Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 7. If \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) be three non-collinear points of the plane determined by \( {ABC} \), then the planes determined by \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) and \( {ABC} \) are identical. | This will come immediately from the two preceding. | No |
Theorem 8. Two lines in the same plane always intersect. | Let \( B \) and \( C \) be two points of the one line, and \( A \) a point of the other, If \( A \) be also a point of \( {BC} \) the theorem is proved. If not, we may use the point \( A \) and the line \( {BC} \) to determine the plane, and our second line must be identical with a line through \( A \) meeting \( {BC} ... | No |
Theorem 12. If \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime } \) be four non-coplanar points of the space determined by \( A, B, C, D \), then the two spaces determined by the two sets of four points are identical. | With regard to the last theorem it is clear that all points of the space determined by \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime } \) lie in that determined by \( A, B, C, D \) . Let us assume that \( {B}^{\prime },{C}^{\prime },{D}^{\prime } \) are points of \( {AB},{AC},{AD} \) respectively. The planes... | Yes |
Theorem 17. If \( A \) and \( C \) be harmonically separated by \( B \) and \( D \), then \( B \) and \( D \) are harmonically separated by \( A \) and \( C \) . | The proof will come immediately from 15 , after drawing two or three lines; we leave the details to the reader. | No |
Theorem 18. A given point has a unique harmonic conjugate with regard to any two points collinear with it. | This is an immediate result of 16. | No |
Theorem 19. If a point \( O \) be connected with four points \( A, B, C, D \) not collinear with it by lines \( {OA},{OB},{OC},{OD} \), and if these lines meet another line in \( {A}^{\prime },{B}^{\prime }.{C}^{\prime },{D}^{\prime } \) respectively, and, lastly, if \( A \) and \( C \) be harmonic conjugates with rega... | We may legitimately assume that the quadrilateral construction which yielded \( A, B, C, D \) was in a plane which did not contain \( O \), for this construction may be effected in any plane which contains \( {AD} \) . Then radiating lines through \( O \) will transfer this quadrilateral construction into another givin... | No |
Theorem 20. If four planes \( \alpha ,\beta ,\gamma ,\delta \) determined by a line \( l \) and four points \( A, B, C, D \) meet another line in four points \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime } \) respectively, and if \( A \) and \( C \) be harmonically separated by \( B \) and \( D \), then \( {... | It is sufficient to draw the line \( A{D}^{\prime } \) and apply 19. | No |
Theorem 22. The laws of separation laid down for points in Axioms III-VII hold equally for coplanar concurrent lines, and coaxal planes. | We have merely to bring the four lines or planes to intersect another line in distinct points, and apply XI. | No |
Theorem 24. If \( A, B, C, D \) be four collinear points, and \( A \) and \( C \) be harmonically separated by \( B \) and \( D \), then \( {AC}\int {BD} \) . | We have merely to observe that our quadrilateral construction for harmonic separation permits us to pass by two projections from \( A, B, C, D \) to \( C, B, A \) , \( D \) respectively, so that if we had \( {AB}\int {CD} \) we should also have \( {CB}\int {AD} \), and vice versa. Hence our theorem. | No |
Theorem 25. \[ {P}_{0}{P}_{n + 1}\int {P}_{n}{P}_{\infty }\text{ if }n > 0. \] | The theorem certainly holds when \( n = 1 \) . Suppose that \( {P}_{0}{P}_{n}\int {P}_{n - 1}{P}_{\infty } \) . We also know that \( {P}_{n - 1}{P}_{n + 1}\int {P}_{n}{P}_{\infty } \) . Hence, clearly \( {P}_{0}{P}_{n + 1}\int {P}_{n}{P}_{\infty } \) . We notice also that \( {P}_{0}{P}_{n + 2}\int {P}_{n}{P}_{\infty } ... | Yes |
Theorem 26. If \( P \) be any point which satisfies the condition \( {P}_{0}P\int {P}_{1}{P}_{\infty } \), then such a positive integer \( n \) may be found that \( {P}_{0}P\int {P}_{n}{P}_{\infty },{P}_{0}{P}_{n + 1}\int P{P}_{\infty } \) . | Let us divide all points of the separation class determined by \( {P}_{0}{P}_{\infty } \) which include \( {P}_{1} \) and \( P \) the positive separation class let us say, into two sub-classes as follows. A point \( A \) shall be assigned to the first class if we may find such a positive integer \( n \) that \( {P}_{0}... | Yes |
Theorem 28. If in two projective one-dimensional forms three elements of one lie in the corresponding elements of the other, then every element of the first lies in the corresponding element of the second. | For we may use these three elements in each case as \( \infty ,0,1 \), and then, remembering the definition of cross ratio, make use of the fact that the construction of the harmonic conjugate of a point with regard to two others is unique. This theorem is known as the fundamental one of projective geometry. \( {}^{102... | Yes |
Theorem 29. If two fundamental one-dimensional forms be connected by a finite number of projections and intersections they are projective. | This comes immediately from 27. | No |
Theorem 30. If two fundamental one-dimensional forms be projective, they may be connected by a finite number of projections and intersections. | It is, in fact, easy to connect them with two other projective forms whereof one contains three, and hence all corresponding members of the other. | No |
Theorem 31. Four elements of a fundamental one-dimensional form determine six cross ratios which bear to one another the relations of the six numbers\n\n\\[ \lambda ,\\;\\frac{1}{\lambda },\\;1 - \\lambda ,\\;\\frac{1}{1 - \\lambda },\\;\\frac{\\lambda - 1}{\\lambda },\\;\\frac{\\lambda }{\\lambda - 1}. \\] | The proof is perfectly straightforward, and is left to the reader. | No |
Theorem 32. If a fundamental one-dimensional form be projectively transformed into itself there will be two distinct or coincident self-corresponding elements. | We have merely to put \( \left( {\rho x}\right) \) for \( \left( {x}^{\prime }\right) \) in (3), and solve the quadratic equation in \( \rho \) obtained by equating to zero the determinant of the two linear homogeneous equations in \( {x}_{0},{x}_{1} \) . | Yes |
