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Theorem 8.30 Let \( G \) be a nontrivial group of odd order.\n\n1) \( {G}^{\prime } < G \)\n\n2) G has a normal subgroup \( K \) of prime index. | Proof. Part 1) can be proved by induction on \( o\left( G\right) \) . If \( o\left( G\right) = 3 \), then \( G \) is abelian and \( {G}^{\prime } = \{ 1\} < G \) . Assume the result holds for groups of odd order less than \( o\left( G\right) \) and let \( o\left( G\right) \) be odd. Then the Feit-Thompson theorem impli... | Yes |
Corollary 8.32 Let \( o\left( G\right) = {nm} \) where \( \left( {n, m}\right) = 1 \) . If \( G \) has a normal (Hall) subgroup \( N \) of order \( n \), then any subgroup \( H \) of \( G \) that has order \( {m}^{\prime } \) dividing \( m \) is contained in some complement of \( N \) . | Proof. The Schur-Zassenhaus Theorem implies that there is a \( K \leq G \) for which \( G = N \rtimes K \) . Then \( \left| {{NH} \cap K}\right| = {m}^{\prime } \) and\n\n\[ N \rtimes H = N \rtimes \left( {{NH} \cap K}\right) \]\n\nHence, the Schur-Zassenhaus Theorem implies that there exists \( a \in G \) for which\n\... | Yes |
Theorem 8.33 Let \( p \) be a prime dividing \( n \) ! and let\n\n\[ n = {a}_{0} + {a}_{1}p + \cdots + {a}_{m}{p}^{m} \]\n\nbe the base-p representation of \( n \), that is, \( 0 \leq {a}_{k} < p \) . The largest exponent \( e \) of \( p \) for which \( {p}^{e} \mid n \) ! is\n\n\[ L\left( m\right) = \mathop{\sum }\lim... | Proof. The number of factors in \( n! = 1 \cdot 2\cdots n \) that are multiples of \( p \) is \( \lfloor n/p\rfloor \) where \( \lfloor x\rfloor \) is the floor of \( x \) . Among these \( \lfloor n/p\rfloor \) factors, there are \( \left\lfloor {n/{p}^{2}}\right\rfloor \) factors that are multiples of \( p \) . Thus,\... | Yes |
Theorem 9.6 (R. Brauer and K. A. Fowler [4], 1955) Let \( G \) be a group of even order with exactly \( n \geq 1 \) involutions. Assume that \( Z\left( G\right) \) has odd order. Then either \( G \) has a subgroup of index 2 (which must be normal) or \( G \) has a proper subgroup \( H \) with\n\n\[ \left( {G : H}\right... | Equation (9.3) implies that for any involution \( b \), which we can assume is \( {x}_{1} \), we have\n\n\[ \frac{n}{\left| G\right| } = \mathop{\sum }\limits_{{i = 1}}^{u}\frac{1}{\left| C\left( {x}_{i}\right) \right| } \geq \frac{1}{\left| C\left( b\right) \right| } \]\n\nthat is,\n\n\[ \frac{\left| G\right| }{n} \le... | Yes |
1) (First \( \Omega \) -isomorphism theorem) Every \( \Omega \) -homomorphism \( \sigma : G \rightarrow H \) induces an \( \Omega \) -embedding \( \bar{\sigma } : G/\ker \left( \sigma \right) \hookrightarrow H \) defined by | \[ \bar{\sigma }\left( {g\ker \left( \sigma \right) }\right) = \sigma \left( g\right) \] and so \[ \frac{G}{\ker \left( \sigma \right) } \approx {}_{\Omega }\operatorname{im}\left( \sigma \right) \] | Yes |
Theorem 10.4 A proper \( \Omega \) -series is an \( \Omega \) -composition series if and only if its factor groups are \( \Omega \) -simple. | Thus, a series\n\n\[ \mathcal{G} : {G}_{0} \vartriangleleft {G}_{1} \vartriangleleft \cdots \vartriangleleft {G}_{n} \]\n\nin \( G \) is a composition series if and only if its factor groups \( {G}_{k + 1}/{G}_{k} \) are simple and \( \mathcal{G} \) is a chief series if and only if each factor group \( {G}_{k + 1}/{G}_... | No |
Theorem 10.8 Let \( G \) be an \( \Omega \) -group and let \( H, K \in {\operatorname{subn}}_{\Omega }\left( G\right) \) . Then\n\n\[ \exists {\operatorname{CompSer}}_{\Omega }\left( {H;K}\right) \; \Leftrightarrow \;{\operatorname{subn}}_{\Omega }\left( {H;K}\right) \text{ has BCC } \] | Proof of the following is left to the reader. | No |
Theorem 10.13 Let\n\n\[ G = {H}_{1} \ltimes \cdots \ltimes {H}_{n} = {K}_{1} \ltimes \cdots \ltimes {K}_{m} \]\n\nwhere the factors \( {H}_{k} \) and \( {K}_{k} \) are indecomposable. If the composition\n\n\[ {\alpha }_{i, j} = \left( {\left. {\pi }_{{H}_{i}}\right| }_{{K}_{j}}\right) \circ \left( {\left. {\kappa }_{{K... | In attempting to show that a composition is an automorphism, we are reminded of Fitting’s lemma. We have assumed that \( {H}_{i} \) is indecomposable. Also, since the restriction and composition of normal maps is normal, \( {\alpha }_{i, j} \) is normal and so has normal higher images. Thus, if we assume that \( G \) h... | Yes |
Theorem 10.15 (The Krull–Remak–Schmidt Theorem) Let \( G \) be a group that has BCC on normal subgroups. Suppose that\n\n\[ G = {H}_{1} \ltimes \cdots \ltimes {H}_{n} = {K}_{1} \ltimes \cdots \ltimes {K}_{m} \]\n\nwhere all factors \( {H}_{i} \) and \( {K}_{j} \) are indecomposable. Then \( n = m \) and there is a rein... | ## True Uniqueness\n\nThe Krull-Remak-Schmidt Theorem gives uniqueness of the terms of a Remak decomposition up to isomorphism. Let us now consider the question of when a group \( G \) has an essentially unique Remak decomposition, that is, a Remak decomposition that is unique up to the order of the factors. First supp... | Yes |
Theorem 10.16 Let \( G \) have \( {BCC} \) on normal subgroups and let\n\n\[ G = {H}_{1} \ltimes \cdots \ltimes {H}_{n} \]\n\nbe a Remak decomposition of \( G \) . The following are equivalent:\n\n1) This Remak decomposition of \( G \) is essentially unique .\n\n2) \( {H}_{k} \) is invariant under all normal endomorphi... | If \( \alpha : {H}_{i} \rightarrow Z\left( G\right) \) is a nonzero homomorphism, then we can build a normal endomorphism \( \lambda : G \rightarrow G \) by specifying that\n\n\[ {\left. \lambda \right| }_{{H}_{k}} = \left\{ \begin{array}{ll} \alpha & \text{ if }k = i \\ 0 & \text{ if }k \neq i \end{array}\right. \]\n\... | Yes |
Theorem 10.17 Let \( G \) have \( {BCC} \) on normal subgroups and let\n\n\[ G = {H}_{1} \ltimes \cdots \ltimes {H}_{n} \]\n\nbe a Remak decomposition of \( G \) .\n\n1) The following are equivalent:\n\na) This Remak decomposition of \( G \) is essentially unique.\n\nb) Every homomorphism \( \alpha : {H}_{i} \rightarro... | Proof. For part 2), if \( G = {G}^{\prime } \), then \( {H}_{i} = {H}_{i}^{\prime } \) for all \( i \) and so if \( \lambda : {H}_{i} \rightarrow Z\left( {H}_{j}\right) \) for \( j \neq i \), then for \( a, b \in {H}_{i} \), \n\n\[ \lambda \left( \left\lbrack {a, b}\right\rbrack \right) = \left\lbrack {{\lambda a},{\la... | Yes |
Theorem 11.1 The following implications hold for a class \( \mathcal{K} \) of groups:\n\n1) subgroup \( \Rightarrow \) intersection\n\n2) quotient \( \Rightarrow \) cojoin\n\n3) seminormal join \( \Rightarrow \) normal join \( \Rightarrow \) direct product\n\n4) subgroup and direct product \( \Rightarrow \) cointersect... | Proof. Part 1) is clear. For part 2), we have\n\n\[ \frac{G}{HK} \approx \frac{G}{H}/\frac{HK}{H} \in \mathcal{K} \]\n\nFor part 3), the direct product \( H \boxplus K \) is the seminormal join of \( H \boxplus \{ 1\} \) and\n\n\( \{ 1\} \boxplus K \), each of which is in \( \mathcal{K} \) if \( H, K \in \mathcal{K} \)... | Yes |
