrow C \]\n\nin \( \widehat{\mathrm{... | No |
Let \( \mathrm{C} \) be a category, and let \( A \) be an object of \( \mathrm{C} \). We are going to define a category \( {\mathrm{C}}_{A} \) whose objects are certain morphisms in \( \mathrm{C} \) and whose morphisms are certain diagrams of \( \mathrm{C} \) (surprise!). | - \( \operatorname{Obj}\left( {\mathrm{C}}_{A}\right) = \) all morphisms from any object of \( \mathrm{C} \) to \( A \) ; thus, an object of \( {\mathrm{C}}_{A} \) is a morphism \( f \in {\operatorname{Hom}}_{\mathrm{C}}\left( {Z, A}\right) \) for some object \( Z \) of \( \mathrm{C} \). Pictorially, an object of \( {\... | Yes |
For the sake of concreteness, let's apply the construction given in Example 3.5 to the category constructed in Example 3.3, say for \( S = \mathbb{Z} \) and \( \sim \) the relation \( \leq \) . Call \( \mathrm{C} \) this category, and choose an object \( A \) of \( \mathrm{C} \) -that is, an integer, for example, \( A ... | \[ \left( {m,3}\right) \rightarrow \left( {n,3}\right) \] if and only if \( m \leq n \) . In this case \( {\mathrm{C}}_{A} \) may be harmlessly identified with the ’subcategory’ of integers \( \leq 3 \), with ’the same’ morphisms as in \( \mathrm{C} \). | Yes |
It is useful to contemplate a few more 'abstract' examples in the style of Examples 3.5 and 3.7. These will be essential ingredients in the promised revisitation of some of the operations mentioned in \( §{1.3} \) . Their definition will appear disappointedly simple-minded to the reader who has mastered Examples 3.5 an... | We will leave to the reader the task of formalizing this rough description. This example is really nothing more than a mixture of \( {\mathrm{C}}_{A} \) and \( {\mathrm{C}}_{B} \), where the two structures interact because of the stringent requirement that the same \( \sigma \) must make both sides of the diagram commu... | No |
Example 3.10. As a final variation on these examples, we conclude by considering the fibered version of \( {\mathrm{C}}_{A, B} \) (and \( {\mathrm{C}}^{A, B} \) ). Take this as a test to see if you have really understood \( {\mathrm{C}}_{A, B} \) -experts would tell you that this looks fairly sophisticated for students... | A solid understanding of Example 3.9 will make this example look just as tame; at this point the reader should have no difficulties formalizing it (that is, explaining how composition works, what identities are, etc.).\n\nAlso left to the reader is the construction of the ’mirror’ example \( {C}^{\alpha ,\beta } \), st... | No |
Proposition 4.2. The inverse of an isomorphism is unique. | Proof. We have to verify that if both \( {g}_{1} \) and \( {g}_{2} : B \rightarrow A \) act as inverses of a given isomorphism \( f : A \rightarrow B \), then \( {g}_{1} = {g}_{2} \) . The standard trick for this kind of verification is to compose \( f \) on the left by one of the morphisms, and on the right by the oth... | Yes |
Proposition 4.3. With notation as above:\n\n- Each identity \( {1}_{A} \) is an isomorphism and is its own inverse.\n\n- If \( f \) is an isomorphism, then \( {f}^{-1} \) is an isomorphism and further \( {\left( {f}^{-1}\right) }^{-1} = f \) .\n\n- If \( f \in {\operatorname{Hom}}_{\mathrm{C}}\left( {A, B}\right), g \i... | Proof. These all 'prove themselves'. For example, it is immediate to verify that \( {f}^{-1}{g}^{-1} \) is a left-inverse of \( {gf} \) : indeed \( {}^{20} \) ,\n\n\[ \left( {{f}^{-1}{g}^{-1}}\right) \left( {gf}\right) = {f}^{-1}\left( {\left( {{g}^{-1}g}\right) f}\right) = {f}^{-1}\left( {{1}_{B}f}\right) = {f}^{-1}f ... | Yes |
