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Example 7.61.\n\nLet \( f\left( {x, y, z}\right) = {x}^{2}{y}^{2} + {xy} \), and let \( G = \left\lbrack {{g}_{1},{g}_{2},{g}_{3}}\right\rbrack \), where\n\n\[ \n{g}_{1} = {y}^{2} + {z}^{2} \n\]\n\n\[ \n{g}_{2} = {x}^{2}y + {yz} \n\]\n\n\[ \n{g}_{3} = {z}^{3} + {xy} \n\]\n\nWe use the degree-lexicographic order on \( {... | On the other hand, let us apply the division algorithm using the 3-tuple \( {G}^{\prime } = \) \( \left\lbrack {{g}_{2},{g}_{1},{g}_{3}}\right\rbrack \) . The first reduction gives \( f\overset{{g}_{2}}{ \rightarrow }{h}^{\prime } \), where\n\n\[ \n{h}^{\prime } = f - \frac{{x}^{2}{y}^{2}}{{x}^{2}y}\left( {{x}^{2}y + {... | Yes |
Proposition 7.62. A set \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) of polynomials is a Gröbner basis of \( I = \left( {{g}_{1},\ldots ,{g}_{m}}\right) \) if and only if, for each \( m \) -tuple \( {G}_{\sigma } = \left\lbrack {{g}_{\sigma \left( 1\right) },\ldots ,{g}_{\sigma \left( m\right) }}\right\rbrack \) (wh... | Proof. Assume there is some permutation \( \sigma \in {S}_{m} \) and some \( f \in I \) whose remainder mod \( {G}_{\sigma } \) is not 0 . Among all such polynomials, choose \( f \) of minimal Degree. Since \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) is a Gröbner basis, \( \operatorname{LT}\left( {g}_{i}\right) \mi... | Yes |
Corollary 7.63. Let \( I = \left( {{g}_{1},\ldots ,{g}_{m}}\right) \) be an ideal, let \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) be a Gröbner basis of \( I \), and let \( G = \left\lbrack {{g}_{1},\ldots ,{g}_{m}}\right\rbrack \) be any \( m \) -tuple formed from the \( {g}_{i} \) . If \( f\left( X\right) \in k\l... | Proof. The division algorithm gives a polynomial \( r \) which is reduced mod \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \), and polynomials \( {a}_{1},\ldots ,{a}_{m} \) with \( f = {a}_{1}{g}_{1} + \cdots + {a}_{m}{g}_{m} + r \) ; clearly, \( f - r = {a}_{1}{g}_{1} + \cdots + {a}_{m}{g}_{m} \in I \). To prove uniqu... | No |
Corollary 7.64. Let \( I = \left( {{g}_{1},\ldots ,{g}_{m}}\right) \) be an ideal, let \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) be a Gröbner basis of \( I \), and let \( G \) be the m-tuple \( G = \left\lbrack {{g}_{1},\ldots ,{g}_{m}}\right\rbrack \). (i) If \( f\left( X\right) \in k\left\lbrack X\right\rbrack ... | Proof. (i) If \( r \) is the remainder of \( f{\;\operatorname{mod}\;G} \), then Corollary 7.63 says that \( r \) is the unique polynomial, reduced \( {\;\operatorname{mod}\;\left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} } \), with \( f - r \in I \) ; similarly, the remainder \( {r}_{\sigma } \) of \( f{\;\operatorname{mod}... | Yes |
Lemma 7.66. Given \( {g}_{1}\left( X\right) ,\ldots ,{g}_{\ell }\left( X\right) \in k\left\lbrack X\right\rbrack \) and monomials \( {c}_{j}{X}^{\alpha \left( j\right) } \), let \( h\left( X\right) = \mathop{\sum }\limits_{{j = 1}}^{\ell }{c}_{j}{X}^{\alpha \left( j\right) }{g}_{j}\left( X\right) . \)\n\nLet \( \delta ... | Proof. Let \( \operatorname{LT}\left( {g}_{j}\right) = {b}_{j}{X}^{\beta \left( j\right) } \), so that \( \operatorname{LT}\left( {{c}_{j}{X}^{\alpha \left( j\right) }{g}_{j}\left( X\right) }\right) = {c}_{j}{b}_{j}{X}^{\delta } \) . The coefficient of \( {X}^{\delta } \) in \( h\left( X\right) \) is thus \( \mathop{\s... | Yes |
Corollary 7.68. If \( I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) \) in \( k\left\lbrack X\right\rbrack \), where each \( {f}_{i} \) is a monomial (that is, if \( I \) is a monomial ideal), then \( \left\{ {{f}_{1},\ldots ,{f}_{s}}\right\} \) is a Gröbner basis of \( I \) . | Proof. By Example 7.65, the \( S \) -polynomial of any pair of monomials is 0 . | Yes |
Theorem 7.69 (Buchberger’s Algorithm). Every ideal \( I = \left( {{f}_{1},\ldots ,{f}_{s}}\right) \) in \( k\left\lbrack X\right\rbrack \) has a Gröbner basis \( {}^{11} \) which can be computed by an algorithm. | Proof. Here is a pseudocode for an algorithm.\n\nInput: \( B = \left\{ {{f}_{1},\ldots ,{f}_{s}}\right\} \;G = \left\lbrack {{f}_{1},\ldots ,{f}_{s}}\right\rbrack \)\n\nOutput: a Gröbner basis \( B = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \)\n\ncontaining \( \left\{ {{f}_{1},\ldots ,{f}_{s}}\right\} \)\n\n\( B \math... | Yes |