Theorem 33. The congruent group is transitive for a sufficiently small restricted region. | This comes at once by reductio ad absurdum. For the tangents to all possible paths which a chosen point might follow would, if 33 were untrue, generate a surface or set of surfaces, or a line or set of lines, and this assemblage of surfaces or lines would be carried into itself by every congruent transformation which l... | No |
Theorem 34. The congruent group depends on six essential parameters. | The number of parameters is certainly finite since the congruent group arises from analytic relations among the fifteen essential parameters of the general collineation group. The transference from a point to a point imposes three restrictions, necessarily distinct, as three independent parameters are needed to determi... | Yes |
Theorem 35. The congruent group is a six-parameter collineation group which leaves invariant a quadric or a conic. | Let us first take cases \( \left( a\right) \) and \( \left( b\right) \) together. The distance must be a continuous function of each cross ratio determined by the two points and the intersections of their line with the quadric. If we call a distance \( d \), and the corresponding cross ratio of this type \( c \), we mu... | No |
Theorem 2. The angle of two intersecting geodesics is an absolute invariant for all congruent transformations. | This comes at once from the fact that\n\n\[ \frac{\mathop{\sum }\limits_{{ij}}^{{1,2,3}}{a}_{ij}d{z}_{i}\delta {z}_{j}}{ds\delta s} \]\n\nis obviously an absolute invariant for all congruent transformations. | Yes |
Corollary 1. Every simultaneous set of linear homogeneous Diophantine equations possesses a set of irreducible solutions, finite in number. | A direct proof without reference to the present lemma is not difficult. \( {}^{4} \) Applied to the given illustration of the above lemma on monomials the above analysis shows that if the prescribed conditions on the exponents are given by (141) then the complete system of monomials is given by\n\n\[ \n{x}_{1}^{2\alpha... | No |
Lemma 2. If \( {u}_{1} + {u}_{2} + {u}_{3} = 0 \), then any product of powers of \( {u}_{1},{u}_{2},{u}_{3} \) of order \( n \) can be expressed linearly in terms of such products as contain one exponent equal to or greater than \( \frac{2}{3}n \) . | We shall obtain this result as a special case of a considerably more general result embodied in a theorem on the representation of a binary form in terms of other binary forms. | No |
Lemma 3. If \( \left( A\right) : {A}_{1},{A}_{2},\ldots ,{A}_{k} \) is a system of binary forms of respective orders \( {a}_{1},{a}_{2},\ldots ,{a}_{k} \), and \( \left( B\right) : {B}_{1},{B}_{2},\ldots ,{B}_{l} \), a system of respective orders by \( {b}_{1},{b}_{2},\ldots ,{b}_{l} \), and if\n\n\[ \phi = {A}_{1}^{{\... | To prove this, assume that any term of \( \tau \) contains \( \rho \) symbols of the forms \( A \) not in second order determinant combinations with a symbol of the \( B \) forms, and \( \sigma \) symbols of the \( B \) ’s not in combination with a symbol of the \( A \) ’s. Then evidently we have for the total number o... | Yes |
Theorem. The system of all concomitants of a binary form \( f = {a}_{x}^{n} = \ldots \) of order \( n \) is finite. | The proof of this theorem can now be readily accomplished in view of the theorems in Paragraphs III, IV of Chapter IV, Section 7, and lemma 3 just proved.\n\nThe system consisting of \( f \) itself is relatively complete modulo \( {\left( ab\right) }^{2} \) . It is a finite system also, and hence it satisfies the hypot... | Yes |
Lemma 6. If the order \( \omega \) of a covariant \( C \) of a binary quadratic form modulo 3 is not divisible by 3, its leading coefficient \( S \) is a linear homogeneous function of the seminvariants \( \left( i\right) ,\left( s\right) \), other than \( 1, I, q \) . | In proof of this lemma we have under the transformation \( {x}_{1} \equiv {x}_{1}\prime + {x}_{2}\prime ,{x}_{2} \equiv \) \( {x}_{2}\prime \)\n\n\[ C = S{x}_{1}^{\omega } + {S}_{1}{x}_{1}^{\omega - 1}{x}_{2} + \ldots \equiv {Sx}{\prime }_{1}^{\omega } + \left( {{S}_{1} + {\omega S}}\right) x{\prime }_{1}^{\omega - 1}x... | Yes |
Lemma 7. The following formula is true:\n\n\[ \n{\Delta }^{n}{D}^{n} \equiv {\left| \begin{array}{lll} \frac{\partial }{\partial {\lambda }_{1}} & \frac{\partial }{\partial {\lambda }_{2}} & \frac{\partial }{\partial {\lambda }_{3}} \\ \frac{\partial }{\partial {\mu }_{1}} & \frac{\partial }{\partial {\mu }_{2}} & \fra... | In proof of this we note that \( {D}^{n} \), expanded by the multinomial theorem, gives\n\n\[ \n{D}^{n} = \mathop{\sum }\limits_{{ij}}\frac{n!}{{i}_{1}!{i}_{2}!{i}_{3}!}{\lambda }_{1}^{{i}_{1}}{\lambda }_{2}^{{i}_{2}}{\lambda }_{3}^{{i}_{3}}{\left( {\mu }_{2}{\nu }_{3} - {\mu }_{3}{\nu }_{2}\right) }^{{i}_{1}}{\left( {... | Yes |
Lemma 8. If \( P \) is a product of \( m \) factors of type \( {\alpha }_{\lambda }, n \) of type \( {\beta }_{\mu } \), and \( p \) of type \( {\gamma }_{\nu } \), then \( {\Delta }^{k}P \) is a sum of a number of monomials, each monomial of which contains \( k \) factors of type \( \left( {\alpha \beta \gamma }\right... | This is easily proved. Let \( P = {ABC} \), where\n\n\[ A = {\alpha }_{\lambda }^{\left( 1\right) }{\alpha }_{\lambda }^{\left( 2\right) }\cdots {\alpha }_{\lambda }^{\left( m\right) },\]\n\n\[ B = {\beta }_{\mu }^{\left( 1\right) }{\beta }_{\mu }^{\left( 2\right) }\cdots {\beta }_{\mu }^{\left( n\right) },\]\n\n\[ C =... | Yes |
To divide a given triangle by lines parallel to its base into parts which have given ratios to one another. | Again in the manner of Proposition 5, suppose it be required to divide the triangle \( {abg} \) into three parts in the ratio \( {ez} : {zt} : {ti} \) . Then determine the points \( h, l \) in \( {ab} \) such that\n\n\[ a{h}^{2} : a{b}^{2} = {ez} : e{i}^{92} \]\n\nand\n\n\[ a{l}^{2} : a{b}^{2} = {et} : {ei}. \]\n\nThen... | Yes |
Corollary 1. If \( \lambda = {\left( \left( \frac{3}{4} + \epsilon \right) \left( n - 2\right) \log \left( n - 2\right) \right) }^{1/2} \), where \( \epsilon \) is any positive constant, then almost all tournaments \( {T}_{n} \) have the property that\n\n\[ \left| {{r}_{ij} - \frac{1}{4}\left( {n - 2}\right) }\right| <... | Proof. Let \( N\left( k\right) \) denote the number of tournaments \( {T}_{n} \) such that the inequality \( \left| {{r}_{ij} - \frac{1}{4}\left( {n - 2}\right) }\right| > \lambda \) holds for exactly \( k \) ordered pairs of distinct nodes \( {p}_{i} \) and \( {p}_{j} \) . Then\n\n\[ 0 \leq \frac{{2}^{\left( \begin{ar... | Yes |