An automorphism of an object \( A \) of a category \( \mathrm{C} \) is an isomorphism from \( A \) to itself. The set of automorphisms of \( A \) is denoted \( {\operatorname{Aut}}_{\mathsf{C}}\left( A\right) \) ; it is a subset of \( {\operatorname{End}}_{\mathsf{C}}\left( A\right) \) . By Proposition 4.3, composition... | - the composition of two elements \( f, g \in {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) is an element \( {gf} \in {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) ;\n\n- composition is associative;\n\n- \( {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) contains the element \( {1}_{A} \), which is an i... | Yes |
Example 4.9. As proven in Proposition 2.3, in the category Set the monomorphisms are precisely the injective functions. | The reader should have by now checked that, likewise, in Set the epimorphisms are precisely the surjective functions (cf. Exercise 2.5). Thus, while the definitions given in \( §{2.6} \) may have looked counterintuitive at first, they work as natural 'categorical counterparts' of the ordinary notions of injective/surje... | No |
In the categories of Example 3.3, every morphism is both a monomorphism and an epimorphism. | Indeed, recall that there is at most one morphism between any two objects in these categories; hence the conditions defining monomorphisms and epimorphisms are vacuous. | Yes |
The category obtained by endowing \( \mathbb{Z} \) with the relation \( \leq \) (see Example 3.3) has no initial or final object. | Indeed, an initial object in this category would be an integer \( i \) such that \( i \leq a \) for all integers \( a \) ; there is no such integer. Similarly, a final object would be an integer \( f \) larger than every integer, and there is no such thing. | Yes |
If \( {I}_{1},{I}_{2} \) are both initial objects in \( \mathrm{C} \), then \( {I}_{1} \cong {I}_{2} \). | Proof. Recall that (by definition of category!) for every object \( A \) of \( \mathrm{C} \) there is at least one element in \( {\operatorname{Hom}}_{\mathrm{C}}\left( {A, A}\right) \), namely the identity \( {1}_{A} \). If \( I \) is initial, then there is a unique morphism \( I \rightarrow I \), which therefore must... | Yes |
Proposition 5.6. The disjoint union is a coproduct in Set. | Proof. Recall (§1.4) that the disjoint union \( A \coprod B \) is defined as the union of two disjoint isomorphic copies \( {A}^{\prime },{B}^{\prime } \) of \( A, B \), respectively; for example, we may let \( {A}^{\prime } = \{ 0\} \times A,{B}^{\prime } = \{ 1\} \times B \) . The functions \( {i}_{A},{i}_{B} \) are ... | Yes |
Since we explicitly require \( G \) to be nonempty, the most economical way to concoct a group is by letting \( G = \{ e\} \) be a singleton. There is only one function \( G \times G \rightarrow G \) in this case, so there is only one possible binary operation on \( G \ ), defined by\n\n\[ e \bullet e \mathrel{\text{:=... | The three axioms trivially hold for this example, so \( \{ e\} \) is equipped with a unique group structure.\n\nThis is usually called the trivial group; purists should call any such group \( a \) trivial group, since every singleton gives rise to one. | Yes |
Proposition 1.6. If \( h \in G \) is an identity of \( G \), then \( h = {e}_{G} \) . | Proof. Using first that \( {e}_{G} \) is an identity and then that \( h \) is an identity, one gets \( {}^{3} \)\n\n\[ \nh = {e}_{G}h = {e}_{G} \]\n\n(Amusingly, this argument only uses that \( {e}_{G} \) is a ’left’ identity and \( h \) is a ’right’ identity.) | Yes |