Proposition 7.72. Let \( k \) be a field and let \( k\left\lbrack X\right\rbrack = k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) have a monomial order for which \( {x}_{1} \succ {x}_{2} \succ \cdots \succ {x}_{n} \) (for example, the lexicographic order) and, for fixed \( p > 1 \), let \( Y = {x}_{p},\ldots ,... | Proof. Recall that \( \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) being a Gröbner basis of \( I = \left( {{g}_{1},\ldots ,{g}_{m}}\right) \) means that for each nonzero \( f \in I \), there is \( {g}_{i} \) with \( \operatorname{LT}\left( {g}_{i}\right) \mid \operatorname{LT}\left( f\right) \) . Let \( f\left( {{x}_{p... | Yes |
Proposition 7.73. Let \( k \) be a field, and let \( {I}_{1},\ldots ,{I}_{t} \) be ideals in \( k\left\lbrack X\right\rbrack \), where \( X = {x}_{1},\ldots ,{x}_{n} \n\n(i) Consider the polynomial ring \( k\left\lbrack {X,{y}_{1},\ldots ,{y}_{t}}\right\rbrack \) in \( n + t \) indeterminates. If \( J \) is the ideal i... | (i) If \( f = f\left( X\right) \in {J}_{X} = J \cap k\left\lbrack X\right\rbrack \), then \( f \in J \), and so there is an equation\n\n\[ \nf\left( X\right) = g\left( {X, Y}\right) \left( {1-\sum {y}_{j}}\right) + \mathop{\sum }\limits_{j}{h}_{j}\left( {X,{y}_{1},\ldots ,{y}_{t}}\right) {y}_{j}{q}_{j}\left( X\right) ,... | Yes |
Hopkins-Levitzki theorem (霍普金斯-列维茨基定理) | 555 | No |
Proposition 1.1. Let \( G \) be a finite group and let \( X \) be a finite \( G \) -set. For every \( g \in G \) , let \( {X}^{g} \subset X \) be the set of elements \( x \) of \( X \) which are fixed under the action of \( g \), and let \( {\chi }_{X}\left( g\right) = \left| {X}^{g}\right| \) . Then :\n\n\[ \left| {X/... | Proof. By splitting \( X \) into orbits, we may assume that \( G \) acts transitively, hence that \( X = G/H \), where \( H \) is a subgroup of \( G \) . If \( \left( {g, x}\right) \in G \times X \), define \( f\left( {g, x}\right) \) to be equal to 1 if \( {gx} = x \), and to 0 if \( {gx} \neq x \) . Let us compute in... | Yes |
Proposition 1.2. Let \( M \) and \( N \) be two normal subgroups of \( G \) such that \( M \cap N = 1 \) . Then \( M \) and \( N \) commute elementwise, i.e., \( {xy} = {yx} \) for every \( x \in M \) and \( y \in N \) . | Indeed, the element \( {xy}{x}^{-1}{y}^{-1} = {xy}{x}^{-1}{y}^{-1} \) belongs to both \( M \) and \( N \), hence is equal to 1 . | Yes |
Proposition 1.3. Every finite group has a Jordan-Hölder filtration. | Proof. If \( G = 1 \), take the trivial Jordan-Hölder filtration with \( n = 0 \) in (1.4); if \( G \) is simple, take \( n = 1 \) in (1.4). Suppose that \( G \) is neither 1 nor simple. Use induction on the order of \( G \) . Let \( N \) be a normal subgroup of \( G \), distinct from \( G \), and of maximal order. The... | Yes |
Corollary 1.5. If \( N \) is a normal subgroup of \( G \), then \( \ell \left( G\right) = \ell \left( N\right) + \ell \left( {G/N}\right) \) . | This has been proved in the argument above when \( \ell \left( G\right) < \infty \) . When \( \ell \left( G\right) = \infty \), then the same is true for either \( N \) or \( G/N \), and we still have \( \ell \left( G\right) = \ell \left( N\right) + \ell \left( {G/N}\right) . | Yes |
Proposition 1.6 (Goursat’s lemma). For \( i = 1,2 \), let \( {N}_{i} \) be a normal subgroup of \( {G}_{i} \) and let \( \varphi : {G}_{1}/{N}_{1} \rightarrow {G}_{2}/{N}_{2} \) be an isomorphism. Let \( {H}_{{N}_{1},{N}_{2},\varphi } \) be the set of all \( \left( {{g}_{1},{g}_{2}}\right) \in {G}_{1} \times {G}_{2} \)... | Proof of \( i \) ). Let \( {g}_{1} \) be an element of \( {G}_{1} \) . Choose \( {g}_{2} \in {G}_{2} \) such that \( \overline{{g}_{2}} = \varphi \left( \overline{{g}_{1}}\right) \), and let \( h = \left( {{g}_{1},{g}_{2}}\right) \) . We have \( h \in H \) and \( {\operatorname{pr}}_{1}\left( h\right) = {g}_{1} \) . Sa... | Yes |