Proposition 2.1. Let \( G \) be a group and let \( H, K \) be two subgroups such that \( H \cap K = e,{HK} = G \), and such that \( {xy} = {yx} \) for all \( x \in H \) and \( y \in K \) . Then the map \[ H \times K \rightarrow G \] such that \( \left( {x, y}\right) \mapsto {xy} \) is an isomorphism. | Proof. It is obviously a homomorphism, which is surjective since \( {HK} = G \) . If \( \left( {x, y}\right) \) is in its kernel, then \( x = {y}^{-1} \), whence \( x \) lies in both \( H \) and \( K \), and \( x = e \) , so that \( y = e \) also, and our map is an isomorphism. | Yes |
Proposition 2.2. Let \( G \) be a group and \( H \) a subgroup. Then\n\n\[ \left( {G : H}\right) \left( {H : 1}\right) = \left( {G : 1}\right) ,\]\n\nin the sense that if two of these indices are finite, so is the third and equality holds as stated. If \( \left( {G : 1}\right) \) is finite, the order of \( H \) divides... | Proof. Note that\n\n\[ H = \mathop{\bigcup }\limits_{i}{x}_{i}K\;\text{ (disjoint), \]\n\n\[ G = \mathop{\bigcup }\limits_{j}{y}_{j}H\;\text{ (disjoint). \]\n\nHence\n\n\[ G = \mathop{\bigcup }\limits_{{i, j}}{y}_{j}{x}_{i}K \]\n\nWe must show that this union is disjoint, i.e. that the \( {y}_{j}{x}_{i} \) represent di... | Yes |
Let \( G \) be a finite group. An abelian tower of \( G \) admits a cyclic refinement. Let \( G \) be a finite solvable group. Then \( G \) admits a cyclic tower, whose last element is \( \{ e\} \) . | The second assertion is an immediate consequence of the first, and it clearly suffices to prove that if \( G \) is finite, abelian, then \( G \) admits a cyclic tower. We use induction on the order of \( G \) . Let \( x \) be an element of \( G \) . We may assume that \( x \neq e \) . Let \( X \) be the cyclic group ge... | Yes |
Theorem 3.2. Let \( G \) be a group and \( H \) a normal subgroup. Then \( G \) is solvable if and only if \( H \) and \( G/H \) are solvable. | Proof. We prove that \( G \) solvable implies that \( H \) is solvable. Let \( G = {G}_{0} \supset {G}_{1} \supset \ldots \supset {G}_{r} = \{ e\} \) be a tower of groups with \( {G}_{i + 1} \) normal in \( {G}_{i} \) and such that \( {G}_{i}/{G}_{i + 1} \) is abelian. Let \( {H}_{i} = H \cap {G}_{i} \) . Then \( {H}_{... | No |
Lemma 3.3. (Butterfly Lemma.) (Zassenhaus) Let \( U, V \) be subgroups of a group. Let \( u, v \) be normal subgroups of \( U \) and \( V \), respectively. Then\n\n\[ u\left( {U \cap v}\right) \;\text{ is normal in }\;u\left( {U \cap V}\right) ,\]\n\n\[ \left( {u \cap V}\right) v\;\text{ is normal in }\;\left( {U \cap ... | Proof. The combination of groups and factor groups becomes clear if one visualizes the following diagram of subgroups (which gives its name to the lemma):\n\n\n\nIn this diagram, we are given \( U, u, V, v \) . All the... | Yes |
Theorem 3.4. (Schreier) Let \( G \) be a group. Two normal towers of subgroups ending with the trivial group have equivalent refinements. | Proof. Let the two towers be as above. For each \( i = 1,\ldots, r - 1 \) and \( j = 1,\ldots, s \) we define\n\n\[ \n{G}_{ij} = {G}_{i + 1}\left( {{H}_{j} \cap {G}_{i}}\right) \n\]\n\nThen \( {G}_{is} = {G}_{i + 1} \), and we have a refinement of the first tower:\n\n\[ \nG = {G}_{11} \supset {G}_{12} \supset \cdots \s... | Yes |
Theorem 3.5. (Jordan-Hölder) Let \( G \) be a group, and let\n\n\[ G = {G}_{1} \supset {G}_{2} \supset \cdots \supset {G}_{r} = \{ e\} \]\n\nbe a normal tower such that each group \( {G}_{i}/{G}_{i + 1} \) is simple, and \( {G}_{i} \neq {G}_{i + 1} \) for \( i = 1,\ldots, r - 1 \) . Then any other normal tower of \( G ... | Proof. Given any refinement \( \left\{ {G}_{ij}\right\} \) as before for our tower, we observe that for each \( i \), there exists precisely one index \( j \) such that \( {G}_{i}/{G}_{i + 1} = {G}_{ij}/{G}_{i, j + 1} \) . Thus the sequence of non-trivial factors for the original tower, or the refined tower, is the sam... | Yes |
Proposition 4.2. Let \( G \) be a cyclic group. Then every subgroup of \( G \) is cyclic. | Proof. If \( G \) is infinite cyclic, it is isomorphic to \( \mathbf{Z} \), and we determined above all subgroups of \( \mathbf{Z} \), finding that they are all cyclic. If \( f : G \rightarrow {G}^{\prime } \) is a homomorphism, and \( a \) is a generator of \( G \), then \( f\left( a\right) \) is obviously a generator... | Yes |
Proposition 5.2. The number of conjugate subgroups to \( H \) is equal to the index of the normalizer of \( H \) . | Proof. Note that \( H \) is contained in its normalizer \( {N}_{H} \), so the index of \( {N}_{H} \) in \( G \) is 1 or 2. If it is 1, then we are done. Suppose it is 2. Let \( G \) operate by conjugation on the set of subgroups. The orbit of \( H \) has 2 elements, and \( G \) operates on this orbit. In this way we ge... | No |
Proposition 5.3. There exists a unique homomorphism \( \varepsilon : {S}_{n} \rightarrow \{ \pm 1\} \) such that for every transposition \( \tau \) we have \( \varepsilon \left( \tau \right) = - 1 \) . | Proof. Let \( \Delta \) be the function \[ \Delta \left( {{x}_{1},\ldots ,{x}_{n}}\right) = \mathop{\prod }\limits_{{i < j}}\left( {{x}_{j} - {x}_{i}}\right) \] the product being taken for all pairs of integers \( i, j \) satisfying \( 1 \leqq i < j \leqq n \) . Let \( \tau \) be a transposition, interchanging the two ... | Yes |
Theorem 5.4. If \( n \geqq 5 \) then \( {S}_{n} \) is not solvable. | Proof. We shall first prove that if \( H, N \) are two subgroups of \( {S}_{n} \) such that \( N \subset H \) and \( N \) is normal in \( H \), if \( H \) contains every 3-cycle, and if \( H/N \) is abelian, then \( N \) contains every 3-cycle. To see this, let \( i, j, k, r, s \) be five distinct integers in \( {J}_{n... | Yes |
Lemma 6.1. Let \( G \) be a finite abelian group of order \( m \), let \( p \) be a prime number dividing \( m \) . Then \( G \) has a subgroup of order \( p \) . | Proof. We first prove by induction that if \( G \) has exponent \( n \) then the order of \( G \) divides some power of \( n \) . Let \( b \in G, b \neq 1 \), and let \( H \) be the cyclic subgroup generated by \( b \) . Then the order of \( H \) divides \( n \) since \( {b}^{n} = 1 \), and \( n \) is an exponent for \... | Yes |