Proposition 1.7. The inverse is also unique: if \( {h}_{1},{h}_{2} \) are both inverses of \( g \) in \( G \) , then \( {h}_{1} = {h}_{2} \) . | Proof. This actually follows from Proposition I.4.2 (by viewing \( G \) as the set of isomorphisms of a groupoid with a single object). The reader should construct a stand-alone proof, using the same trick, but carefully hiding any reference to morphims. | No |
Proposition 1.8. Let \( G \) be a group. Then \( \forall a, g, h \in G \)\n\n\[ \n{ga} = {ha} \Rightarrow g = h,\;{ag} = {ah} \Rightarrow g = h.\n\] | Proof. Both statements are proven by multiplying (on the appropriate side) by \( {a}^{-1} \) and applying associativity. For example,\n\n\[ \n{ga} = {ha} \Longrightarrow \left( {ga}\right) {a}^{-1} = \left( {ha}\right) {a}^{-1} \Longrightarrow g\left( {a{a}^{-1}}\right) = h\left( {a{a}^{-1}}\right) \Longrightarrow g{e}... | Yes |
Lemma 1.10. If \( {g}^{n} = e \) for some positive integer \( n \), then \( \left| g\right| \) is a divisor of \( n \) . | Proof. As observed, \( n \geq \left| g\right| \) by definition of order, that is, \( n - \left| g\right| \geq 0 \) . There must then exist \( {}^{7} \) a positive integer \( m \) such that\n\n\[ r = n - \left| g\right| \cdot m \geq 0\;\text{ and }\;n - \left| g\right| \cdot \left( {m + 1}\right) < 0, \]\n\nthat is, \( ... | Yes |
Proposition 1.13. Let \( g \in G \) be an element of finite order. Then \( {g}^{m} \) has finite order \( \forall m \geq 0 \), and in fact \( {}^{8} \n\n\[ \n\left| {g}^{m}\right| = \frac{\operatorname{lcm}\left( {m,\left| g\right| }\right) }{m} = \frac{\left| g\right| }{\gcd \left( {m,\left| g\right| }\right) }\n\] | Proof. The equality of the two numbers \( \frac{\text{lcm}\left( {m,\left| g\right| }\right) }{m} \) and \( \frac{\left| g\right| }{\gcd \left( {m,\left| g\right| }\right) } \) follows from elementary properties of gcd and lcm: \( \operatorname{lcm}\left( {a, b}\right) = {ab}/\gcd \left( {a, b}\right) \) for all \( a \... | Yes |
Proposition 1.14. If \( {gh} = {hg} \), then \( \left| {gh}\right| \) divides \( \operatorname{lcm}\left( {\left| g\right| ,\left| h\right| }\right) \) . | Proof. Let \( \left| g\right| = m,\left| h\right| = n \) . If \( N \) is any common multiple of \( m \) and \( n \), then \( {g}^{N} = {h}^{N} = e \) by Corollary 1.11. Since \( g \) and \( h \) commute,\n\n\[{\left( gh\right) }^{N} = \underset{N\text{ times }}{\underbrace{\left( {gh}\right) \left( {gh}\right) \cdots \... | Yes |
Lemma 2.2. If \( a \equiv {a}^{\prime }{\;\operatorname{mod}\;n} \) and \( b \equiv {b}^{\prime }{\;\operatorname{mod}\;n} \), then\n\n\[ \left( {a + b}\right) \equiv \left( {{a}^{\prime } + {b}^{\prime }}\right) {\;\operatorname{mod}\;n}. \]\n | Proof. By hypothesis \( n \mid \left( {{a}^{\prime } - a}\right) \) and \( n \mid \left( {{b}^{\prime } - b}\right) \) ; therefore \( \exists k,\ell \in \mathbb{Z} \) such that\n\n\[ \left( {{a}^{\prime } - a}\right) = {kn},\;\left( {{b}^{\prime } - b}\right) = \ell n. \]\n\nThen\n\n\[ \left( {{a}^{\prime } + {b}^{\pri... | Yes |
Proposition 2.3. The order of \( {\left\lbrack m\right\rbrack }_{n} \) in \( \mathbb{Z}/n\mathbb{Z} \) is 1 if \( n \mid m \), and more generally\n\n\[ \left| {\left\lbrack m\right\rbrack }_{n}\right| = \frac{n}{\gcd \left( {m, n}\right) }.\] | Proof. If \( n \mid m \), then \( {\left\lbrack m\right\rbrack }_{n} = {\left\lbrack 0\right\rbrack }_{n} \) . If \( n \) does not divide \( m \), observe again that \( {\left\lbrack m\right\rbrack }_{n} = m{\left\lbrack 1\right\rbrack }_{n} \) and apply Proposition 1.13. | Yes |