Proposition 1.8 (Ribet). Let \( H \) be a subgroup of \( {G}_{1} \times \cdots \times {G}_{n} \). Assume that \( {\operatorname{pr}}_{ij}\left( H\right) = {G}_{i} \times {G}_{j} \) for every \( i, j \), and that the \( {G}_{i} \), with at most two exceptions, can be generated by commutators (i.e., elements of the form ... | Proof. Use induction on \( n \), starting with \( n = 2 \), which is obvious. Suppose that \( n \geq 3 \). We may suppose that \( {G}_{1} \) is generated by commutators. The induction assumption shows that the projection \( H \rightarrow {G}_{2} \times \cdots \times {G}_{n} \) is surjective. Hence we only have to show ... | Yes |
Proposition 2.2. Let \( H \) be a subgroup of \( G \) and let \( S \) be a p-Sylow subgroup of \( G \) . There exists \( g \in G \) such that \( H \cap {gS}{g}^{-1} \) is a p-Sylow subgroup of \( H \) . | Proof. Let \( X = G/S \) . The group \( H \) acts on \( X \) by left multiplication. The \( H \) -stabilizers of the points of \( X \) are of the form \( H \cap {gS}{g}^{-1} \), with \( g \in G \) . Since \( S \) is a \( p \) -Sylow subgroup of \( G \), we have \( p \nmid \left| X\right| \) ; hence there is at least on... | Yes |
Lemma 2.4. \( \left( \begin{matrix} {qm} \\ q \end{matrix}\right) \equiv m\left( {\;\operatorname{mod}\;p}\right) \) . | Proof. Let \( R = {\mathbf{F}}_{p}\left\lbrack t\right\rbrack \) be the polynomial ring in an indeterminate \( t \) over \( {\mathbf{F}}_{p} \) . Since \( q \) is a power of \( p \), we have \( {\left( 1 + t\right) }^{q} = 1 + {t}^{q} \) in \( R \), hence \( {\left( 1 + t\right) }^{qm} = 1 + m{t}^{q} + \ldots \) The co... | Yes |
Corollary 2.5 (Cauchy). If \( p \) divides the order of \( G \), then \( G \) contains an element of order \( p \) . | Proof. Let \( S \) be a \( p \) -Sylow subgroup of \( G \) . Since \( p \) divides \( \left| G\right|, S \) is nontrivial. Let \( x \) be an element of \( S - \{ 1\} \) . The order of \( x \) is \( {p}^{m} \) for some \( m \geq 1 \) . Then \( {x}^{{p}^{m - 1}} \) has order \( p \) . | Yes |
Lemma 2.7. Let \( P \) be a p-group acting on a finite set \( X \) . Let \( {X}^{P} \) be the set of elements of \( X \) fixed by \( P \) . Then \( \left| X\right| \equiv \left| {X}^{P}\right| \left( {\;\operatorname{mod}\;p}\right) \) . | Indeed, every orbit of \( P \) in \( X - {X}^{P} \) has order divisible by \( p \) ; hence \( \left| {X - {X}^{P}}\right| \) is divisible by \( p \) . | Yes |
Lemma 2.8. Let \( S \) and \( {S}^{\prime } \) be two p-Sylow subgroups of \( G \) . If \( {S}^{\prime } \) normalizes \( S \), then \( {S}^{\prime } = S \) . | Proof of the lemma. The groups \( S \) and \( {S}^{\prime } \) are \( p \) -Sylow subgroups of \( {N}_{G}\left( S\right) \) . Since \( S \) is normal in \( {N}_{G}\left( S\right) \), it is the unique \( p \) -Sylow subgroup of \( {N}_{G}\left( S\right) \) . Hence \( {S}^{\prime } = S \) . | Yes |
Corollary 2.9. If \( S \) is a \( p \) -Sylow subgroup of \( G \), then \( \left( {G : {N}_{G}\left( S\right) }\right) \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) . | Proof. Part (2) of th.2.6 imply that \( \left( {G : {N}_{G}\left( S\right) }\right) \) is the number of the \( p \) -Sylow subgroups of \( G \) . By part (3), that number is \( \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) . | Yes |
Proposition 2.10. Let H be a subgroup of \( G \) and let \( P \) be a p-Sylow subgroup of H. There exists a p-Sylow subgroup \( S \) of \( G \) such that \( P = S \cap H \) . | Proof. This follows from prop.2.2. | No |
Proposition 2.11. Let \( H \) be a normal subgroup of \( G \) and let \( S \) be a p-Sylow subgroup of \( G \) . Then :\n\n(1) \( S \cap H \) is a p-Sylow subgroup of \( H \) . | (1) By prop.2.10 there exists a \( p \) -Sylow \( {S}^{\prime } \) of \( G \) such that \( {S}^{\prime } \cap H \) is a \( p \) -Sylow of \( H \) . Since \( S \) and \( {S}^{\prime } \) are conjugate, and \( H \) is normal, \( {S}^{\prime } \cap H \) and \( S \cap H \) are \( G \) -conjugate, hence have the same order.... | Yes |
Corollary 2.12. If \( f : G \rightarrow {G}^{\prime } \) is a surjective homomorphism of finite groups, the p-Sylow subgroups of \( {G}^{\prime } \) are the images of the p-Sylow subgroups of \( G. \) | This is just a reformulation of (2) and (3). | No |