Theorem 6.2. Let \( G \) be a finite group and \( {pa} \) prime number dividing the order of \( G \) . Then there exists a p-Sylow subgroup of \( G \) . | Proof. By induction on the order of \( G \) . If the order of \( G \) is prime, our assertion is obvious. We now assume given a finite group \( G \), and assume the theorem proved for all groups of order smaller than that of \( G \) . If there exists a proper subgroup \( H \) of \( G \) whose index is prime to \( p \),... | Yes |
Lemma 6.3. Let \( H \) be a p-group acting on a finite set \( S \) . Then:\n\n(a) The number of fixed points of \( H \) is \( \equiv \# \left( S\right) {\;\operatorname{mod}\;p} \) .\n\n(b) If \( H \) has exactly one fixed point, then \( \# \left( S\right) \equiv 1{\;\operatorname{mod}\;p} \) .\n\n(c) If \( p \mid \# \... | Proof. We repeatedly use the orbit formula\n\n\[ \n\# \left( S\right) = \sum \left( {H : {H}_{{s}_{i}}}\right) \n\]\n\nFor each fixed point \( {s}_{i} \) we have \( {H}_{{s}_{i}} = H \) . For \( {s}_{i} \) not fixed, the index \( \left( {H : {H}_{{s}_{i}}}\right) \) is divisible by \( p \), so (a) follows at once. Part... | Yes |
Theorem 6.4. Let \( G \) be a finite group.\n\n(i) If \( H \) is a p-subgroup of \( G \), then \( H \) is contained in some p-Sylow subgroup.\n\n(ii) All p-Sylow subgroups are conjugate.\n\n(iii) The number of p-Sylow subgroups of \( G \) is \( \equiv 1{\;\operatorname{mod}\;p} \) . | Proof. Let \( P \) be a \( p \) -Sylow subgroup of \( G \) . Suppose first that \( H \) is contained in the normalizer of \( P \) . We prove that \( H \subset P \) . Indeed, \( {HP} \) is then a subgroup of the normalizer, and \( P \) is normal in \( {HP} \) . But\n\n\[ \n\left( {{HP} : P}\right) = \left( {H : H \cap P... | Yes |
Theorem 6.5. Let \( G \) be a finite p-group. Then \( G \) is solvable. If its order is \( > 1 \), then \( G \) has a non-trivial center. | Proof. The first assertion follows from the second, since if \( G \) has center \( Z \), and we have an abelian tower for \( G/Z \) by induction, we can lift this abelian tower to \( G \) to show that \( G \) is solvable. To prove the second assertion, we use the class equation\n\n\[ \left( {G : 1}\right) = \operatorna... | Yes |
Corollary 6.6. Let \( G \) be a p-group which is not of order 1 . Then there exists a sequence of subgroups\n\n\[ \{ e\} = {G}_{0} \subset {G}_{1} \subset {G}_{2} \subset \cdots \subset {G}_{n} = G \]\n\nsuch that \( {G}_{i} \) is normal in \( G \) and \( {G}_{i + 1}/{G}_{i} \) is cyclic of order \( p \) . | Proof. Since \( G \) has a non-trivial center, there exists an element \( a \neq e \) in the center of \( G \), and such that \( a \) has order \( p \) . Let \( H \) be the cyclic group generated by \( a \) . By induction, if \( G \neq H \), we can find a sequence of subgroups as stated above in the factor group \( G/H... | No |
Lemma 6.7. Let \( G \) be a finite group and let \( p \) be the smallest prime dividing the order of \( G \) . Let \( H \) be a subgroup of index \( p \) . Then \( H \) is normal. | Proof. Let \( N\left( H\right) = N \) be the normalizer of \( H \) . Then \( N = G \) or \( N = H \) . If \( N = G \) we are done. Suppose \( N = H \) . Then the orbit of \( H \) under conjugation has \( p = \left( {G : H}\right) \) elements, and the representation of \( G \) on this orbit gives a homomorphism of \( G ... | Yes |
Proposition 6.8. Let \( p, q \) be distinct primes and let \( G \) be a group of order pq. Then \( G \) is solvable. | Proof. Say \( p < q \) . Let \( Q \) be a Sylow subgroup of order \( q \) . Then \( Q \) has index \( p \), so by the lemma, \( Q \) is normal and the factor group has order \( p \) . But a group of prime order is cyclic, whence the proposition follows. | Yes |
Proposition 7.1. Let \( \\left\\{ {{f}_{i} : {A}_{i} \rightarrow B}\\right\\} \) be a family of homomorphisms into an abelian group \( B \) . Let \( A = \\oplus {A}_{i} \) . There exists a unique homomorphism\n\n\[ f : A \\rightarrow B \]\n\n such that \( f \\circ {\\lambda }_{j} = {f}_{j} \) for all \( j \) . | Proof. We can define a map \( f : A \\rightarrow B \) by the rule\n\n\[ f\\left( {\\left( {x}_{i}\\right) }_{i \\in I}\\right) = \\mathop{\\sum }\\limits_{{i \\in I}}{f}_{i}\\left( {x}_{i}\\right) \]\n\nThe sum on the right is actually finite since all but a finite number of terms are 0 . It is immediately verified tha... | Yes |
Lemma 7.2. Let \( A\overset{f}{ \rightarrow }{A}^{\prime } \) be a surjective homomorphism of abelian groups, and assume that \( {A}^{\prime } \) is free. Let \( B \) be the kernel of \( f \) . Then there exists a subgroup \( C \) of \( A \) such that the restriction of \( f \) to \( C \) induces an isomorphism of \( C... | Proof. Let \( {\left\{ {x}_{i}^{\prime }\right\} }_{i \in I} \) be a basis of \( {A}^{\prime } \), and for each \( i \in I \), let \( {x}_{i} \) be an element of \( A \) such that \( f\left( {x}_{i}\right) = {x}_{i}^{\prime } \) . Let \( C \) be the subgroup of \( A \) generated by all elements \( {x}_{i}, i \in I \) .... | Yes |
Theorem 7.3. Let \( A \) be a free abelian group, and let \( B \) be a subgroup. Then \( B \) is also a free abelian group, and the cardinality of a basis of \( B \) is \( \leqq \) the cardinality of a basis for \( A \) . Any two bases of \( B \) have the same cardinality. | Proof. We shall give the proof only when \( A \) is finitely generated, say by a basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \left( {n \geqq 1}\right) \), and give the proof by induction on \( n \) . We have an expression of \( A \) as direct sum:\n\n\[ A = \mathbf{Z}{x}_{1} \oplus \cdots \oplus \mathbf{Z}{x}_{n... | Yes |
Theorem 8.1 Let \( A \) be a torsion abelian group. Then \( A \) is the direct sum of its subgroups \( A\left( p\right) \) for all primes \( p \) such that \( A\left( p\right) \neq 0 \) . | Proof. There is a homomorphism\n\n\[ \n{\bigoplus }_{p}A\left( p\right) \rightarrow A \n\] \n\nwhich to each element \( \left( {x}_{p}\right) \) in the direct sum associates the element \( \sum {x}_{p} \) in \( A \) . We prove that this homomorphism is both surjective and injective. Suppose \( x \) is in the kernel, so... | Yes |