Proposition 2.6. Multiplication makes \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) into a group. | Proof. Simple properties of gcd’s show that if \( \gcd \left( {{m}_{1}, n}\right) = \gcd \left( {{m}_{2}, n}\right) = 1 \), then \( \gcd \left( {{m}_{1}{m}_{2}, n}\right) = 1 \) . (For example, if a prime integer divided both \( n \) and \( {m}_{1}{m}_{2} \), then it would necessarily divide \( {m}_{1} \) or \( {m}_{2}... | No |
Proposition 3.2. Let \( \varphi : G \rightarrow H \) be a group homomorphism. Then\n\n- \( \varphi \left( {e}_{G}\right) = {e}_{H} \) ;\n\n- \( \forall g \in G,\varphi \left( {g}^{-1}\right) = \varphi {\left( g\right) }^{-1} \) . | Proof. The first item follows from the definition of homomorphism and cancellation: since \( {e}_{H} = {e}_{H} \cdot {e}_{H} \), \n\n\[ \n{e}_{H} \cdot \varphi \left( {e}_{G}\right) = \varphi \left( {e}_{G}\right) = \varphi \left( {{e}_{G} \cdot {e}_{G}}\right) = \varphi \left( {e}_{G}\right) \cdot \varphi \left( {e}_{... | Yes |
Proposition 3.3. Trivial groups are both initial and final in Grp. | Proof. It should be clear that trivial groups are final: there is only one function from a set to a singleton, that is, the constant function; this is vacuously a group homomorphism.\n\nTo see that trivial groups are initial, let \( T = \{ e\} \) be a trivial group; for any group \( G \), define \( \varphi : T \rightar... | Yes |
Proposition 3.4. With operation defined componentwise, \( G \times H \) is a product \( {in}\mathsf{{Grp}} \) . | Proof. Recall (§I.5.4) that this means that \( G \times H \) satisfies the following universal property: for any group \( A \) and any choice of group homomorphisms \( {\varphi }_{G} : A \rightarrow G \) , \( {\varphi }_{H} : A \rightarrow H \), there exists a unique group homomorphism \( {\varphi }_{G} \times {\varphi... | Yes |
Proposition 4.1. Let \( \varphi : G \rightarrow H \) be a group homomorphism, and let \( g \in G \) be an element of finite order. Then \( \left| {\varphi \left( g\right) }\right| \) divides \( \left| g\right| \) . | Proof. As observed, \( \varphi {\left( g\right) }^{\left| g\right| } = {e}_{H} \) ; applying Lemma 1.10 gives the statement. | No |
There are no nontrivial homomorphisms \( \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z} \) : indeed, the image of every element of \( \mathbb{Z}/n\mathbb{Z} \) must have finite order, and the only element with finite order in \( \left( {\mathbb{Z}, + }\right) \) is 0 . | Indeed, the image of every element of \( \mathbb{Z}/n\mathbb{Z} \) must have finite order, and the only element with finite order in \( \left( {\mathbb{Z}, + }\right) \) is 0. | Yes |
Proposition 4.3. Let \( \varphi : G \rightarrow H \) be a group homomorphism. Then \( \varphi \) is an isomorphism of groups if and only if it is a bijection. | Proof. One implication is immediate, as pointed out above. For the other implication, assume \( \varphi : G \rightarrow H \) is a bijective group homomorphism. As a bijection, \( \varphi \) has an inverse in Set:\n\n\[{\varphi }^{-1} : H \rightarrow G\]\n\nwe simply need to check that this is a group homomorphism. Let ... | Yes |
Proposition 4.8. Let \( \varphi : G \rightarrow H \) be an isomorphism.\n\n- \( \left( {\forall g \in G}\right) : \left| {\varphi \left( g\right) }\right| = \left| g\right| \) ;\n\n- \( G \) is commutative if and only if \( H \) is commutative. | Proof. The first assertion follows from Proposition 4.1: the order of \( \varphi \left( g\right) \) divides the order of \( g \), and on the other hand the order of \( g = {\varphi }^{-1}\left( {\varphi \left( g\right) }\right) \) must divide the order of \( \varphi \left( g\right) \) ; thus the two orders must be equa... | No |