Corollary 2.13. Let \( S \) be a p-Sylow subgroup of \( G \) and let \( H \) be a subgroup of \( G \) containing \( {N}_{G}\left( S\right) \) . Then \( {N}_{G}\left( H\right) = H \) . In particular \( {N}_{G}\left( {{N}_{G}\left( S\right) }\right) = {N}_{G}\left( S\right) \) . | Proof. Let \( K = {N}_{G}\left( H\right) \) . Since \( H \) is normal in \( K \), we may apply part (4) of prop.2.11 to \( K \) . We obtain \( K = H{N}_{K}\left( S\right) \), hence \( K = H \) since \( {N}_{K}\left( S\right) \subset {N}_{G}\left( S\right) \subset H \) . | Yes |
Proposition 2.14. Let \( S \) be a p-Sylow subgroup of \( G \), and let \( {\left( {A}_{i},{B}_{i}\right) }_{i \in I} \) be two families of \( S \) -normal subsets of \( S \) . Let \( g \in G \) be such that \( g{A}_{i}{g}^{-1} = {B}_{i} \) for every \( i \) . Then there exists \( n \in {N}_{G}\left( S\right) \) such t... | Proof. Let \( {N}_{A} = \mathop{\bigcap }\limits_{{i \in I}}{N}_{G}\left( {A}_{i}\right) \) and \( {N}_{B} = \mathop{\bigcap }\limits_{{i \in I}}{N}_{G}\left( {B}_{i}\right) \) . We have \( g{N}_{A}{g}^{-1} = {N}_{B} \) . The hypothesis of \( S \) -normality on the \( {A}_{i} \) and \( {B}_{i} \) means that \( S \subse... | Yes |
Theorem 2.16 (Burnside). Let \( X \) and \( Y \) be two subsets of the center \( Z\left( S\right) \) of \( S \) . Let \( g \in G \) be such that \( {gX}{g}^{-1} = Y \) . Then there exists \( n \in {N}_{G}\left( S\right) \) such that \( {nx}{n}^{-1} = {gx}{g}^{-1} \) for every \( x \in X \) . In particular, \( {nX}{n}^{... | Proof.\n\nWe apply prop.2.14 with \( I = X,{A}_{x} = \{ x\} ,{B}_{x} = \left\{ {{gx}{g}^{-1}}\right\} \) . Since \( {A}_{x} \) and \( {B}_{x} \) are contained in the center of \( S \), they are \( S \) -normal. We thus obtain \( n \in {N}_{G}\left( S\right) \) with \( {nx}{n}^{-1} = {gx}{g}^{-1} \) for every \( x \in X... | Yes |
Theorem 2.19. If \( x, y \in S \) are conjugate in \( G \), then there exists a sequence of elements \( {a}_{0},\ldots ,{a}_{n} \in S \) such that :\n\n(1) \( {a}_{0} = x \) and \( {a}_{n} = y \) .\n\n(2) \( {a}_{i} \) is locally conjugate to \( {a}_{i + 1} \) for \( 0 \leq i \leq n - 1 \) . | Theorem 2.19 follows from the following more general result (in which we write \( {A}^{g} \) instead of \( {g}^{-1}{Ag} \ | No |
Proposition 3.2. Let \( G \) be a group and let \( n \) be an integer \( \geq 1 \) . The following properties are equivalent :\n\n(1) \( G \) is solvable with \( d\ell \left( G\right) \leq n \) .\n\n(2) There exists a sequence \( G = {G}_{0} \supset {G}_{1} \supset \cdots \supset {G}_{n} = 1 \) of characteristic subgro... | Proof. Use induction on \( n \), the case \( n = 1 \) being trivial.\n\n\( \left( 1\right) \Rightarrow \left( 2\right) \) : For \( i \geq 0 \), set \( {G}_{i} = {D}^{i}G \) .\n\n\( \left( 2\right) \Rightarrow \left( {2}^{\prime }\right) \) : This is clear.\n\n\( \left( {2}^{\prime }\right) \Rightarrow \left( 1\right) \... | Yes |
Corollary 3.3. Assume that \( G \) is solvable, finite and nontrivial.\n\n(i) Then there exists a prime number \( p \) and a characteristic nontrivial p-subgroup \( N \) of \( G \) which is elementary abelian. | Proof of \( \left( i\right) \) . If \( d\ell \left( G\right) = 1, G \) is abelian; if \( p \) is a prime divisor of \( \left| G\right| \), choose for \( N \) the set of elements \( x \in G \) such that \( {x}^{p} = 1 \) . If \( d\ell \left( G\right) > 1 \), choose an abelian subgroup \( A \) of \( G \) having property ... | No |
Proposition 3.4. Let \( N \) be a normal subgroup of \( G \). If \( N \) and \( G/N \) are solvable, so is \( G \), and \( d\ell \left( G\right) \leq d\ell \left( N\right) + d\ell \left( {G/N}\right) \) . | Proof. Let \( i = d\ell \left( N\right) \) and \( j = d\ell \left( {G/N}\right) \). We have \( {D}^{j}\left( G\right) \subset N \), and \( {D}^{i}\left( N\right) = 1 \), hence \( {D}^{i + j}\left( G\right) = {D}^{i}\left( {{D}^{j}G}\right) = 1 \). The proposition follows. | Yes |
Proposition 3.5. Let \( G \) be a finite group and let \( G = {G}_{0} \supset {G}_{1} \supset \cdots \supset {G}_{n} = 1 \) be a Jordan-Hölder filtration of \( G \) . The group \( G \) is solvable if and only if \( {G}_{i}/{G}_{i + 1} \) is a cyclic group of prime order, for \( 0 \leq i \leq n - 1 \) . | Proof. Note that \( {G}_{i}/{G}_{i + 1} \) is both simple and solvable, hence its derived group is trivial, which implies that it is cyclic of prime order. | No |