Theorem 8.2. Every finite abelian p-group is isomorphic to a product of cyclic p-groups. If it is of type \( \left( {{p}^{{r}_{1}},\ldots ,{p}^{{r}_{s}}}\right) \) with\n\n\[ \n{r}_{1} \geqq {r}_{2} \geqq \cdots \geqq {r}_{s} \geqq 1 \n\]\n\nthen the sequence of integers \( \left( {{r}_{1},\ldots ,{r}_{s}}\right) \) is... | Proof. We shall prove the existence of the desired product by induction. Let \( {a}_{1} \in A \) be an element of maximal period. We may assume without loss of generality that \( A \) is not cyclic. Let \( {A}_{1} \) be the cyclic subgroup generated by \( {a}_{1} \) , say of period \( {p}^{{r}_{1}} \) . We need a lemma... | No |
Lemma 8.3. Let \( \bar{b} \) be an element of \( A/{A}_{1} \), of period \( {p}^{r} \) . Then there exists a representative a of \( \bar{b} \) in \( A \) which also has period \( {p}^{r} \) . | Proof. Let \( b \) be any representative of \( \bar{b} \) in \( A \) . Then \( {p}^{r}b \) lies in \( {A}_{1} \), say \( {p}^{r}b = n{a}_{1} \) with some integer \( n \geqq 0 \) . We note that the period of \( \bar{b} \) is \( \leqq \) the period of \( b \) . If \( n = 0 \) we are done. Otherwise write \( n = {p}^{k}\m... | Yes |
Theorem 8.4. Let \( A \) be a finitely generated torsion-free abelian group. Then A is free. | Proof. Assume \( A \neq 0 \) . Let \( S \) be a finite set of generators, and let \( {x}_{1},\ldots ,{x}_{n} \) be a maximal subset of \( S \) having the property that whenever \( {v}_{1},\ldots ,{v}_{n} \) are integers such that\n\n\[ \n{v}_{1}{x}_{1} + \cdots + {v}_{n}{x}_{n} = 0 \n\]\n\nthen \( {v}_{j} = 0 \) for al... | Yes |
Theorem 8.5. Let \( A \) be a finitely generated abelian group, and let \( {A}_{\text{tor }} \) be the subgroup consisting of all elements of \( A \) having finite period. Then \( {A}_{\text{tor }} \) is finite, and \( A/{A}_{\text{tor }} \) is free. There exists a free subgroup \( B \) of \( A \) such that \( A \) is ... | Proof. We recall that a finitely generated torsion abelian group is obviously finite. Let \( A \) be finitely generated by \( n \) elements, and let \( F \) be the free abelian group on \( n \) generators. By the universal property, there exists a surjective homomorphism \[ F\overset{\varphi }{ \rightarrow }A \] of \( ... | Yes |
Theorem 9.2. Let \( A \times {A}^{\prime } \rightarrow C \) be a bilinear map of two abelian groups into a cyclic group \( C \) of order \( m \) . Let \( B,{B}^{\prime } \) be its respective kernels on the left and right. Assume that \( {A}^{\prime }/{B}^{\prime } \) is finite. Then \( A/B \) is finite, and \( {A}^{\pr... | Proof. The injection of \( A/B \) into \( \operatorname{Hom}\left( {{A}^{\prime }/{B}^{\prime }, C}\right) \) shows that \( A/B \) is finite. Furthermore, we get the inequalities\n\n\[ \text{ord}A/B \leqq \operatorname{ord}{\left( {A}^{\prime }/{B}^{\prime }\right) }^{ \land } = \operatorname{ord}{A}^{\prime }/{B}^{\pr... | Yes |
Corollary 9.3. Let \( A \) be a finite abelian group, \( B \) a subgroup, \( {A}^{ \land } \) the dual group, and \( {B}^{ \bot } \) the set of \( \varphi \in {A}^{ \land } \) such that \( \varphi \left( B\right) = 0 \) . Then we have a natural isomorphism of \( {A}^{ \land }/{B}^{ \bot } \) with \( {B}^{ \land } \) . | Proof. This is a special case of Theorem 9.2. | Yes |
Proposition 12.1. Let \( S \) be a set. Then there exists a free group \( \left( {F, f}\right) \) determined by \( S \) . Furthermore, \( f \) is injective, and \( F \) is generated by the image of \( f \) . | Proof. (I owe this proof to J. Tits.) We begin with a lemma. | No |
Lemma 12.2. There exists a set \( I \) and a family of groups \( {\left\{ {G}_{i}\right\} }_{i \in I} \) such that, if \( g : S \rightarrow G \) is a map of \( S \) into a group \( G \), and \( g \) generates \( G \), then \( G \) is isomorphic to some \( {G}_{i} \) . | Proof. This is a simple exercise in cardinalities, which we carry out. If \( S \) is finite, then \( G \) is finite or denumerable. If \( S \) is infinite, then the cardinality of \( G \) is \( \leqq \) the cardinality of \( S \) because \( G \) consists of finite products of elements of \( g\left( S\right) \) . Let \(... | Yes |
Proposition 12.3. Coproducts exist in the category of groups. | Proof. Let \( {\left\{ {G}_{i}\right\} }_{i \in I} \) be a family of groups. We let \( \mathcal{C} \) be the category whose objects are families of group-homomorphisms\n\n\[ \n{\left\{ {g}_{i} : {G}_{i} \rightarrow G\right\} }_{i \in I} \n\] \nand whose morphisms are the obvious ones. We must find a universal element i... | Yes |
Proposition 12.4. Let \( G \) be a group and \( {\left\{ {G}_{i}\right\} }_{i \in I} \) a family of subgroups. Assume:\n\n(a) The family generates \( G \) .\n\n(b) If\n\n\[ x = {x}_{{i}_{1}}\cdots {x}_{{i}_{n}}\;\text{ with }{x}_{{i}_{\nu }} \in {G}_{{i}_{\nu }},{x}_{{i}_{\nu }} \neq e\text{ and }{i}_{\nu } \neq {i}_{\... | Proof. The homomorphism from the coproduct into \( G \) is surjective by the assumption that the family generates \( G \) . Suppose an element is in the kernel. Then such an element has a representation\n\n\[ {x}_{{i}_{1}}\cdots {x}_{{i}_{n}} \]\n\nas in (b), mapping to the identity in \( G \), so all \( {x}_{{i}_{\nu ... | Yes |
Corollary 12.6. Let \( F\\left( S\\right) \) be the free group on a set \( S \), and let \( {x}_{1},\\ldots ,{x}_{n} \) be distinct elements of \( S \) . Let \( {\\nu }_{1},\\ldots ,{\\nu }_{r} \) be integers \( \\neq 0 \) and let \( {i}_{1},\\ldots ,{i}_{r} \) be integers,\n\n\[ \n1 \\leqq {i}_{1},\\ldots ,{i}_{r} \\l... | Proof. Let \( {G}_{1},\\ldots ,{G}_{n} \) be the cyclic groups generated by \( {x}_{1},\\ldots ,{x}_{n} \) . Let \( G = {G}_{1} \\circ \\cdots \\circ {G}_{n} \) . Let\n\n\[ \nF\\left( S\\right) \\rightarrow G \n\]\n\nbe the homomorphism sending each \( {x}_{i} \) on \( {x}_{i} \), and all other elements of \( S \) on t... | Yes |
Corollary 12.7. Let \( S \) be a set with \( n \) elements \( {x}_{1},\ldots ,{x}_{n}, n \geqq 1 \) . Let \( {G}_{1} \) , \( \ldots ,{G}_{n} \) be the infinite cyclic groups generated by these elements. Then the map\n\n\[ F\left( S\right) \rightarrow {G}_{1} \circ \cdots \circ {G}_{n} \]\n\nsending each \( {x}_{i} \) o... | Proof. It is obviously surjective and injective. | No |
Corollary 12.8. Let \( {G}_{1},\ldots ,{G}_{n} \) be groups with \( {G}_{i} \cap {G}_{j} = \{ 1\} \) if \( i \neq j \) . The homomorphism\n\n\[ \n{G}_{1} \coprod \cdots \coprod {G}_{n} \rightarrow {G}_{1} \circ \cdots \circ {G}_{n} \n\]\n\nof their coproduct into \( {G}_{1} \circ \cdots \circ {G}_{n} \) induced by the ... | Proof. Again, it is obviously injective and surjective. | No |