Lemma 5.1. If \( w \in W\left( A\right) \) has length \( n \), then \( {}^{22}{r}^{\lfloor \frac{n}{2}\rfloor }\left( w\right) \) is a reduced word. | Proof. Indeed, either \( r\left( w\right) = w \) or the length of \( r\left( w\right) \) is less than the length of \( w \) ; but one cannot decrease the length of \( w \) more than \( n/2 \) times, since each nonidentity application of \( r \) decreases the length by two. | Yes |
Proposition 5.2. The pair \( \left( {j, F\left( A\right) }\right) \) satisfies the universal property for free groups on \( A \) . | Proof. This is also essentially evident, once one has absorbed all the notation. Any function \( f : A \rightarrow G \) to a group extends uniquely to a map \( \varphi : F\left( A\right) \rightarrow \) \( G \), determined by the homomorphism condition and by the requirement that the diagram commutes, which fixes its va... | Yes |
Example 5.3. It is easy to ’visualize’ \( F\left( {\{ a\} }\right) \cong \mathbb{Z} \) ; but it is already somewhat challenging for the free group on two generators, \( F\left( {\{ x, y\} }\right) \) . The best we can do is the following: behold the infinite graph \( {}^{23} \) | This is an example of the Cayley graph of a group (cf. Exercise 8.6): a graph whose vertices correspond to the elements of the group and whose edges connect vertices according to the action of generators.\n\nobtained by starting at a point (the center of the picture), then branching out in four directions by a length o... | No |
Proposition 5.6. For every set \( A,{F}^{ab}\left( A\right) \cong {\mathbb{Z}}^{\oplus A} \) . | Proof. The key point is again that every element of \( {\mathbb{Z}}^{\oplus A} \) may be written uniquely as a finite sum\n\n\[ \mathop{\sum }\limits_{{a \in A}}{m}_{a}j\left( a\right) ,\;{m}_{a} \neq 0\text{ for only finitely many }a; \]\n\nonce this is understood, the argument is precisely the same as for Claim 5.4. | No |
Proposition 6.2. A nonempty subset \( H \) of a group \( G \) is a subgroup if and only if\n\n\[ \left( {\forall a, b \in H}\right) : \;a{b}^{-1} \in H. \] | Proof. It is clear that if \( H \) is a subgroup, then the stated condition holds: indeed, if \( b \in H \), then the inverse of \( b \) must also be in \( H \) and \( H \) is closed under the operation of \( G \) .\n\nConversely, assume the stated condition holds; we have to check that \( H \) is closed under the oper... | Yes |
Lemma 6.3. If \( {\left\{ {H}_{\alpha }\right\} }_{\alpha \in A} \) is any family of subgroups of a group \( G \), then\n\n\[ H = \mathop{\bigcap }\limits_{{\alpha \in A}}{H}_{i} \]\n\nis a subgroup of \( G \) . | Proof. This follows right away from Proposition 6.2: \( H \) is nonempty, because \( e \in {H}_{\alpha } \) for all \( \alpha \), so \( e \in H \) ; and\n\n\[ a, b \in H \Rightarrow \left( {\forall \alpha \in A}\right) : a, b \in {H}_{\alpha } \Rightarrow \left( {\forall \alpha \in A}\right) : a{b}^{-1} \in {H}_{\alpha... | Yes |
Lemma 6.4. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a group homomorphism, and let \( {H}^{\prime } \) be a subgroup of \( {G}^{\prime } \) . Then \( {\varphi }^{-1}\left( {H}^{\prime }\right) \) is a subgroup of \( G \) . | Proof. Recall (end of §I.2.5) that \( {\varphi }^{-1}\left( {H}^{\prime }\right) \) consists of all \( g \in G \) such that \( \varphi \left( g\right) \in \) \( {H}^{\prime } \) . Since \( \varphi \left( {e}_{G}\right) = {e}_{{G}^{\prime }} \in {H}^{\prime } \), this set is nonempty. If \( a, b \in {\varphi }^{-1}\left... | Yes |