Theorem 11.3 Let \( \mathcal{K} \) be a class of groups.\n\n1) (Subgroup) If \( \mathcal{K} \) is closed under subgroup, then the \( {\mathcal{K}}_{s} \) and \( {\mathcal{K}}_{n} \) classes are closed under subgroup. | Proof. For 1), if \( \mathcal{K} \) is closed under subgroup, then (normal) \( \mathcal{K} \) -series are closed under intersection and so the \( {\mathcal{K}}_{s} \) and \( {\mathcal{K}}_{n} \) classes are closed under subgroup. | Yes |
Theorem 11.7 Let \( G \) be nilpotent, with higher centers \( {\zeta }^{k}\left( 1\right) \).\n\n1) If \( H \leq G \), then\n\n\[ H{\zeta }^{k}\left( 1\right) \trianglelefteq H{\zeta }^{k + 1}\left( 1\right) \]\n\n2) If \( N \trianglelefteq G \), then\n\n\[ N \cap {\zeta }^{k}\left( 1\right) = \{ 1\} \; \Rightarrow \;N... | Proof. For part 1), since \( \left\lbrack {G,{\zeta }^{k + 1}\left( 1\right) }\right\rbrack \leq {\zeta }^{k}\left( 1\right) \), Theorem 3.41 implies that\n\n\[ \left\lbrack {H{\zeta }^{k}\left( 1\right), H{\zeta }^{k + 1}\left( 1\right) }\right\rbrack = \left\lbrack {H{\zeta }^{k}\left( 1\right), H}\right\rbrack {\lef... | Yes |
Theorem 11.8 The following are equivalent for a finite group \( G \):\n\n1) \( G \) is nilpotent.\n\n2) Every Sylow subgroup of \( G \) is normal.\n\n3) \( G \) is the direct product of its Sylow p-subgroups\n\n\[ G = \underset{p \in \mathcal{P}}{ \ltimes }{Y}_{p} \]\n\n4) If \( H \leq G \), then\n\n\[ H = \mathop{\bow... | Proof. Theorem 8.11 states that 2)-10) are equivalent. Moreover, Theorem 7.10 implies that a finite \( p \) -group is nilpotent and therefore so is any direct product of finite \( p \) -groups. Hence,6) implies 1). Theorem 11.7 shows that 1) implies 7). \( ▱ \) | No |
Theorem 11.11 Let \( G \) be nilpotent.\n\n1) If \( H \leq G \), then\n\n\[ \operatorname{nilclass}\left( H\right) \leq \operatorname{nilclass}\left( G\right) \]\n\n2) If \( N \trianglelefteq G \), then\n\n\[ \operatorname{nilclass}\left( {G/N}\right) \leq \operatorname{nilclass}\left( G\right) \]\n\n3) (Fitting's theo... | Proof. The first two parts follow from Theorem 11.10. For part 3), Theorem 11.10 implies that\n\n\[ {\Gamma }_{HK}^{k}\left( {HK}\right) = \mathop{\prod }\limits_{{{A}_{1},\ldots ,{A}_{k}, B \in \{ H, K\} }}{\Gamma }_{{A}_{1}}\cdots {\Gamma }_{{A}_{k}}\left( B\right) \]\n\nfor all \( k \geq 1 \) . Now, suppose that nil... | Yes |
Theorem 11.13 Let \( G \) be a group and let \( N \trianglelefteq G \) . Then\n\n\[{\left( \frac{G}{N}\right) }^{\left( n\right) } = \frac{{G}^{\left( n\right) }N}{N}\]\n\nfor all \( n \geq 0 \) . | Proof. For any \( N \leq H \trianglelefteq G \), we have\n\n\[{\left( \frac{H}{N}\right) }^{\prime } = \left\lbrack {\frac{H}{N},\frac{H}{N}}\right\rbrack = \frac{\left\lbrack {H, H}\right\rbrack N}{N} = \frac{{H}^{\prime }N}{N}\]\n\nIn particular, for \( H = G \), we have\n\n\[{\left( \frac{G}{N}\right) }^{\prime } = ... | Yes |
Theorem 11.15 Let \( G \) be solvable. Any minimal normal subgroup \( N \) of \( G \) is abelian. Moreover, if \( N \) contains a nontrivial element of finite order, then \( N \) is elementary abelian. | Proof. For the final statement, \( N \) has an element of prime order \( p \) and since\n\n\[ \n{N}_{p} \mathrel{\text{:=}} \left\{ {x \in N \mid {x}^{p} = 1}\right\} \trianglelefteq N \n\]\n\nit follows that \( N = {N}_{p} \) is an elementary abelian group. \( ▱ \) | No |
Theorem 11.16 The following are equivalent for a group \( G \) that has a composition series.\n\n1) \( G \) is solvable.\n\n2) Every composition series for \( G \) has prime order factor groups.\n\n3) G has a cyclic series in which each factor group has prime order.\n\n4) \( G \) has a cyclic series, that is, \( G \) i... | Proof. We have seen that 1) implies 2) and it is clear that 2) implies 3), that 3) implies 4) and that 4) implies 1). Thus, 1)-4) are equivalent.\n\nIt is clear that 5) implies 1). If \( G \) is solvable, the factor groups \( {G}_{k + 1}/{G}_{k} \) of a chief series are minimal normal in the solvable group \( G/{G}_{k}... | Yes |
1) (Feit-Thompson Theorem) Any group of odd order is solvable; equivalently, every finite nonabelian simple group has even order. | Proof. The equivalence in part 1) is left as an exercise. \( ▱ \) | No |
Theorem 11.18 (Hall’s theorem,1928) Let \( G \) be a finite solvable group with \( o\left( G\right) = {ab} \), where \( \left( {a, b}\right) = 1 \) . Then \( G \) has a Hall subgroup of order \( a \) and all subgroups of order a are conjugate. | Proof. We may assume that \( a, b > 1 \) . The proof is by induction on \( o\left( G\right) \) . If \( o\left( G\right) = 1 \), the result holds trivially. Assume that it holds for all groups of order less than \( o\left( G\right) \) . If \( G \) does not have a minimal normal subgroup, then \( G \) is simple and solva... | Yes |
Theorem 11.19 If a finite group \( G \) has a Hall \( {p}^{\prime } \) -subgroup for every prime \( p \) dividing \( o\left( G\right) \), then \( G \) is solvable. | Proof. Assume that the theorem is false and let \( G \) be a counterexample of smallest order. If \( G \) is not simple, then let \( N \) be a nontrivial proper normal subgroup of \( G \) . We leave it as an exercise to show that \( N \cap H \) is a Hall \( {p}^{\prime } \) - subgroup of \( N \) and \( {HN}/N \) is a H... | No |
Theorem 12.1 Let \( X \) be a nonempty set. If \( \left( {F,\kappa }\right) \) and \( \left( {G,\lambda }\right) \) are universal for \( X \), then there is an isomorphism \( \sigma : F \approx G \) connecting the universal maps, that is, for which | Proof. There are unique mediating morphisms \( {\tau }_{\lambda } : F \rightarrow G \) and \( {\tau }_{\kappa } : G \rightarrow F \) for which\n\n\[{\tau }_{\lambda } \circ \kappa = \lambda \;\text{ and }\;{\tau }_{\kappa } \circ \lambda = \kappa\]\n\nand so\n\n\[{\tau }_{\lambda } \circ {\tau }_{\kappa } \circ \lambda... | Yes |
Theorem 12.2 Let \( X \) be a nonempty set and let \( w\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) be a word over \( {X}^{\prime } \) . If \( \left( {{K}_{X},\kappa }\right) \) is universal on \( X \), then the following are equivalent:\n\n1) \( w\left( {\kappa {x}_{1},\ldots ,\kappa {x}_{n}}\right) = 1 \) in \( {F}_{X}... | Proof. Let \( G \) be a group and let \( {a}_{i} \in G \) for \( i = 1,\ldots, n \) . The function sending \( {x}_{i} \) to \( {a}_{i} \) can be lifted to a unique homomorphism \( \sigma : {F}_{X} \rightarrow G \) for which \( {\sigma \kappa }{x}_{i} = {a}_{i} \) and so\n\n\[ w\left( {{a}_{1},\ldots ,{a}_{n}}\right) = ... | No |