Proposition 3.6. We have \( \left( {{C}^{i}G,{C}^{j}G}\right) \subset {C}^{i + j}G \) for all \( i \geq 1 \) and \( j \geq 1 \) . | Proof. Use induction on \( i \) . The proposition is clear when \( i = 1 \) and \( j \geq 1 \) . Fix \( j \geq 1 \) . We have \( \left( {{C}^{i}G,{C}^{j}G}\right) = \left( {\left( {G,{C}^{i - 1}G}\right) ,{C}^{j}G}\right) \) . But \( \left( {\left( {{C}^{i - 1}G,{C}^{j}G}\right), G}\right) \) is contained in \( \left( ... | Yes |
Lemma 3.7. If \( X, Y \) and \( Z \) are normal subgroups of \( G \) and if \( H \) is a subgroup of \( G \) containing \( \left( {\left( {Y, Z}\right), X}\right) \) and \( \left( {\left( {Z, X}\right), Y}\right) \), then \( H \) contains \( \left( {\left( {X, Y}\right), Z}\right) \) . | Proof. After dividing \( G \) by the subgroup generated by \( \left( {\left( {Y, Z}\right), X}\right) \) and \( \left( {\left( {Z, X}\right), Y}\right) \) , we may assume that \( \left( {\left( {Y, Z}\right), X}\right) = 1 = \left( {\left( {Z, X}\right), Y}\right) \), and we have to prove that \( \left( {\left( {X, Y}\... | Yes |
Proposition 3.8. A nilpotent group is solvable. | Proof. Indeed, for every \( n \geq 0 \), we have \( {D}^{n}G \subset {C}^{{2}^{n}}G \) ; hence \( {D}^{n}G = 1 \) when \( n \) is large enough. | No |
Proposition 3.9. A group \( G \) is nilpotent of nilpotency class \( \leq n \), with \( n > 0 \), if and only if \( G \) is a central extension of a nilpotent group \( \Gamma \) of nilpotency class \( \leq n - 1 \) . | Proof. If \( G \) has nilpotency class \( \leq n \), then \( {C}^{n + 1}G = 1 \), i.e., \( {C}^{n}G \) is contained in the center of \( G \) . Define \( \Gamma = G/{C}^{n}G \) . We have \( {C}^{n}\Gamma = 1 \) ; therefore, \( \Gamma \) is nilpotent of nilpotency class \( \leq n - 1 \) .\n\nConversely, if such an exact ... | Yes |
Corollary 3.12. Let \( G \) be a nilpotent group and let \( H \) be a proper subgroup of \( G \). Then : a) \( {N}_{G}\left( H\right) \neq H \). b) The following properties are equivalent : (i) \( H \) is maximal among the proper subgroups of \( G \). (ii) The index of \( H \) in \( G \) is a prime number. (iii) \( H \... | Proof of a). Use induction on the nilpotency class \( n \) of \( G \). If \( n = 1 \), then \( G \) is abelian and \( {N}_{G}\left( H\right) = G \), hence \( {N}_{G}\left( H\right) \neq H \). If \( n \geq 2 \), let \( A = Z\left( G\right) \). The group \( G/A \) is nilpotent of nilpotency class \( n - 1 \). We have \( ... | Yes |
Proposition 3.13. Let \( G \) be a group. The following properties are equivalent :\n\n(1) \( G \) is nilpotent.\n\n(2) \( G \) has a central filtration. | Proof. We have \( \left( 1\right) \Rightarrow \left( 2\right) \) by taking \( {G}_{i} = {C}^{i}G \) . Conversely, if \( \left( {G}_{i}\right) \) is a central filtration of \( G \), we have \( {C}^{i}G \subset {G}_{i} \) for all \( i \), hence \( {C}^{i}G = 1 \) for large enough \( i \) . | No |
Theorem 3.14. Let \( G \) be a nilpotent group.\n\n(a) Let \( \varphi : H \rightarrow G \) be a homomorphism of a group \( H \) into \( G \) . If \( H\overset{\varphi }{ \rightarrow }G \rightarrow {G}^{\text{ab }} \) is surjective, then \( \varphi \) is surjective. | Proof of \( \left( a\right) \) . Define \( {\operatorname{gr}}_{n}\left( G\right) = {C}^{n}G/{C}^{n + 1}G \) and \( {\operatorname{gr}}_{n}\left( H\right) = {C}^{n}H/{C}^{n + 1}H \), as above. The map \( \varphi : H \rightarrow G \) defines for every \( n \) a homomorphism \( {\varphi }_{n} : {C}^{n}H \rightarrow {C}^{... | Yes |
Lemma 3.15. Let \( N \) be a normal subgroup of a nilpotent group \( G \) . Then \( N \cap Z\left( G\right) \neq 1 \) if \( N \neq 1 \) . | Proof of lemma 3.15. Let \( n \) be the largest integer such that the group \( {N}_{n} = N \cap {C}^{n}G \) is nontrivial. The group \( \left( {G,{N}_{n}}\right) \) is contained in \( N \cap {C}^{n + 1}G \), hence is trivial. This shows that \( {N}_{n} \) is contained in the center of \( G \), hence \( N \cap Z\left( G... | Yes |