Proposition 6.6. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a homomorphism. Then the inclusion \( i \) : \( \ker \varphi \hookrightarrow G \) is final in the category \( {}^{26} \) of group homomorphisms \( \alpha : K \rightarrow G \) such that \( \varphi \circ \alpha \) is the trivial map. | Proof. If \( \alpha : K \rightarrow G \) is such that \( \varphi \circ \alpha \) is the trivial map, then \( \forall k \in K \n\n\[ \n\varphi \circ \alpha \left( k\right) = \varphi \left( {\alpha \left( k\right) }\right) = {e}_{G}, \n\]\n\nthat is, \( \alpha \left( k\right) \in \ker \varphi \) . We can (and must) then ... | No |
Proposition 6.9. Let \( G \subseteq \mathbb{Z} \) be a subgroup. Then \( G = d\mathbb{Z} \) for some \( d \geq 0 \) . | Proof of Proposition 6.9. If \( G = \{ 0\} \), then \( G = 0\mathbb{Z} \) . If not, note that \( G \) must contain positive integers: indeed, if \( a \in G \) and \( a < 0 \), then \( - a \in G \) and \( - a > 0 \) . We can then let \( d \) be the smallest positive integer \( {}^{29} \) in \( G \), and we claim \( G = ... | Yes |
Proposition 6.11. Let \( n > 0 \) be an integer and let \( G \subseteq \mathbb{Z}/n\mathbb{Z} \) be a subgroup. Then \( G \) is the cyclic subgroup of \( \mathbb{Z}/n\mathbb{Z} \) generated by \( {\left\lbrack d\right\rbrack }_{n} \) , for some divisor \( d \) of \( n \) . | Proof. Let \( {\pi }_{n} : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} \) be the quotient map, and consider \( {G}^{\prime } \mathrel{\text{:=}} {\pi }_{n}^{-1}\left( G\right) \) . By Lemma 6.4, \( {G}^{\prime } \) is a subgroup of \( \mathbb{Z} \) ; by Proposition 6.9, \( {G}^{\prime } \) is a cyclic subgroup of \( ... | Yes |
Proposition 6.12. The following are equivalent:\n\n(a) \( \varphi \) is a monomorphism;\n\n(b) \( \ker \varphi = \left\{ {e}_{G}\right\} \) ;\n\n(c) \( \varphi : G \rightarrow {G}^{\prime } \) is injective (as a set-function). | Proof. (a) \( \Rightarrow \) (b): Assume (a) holds, and consider the two parallel compositions\n\n\[ \ker \varphi \overrightarrow{\overset{i}{ \rightarrow }}G\overset{\varphi }{ \rightarrow }{G}^{\prime } \]\nwhere \( i \) is the inclusion and \( e \) is the trivial map. Both \( \varphi \circ i \) and \( \varphi \circ ... | Yes |
Lemma 7.2. If \( \varphi : G \rightarrow {G}^{\prime } \) is any group homomorphism, then \( \ker \varphi \) is a normal subgroup of \( G \) . | Proof. We already know that \( \ker \varphi \) is a subgroup of \( G \) ; to verify it is normal note that \( \forall g \in G,\forall n \in \ker \varphi \)\n\n\[ \varphi \left( {{gn}{g}^{-1}}\right) = \varphi \left( g\right) \varphi \left( n\right) \varphi \left( {g}^{-1}\right) = \varphi \left( g\right) {e}_{{G}^{\pri... | Yes |
Proposition 7.4. Let \( \sim \) be an equivalence relation on a group \( G \), satisfying \( \left( \dagger \right) \) . Then\n\n- the equivalence class of \( {e}_{G} \) is a subgroup \( H \) of \( G \) ; and\n\n- \( a \sim b \Leftrightarrow {a}^{-1}b \in H \Leftrightarrow {aH} = {bH} \) . | Proof. Let \( H \subseteq G \) be the equivalence class of the identity; \( H \neq \varnothing \) as \( {e}_{G} \in H \) . For \( a, b \in H \), we have \( {e}_{G} \sim b \) and hence \( {b}^{-1} \sim {e}_{G} \) (applying \( \left( \dagger \right) \), multiplying on the left by \( \left. {b}^{-1}\right) \) ; hence \( a... | Yes |
Proposition 7.6. If \( H \) is any subgroup of a group \( G \), the relation \( { \sim }_{L} \) defined by\n\n\[ \n\left( {\forall a, b \in G}\right) : \;a{ \sim }_{L}b \Leftrightarrow {a}^{-1}b \in H \n\]\n\nis an equivalence relation satisfying \( \left( \dagger \right) \) . | Proof. This is straightforward and is mostly left to the reader (Exercise 7.8). To see that the relation satisfies \( \left( \dagger \right) \), note that\n\n\[ \na{ \sim }_{L}b \Longrightarrow {a}^{-1}b \in H \Longrightarrow {a}^{-1}\left( {{g}^{-1}g}\right) b \in H \Longrightarrow {\left( ga\right) }^{-1}\left( {gb}\... | No |