Theorem 12.4 Let \( r \in W \) be a word that does not contain a subword of the form \( x{x}^{-1} \) or \( {x}^{-1}x \) for \( x \in X \) . If \( w \equiv r \), then there is a reduction from \( w \) to \( r \) that involves only removal rules. | Proof. Among all reductions from \( w \) to \( r \), select a reduction with the fewest number of steps and suppose that there is at least one insertion step. Denote the steps by \( {s}_{1},{s}_{2},\ldots ,{s}_{m} \) and suppose that step \( {s}_{k} \) results in the word \( {u}_{k} \) . Let \( {s}_{k} \) be the last i... | Yes |
Theorem 12.5 Let \( {F}_{X} \) be the concrete free group on a set \( X \) and let \( j : X \rightarrow {F}_{X} \) be the inclusion map. Then the pair \( \left( {{F}_{X}, j}\right) \) is universal for \( X \) and so \( {F}_{X} \) is a free group on \( X \) . | Proof. Let \( f : X \rightarrow G \) . If \( \tau : {F}_{X} \rightarrow G \) is defined by \( {\tau \epsilon } = 1 \) and\n\n\[ \tau \left( {{x}_{1}^{{e}_{1}}\cdots {x}_{n}^{{e}_{n}}}\right) = f{\left( {x}_{1}\right) }^{{e}_{1}}\cdots f{\left( {x}_{n}\right) }^{{e}_{n}} \]\n\nfor \( {x}_{1}^{{e}_{1}}\cdots {x}_{n}^{{e}... | Yes |
Theorem 12.6 Let \( X \) be a nonempty set. If \( \left( {K,\kappa }\right) \) and \( \left( {G,\lambda }\right) \) are \( \mathcal{K} \) -universal for \( X \), then there is an isomorphism \( \sigma : K \approx G \) for which | \[ \sigma \circ \kappa = \lambda \] | No |
Theorem 12.7 Let \( \mathcal{K} \) be a class of groups and let \( X \) be an infinite set. Then \( {\mathcal{L}}_{X}\left( \mathcal{K}\right) \) is a fully invariant subgroup of the free group \( {F}_{X} \) . | Proof. It is clear that the product of two laws of \( \mathcal{K} \) -groups is a law of \( \mathcal{K} \) -groups, as is the inverse of a law of \( \mathcal{K} \) -groups. Also, if \( \sigma \in \operatorname{End}\left( {F}_{X}\right) \), then for any \( w\left( {{x}_{1},\ldots ,{x}_{n}}\right) \in \mathcal{L}\left( \... | Yes |
Theorem 12.9 Let \( \mathcal{K} = \mathcal{E}\left( \mathcal{L}\right) \) be an equational class of groups with laws \( \mathcal{L} \). 1) For any group \( G \), the verbal subgroup \( \mathcal{L}\left( G\right) = \left\langle {w\left( {{a}_{1},\ldots ,{a}_{n}}\right) \mid w \in \mathcal{L},{a}_{i} \in G}\right\rangle ... | Proof. Write \( F = {F}_{X} \). For part 1), if \( \sigma \in \operatorname{End}\left( G\right) \), then for any \( {a}_{i} \in G \), \( {\sigma w}\left( {{a}_{1},\ldots ,{a}_{n}}\right) = w\left( {\sigma {a}_{1},\ldots ,\sigma {a}_{n}}\right) \in \mathcal{L}\left( G\right) \) and so \( \mathcal{L}\left( G\right) \) is... | Yes |
Let \( \mathcal{K} \) be the class of all finitely-generated groups. If \( w\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is a law of \( \mathcal{K} \), then \( w\left( {{x}_{1},\ldots ,{x}_{n}}\right) = 1 \) in all finitely-generated groups and therefore in all groups. Hence, \( \mathcal{E}\left( {\mathcal{L}\left( \math... | \( ▱ \) | No |
Theorem 12.12 (Birkhoff) The following are equivalent for a class \( \mathcal{K} \) of groups:\n\n1) \( \mathcal{K} \) is an equational class\n\n2) \( \mathcal{K} \) is closed under subgroup, quotient and direct product\n\n3) \( \mathcal{K} \) is closed under quotient and subdirect product. | Proof. It is clear that 1) implies 2), which implies 3). To show that 3) implies 1), we show that \( \mathcal{E}\left( {\mathcal{L}\left( \mathcal{K}\right) }\right) \subseteq \mathcal{K} \) . Let \( G \in \mathcal{E}\left( {\mathcal{L}\left( \mathcal{K}\right) }\right) \), that is, \( G \) satisfies the laws of \( \ma... | Yes |
Theorem 12.14 If \( X \) is a nonempty set, then the free abelian group \( {A}_{X} \) satisfies\n\n\[ \n{A}_{X} \approx \underset{x \in X}{ \boxplus }\langle x\rangle \approx \underset{x \in X}{ \boxplus }\mathbb{Z} \n\] | Proof. The function \( f : X \rightarrow \boxplus \langle x\rangle \) defined by\n\n\[ \nf\left( x\right) \left( y\right) = \left\{ \begin{array}{ll} x & \text{ if }y = x \\ 1 & \text{ if }y \neq x \end{array}\right.\n\]\n\nfor all \( y \in X \) can be lifted uniquely to a homomorphism \( \tau : {A}_{X} \rightarrow \bo... | Yes |
Theorem 12.15 Let \( X \) be a nonempty subset of an abelian group \( A \) . The following are equivalent:\n\n1) A is a free abelian group with basis \( X \)\n\n2) \( X \) is independent in \( A \) and generates \( A \)\n\n3) Except for the order of the factors, every nonidentity element \( a \in A \) has a unique expr... | Proof. If 1) holds, we have seen that \( X \) generates \( A \) and if\n\n\[ w = {x}_{1}^{{e}_{1}}\cdots {x}_{n}^{{e}_{n}} = \epsilon \]\n\nfor \( {x}_{i} \neq {x}_{j} \), then \( {e}_{i} = 0 \) for all \( i \), since otherwise \( w \) would be a reduced word equivalent to the shorter word \( \epsilon \) . Hence,2) hol... | Yes |
Theorem 12.17 Let \( {A}_{X} \) be free abelian on \( X \) . Then all independent sets have cardinality at most \( \left| X\right| \) . | Proof. It is sufficient to prove the result for \( A = { \boxplus }_{x \in X}{\mathbb{Z}}_{x} \) where \( {\mathbb{Z}}_{x} = \mathbb{Z} \) for all \( x \) . The set \( V = { \boxplus }_{x \in X}{\overline{\mathbb{Q}}}_{x} \) is a vector space over the rational field \( \mathbb{Q} \) and it is easy to see that a subset\... | Yes |
Theorem 12.19 If \( \sigma : G \rightarrow {F}_{X} \) is an epimorphism, where \( {F}_{X} \) is free on \( X \), then \( K = \ker \left( \sigma \right) \) is complemented in \( G \) . | Proof. Define a function \( \tau : X \rightarrow G \) by letting \( {\tau x} \) be a fixed member of \( {\sigma }^{-1}\left( x\right) \) . Since \( {F}_{X} \) is free on \( X \), there is a unique homomorphism \( \tau : {F}_{X} \rightarrow G \) that extends \( \tau \) on \( X \) . Since for any \( x \in X \) ,\n\n\[ \s... | Yes |
Theorem 12.20 Let \( X = \{ x, y\} \) and let \( {F}_{X} \) be the 2-generated free group on \( X \) . Let \( G = \langle S\rangle \), where\n\n\[ S = \left\{ {{y}^{k}x{y}^{-k} \mid k > 0}\right\} \]\n\nThen \( G \) is isomorphic to the free group \( {F}_{Z} \) on a countably infinite set \( Z = \left\{ {{z}_{1},{z}_{2... | Proof. Consider the function \( f : Z \rightarrow G \) defined by \( f\left( {z}_{k}\right) = {y}^{k}x{y}^{-k} \) . Then there is a unique mediating morphism \( \tau : {F}_{Z} \rightarrow G \) for which \( \tau \left( {z}_{k}\right) = {y}^{k}x{y}^{-k} \) . It is clear that \( \tau \) is surjective.\n\nIn addition, if\n... | Yes |
Theorem 12.21 Let \( {F}_{X} \) be the free group on \( X = \{ x, y\} \) . Then \( {F}_{X} \) has a subgroup \( H \) for which \( {xH}{x}^{-1} < H \) . | Proof. Let \( H \) consist of the empty word \( \epsilon \) and the set of all words of the form\n\n\[ w = {x}^{{n}_{1}}{y}^{{k}_{1}}{x}^{{n}_{2}}{y}^{{k}_{2}}\cdots {x}^{{n}_{r}}{y}^{{k}_{r}}{x}^{{n}_{r + 1}} \]\n\nwith \( r \geq 1,{k}_{i} \neq 0,{n}_{1} \geq 0,{n}_{r + 1} \leq 0,{n}_{i} \neq 0 \) for \( 2 \leq i \leq... | No |