Theorem 3.16 (E. Kolchin). Let \( V \) be a finite dimensional vector space over a field \( K \) and let \( G \) be a subgroup of \( \mathbf{GL}\left( V\right) \). Suppose that \( g - 1 \) is nilpotent for every \( g \in G^2 \). Then there exists a complete flag (see example 3 of §3.2) of \( V \) such that \( G \) is c... | Proof. Use induction on the dimension \( n \) of \( V \). The case \( n = 0 \) is trivial. Suppose that \( n \geq 1 \) and let us show first that there exists a nonzero \( x \in V \) which is \( G \)-invariant. Finding such an \( x \) means solving a family of linear equations with coefficients in \( K \); it is well k... | Yes |
Proposition 3.17. Every p-group is nilpotent. | First proof. Let \( P \) be a \( p \) -group. Then we can embed \( P \) in \( {\mathbf{{GL}}}_{n}\left( {\mathbf{Z}/p\mathbf{Z}}\right) \) for some sufficiently large integer \( n \) . Thus, \( P \) is contained in a \( p \) -Sylow subgroup of \( {\mathbf{{GL}}}_{n}\left( {\mathbf{Z}/p\mathbf{Z}}\right) \) ; as we have... | Yes |
Theorem 3.18. Let \( G \) be a finite p-group.\n\n(1) If \( H \) is a proper subgroup of \( G \), then \( {N}_{G}\left( H\right) \neq H \) . | Proof. (1) follows from the fact that \( G \) is nilpotent, cf. cor.3.12; it implies (2),(3) and (4). As for (5), it follows from cor.3.10. | No |
Theorem 3.20. Let \( G \) be a finite group. The following statements are equivalent :\n\n(1) \( G \) is nilpotent.\n\n(2) \( G \) is a product of p-groups, for a suitable finite set of primes.\n\n(3) For every prime \( p, G \) has a unique p-Sylow subgroup.\n\n(4) Let \( p \) and \( {p}^{\prime } \) be two distinct pr... | Proof.\n\n\( \\left( 2\\right) \\Rightarrow \\left( 1\\right) \) : This is corollary 3.19 proved above.\n\n\( \\left( 1\\right) \\Rightarrow \\left( 3\\right) \) : Let \( S \) be a \( p \) -Sylow subgroup of \( G \) and let \( N = {N}_{G}\\left( S\\right) \) . Then \( N \) is its own normalizer (cf. cor.2.13). Since \(... | Yes |
Corollary 3.21. If \( G \) is nilpotent, and if \( d \) is a divisor of \( \left| G\right| \), there exists a subgroup of \( G \) of order \( d \) . | Proof. Because of (2), it is enough to prove this when \( G \) is a \( p \) -group, in which case it follows from part (4) of th.3.18. | No |
Theorem 3.22. Every finite abelian p-group is a product of cyclic groups. | Note that, if \( G \) is a finite abelian \( p \) -group, there exists an integer \( n \) such that \( {p}^{n}x = 0 \) for all \( x \in G \) ; hence \( G \) can be viewed as a \( \mathbf{Z}/{p}^{n}\mathbf{Z} \) -module. Theorem 3.22 is thus a special case of the following one, where there is no finiteness assumption: | No |
Theorem 3.23. Every \( \mathbf{Z}/{p}^{n}\mathbf{Z} \) -module is a direct sum of cyclic modules (isomorphic to \( \mathbf{Z}/{p}^{i}\mathbf{Z} \) for some \( i \leq n \) ). | Proof. Let \( M \) be a \( \mathbf{Z}/{p}^{n}\mathbf{Z} \) -module and let \( V \) be the \( {\mathbf{F}}_{p} \) -vector space \( M/{pM} \) .\n\nFor \( i = 0,\ldots, n \), let \( {M}_{i} \) be the set \( \left\{ {x \in M \mid {p}^{i}x = 0}\right\} \) . We have:\n\n\[ \n{M}_{0} = 0 \subset {M}_{1} \subset \cdots \subset... | Yes |
Proposition 3.24. Let \( S \subset G \) and let \( H \) be the subgroup of \( G \) generated by \( S \) . We have \( H = G \) if and only if \( H.\Phi \left( G\right) = G \), i.e., if \( S \) generates \( G/\Phi \left( G\right) \) . | Proof. One direction is obvious. For the nontrivial direction, if \( H.\Phi \left( G\right) = G \) and \( H \neq G \) , there exists a maximal subgroup \( {H}^{\prime } \) of \( G \) containing \( H \) ; by definition, \( \Phi \left( G\right) \subset {H}^{\prime } \) . It follows that \( G \subset {H}^{\prime } \), whi... | Yes |
Theorem 3.25. The group \( \Phi \left( G\right) \) is nilpotent. | Proof. Let \( p \) be a prime number and let \( S \) be a \( p \) -Sylow subgroup of \( \Phi \left( G\right) \) . By part (4) of prop.2.11, we have \( G = \Phi \left( G\right) .{N}_{G}\left( S\right) \) . According to prop.3.24, this implies \( {N}_{G}\left( S\right) = G \) ; hence \( S \) is normal in \( G \), and a f... | Yes |