Proposition 7.8. There is a one-to-one correspondence between subgroups of \( G \) and equivalence relations on \( G \) satisfying \( \left( {\dagger \dagger }\right) \) ; for the relation \( { \sim }_{R} \) corresponding to a subgroup \( H, G/{ \sim }_{R} \) may be described as the set of right-cosets \( {Ha} \) of \(... | The relation corresponding to \( H \) in this second way is defined by\n\n\[ a{ \sim }_{R}b \Leftrightarrow a{b}^{-1} \in H \Leftrightarrow {Ha} = {Hb}. \] | Yes |
Let \( G = {S}_{3} \), and let \( H \) be the subgroup consisting of the identity and the \( 1 \leftrightarrow 2 \) switch:\n\n\[ H = \left\{ {\left( \begin{array}{lll} 1 & 2 & 3 \\ 1 & 2 & 3 \end{array}\right) ,\left( \begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 3 \end{array}\right) }\right\} \]\n\nThen\n\n\[ \left( \begin... | This state of affairs simply reflects the fact that the two conditions \( \left( \dagger \right) \) and \( \left( {\dagger \dagger }\right) \) are different: there is no reason to expect that if one holds, the other one should also hold (unless \( G \) is commutative, of course). Once more, keep in mind that both have ... | Yes |
Proposition 7.10. The relations \( { \sim }_{L},{ \sim }_{R} \) corresponding to a subgroup \( H \) coincide if and only if \( H \) is normal. | Proof. Two relations coincide if the corresponding partitions agree. Therefore\n\n\( { \sim }_{L} = { \sim }_{R} \Leftrightarrow \) left- and right-cosets of \( H \) coincide \( \Leftrightarrow \left( {\forall g \in G}\right) : {gH} = {Hg}. \)\n\nBut this is one of the equivalent conditions defining the notion of norma... | Yes |
Theorem 7.12. Let \( H \) be a normal subgroup of a group \( G \) . Then for every group homomorphism \( \varphi : G \rightarrow {G}^{\prime } \) such that \( H \subseteq \ker \varphi \) there exists a unique group homomorphism \( \widetilde{\varphi } : G/H \rightarrow {G}^{\prime } \) so that the diagram\n\n![ed3b132a... | Proof. We only need to match the stated universal property with the one we proved in Proposition 7.3, and indeed,\n\n\[ H \subseteq \ker \varphi \Leftrightarrow \left( {\forall h \in H}\right) : \varphi \left( h\right) = {e}_{{G}^{\prime }} \]\n\nis equivalent to\n\n\[ \left( {\forall a, b \in G}\right) : a{b}^{-1} \in... | Yes |
Theorem 8.1. Every group homomorphism \( \varphi : G \rightarrow {G}^{\prime } \) may be decomposed as follows: \n\nwhere the isomorphism \( \widetilde{\varphi } \) in the middle is the homomorphism induced by \( \va... | It is important that the reader agree that we have already proved anything that deserves to be proven here. We know that the projection on the left and the inclusion on the right are homomorphisms and \( \widetilde{\varphi } \) comes from Theorem 7.12. The decomposition is the same one obtained at the level of set-func... | No |
Corollary 8.2. Suppose \( \varphi : G \rightarrow {G}^{\prime } \) is a surjective group homomorphism. Then\n\n\[ \n{G}^{\prime } \cong \frac{G}{\ker \varphi } \n\] | Proof. \( {im\varphi } = {G}^{\prime } \) in Theorem 8.1. | No |
If \( {H}_{1} \subseteq {G}_{1} \) and \( {H}_{2} \subseteq {G}_{2} \) are normal subgroups, then \( {H}_{1} \times {H}_{2} \) is a normal subgroup of the group \( {G}_{1} \times {G}_{2} \) and\n\n\[ \frac{{G}_{1} \times {G}_{2}}{{H}_{1} \times {H}_{2}} \cong \frac{{G}_{1}}{{H}_{1}} \times \frac{{G}_{2}}{{H}_{2}} \] | Indeed, composing the projections\n\n\[ {\pi }_{1} : {G}_{1} \times {G}_{2} \rightarrow {G}_{1},\;{\pi }_{2} : {G}_{1} \times {G}_{2} \rightarrow {G}_{2} \]\n\nwith the morphisms to the quotients gives surjective homomorphisms\n\n\[ {\pi }_{1} : {G}_{1} \times {G}_{2} \rightarrow \frac{{G}_{1}}{{H}_{1}},\;{\pi }_{2} : ... | Yes |