Theorem 12.23 If \( G \approx \langle X \mid \mathcal{R}\rangle \), then \( G = \langle Y \mid \mathcal{S}\rangle \), where \( \left| Y\right| \leq \left| X\right| \) and \( \left| \mathcal{R}\right| \leq \left| \mathcal{S}\right| \) . In particular, \( G \) has a finite presentation if and only if it has a finite conc... | Proof. Let \( \mu : {F}_{X} \rightarrow G \) be a free presentation of \( G \) with kernel \( N = \langle \mathcal{R}{\rangle }_{\text{nor }} \) . Then the set \( Y = {\mu X} \) generates \( G \) . Let \( \sigma : {F}_{Y} \rightarrow G \) be the change of context map. If \( \mathcal{S} \) is the set of relators in \( Y... | Yes |
Theorem 12.24 If \( G \) has a finite presentation and if \( X \) is a generating set for \( G \) , then \( G \) has a finite presentation of the form\n\n\[ \left\langle {{x}_{1},\ldots ,{x}_{n} \mid {r}_{1},\ldots ,{r}_{m}}\right\rangle \]\n\nwhere \( {x}_{i} \in X \) . | Proof. Let \( \langle Y \mid \mathcal{S}\rangle \) be a finite concrete presentation of \( G \), with\n\n\[ Y = \left\{ {{y}_{1},\ldots ,{y}_{u}}\right\} \;\text{ and }\;\mathcal{S} = \left\{ {{s}_{i}\left( {{y}_{1},\ldots ,{y}_{u}}\right) \mid i = 1,\ldots, v}\right\} \]\n\nThen there is a finite subset \( {X}_{0} = \... | Yes |
Theorem 12.26 The word problem is solvable for the class of all finitely-presented residually finite groups. | Proof. Let \( G = \langle X \mid \mathcal{R}\rangle \) be a finite presentation of a residually finite group \( G \) and let \( w \in G \) . It is possible to enumerate all finite groups by constructing multiplication tables. Also, for a given finite group \( F \), there are only a finite number of group homomorphisms ... | Yes |
Theorem 12.27 Suppose that a group with presentation \( \langle X \mid \mathcal{R}\rangle \) has order at most \( n < \infty \) . Then any group \( G \) of order \( n \) generated by \( X \) and satisfying the relations in \( \mathcal{R} \) has presentation \( \langle X \mid \mathcal{R}\rangle \) . | Proof. Let \( \sigma : {F}_{X} \rightarrow G \) be the change of context epimorphism. Since \( \langle \mathcal{R}{\rangle }_{\text{nor }} \leq \ker \left( \sigma \right) \), we have\n\n\[ n \geq \left( {{F}_{X} : \langle \mathcal{R}{\rangle }_{\text{nor }}}\right) \geq \left( {{F}_{X} : \ker \left( \sigma \right) }\ri... | Yes |
Theorem 13.1 The torsion subgroup of an abelian group need not be complemented. | Proof. Let \( A = \boxtimes {\mathbb{Z}}_{p} \) be the external direct product of the abelian groups \( {\mathbb{Z}}_{p} \) , taken over all primes \( p \) . For \( a \in A \), we use the notation \( {a}_{p} \) in place of \( a\left( p\right) \) . The torsion subgroup \( {A}_{\text{tor }} \) is the subgroup of all elem... | Yes |
Theorem 13.3 A finitely-generated abelian group \( A \) is torsion free if and only if it is free. | Proof. We leave proof that if \( A \) is free, then it is torsion free as an exercise. For the converse, let \( S = \left\{ {{v}_{1},\ldots ,{v}_{n}}\right\} \) be a generating set for the torsion-free abelian group \( A \) . The proof is based on the fact that since \( A \) is torsion free, it is a torsion-free \( \ma... | No |
Theorem 13.5 Let \( G \) be an abelian group. The following are equivalent:\n\n1) \( G \) is a free abelian group.\n\n2) \( G \) is projective.\n\n3) \( G \) has the right-inverse property, that is, any surjection \( \tau : A \rightarrow G \), where \( A \) is abelian, has a right inverse.\n\n4) If \( \sigma : A \right... | Proof. We have seen that 3) and 4) are equivalent. Assume that 1) holds and let \( G = {F}_{X} \) be free on \( X \) . Let \( \sigma : A \rightarrow B \) be surjective and let \( \tau : {F}_{X} \rightarrow B \) . Then for\neach \( x \in X \), there is an \( {a}_{x} \in A \) for which \( \sigma {a}_{x} = {\tau x} \) . D... | Yes |
Theorem 5.1 (Cayley). Every group is isomorphic to a collection of permutations. | Proof. We have just seen that from the columns of any group's multiplication table, we can create a permutation for each element of the group, as Figure 5.31 exemplifies. We can also make a multiplication table out of those permutations. This proof explains why such a multiplication table must behave the same as the or... | Yes |
Theorem 6.7. If \( H \) is a subgroup of \( G \), then each element of \( G \) belongs to exactly one left coset of \( H \) . | Proof. Suppose the element \( g \) of \( G \) appears to belong to two different left cosets, say \( {aH} \) and \( {bH} \) . In such a case, \( {aH} \) and \( {bH} \) must actually be two different names for the same coset (in the sense of Observation 6.5); here's why.\n\nSince \( g \) is in \( {aH} \), we can conclud... | Yes |
Theorem 6.8 (Lagrange’s Theorem). If \( H < G \), then the order \( \left| H\right| \) of the subgroup divides the order \( \left| G\right| \) of the larger group. | Proof. Theorem 6.7 proved that the group \( G \) is partitioned into copies of \( H \) . So the size of \( G \) can be determined just by counting how many copies of \( H \) there are and multiplying that number by the size of each one, the number \( \left| H\right| \) . So if \( n \) is the number of left cosets (incl... | Yes |
Theorem 7.6. If \( H < G \), then a quotient group \( \frac{G}{H} \) can be constructed just when \( H \vartriangleleft G \) . | Proof. The quotient process from Definition 7.5 succeeds just when the resulting diagram is a valid Cayley diagram. Most aspects of valid Cayley diagrams are guaranteed by the quotient process. For instance, because we begin with a diagram that has an arrow of every color exiting every node, our resulting diagram has t... | No |
Theorem 8.5 (Fundamental Homomorphism Theorem). If \( \phi : G \rightarrow H \) is a homomorphism, then \( \operatorname{Im}\left( \phi \right) \cong \frac{G}{\operatorname{Ker}\left( \phi \right) } \) . | Proof. For any homomorphism \( \phi : G \rightarrow H \), we know from Observations 8.2 to 8.4 that \( \operatorname{Ker}\left( \phi \right) \vartriangleleft G \), so we know that we can take a quotient \( \frac{G}{\operatorname{Ker}\left( \phi \right) } \) and obtain a group. Now I must explain why that group is isomo... | Yes |
Theorem 8.7. \( {C}_{n} \times {C}_{m} \cong {C}_{nm} \) if and only if \( n \) and \( m \) are relatively prime. | Proof. If \( {C}_{n} \times {C}_{m} \) is cyclic, it must be generated by one of its elements; let’s call it \( \left( {a, b}\right) \) because we do not know specifically which \( a \in {C}_{n} \) or \( b \in {C}_{m} \) it involves. Because the orbit of \( \left( {a, b}\right) \) includes every element in \( {C}_{n} \... | Yes |