Theorem 2.1. (Chinese Remainder Theorem). Let \( {a}_{1},\ldots ,{a}_{n} \) be ideals of \( A \) such that \( {\mathfrak{a}}_{i} + {\mathfrak{a}}_{j} = A \) for all \( i \neq j \) . Given elements \( {x}_{1},\ldots ,{x}_{n} \in A \), there exists \( x \in A \) such that \( x \equiv {x}_{i}\left( {\;\operatorname{mod}\;... | Proof. If \( n = 2 \), we have an expression\n\n\[ 1 = {a}_{1} + {a}_{2} \]\n\nfor some elements \( {a}_{i} \in {\mathfrak{a}}_{i} \), and we let \( x = {x}_{2}{a}_{1} + {x}_{1}{a}_{2} \) .\n\nFor each \( i \geqq 2 \) we can find elements \( {a}_{i} \in {\mathfrak{a}}_{1} \) and \( {b}_{i} \in {\mathfrak{a}}_{i} \) suc... | Yes |
Corollary 2.2. Let \( {\mathfrak{a}}_{1},\ldots ,{\mathfrak{a}}_{n} \) be ideals of \( A \) . Assume that \( {\mathfrak{a}}_{i} + {\mathfrak{a}}_{j} = A \) for \( i \neq j \) . Let\n\n\[ f : A \rightarrow \mathop{\prod }\limits_{{i = 1}}^{n}A/{\mathfrak{a}}_{i} = \left( {A/{\mathfrak{a}}_{1}}\right) \times \cdots \time... | Proof. That the kernel of \( f \) is what we said it is, is obvious. The surjectivity follows from the theorem. | No |
Theorem 2.3. Let \( A \) be a cyclic group of order \( n \) . For each \( k \in \mathbf{Z} \) let \( {f}_{k} : A \rightarrow A \) be the endomorphism \( x \mapsto {kx} \) (writing \( A \) additively). Then \( k \mapsto {f}_{k} \) induces a ring isomorphism \( \mathbf{Z}/n\mathbf{Z} \approx \operatorname{End}\left( A\ri... | Proof. Recall that the additive group structure on \( \operatorname{End}\left( A\right) \) is simply addition of mappings, and the multiplication is composition of mappings. The fact that \( k \mapsto {f}_{k} \) is a ring-homomorphism is then a restatement of the formulas\n\n\[ \n{1a} = a,\;\left( {k + {k}^{\prime }}\r... | Yes |
Proposition 3.1. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a homomorphism of monoids. Then there exists a unique homomorphism \( h : A\left\lbrack G\right\rbrack \rightarrow A\left\lbrack {G}^{\prime }\right\rbrack \) such that \( h\left( x\right) = \) \( \varphi \left( x\right) \) for all \( x \in G \) and \(... | Proof. In fact, let \( \alpha = \sum {a}_{x}x \in A\left\lbrack G\right\rbrack \) . Define\n\n\[ h\left( \alpha \right) = \sum {a}_{x}\varphi \left( x\right) \]\n\nThen \( h \) is immediately verified to be a homomorphism of abelian groups, and \( h\left( x\right) = \varphi \left( x\right) \) . Let \( \beta = \sum {b}_... | Yes |
Proposition 3.2. Let \( G \) be a monoid and let \( f : A \rightarrow B \) be a homomorphism of commutative rings. Then there is a unique homomorphism | Proof. Since every element of \( A\left\lbrack G\right\rbrack \) has a unique expression as a sum \( \sum {a}_{x}x \), the formula giving \( h \) gives a well-defined map from \( A\left\lbrack G\right\rbrack \) into \( B\left\lbrack G\right\rbrack \) . This map is obviously a homomorphism of abelian groups. As for mult... | Yes |
Proposition 5.1. Let \( A \) be a principal entire ring and \( a, b \in A, a, b \neq 0 \). Let \( \left( {a, b}\right) = \left( c\right) \). Then \( c \) is a greatest common divisor of \( a \) and \( b \). | Proof. Since \( b \) lies in the ideal \( \left( c\right) \), we can write \( b = {xc} \) for some \( x \in A \), so that \( c \mid b \). Similarly, \( c \mid a \). Let \( d \) divide both \( a \) and \( b \), and write \( a = {dy} \), \( b = {dz} \) with \( y, z \in A \). Since \( c \) lies in \( \left( {a, b}\right) ... | Yes |
Theorem 5.2. Let \( A \) be a principal entire ring. Then \( A \) is factorial. | Proof. We first prove that every non-zero element of \( A \) has a factorization into irreducible elements. Let \( S \) be the set of principal ideals \( \neq 0 \) whose generators do not have a factorization into irreducible elements, and suppose \( S \) is not empty. Let \( \left( {a}_{1}\right) \) be in \( S \) . Co... | Yes |
Proposition 2.1. A sequence\n\n\[ \n{X}^{\prime }\overset{\lambda }{ \rightarrow }X \rightarrow {X}^{\prime \prime } \rightarrow 0 \n\]\n\nis exact if and only if the sequence\n\n\[ \n{\operatorname{Hom}}_{A}\left( {{X}^{\prime }, Y}\right) \leftarrow {\operatorname{Hom}}_{A}\left( {X, Y}\right) \leftarrow {\operatorna... | Proof. This is an important fact, whose proof is easy. For instance, suppose the first sequence is exact. If \( g : {X}^{\prime \prime } \rightarrow Y \) is an \( A \) -homomorphism, its image in \( {\operatorname{Hom}}_{A}\left( {X, Y}\right) \) is obtained by composing \( g \) with the surjective map of \( X \) on \(... | No |
Proposition 2.2. A sequence\n\n\[ 0 \rightarrow {Y}^{\prime } \rightarrow Y \rightarrow {Y}^{\prime \prime } \]\n\nis exact if and only if\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{A}\left( {X,{Y}^{\prime }}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {X, Y}\right) \rightarrow {\operatorname{Hom}}_{A}\left( {X,{Y... | The verification will be left to the reader. It follows at once from the definitions. | No |
Proposition 3.1. Let \( M \) be an \( A \) -module and \( n \) an integer \( \geqq 1 \) . For each \( i = 1,\ldots, n \) let \( {\varphi }_{i} : M \rightarrow M \) be an \( A \) -homomorphism such that\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{\varphi }_{i} = \mathrm{{id}}\;\text{ and }\;{\varphi }_{i} \circ {\varphi }... | Proof. For each \( j \), we have\n\n\[ {\varphi }_{j} = {\varphi }_{j} \circ \mathrm{{id}} = {\varphi }_{j} \circ \mathop{\sum }\limits_{{i = 1}}^{n}{\varphi }_{i} = {\varphi }_{j} \circ {\varphi }_{j} = {\varphi }_{j}^{2}, \]\n\nthereby proving the first assertion. It is clear that \( \varphi \) is an \( A \) -homomor... | Yes |
Proposition 3.2. Let \( 0 \rightarrow {M}^{\prime }\overset{f}{ \rightarrow }M\overset{g}{ \rightarrow }{M}^{\prime \prime } \rightarrow 0 \) be an exact sequence of modules. The following conditions are equivalent:\n\n1. There exists a homomorphism \( \varphi : {M}^{\prime \prime } \rightarrow M \) such that \( g \cir... | Proof. Let us write the homomorphisms on the right:\n\n\[ M\underset{\varphi }{\overset{g}{ \rightleftarrows }}{M}^{\prime \prime } \rightarrow 0 \]\n\nLet \( x \in M \). Then\n\n\[ x - \varphi \left( {g\left( x\right) }\right) \]\n\nis in the kernel of \( g \), and hence \( M = \operatorname{Ker}g + \operatorname{Im}\... | No |