As a particular case of Claim 8.4, take \( {H}_{1} = \left\{ {e}_{{G}_{1}}\right\} \subseteq {G}_{1} \) and \( {H}_{2} = {G}_{2} \subseteq {G}_{2} \) : | \[ \frac{{G}_{1} \times {G}_{2}}{{G}_{2}} \cong \frac{{G}_{1}}{\left\{ {e}_{{G}_{1}}\right\} } \times \frac{{G}_{2}}{{G}_{2}} \cong {G}_{1} \] where on the left we identify \( {G}_{2} \) with the subgroup \( \left\{ {e}_{{G}_{1}}\right\} \times {G}_{2} \) . | Yes |
The cyclic group \( {C}_{3} \) may be viewed as a subgroup of the dihedral group \( {D}_{6} \) : the rotations of a triangle give a copy of \( {C}_{3} \) inside \( {D}_{6} \) . Then \( {C}_{3} \) is normal in \( {D}_{6} \), and \[ \frac{{D}_{6}}{{C}_{3}} \cong {C}_{2} \] | This can of course be checked 'by hand'. But note that there is an evident surjective homomorphism \( {D}_{6} \rightarrow {C}_{2} \), whose kernel is \( {C}_{3} \) : map an element \( \sigma \) of \( {D}_{6} \) to the identity in \( {C}_{2} \) if it does not flip the triangle (that is, precisely when \( \sigma \in {C}_... | Yes |
One can give a circle (denoted \( {S}^{1} \) ) a group structure by identifying its points with rotations of a plane about a point and adding them accordingly. The function\n\n\[ \rho : {\mathbb{R}}^{1} \rightarrow {S}^{1} \]\n\nmapping a number \( r \) to the result of a rotation by \( {2\pi r} \) radians is then a su... | By Corollary 8.2, therefore,\n\n\[ \frac{\mathbb{R}}{\mathbb{Z}} \cong {S}^{1} \]\n\n(Cf. Exercise I.1.6.) Geometrically, this amounts to ’wrapping’ \( \mathbb{R} \) infinitely many times around the circle, realizing \( \mathbb{R} \) as the ’universal cover’ of \( {S}^{1} \) ; here, \( \mathbb{Z} \) plays the role of ’... | No |
Here is the effect of this operation on the lattice of subgroups of \( {C}_{12} \cong \mathbb{Z}/{12}\mathbb{Z} \) (labeled by generators; cf. §6.4), after quotienting by \( H = \langle \left\lbrack 6\right\rbrack \rangle \cong {C}_{2} \). | Here is why this works. First note that if \( H \subseteq K \) are subgroups of a group \( G \) and \( H \) is normal in \( G \), then \( H \) is normal in \( K \). | No |
Proposition 8.9. Let \( H \) be a normal subgroup of a group \( G \). Then for every subgroup \( K \) of \( G \) containing \( H \), \( K/H \) may be identified with a subgroup of \( G/H \). The function \[ u : \{ \text{subgroups}\;K\;\text{of}\;G\;\text{containing}\;H\} \rightarrow \{ \text{subgroups}\;\text{of}\;G/H\... | Proof. The group \( K/H \) consists of the cosets \( {aH} \in G/H \) with \( a \in K \), and in this sense it is a subset (and clearly a subgroup) of \( G/H \). It is also clear that if \( H \subseteq K \subseteq L \), then \( u\left( K\right) = K/H \subseteq L/H = u\left( L\right) \); that is, \( u \) preserves inclus... | Yes |
Proposition 8.10. Let \( H \) be a normal subgroup of a group \( G \), and let \( N \) be a subgroup of \( G \) containing \( H \) . Then \( N/H \) is normal in \( G/H \) if and only if \( N \) is normal in \( G \), and in this case \[ \frac{G/H}{N/H} \cong \frac{G}{N} \] | Proof. If \( N \) is normal, then consider the projection \[ G \rightarrow \frac{G}{N} : \] the subgroup \( H \) is contained in \( N \), which is the kernel of this homomorphism, so we get (by the universal property of quotients, Theorem 7.12) an induced homomorphism \[ \frac{G}{H} \rightarrow \frac{G}{N} \] The subgr... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.