Theorem 9.4 (Orbit-Stabilizer Theorem). The size of an element's orbit times the size of its stabilizer is the size of the group. | Proof. Because Stab \( \left( s\right) \) is a subgroup of \( G \), the definition of subgroup index (Definition 6.9) tells us that\n\n\[ \underset{\text{size of subgroup }}{\underbrace{\left| \operatorname{Stab}\left( S\right) \right| }} \cdot \underset{\text{number of cosets }}{\underbrace{\left\lbrack G : \operatorn... | Yes |
Theorem 9.5. If a group \( G \) of prime order \( p \) acts on a set \( S \), then the order of \( S \) and the number of stable elements in \( S \) are congruent mod \( p \) . | Proof. The Orbit-Stabilizer Theorem tells us that the size of each \( \operatorname{Orb}\left( s\right) \) is a factor of \( \left| G\right| \), so when \( \left| G\right| \) is a prime \( p \), all orbits have size 1 or \( p \) . The stable elements are in orbits of size 1 (each by itself) and the rest of \( S \) is p... | Yes |
Theorem 9.6 (Cauchy’s Theorem). If \( p \) is a prime number that divides \( \left| G\right| \), then \( G \) has an element \( g \) of order \( p \), and therefore a subgroup \( \langle g\rangle \) of order \( p \) . | Proof. Because \( p \) is prime, if I find some \( g \neq e \) satisfying \( {g}^{p} = e \), then \( g \) must have order \( p \) . Exercise 9.15 asks you to explain why this is so, but for now I use the fact without justification.\n\nIn most groups there are lots of ways to multiply \( p \) group elements together and... | No |
Theorem 9.8. If a p-group \( G \) acts on a set \( S \), then the order of \( S \) and the number of stable elements in \( S \) are congruent mod \( p \) . | Proof. As in the proof of Theorem 9.5, the size of each orbit must divide the order of \( G \) . In this case, only powers of \( p \) divide \( \left| G\right| \), so the orbits are therefore of various sizes including \( 1, p,{p}^{2},{p}^{3} \), up to at most \( {p}^{n} \) . So simply modify the illustration for Theor... | No |
Theorem 9.9. If \( H \) is a p-subgroup of \( G \), then \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack { \equiv }_{p}\left\lbrack {G : H}\right\rbrack \) . | Proof. Let \( S \) be the left cosets of \( H \) in \( G \) and consider the group \( H \) acting on \( S \) by the interpretation homomorphism \( \phi : G \rightarrow \operatorname{Perm}\left( S\right) \) defined by\n\n\[ \phi \left( h\right) = \text{the permutation that sends a coset}{gH}\text{to the coset}{hgH}\text... | No |
Theorem 9.10 (First Sylow Theorem). If \( G \) is a group and \( {p}^{n} \) is the highest power of \( p \) dividing \( \left| G\right| \), then there are subgroups of \( G \) of every order \( 1, p,{p}^{2},{p}^{3} \), up to \( {p}^{n} \) . Also, every p-subgroup with fewer than \( {p}^{n} \) elements is inside one of ... | Proof. It is easy to find a \( p \) -subgroup of order 1 (which is \( {p}^{0} \) ) because it is obviously \( \{ e\} \) . We also know that there is a \( p \) -subgroup of order \( p \) (which is \( {p}^{1} \) ) from Cauchy’s Theorem (as long as \( \left| G\right| > 1 \) ). The main job of this proof is showing the exi... | Yes |
Theorem 9.12 (Second Sylow Theorem). Any two Sylow p-subgroups are conjugates. | Proof. I use again the strategy described after the statement of Theorem 9.9. Let \( S \) be the left cosets of some Sylow \( p \) -subgroup \( H < G \) and have another Sylow \( p \) -subgroup \( K \) act on \( S \) by left multiplication (as in the proof of Theorem 9.9).\n\nA stable element of this action is a left c... | Yes |
Theorem 9.13 (Third Sylow Theorem). The number \( n \) of Sylow p-subgroups of \( G \) obeys the following two restrictions.\nn divides \( \left| G\right| \)\n\[n{ \equiv }_{p}1\] | Proof. The first of the two restrictions is the easier to prove. Let \( H \) be one of the \( n \) Sylow \( p \) -subgroups of \( G \) . We know that the set of Sylow \( p \) -subgroups is the set of conjugates of \( H \) . Another way to say this is that if \( G \) acts on its subgroups by conjugation, then the set of... | Yes |
Theorem 10.5. The degree of an extension \( \mathbb{Q}\left( r\right) \) always matches the degree of the irreducible polynomial to which \( r \) is a root. | For example, \( \mathbb{Q}\left( \sqrt{2}\right) \) is a degree-2 extension and \( \sqrt{2} \) is a root of the degree-2 irreducible polynomial \( {x}^{2} - 2 \) . | "No" |
Theorem 10.7. Successive extensions multiply degrees. | \[ \left\lbrack {\mathbb{Q}\left( {a, b}\right) : \mathbb{Q}}\right\rbrack = \left\lbrack {\mathbb{Q}\left( {a, b}\right) : \mathbb{Q}\left( a\right) }\right\rbrack \left\lbrack {\mathbb{Q}\left( a\right) : \mathbb{Q}}\right\rbrack \] This theorem is best shown in a diagram like Figure 10.10. On the bottom is the field... | Yes |
Theorem 10.8. Take any polynomial irreducible over \( \mathbb{Q} \) and any two of its roots, \( {r}_{1} \) and \( {r}_{2} \). | 1. There is an isomorphism \( \phi : \mathbb{Q}\left( {r}_{1}\right) \rightarrow \mathbb{Q}\left( {r}_{2}\right) \) that replaces every \( {r}_{1} \) with \( {r}_{2} \), but fixes \( \mathbb{Q} \) . For example, in a degree-2 extension, \( \phi \) would look like this.\n\n\[ \phi \left( {a + b{r}_{1}}\right) = a + b{r}... | Yes |
Theorem 10.9. The degree of a normal extension equals the order of its Galois group. | This theorem makes it easier to compute the Galois group of \( \mathbb{Q}\left( {{r}_{1},{r}_{2},{r}_{3}}\right) \) . Because there are three roots, the Galois group (which permutes those roots) must be isomorphic to some subgroup of \( {S}_{3} \) . Theorem 10.9 says that the order of the Galois group must match the de... | Yes |
Proposition 2.1. If a binary operation \( \cdot : M \times M \rightarrow M \) is associative and if \( {e}^{\prime } \in M \) is a left identity and \( {e}^{\prime \prime } \in M \) is a right identity, which means that\n\n\[ \n{e}^{\prime } \cdot a = a\;\text{ for all }\;a \in M \n\]\n\n(G21)\n\nand\n\n\[ \na \cdot {e... | Proof. If we let \( a = {e}^{\prime \prime } \) in equation (G2l), we get\n\n\[ \n{e}^{\prime } \cdot {e}^{\prime \prime } = {e}^{\prime \prime } \n\]\n\nand if we let \( a = {e}^{\prime } \) in equation \( \left( \mathrm{{G2r}}\right) \), we get\n\n\[ \n{e}^{\prime } \cdot {e}^{\prime \prime } = {e}^{\prime } \n\]\n\n... | Yes |
In a monoid \( M \) with identity element \( e \), if some element \( a \in M \) has some left inverse \( {a}^{\prime } \in M \) and some right inverse \( {a}^{\prime \prime } \in M \), which means that\n\n\[ \n{a}^{\prime } \cdot a = e \n\]\n\n(G31)\n\nand\n\n\[ \na \cdot {a}^{\prime \prime } = e, \n\]\n\n\( \left( {\... | Proof. Using (G3l) and the fact that \( e \) is an identity element, we have\n\n\[ \n\left( {{a}^{\prime } \cdot a}\right) \cdot {a}^{\prime \prime } = e \cdot {a}^{\prime \prime } = {a}^{\prime \prime }. \n\]\n\nSimilarly, Using (G3r) and the fact that \( e \) is an identity element, we have\n\n\[ \n{a}^{\prime } \cdo... | Yes |