Theorem 4.1. Let \( A \) be a ring and \( M \) a module over \( A \) . Let \( I \) be a non-empty set, and let \( {\left\{ {x}_{i}\right\} }_{i \in I} \) be a basis of \( M \) . Let \( N \) be an \( A \) -module, and let \( {\left\{ {y}_{i}\right\} }_{i \in I} \) be a family of elements of \( N \) . Then there exists a... | Proof. Let \( x \) be an element of \( M \) . There exists a unique family \( {\left\{ {a}_{i}\right\} }_{i \in I} \) of elements of \( A \) such that\n\n\[ x = \mathop{\sum }\limits_{{i \in I}}{a}_{i}{x}_{i} \]\n\nWe define\n\n\[ f\left( x\right) = \sum {a}_{i}{y}_{i} \]\n\nIt is then clear that \( f \) is a homomorph... | Yes |
Corollary 4.2. Let the notation be as in the theorem, and assume that \( {\left\{ {y}_{i}\right\} }_{i \in I} \) is a basis of \( N \) . Then the homomorphism \( f \) is an isomorphism, i.e. a module-isomorphism. | Proof. By symmetry, there exists a unique homomorphism\n\n\[ g : N \rightarrow M \]\n\nsuch that \( g\left( {y}_{i}\right) = {x}_{i} \) for all \( i \), and \( f \circ g \) and \( g \circ f \) are the respective identity mappings. | Yes |
Corollary 4.3. Two modules having bases whose cardinalities are equal are isomorphic. | Proof. Clear. | No |
Theorem 5.1. Let \( V \) be a vector space over a field \( K \), and assume that \( V \neq \{ 0\} \) . Let \( \Gamma \) be a set of generators of \( V \) over \( K \) and let \( S \) be a subset of \( \Gamma \) which is linearly independent. Then there exists a basis \( \mathcal{B} \) of \( V \) such that \( S \subset ... | Proof. Let \( \mathfrak{T} \) be the set whose elements are subsets \( T \) of \( \Gamma \) which contain \( S \) and are linearly independent. Then \( \mathfrak{T} \) is not empty (it contains \( S \) ), and we contend that \( \mathfrak{T} \) is inductively ordered. Indeed, if \( \left\{ {T}_{i}\right\} \) is a totall... | Yes |
Theorem 5.2. Let \( V \) be a vector space over a field \( K \). Then two bases of \( V \) over \( K \) have the same cardinality. | Proof. Let us first assume that there exists a basis of \( V \) with a finite number of elements, say \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\}, m \geqq 1 \). We shall prove that any other basis must also have \( m \) elements. For this it will suffice to prove: If \( {w}_{1},\ldots ,{w}_{n} \) are elements of \( V ... | No |
Theorem 5.3. Let \( V \) be a vector space over a field \( K \), and let \( W \) be a subspace. Then\n\n\[{\dim }_{K}V = {\dim }_{K}W + {\dim }_{K}V/W\] | Proof. The first statement is a special case of the second, taking for \( f \) the canonical map. Let \( {\left\{ {u}_{i}\right\} }_{i \in I} \) be a basis of \( \operatorname{Im}f \), and let \( {\left\{ {w}_{j}\right\} }_{j \in J} \) be a basis of Ker \( f \) . Let \( {\left\{ {v}_{i}\right\} }_{i \in I} \) be a fami... | Yes |
Corollary 5.4. Let \( V \) be a vector space and \( W \) a subspace. Then\n\n\[ \n\dim W \leqq \dim V \n\]\n\nIf \( V \) is finite dimensional and \( \dim W = \dim V \) then \( W = V \) . | Proof. Clear. | No |
Theorem 6.1. Let \( E \) be a finite free module over the commutative ring \( A \) , of finite dimension \( n \) . Then \( {E}^{ \vee } \) is also free, and \( \dim {E}^{ \vee } = n \) . If \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is a basis for \( E \), and \( {f}_{i} \) is the functional such that \( {f}_{i}\l... | Proof. Let \( f \in {E}^{ \vee } \) and let \( {a}_{i} = f\left( {x}_{i}\right) \left( {i = 1,\ldots, n}\right) \) . We have\n\n\[ f\left( {{c}_{1}{x}_{1} + \cdots + {c}_{n}{x}_{n}}\right) = {c}_{1}f\left( {x}_{1}\right) + \cdots + {c}_{n}f\left( {x}_{n}\right) . \]\n\nHence \( f = {a}_{1}{f}_{1} + \cdots + {a}_{n}{f}_... | Yes |
Corollary 6.2. When \( E \) is free finite dimensional, then the map \( E \rightarrow {E}^{\vee \vee } \) which to each \( x \in V \) associates the functional \( f \mapsto \langle x, f\rangle \) on \( {E}^{ \vee } \) is an isomorphism of \( E \) onto \( {E}^{\vee \vee } \) . | Proof. Note that since \( \left\{ {{f}_{1},\ldots ,{f}_{n}}\right\} \) is a basis for \( {E}^{ \vee } \), it follows from the definitions that \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) is the dual basis in \( E \), so \( E = {E}^{\vee \vee } \) . | No |
Theorem 6.3. Let \( U, V, W \) be finite free modules over the commutative ring \( A \), and let\n\n\[ 0 \rightarrow W\overset{\lambda }{ \rightarrow }V\overset{\varphi }{ \rightarrow }U \rightarrow 0 \]\n\nbe an exact sequence of A-homomorphisms. Then the induced sequence\n\n\[ 0 \rightarrow {\operatorname{Hom}}_{A}\l... | Proof. This is a consequence of \( \mathbf{P}\mathbf{2} \), because a free module is projective. | No |
Theorem 6.4. Let \( V \times {V}^{\prime } \rightarrow K \) be a bilinear map, let \( W,{W}^{\prime } \) be its kernels on the left and right respectively, and assume that \( {V}^{\prime }/{W}^{\prime } \) is finite dimensional. Then the induced homomorphism \( {V}^{\prime }/{W}^{\prime } \rightarrow {\left( V/W\right)... | Proof. By symmetry, we have an induced homomorphism\n\n\[ V/W \rightarrow {\left( {V}^{\prime }/{W}^{\prime }\right) }^{ \vee } \]\n\nwhich is injective. Since\n\n\[ \dim {\left( {V}^{\prime }/{W}^{\prime }\right) }^{ \vee } = \dim {V}^{\prime }/{W}^{\prime } \]\n\nit follows that \( V/W \) is finite dimensional. From ... | Yes |
Theorem 7.1. Let \( F \) be a free module, and \( M \) a submodule. Then \( M \) is free, and its dimension is less than or equal to the dimension of \( F \) . | Proof. For simplicity, we give the proof when \( F \) has a finite basis \( \left\{ {x}_{i}\right\} \) , \( i = 1,\ldots, n \) . Let \( {M}_{r} \) be the intersection of \( M \) with \( \left( {{x}_{1},\ldots ,{x}_{r}}\right) \), the module generated by \( {x}_{1},\ldots ,{x}_{r} \) . Then \( {M}_{1} = M \cap \left( {x... | Yes |
Corollary 7.2. Let \( E \) be a finitely generated module and \( {E}^{\prime } \) a submodule.\n\nThen \( {E}^{\prime } \) is finitely generated. | Proof. We can represent \( E \) as a factor module of a free module \( F \) with a finite number of generators: If \( {v}_{1},\ldots ,{v}_{n} \) are generators of \( E \), we take a free module \( F \) with basis \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) and map \( {x}_{i} \) on \( {v}_{i} \) . The inverse image ... | Yes |