Theorem 3.26. If \( G \) is nilpotent, \( \Phi \left( G\right) \) is the intersection of the normal subgroups of \( G \) of prime index. | Proof. This follows from the characterization of the maximal subgroups given in cor.3.12. | No |
Theorem 3.27. If \( G \) is a p-group, then \( \Phi \left( G\right) = \left( {G, G}\right) .{G}^{p} \) ; it is the intersection of the normal subgroups of \( G \) of index \( p \) . | Proof. The second assertion is a special case of the theorem above. As for the first one, the inclusion \( \left( {G, G}\right) .{G}^{p} \subset \Phi \left( G\right) \) is clear. Conversely, if \( g \notin \left( {G, G}\right) .{G}^{p} \), its image \( v \) in the \( {\mathbf{F}}_{p} \) -vector space \( V = G/\left( {G... | Yes |
Proposition 3.29. (1) \( {P}_{G} \) is a p-group.\n\n(2) Let \( {p}^{m} \) be the order of \( \Phi \left( G\right) \) . We have \( {s}^{{p}^{m}} = 1 \) for every \( s \in {P}_{G} \) . | Proof. It is enough to prove (2). We do it by induction on \( m \) . If \( m = 0 \), then \( G = V \) and \( s = 1 \) by assumption. Assume that \( m > 0 \), i.e., \( \Phi \left( G\right) \neq 1 \) . By lemma 3.15, the abelian group \( A = \Phi \left( G\right) \cap Z\left( G\right) \) is nontrivial. Let \( B \) be the ... | Yes |
Corollary 3.30. If \( C \) is a subgroup of \( \operatorname{Aut}\left( G\right) \) of order prime to \( p \), then \( C \) acts faithfully on \( {V}_{G} = G/\Phi \left( G\right) \), and its order divides \( \mathop{\prod }\limits_{{i = 1}}^{r}\left( {{p}^{i} - 1}\right) \) . | Proof. The first assertion follows from \( C \cap {P}_{G} = 1 \) . It implies that \( \left| C\right| \) divides the order of \( \operatorname{GL}\left( {V}_{G}\right) \), which is \( {p}^{r\left( {r - 1}\right) /2}\mathop{\prod }\limits_{{i = 1}}^{r}\left( {{p}^{i} - 1}\right) \) ; since \( \left| C\right| \) is prime... | Yes |
Proposition 3.31. Let \( G \) be a group (resp. a finite group). Suppose that every subgroup of \( G \) generated by two elements is abelian (resp. nilpotent). Then \( G \) is abelian (resp. nilpotent). | Proof. Let \( x, y \in G \) . The subgroup \( \langle x, y\rangle \) generated by \( x \) and \( y \) is abelian, therefore \( {xy} = {yx} \), hence \( G \) is abelian. If \( G \) is a finite group and \( \langle x, y\rangle \) is nilpotent for every \( x, y \) , characterization (5) of th.3.20 implies that \( G \) is ... | Yes |
Lemma 3.32. If a finite group \( G \) is not solvable, there exist a subgroup \( H \) of \( G \) and a normal subgroup \( K \) of \( H \) such that \( H/K \) is a minimal simple group. | Proof of the lemma. Let \( H \) be a minimal nonsolvable subgroup of \( G \) and let \( K \) be a maximal normal subgroup of \( H \) distinct from \( H \) . The group \( K \) is solvable, since it is strictly contained in \( H \) . The quotient \( H/K \) is simple, because \( K \) is maximal among the proper normal sub... | Yes |
Proposition 3.33. The following two statements are equivalent :\n\n(1) Every minimal simple group can be generated by two elements.\n\n(2) Every finite group, all of whose 2-generated subgroups are solvable, is also solvable. | Proof of (2) \( \Rightarrow \) (1). Suppose that (2) holds. Let \( G \) be a minimal simple group. If \( \langle x, y\rangle \neq G \) for all pairs \( \left( {x, y}\right) \in G \times G \), then \( \langle x, y\rangle \) is solvable since it is a proper subgroup of \( G \) . Now, according to (2), \( G \) is solvable... | Yes |
Theorem 4.1 (Basic Formula). We have \( d \circ d = 0 \) . In other words, the following composition is zero : | Proof. We prove this for \( n = 0 \) and \( n = 1 \) ; for the general case, see exerc.1. For \( n = 0 \) , and \( a \in A = {C}^{0}\left( {G, A}\right) \), we have \( {da}\left( s\right) = {sa} - a \) . It follows that\n\n\[ {dda}\left( {s, t}\right) = {sda}\left( t\right) - {da}\left( {st}\right) + {da}\left( s\right... | No |
Lemma 4.2. Every factor set is cohomologous to a normalized factor set. | Proof. Let \( f \) be a factor set. Choose \( c : G \rightarrow A \) such that \( c\left( 1\right) = - f\left( {1,1}\right) \) . Then \( f + {dc} \) is a normalized factor set which is cohomologous to \( f \) . | Yes |