Proposition 2.3. Given a group \( G \), the translations \( {L}_{g} \) and \( {R}_{g} \) are bijections. | Proof. We show this for \( {L}_{g} \), the proof for \( {R}_{g} \) being similar.\n\nIf \( {L}_{g}\left( a\right) = {L}_{g}\left( b\right) \), then \( {ga} = {gb} \), and multiplying on the left by \( {g}^{-1} \), we get \( a = b \), so \( {L}_{g} \) injective. For any \( b \in G \), we have \( {L}_{g}\left( {{g}^{-1}b... | Yes |
Proposition 2.5. Given a finite group \( G \), a subset \( H \subseteq G \) is a subgroup of \( G \) iff\n\n(1) \( e \in H \) ;\n\n(2) \( H \) is closed under multiplication. | Proof. We just have to prove that Condition (3) of Definition 2.4 holds. For any \( a \in H \) , since the left translation \( {L}_{a} \) is bijective, its restriction to \( H \) is injective, and since \( H \) is finite, it is also bijective. Since \( e \in H \), there is a unique \( b \in H \) such that \( {L}_{a}\le... | Yes |
Proposition 2.6. Given a group \( G \) and any subgroup \( H \) of \( G \), we have \( {g}_{1}H = {g}_{2}H \) iff \( {g}_{2}^{-1}{g}_{1}H = H \) iff \( {g}_{2}^{-1}{g}_{1} \in H \), for all \( {g}_{1},{g}_{2} \in G \) . | Proof. If we apply the bijection \( {L}_{{g}_{2}^{-1}} \) to both \( {g}_{1}H \) and \( {g}_{2}H \) we get \( {L}_{{g}_{2}^{-1}}\left( {{g}_{1}H}\right) = {g}_{2}^{-1}{g}_{1}H \) and \( {L}_{{g}_{2}^{-1}}\left( {{g}_{2}H}\right) = H \), so \( {g}_{1}H = {g}_{2}H \) iff \( {g}_{2}^{-1}{g}_{1}H = H \) . If \( {g}_{2}^{-1... | Yes |
Proposition 2.7. (Lagrange) For any finite group \( G \) and any subgroup \( H \) of \( G \), the order \( h \) of \( H \) divides the order \( n \) of \( G \) . | \[ \left| G\right| = \left( {G : H}\right) \left| H\right| \] | No |
Proposition 2.8. If \( \varphi : G \rightarrow {G}^{\prime } \) is a homomorphism of groups, then \( \varphi : G \rightarrow {G}^{\prime } \) is injective iff \( \operatorname{Ker}\varphi = \{ e\} \) . (We also write \( \operatorname{Ker}\varphi = \left( 0\right) \) .) | Proof. Assume \( \varphi \) is injective. Since \( \varphi \left( e\right) = {e}^{\prime } \), if \( \varphi \left( g\right) = {e}^{\prime } \), then \( \varphi \left( g\right) = \varphi \left( e\right) \), and by injectivity of \( \varphi \) we must have \( g = e \), so \( \operatorname{Ker}\varphi = \{ e\} \) .\n\nCo... | Yes |
Proposition 2.10. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a group homomorphism. Then \( H = \operatorname{Ker}\varphi \) satisfies Property \( \left( {* * }\right) \), and thus Property \( \left( *\right) \) . | Proof. We have\n\n\[ \varphi \left( {{gh}{g}^{-1}}\right) = \varphi \left( g\right) \varphi \left( h\right) \varphi \left( {g}^{-1}\right) = \varphi \left( g\right) {e}^{\prime }\varphi {\left( g\right) }^{-1} = \varphi \left( g\right) \varphi {\left( g\right) }^{-1} = {e}^{\prime }, \]\n\nfor all \( h \in H = \operato... | Yes |
Given a homomorphism of groups \( \varphi : G \rightarrow {G}^{\prime } \), the groups \( G/\operatorname{Ker}\varphi \) and \( \operatorname{Im}\varphi = \varphi \left( G\right) \) are isomorphic. | Since \( \varphi \) is surjective onto its image, we may assume that \( \varphi \) is surjective, so that \( {G}^{\prime } = \operatorname{Im}\varphi \) . We define a map \( \bar{\varphi } : G/\operatorname{Ker}\varphi \rightarrow {G}^{\prime } \) as follows:\n\n\[ \bar{\varphi }\left( \bar{g}\right) = \varphi \left( g... | Yes |
Given an abelian group \( G \), if \( {H}_{1} \) and \( {H}_{2} \) are any subgroups of \( G \) such that \( {H}_{1} \cap {H}_{2} = \{ 0\} \), then the map a is an isomorphism | The map is surjective by definition, so we just have to check that it is injective. For this, we show that \( \operatorname{Ker}a = \{ \left( {0,0}\right) \} \) . We have \( a\left( {{a}_{1},{a}_{2}}\right) = 0 \) iff \( {a}_{1} + {a}_{2} = 0 \) iff \( {a}_{1} = - {a}_{2} \) . Since \( {a}_{1} \in {H}_{1} \) and \( {a}... | Yes |
Proposition 2.14. Every subgroup \( H \) of \( \mathbb{Z} \) is of the form \( H = n\mathbb{Z} \) for some \( n \in \mathbb{N} \) . | Proof. If \( H \) is the trivial group \( \{ 0\} \), then let \( n = 0 \) . If \( H \) is nontrivial, for any nonzero element \( m \in H \), we also have \( - m \in H \) and either \( m \) or \( - m \) is positive, so let \( n \) be the smallest positive integer in \( H \) . By Proposition 2.13, \( n\mathbb{Z} \) is th... | Yes |
Proposition 2.16. Given any integer \( n \geq 1 \), for any \( a \in \mathbb{Z} \), the residue class \( \bar{a} \in \mathbb{Z}/n\mathbb{Z} \) is invertible with respect to multiplication iff \( \gcd \left( {a, n}\right) = 1 \) . | Proof. If \( \bar{a} \) has inverse \( \bar{b} \) in \( \mathbb{Z}/n\mathbb{Z} \), then \( \bar{a}\bar{b} = 1 \), which means that\n\n\[ \n{ab} \equiv 1\;\left( {\;\operatorname{mod}\;n}\right) \n\]\n\nthat is \( {ab} = 1 + {nk} \) for some \( k \in \mathbb{Z} \), which is the Bezout identity\n\n\[ \n{ab} - {nk} = 1 \n... | Yes |
Proposition 3.1. For any \( u \in E \) and any \( \lambda \in K \), if \( \lambda \neq 0 \) and \( \lambda \cdot u = 0 \), then \( u = 0 \) . | Proof. Indeed, since \( \lambda \neq 0 \), it has a multiplicative inverse \( {\lambda }^{-1} \), so from \( \lambda \cdot u = 0 \), we get\n\n\[ \n{\lambda }^{-1} \cdot \left( {\lambda \cdot u}\right) = {\lambda }^{-1} \cdot 0.\n\]\n\nHowever, we just observed that \( {\lambda }^{-1} \cdot 0 = 0 \), and from (V3) and ... | Yes |
Proposition 3.5. Given any vector space \( E \), if \( S \) is any nonempty subset of \( E \), then the smallest subspace \( \langle S\rangle \) (or \( \operatorname{Span}\left( S\right) \) ) of \( E \) containing \( S \) is the set of all (finite) linear combinations of elements from \( S \) . | Proof. We prove that the set \( \operatorname{Span}\left( S\right) \) of all linear combinations of elements of \( S \) is a subspace of \( E \), leaving as an exercise the verification that every subspace containing \( S \) also contains \( \operatorname{Span}\left( S\right) \) .\n\nFirst, Span \( \left( S\right) \) i... | No |
Lemma 3.6. Given a linearly independent family \( {\left( {u}_{i}\right) }_{i \in I} \) of elements of a vector space \( E \), if \( v \in E \) is not a linear combination of \( {\left( {u}_{i}\right) }_{i \in I} \), then the family \( {\left( {u}_{i}\right) }_{i \in I}{ \cup }_{k}\left( v\right) \) obtained by adding ... | Proof. Assume that \( {\mu v} + \mathop{\sum }\limits_{{i \in I}}{\lambda }_{i}{u}_{i} = 0 \), for any family \( {\left( {\lambda }_{i}\right) }_{i \in I} \) of scalars in \( K \) . If \( \mu \neq 0 \), then \( \mu \) has an inverse (because \( K \) is a field), and thus we have \( v = - \mathop{\sum }\limits_{{i \in I... | Yes |