Theorem 7.3. Let \( E \) be finitely generated. Then \( E/{E}_{\text{tor }} \) is free. There exists a free submodule \( F \) of \( E \) such that \( E \) is a direct sum\n\n\[ E = {E}_{\text{tor }} \oplus F. \]\n\nThe dimension of such a submodule \( F \) is uniquely determined. | Proof. We first prove that \( E/{E}_{\text{tor }} \) is torsion free. If \( x \in E \), let \( \bar{x} \) denote its residue class \( {\;\operatorname{mod}\;{E}_{\text{tor }}} \) . Let \( b \in R, b \neq 0 \) be such that \( b\bar{x} = 0 \) . Then \( {bx} \in {E}_{\text{tor }} \) , and hence there exists \( c \in R, c ... | No |
Lemma 7.4. Let \( E,{E}^{\prime } \) be modules, and assume that \( {E}^{\prime } \) is free. Let \( f : E \rightarrow {E}^{\prime } \) be a surjective homomorphism. Then there exists a free submodule \( F \) of \( E \) such that the restriction off to \( F \) induces an isomorphism of \( F \) with \( {E}^{\prime } \),... | Proof. Let \( {\left\{ {x}_{i}^{\prime }\right\} }_{i \in I} \) be a basis of \( {E}^{\prime } \) . For each \( i \), let \( {x}_{i} \) be an element of \( E \) such that \( f\left( {x}_{i}\right) = {x}_{i}^{\prime } \) . Let \( F \) be the submodule of \( E \) generated by all the elements \( {x}_{i} \) , \( i \in I \... | Yes |
Theorem 7.5. Let \( E \) be a finitely generated torsion module \( \neq 0 \) . Then \( E \) is the direct sum\n\n\[ E = {\bigoplus }_{p}E\left( p\right) \]\n\ntaken over all primes \( p \) such that \( E\left( p\right) \neq 0 \) . Each \( E\left( p\right) \) can be written as a direct sum\n\n\[ E\left( p\right) = R/\le... | Proof. Let \( a \) be an exponent for \( E \), and suppose that \( a = {bc} \) with \( \left( {b, c}\right) = \left( 1\right) \) . Let \( x, y \in R \) be such that\n\n\[ 1 = {xb} + {yc}\text{.} \]\n\nWe contend that \( E = {E}_{b} \oplus {E}_{c} \) . Our first assertion then follows by induction, expressing \( a \) as... | Yes |
Lemma 7.6. Let \( E \) be a torsion module of exponent \( {p}^{r}\left( {r \geqq 1}\right) \) for some prime element \( p \) . Let \( {x}_{1} \in E \) be an element of period \( {p}^{r} \) . Let \( \bar{E} = E/\left( {x}_{1}\right) \) . Let \( {\bar{y}}_{1},\ldots ,{\bar{y}}_{m} \) be independent elements of \( \bar{E}... | Proof. Let \( \bar{y} \in \bar{E} \) have period \( {p}^{n} \) for some \( n \geqq 1 \) . Let \( y \) be a representative of \( \bar{y} \) in \( E \) . Then \( {p}^{n}y \in \left( {x}_{1}\right) \), and hence\n\n\[ \n{p}^{n}y = {p}^{s}c{x}_{1},\;c \in R, p \nmid c, \n\]\n\nfor some \( s \leqq r \) . If \( s = r \), we ... | Yes |
Theorem 7.7. Let \( E \) be a finitely generated torsion module, \( E \neq 0 \) . Then \( E \) is isomorphic to a direct sum of non-zero factors\n\n\[ R/\left( {q}_{1}\right) \oplus \cdots \oplus R/\left( {q}_{r}\right) \]\n\nwhere \( {q}_{1},\ldots ,{q}_{r} \) are non-zero elements of \( R \), and \( {q}_{1}\left| {q}... | Proof. Using Theorem 7.5, decompose \( E \) into a direct sum of \( p \) -submodules, say \( E\left( {p}_{1}\right) \oplus \cdots \oplus E\left( {p}_{l}\right) \), and then decompose each \( E\left( {p}_{i}\right) \) into a direct sum of cyclic submodules of periods \( {p}_{i}^{{r}_{ij}} \) . We visualize these symboli... | Yes |
Theorem 7.9. Assume that the elementary matrices in \( R \) generate \( G{L}_{n}\left( R\right) \) . Let \( \left( {x}_{ij}\right) \) be a non-zero matrix with components in \( R \) . Then with a finite number of row and column operations, it is possible to bring the matrix to the form\n\n\[ \left( \begin{matrix} {a}_{... | We leave the proof for the reader. Either Theorem 7.9 can be viewed as equivalent to Theorem 7.8, or a direct proof may be given. | No |
Theorem 8.1. Let \( \varphi \) be a rule which to each simple module associates an element of a commutative group \( \Gamma \), and such that if \( M \approx {M}^{\prime } \) then\n\n\[ \varphi \left( M\right) = \varphi \left( {M}^{\prime }\right) \]\n\nThen \( \varphi \) has a unique extension to an Euler-Poincaré map... | Proof. Given a simple filtration\n\n\[ M = {M}_{1} \supset {M}_{2} \supset \cdots \supset {M}_{r} = 0 \]\n\nwe define\n\n\[ \varphi \left( M\right) = \mathop{\sum }\limits_{{i = 1}}^{{r - 1}}\varphi \left( {{M}_{i}/{M}_{i + 1}}\right) \]\n\nThe Jordan-Hölder theorem shows immediately that this is well-defined, and that... | Yes |
Lemma 9.1. (Snake Lemma). Given a snake diagram as above, the map\n\n\\[ \n\\delta : \\operatorname{Ker}{d}^{\\prime \\prime } \\rightarrow \\operatorname{Coker}{d}^{\\prime }\n\\]\n\ngiven by \\( \\delta {z}^{\\prime \\prime } = {f}^{-1} \\circ d \\circ {g}^{-1}{z}^{\\prime \\prime } \\) is well defined, and we have a... | Proof. It is a routine verification that the class of \\( {z}^{\\prime }{\\operatorname{mod}\\operatorname{Im}}{d}^{\\prime } \\) is independent of the choices made when taking inverse images, whence defining the map \\( \\delta \\) . The proof of the exactness of the sequence is then routine, and consists in chasing a... | No |
Theorem 10.1. Direct limits exist in the category of abelian groups, or more generally in the category of modules over a ring. | Proof. Let \( \left\{ {M}_{i}\right\} \) be a directed system of modules over a ring. Let \( M \) be their direct sum. Let \( N \) be the submodule generated by all elements\n\n\[ \n{x}_{ij} = \left( {\ldots ,0, x,0,\ldots , - {f}_{j}^{i}\left( x\right) ,0,\ldots }\right) \n\]\n\nwhere, for a given pair of indices \( \... | No |
Theorem 10.2. Inverse limits exist in the category of groups, in the category of modules over a ring, and also in the category of rings. | Proof. Let \( \left\{ {G}_{i}\right\} \) be a directed family of groups, for instance, and let \( \Gamma \) be their inverse limit as defined in Chapter I,§10. Let \( {p}_{i} : \Gamma \rightarrow {G}_{i} \) be the projection (defined as the restriction from the projection of the direct product, since \( \Gamma \) is a ... | No |
Proposition 10.3. Assume that \( \left( {A}_{n}\right) \) satisfies ML. Given an exact sequence\n\n\[ 0 \rightarrow \left( {A}_{n}\right) \rightarrow \left( {B}_{n}\right) \overset{g}{ \rightarrow }\left( {C}_{n}\right) \rightarrow 0 \]\n\nof inverse systems, then\n\n\[ 0 \rightarrow \mathop{\lim }\limits_{ \rightarrow... | Proof. The only point is to prove the surjectivity on the right. Let \( \left( {c}_{n}\right) \) be an element of the inverse limit. Then each inverse image \( {g}^{-1}\left( {c}_{n}\right) \) is a coset of \( {A}_{n} \), so in bijection with \( {A}_{n} \) . These inverse images form an inverse system, and the ML condi... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.