Theorem 3.7. Given any finite family \( S = {\left( {u}_{i}\right) }_{i \in I} \) generating a vector space \( E \) and any linearly independent subfamily \( L = {\left( {u}_{j}\right) }_{j \in J} \) of \( S \) (where \( J \subseteq I \) ), there is a basis \( B \) of \( E \) such that \( L \subseteq B \subseteq S \) . | Proof. Consider the set of linearly independent families \( B \) such that \( L \subseteq B \subseteq S \) . Since this set is nonempty and finite, it has some maximal element (that is, a subfamily \( B = {\left( {u}_{h}\right) }_{h \in H} \) of \( S \) with \( H \subseteq I \) of maximum cardinality), say \( B = {\lef... | Yes |
Proposition 3.8. Given a vector space \( E \), for any family \( B = {\left( {v}_{i}\right) }_{i \in I} \) of vectors of \( E \), the following properties are equivalent:\n\n(1) \( B \) is a basis of \( E \) .\n\n(2) \( B \) is a maximal linearly independent family of \( E \) .\n\n(3) \( B \) is a minimal generating fa... | Proof. We will first prove the equivalence of (1) and (2). Assume (1). Since \( B \) is a basis, it is a linearly independent family. We claim that \( B \) is a maximal linearly independent family. If \( B \) is not a maximal linearly independent family, then there is some vector \( w \in E \) such that the family \( {... | Yes |
Theorem 3.11. Let \( E \) be a finitely generated vector space. Any family \( {\left( {u}_{i}\right) }_{i \in I} \) generating \( E \) contains a subfamily \( {\left( {u}_{j}\right) }_{j \in J} \) which is a basis of \( E \) . Any linearly independent family \( {\left( {u}_{i}\right) }_{i \in I} \) can be extended to a... | Proof. The first part follows immediately by applying Theorem 3.7 with \( L = \varnothing \) and \( S = \) \( {\left( {u}_{i}\right) }_{i \in I} \) . For the second part, consider the family \( {S}^{\prime } = {\left( {u}_{i}\right) }_{i \in I} \cup {\left( {v}_{h}\right) }_{h \in H} \), where \( {\left( {v}_{h}\right)... | Yes |
The map \( f : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) defined such that\n\n\[ \n{x}^{\prime } = x - y \n\]\n\n\[ \n{y}^{\prime } = x + y \n\]\n\nis a linear map. | The reader should check that it is the composition of a rotation by \( \pi /4 \) with a magnification of ratio \( \sqrt{2} \) . | No |
Given a linear map \( f : E \rightarrow F \), the set \( \operatorname{Im}f \) is a subspace of \( F \) and the set \( \operatorname{Ker}f \) is a subspace of \( E \). The linear map \( f : E \rightarrow F \) is injective iff \( \operatorname{Ker}f = \left( 0\right) \) (where \( \left( 0\right) \) is the trivial subspa... | Proof. Given any \( x, y \in \operatorname{Im}f \), there are some \( u, v \in E \) such that \( x = f\left( u\right) \) and \( y = f\left( v\right) \) , and for all \( \lambda ,\mu \in K \), we have\n\n\[ f\left( {{\lambda u} + {\mu v}}\right) = {\lambda f}\left( u\right) + {\mu f}\left( v\right) = {\lambda x} + {\mu ... | Yes |
Proposition 3.15. Given any two vector spaces \( E \) and \( F \), given any basis \( {\left( {u}_{i}\right) }_{i \in I} \) of \( E \) , given any other family of vectors \( {\left( {v}_{i}\right) }_{i \in I} \) in \( F \), there is a unique linear map \( f : E \rightarrow F \) such that \( f\left( {u}_{i}\right) = {v}... | Proof. If such a linear map \( f : E \rightarrow F \) exists, since \( {\left( {u}_{i}\right) }_{i \in I} \) is a basis of \( E \), every vector \( x \in E \) can written uniquely as a linear combination\n\n\[ x = \mathop{\sum }\limits_{{i \in I}}{x}_{i}{u}_{i} \]\n\nand by linearity, we must have\n\n\[ f\left( x\right... | Yes |
Proposition 3.16. Given any set \( I \), for any vector space \( F \), and for any function \( f : I \rightarrow F \) , there is a unique linear map \( \bar{f} : {K}^{\left( I\right) } \rightarrow F \), such that\n\n\[ f = \bar{f} \circ \iota \] | Proof. If such a linear map \( \bar{f} : {K}^{\left( I\right) } \rightarrow F \) exists, since \( f = \bar{f} \circ \iota \), we must have\n\n\[ f\left( i\right) = \bar{f}\left( {\iota \left( i\right) }\right) = \bar{f}\left( {e}_{i}\right) \]\n\nfor every \( i \in I \) . However, the family \( {\left( {e}_{i}\right) }... | Yes |
Proposition 3.17. Given any two vector spaces \( E \) and \( F \), with \( F \) nontrivial, given any family \( {\left( {u}_{i}\right) }_{i \in I} \) of vectors in \( E \), the following properties hold:\n\n(1) The family \( {\left( {u}_{i}\right) }_{i \in I} \) generates \( E \) iff for every family of vectors \( {\le... | Proof. (1) If there is any linear map \( f : E \rightarrow F \) such that \( f\left( {u}_{i}\right) = {v}_{i} \) for all \( i \in I \), since \( {\left( {u}_{i}\right) }_{i \in I} \) generates \( E \), every vector \( x \in E \) can be written as some linear combination\n\n\[ x = \mathop{\sum }\limits_{{i \in I}}{x}_{i... | Yes |
Proposition 3.18. Let \( E \) be a vector space of finite dimension \( n \geq 1 \) and let \( f : E \rightarrow E \) be any linear map. The following properties hold:\n\n(1) If \( f \) has a left inverse \( g \), that is, if \( g \) is a linear map such that \( g \circ f = \mathrm{{id}} \), then \( f \) is an isomorphi... | Proof. (1) The equation \( g \circ f = \) id implies that \( f \) is injective; this is a standard result about functions (if \( f\left( x\right) = f\left( y\right) \), then \( g\left( {f\left( x\right) }\right) = g\left( {f\left( y\right) }\right) \), which implies that \( x = y \) since \( g \circ f = \mathrm{{id}} \... | Yes |
Proposition 3.19. Given any vector space \( E \) and any subspace \( M \) of \( E \), the relation \( { \equiv }_{M} \) is an equivalence relation with the following two congruential properties:\n\n1. If \( {u}_{1}{ \equiv }_{M}{v}_{1} \) and \( {u}_{2}{ \equiv }_{M}{v}_{2} \), then \( {u}_{1} + {u}_{2}{ \equiv }_{M}{v... | Proof. It is obvious that \( { \equiv }_{M} \) is an equivalence relation. Note that \( {u}_{1}{ \equiv }_{M}{v}_{1} \) and \( {u}_{2}{ \equiv }_{M}{v}_{2} \) are equivalent to \( {u}_{1} - {v}_{1} = {w}_{1} \) and \( {u}_{2} - {v}_{2} = {w}_{2} \), with \( {w}_{1},{w}_{2} \in M \), and thus,\n\n\[ \n\\left( {{u}_{1} +... | Yes |
Given any differentiable function \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \), by definition, for any \( x \in {\mathbb{R}}^{n} \) , the total derivative \( d{f}_{x} \) of \( f \) at \( x \) is the linear form \( d{f}_{x} : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) defined so that for all \( u = \left( {{u}_{1},... | \[ d{f}_{x}\left( u\right) = \left( \begin{array}{lll} \frac{\partial f}{\partial {x}_{1}}\left( x\right) & \cdots & \frac{\partial f}{\partial {x}_{n}}\left( x\right) \end{array}\right) \left( \begin{matrix} {u}_{1} \\ \vdots \\ {u}_{n} \end{matrix}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\frac{\partial f}{\partia... | Yes |
Theorem 3.20. (Existence of dual bases) Let \( E \) be a vector space of dimension \( n \) . The following properties hold: For every basis \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) of \( E \), the family of coordinate forms \( \left( {{u}_{1}^{ * },\ldots ,{u}_{n}^{ * }}\right) \) is a basis of \( {E}^{ * } \) (ca... | Proof. (a) If \( {v}^{ * } \in {E}^{ * } \) is any linear form, consider the linear form\n\n\[ \n{f}^{ * } = {v}^{ * }\left( {u}_{1}\right) {u}_{1}^{ * } + \cdots + {v}^{ * }\left( {u}_{n}\right) {u}_{n}^{ * }.\n\]\n\nObserve that because \( {u}_{i}^{ * }\left( {u}_{j}\right) = {\delta }_{ij} \),\n\n\[ \n{f}^{ * }\left... | Yes |
Problem 3.9. Consider the following Haar matrix\n\n\[ \nH = \left( \begin{matrix} 1 & 1 & 1 & 0 \\ 1 & 1 & - 1 & 0 \\ 1 & - 1 & 0 & 1 \\ 1 & - 1 & 0 & - 1 \end{matrix}\right) \n\]\n\nProve that the columns of \( H \) are linearly independent. | Hint. Compute the product \( {H}^{\top }H \) . | No |
Problem 3.10. Consider the following Hadamard matrix\n\n\[ \n{H}_{4} = \left( \begin{matrix} 1 & 1 & 1 & 1 \\ 1 & - 1 & 1 & - 1 \\ 1 & 1 & - 1 & - 1 \\ 1 & - 1 & - 1 & 1 \end{matrix}\right) \]\n\nProve that the columns of \( {H}_{4} \) are linearly independent. | Hint. Compute the product \( {H}_{4}^{\top }{H}_{4} \) . | No |
Given any matrices \( A \in {\mathrm{M}}_{m, n}\left( K\right), B \in {\mathrm{M}}_{n, p}\left( K\right) \), and \( C \in {\mathrm{M}}_{p, q}\left( K\right) \), we have\n\n\[ \left( {AB}\right) C = A\left( {BC}\right) \]\n\nthat is, matrix multiplication is associative. | Proof. (1) Every \( m \times n \) matrix \( A = \left( {a}_{ij}\right) \) defines the function \( {f}_{A} : {K}^{n} \rightarrow {K}^{m} \) given by\n\n\[ {f}_{A}\left( x\right) = {Ax} \]\n\nfor all \( x \in {K}^{n} \). It is immediately verified that \( {f}_{A} \) is linear and that the matrix \( M\left( {f}_{A}\right)... | Yes |
Proposition 4.2. Given three vector spaces \( E, F, G \), with respective bases \( \left( {{u}_{1},\ldots ,{u}_{p}}\right) \) , \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \), and \( \left( {{w}_{1},\ldots ,{w}_{m}}\right) \), the mapping \( M : \operatorname{Hom}\left( {E, F}\right) \rightarrow {\mathrm{M}}_{n, p} \) t... | Proof. That \( M\left( {g\left( x\right) }\right) = M\left( g\right) M\left( x\right) \) was shown by Definition 4.2 or equivalently by Formula (1). The identities \( M\left( {g + h}\right) = M\left( g\right) + M\left( h\right) \) and \( M\left( {\lambda g}\right) = {\lambda M}\left( g\right) \) are straightforward, an... | Yes |
Proposition 4.3. Let \( E \) be a vector space, and let \( \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) be a basis of \( E \) . For every family \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \), let \( P = \left( {a}_{ij}\right) \) be the matrix defined such that \( {v}_{j} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ij}{u}_{... | Proof. Note that we have \( P = M\left( f\right) \), the matrix associated with the unique linear map \( f : E \rightarrow E \) such that \( f\left( {u}_{i}\right) = {v}_{i} \) . By Proposition 3.15, \( f \) is bijective iff \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) is a basis of \( E \) . Furthermore, it is obviou... | Yes |
Let \( E = F = {\mathbb{R}}^{2} \), with \( {u}_{1} = \left( {1,0}\right) ,{u}_{2} = \left( {0,1}\right) ,{v}_{1} = \left( {1,1}\right) \) and \( {v}_{2} = \left( {-1,1}\right) \) . The change of basis matrix \( P \) from the basis \( \mathcal{U} = \left( {{u}_{1},{u}_{2}}\right) \) to the basis \( \mathcal{V} = \left(... | \[ P = \left( \begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}\right) \] and its inverse is \[ {P}^{-1} = \left( \begin{matrix} 1/2 & 1/2 \\ - 1/2 & 1/2 \end{matrix}\right) \] The old coordinates \( \left( {{x}_{1},{x}_{2}}\right) \) with respect to \( \left( {{u}_{1},{u}_{2}}\right) \) are expressed in terms of the new co... | Yes |
Example 4.2. Let \( E = F = \mathbb{R}{\left\lbrack X\right\rbrack }_{3} \) be the set of polynomials of degree at most 3, and consider the bases \( \mathcal{U} = \left( {1, x,{x}^{2},{x}^{3}}\right) \) and \( \mathcal{V} = \left( {{B}_{0}^{3}\left( x\right) ,{B}_{1}^{3}\left( x\right) ,{B}_{2}^{3}\left( x\right) ,{B}_... | By expanding the Bernstein polynomials, we find that the change of basis matrix \( {P}_{\mathcal{V},\mathcal{U}} \) is given by\n\n\[ \n{P}_{\mathcal{V},\mathcal{U}} = \left( \begin{matrix} 1 & 0 & 0 & 0 \\ - 3 & 3 & 0 & 0 \\ 3 & - 6 & 3 & 0 \\ - 1 & 3 & - 3 & 1 \end{matrix}\right) \n\]\n\nWe also find that the inverse... | Yes |
Proposition 4.4. Let \( E \) and \( F \) be vector spaces, let \( \mathcal{U} = \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) and \( {\mathcal{U}}^{\prime } = \left( {{u}_{1}^{\prime },\ldots ,{u}_{n}^{\prime }}\right) \) be two bases of \( E \), and let \( \mathcal{V} = \left( {{v}_{1},\ldots ,{v}_{m}}\right) \) and \( {... | Proof. Since \( f : E \rightarrow F \) can be written as \( f = {\operatorname{id}}_{F} \circ f \circ {\operatorname{id}}_{E} \), since \( P \) is the matrix of \( {\operatorname{id}}_{E} \) w.r.t. the bases \( \left( {{u}_{1}^{\prime },\ldots ,{u}_{n}^{\prime }}\right) \) and \( \left( {{u}_{1},\ldots ,{u}_{n}}\right)... | Yes |
Corollary 4.5. Let \( E \) be a vector space, and let \( \mathcal{U} = \left( {{u}_{1},\ldots ,{u}_{n}}\right) \) and \( {\mathcal{U}}^{\prime } = \left( {{u}_{1}^{\prime },\ldots ,{u}_{n}^{\prime }}\right) \) be two bases of \( E \) . Let \( P = {P}_{{\mathcal{U}}^{\prime },\mathcal{U}} \) be the change of basis matri... | or more explicitly, \[ {M}_{{\mathcal{U}}^{\prime }}\left( f\right) = {P}_{{\mathcal{U}}^{\prime },\mathcal{U}}^{-1}{M}_{\mathcal{U}}\left( f\right) {P}_{{\mathcal{U}}^{\prime },\mathcal{U}} = {P}_{\mathcal{U},{\mathcal{U}}^{\prime }}{M}_{\mathcal{U}}\left( f\right) {P}_{{\mathcal{U}}^{\prime },\mathcal{U}}. \] | Yes |
Example 4.3. Let \( E = {\mathbb{R}}^{2},\mathcal{U} = \left( {{e}_{1},{e}_{2}}\right) \) where \( {e}_{1} = \left( {1,0}\right) \) and \( {e}_{2} = \left( {0,1}\right) \) are the canonical basis vectors, let \( \mathcal{V} = \left( {{v}_{1},{v}_{2}}\right) = \left( {{e}_{1},{e}_{1} - {e}_{2}}\right) \), and let\n\n\[ ... | Therefore, in the basis \( \mathcal{V} \), the matrix representing the linear map \( f \) defined by \( A \) is\n\n\[ {A}^{\prime } = {P}^{-1}{AP} = {PAP} = \left( \begin{matrix} 1 & 1 \\ 0 & - 1 \end{matrix}\right) \left( \begin{array}{ll} 2 & 1 \\ 0 & 1 \end{array}\right) \left( \begin{matrix} 1 & 1 \\ 0 & - 1 \end{m... | Yes |
Show that the Bernstein polynomials \( {B}_{0}^{2}\left( t\right) ,{B}_{1}^{2}\left( t\right) ,{B}_{2}^{2}\left( t\right) \) are expressed as linear combinations of the basis \( \left( {1, t,{t}^{2}}\right) \) of the vector space of polynomials of degree at most 2 as follows:\n\[ \left( \begin{array}{l} {B}_{0}^{2}\lef... | Prove that\n\[ {B}_{0}^{2}\left( t\right) + {B}_{1}^{2}\left( t\right) + {B}_{2}^{2}\left( t\right) = 1. \] | No |
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