Theorem 4.3. If \( x \in {H}^{n}\left( {G, A}\right) \) with \( n \geq 1 \), then \( {mx} = 0 \) . | Proof. Let \( f \in {Z}^{n}\left( {G, A}\right) \) be an \( n \) -cocycle in the class \( x \) . Let us show that \( {mf} \) is a coboundary. Since \( f \) is a cocycle, we have:\n\n\[ 0 = {s}_{1}f\left( {{s}_{2},\ldots ,{s}_{n + 1}}\right) - f\left( {{s}_{1}{s}_{2},{s}_{3},\ldots ,{s}_{n + 1}}\right) + \cdots + {\left... | Yes |
Corollary 4.4. If the map : \( a \mapsto {ma} \) is an automorphism of \( A \), then \( {H}^{n}\left( {G, A}\right) = 0 \) for all \( n \geq 1 \) . | Proof. Indeed, \( x \mapsto {mx} \) is an automorphism of \( {C}^{n}\left( {G, A}\right) \) that commutes with \( d \) . Therefore, we obtain an automorphism of \( {H}^{n}\left( {G, A}\right) \) by passing to the quotient. But the map is zero; hence, \( {H}^{n}\left( {G, A}\right) = 0 \) . | Yes |
Corollary 4.5. If \( A \) is finite of order prime to \( m \), then \( {H}^{n}\left( {G, A}\right) = 0 \) for all \( n \geq 1 \) . | Proof. This follows from cor.4.4. | No |
Corollary 4.6. If \( A \) is finitely generated, then \( {H}^{n}\left( {G, A}\right) \) is finite for all \( n \geq 1 \) . | Proof. The cochain groups \( {C}^{n}\left( {G, A}\right) \) are finitely generated, hence the same is true for the \( {H}^{n}\left( {G, A}\right) \) . For \( n > 0 \), we have \( m.{H}^{n}\left( {G, A}\right) = 0 \) ; hence, the \( {H}^{n}\left( {G, A}\right) \) are finitely generated \( \mathbf{Z}/m\mathbf{Z} \) -modu... | Yes |
Proposition 4.7. The map \( {f}_{h} \) is a 2-cocycle on \( G \) with values in \( A \) . | Proof. Let us check that \( d{f}_{h} = 0 \) . In multiplicative notation, this means \( d{f}_{h}\left( {x, y, z}\right) = 1 \) for all \( x, y, z \in G \) . Let us write \( h\left( x\right) h\left( y\right) h\left( z\right) \) in two ways:\n\n\[ \nh\left( x\right) h\left( y\right) .h\left( z\right) = {f}_{h}\left( {x, ... | Yes |
Proposition 4.8. If \( h \) and \( {h}^{\prime } \) are two sections of \( \pi \), the cocycles \( {f}_{h} \) and \( {f}_{{h}^{\prime }} \) are cohomologous. Conversely, every 2-cocycle which is cohomologous to \( {f}_{h} \) is equal to some \( {f}_{{h}^{\prime }} \) . | Proof. For every \( x \in G \) we have \( {h}^{\prime }\left( x\right) = \ell \left( x\right) h\left( x\right) \), where \( \ell \) is a map of \( G \) into \( A \) . Let us compute \( {f}_{{h}^{\prime }} \) in terms of \( \ell \) and \( {f}_{h} \) . We have\n\n\[ \n{h}^{\prime }\left( x\right) {h}^{\prime }\left( y\ri... | Yes |
Proposition 4.9. Two extensions \( {E}_{1} \) and \( {E}_{2} \) of \( G \) by \( A \) are isomorphic if and only if \( \left\lbrack {E}_{1}\right\rbrack = \left\lbrack {E}_{2}\right\rbrack \) in \( {H}^{2}\left( {G, A}\right) \) . | Proof. The \ | No |
Proposition 4.10. Every element of \( {H}^{2}\left( {G, A}\right) \) is equal to \( \left\lbrack E\right\rbrack \), for a suitable extension \( E \) of \( G \) by \( A \) . | Proof. Here it is convenient to write \( A \) additively. Let \( \alpha \) be an element of \( {H}^{2}\left( {G, A}\right) \) , and let \( f \) be a normalized 2-cocycle representing \( \alpha \) . Let \( E = A \times G \) and define a multiplication law on \( E \) by\n\n\[ \left( {a, x}\right) \cdot \left( {b, y}\righ... | Yes |
Corollary 4.12. If \( G \) and the \( G \) -module \( A \) are finite of relatively prime orders, then every extension of \( G \) by \( A \) splits. | Proof. Indeed, we have \( {H}^{2}\left( {G, A}\right) = 0 \), cf. cor.4.5. | No |
Theorem 4.13. The conjugacy classes (by elements of \( A \) or, equivalently, of \( G \) ) of splittings of \( E \) are in bijective correspondence with the elements of the cohomology group \( {H}^{1}\left( {G, A}\right) \) . | [Note that the correspondence depends on the choice of the splitting \( h \) . A more precise statement is that the splittings make up a \( {Z}^{1}\left( {G, A}\right) \) -torsor, and their conjugacy classes make up an \( {H}^{1}\left( {G, A}\right) \) -torsor.] | Yes |
Corollary 4.15. If \( G \) and \( A \) are finite of relatively prime orders, and if \( E \) is an extension of \( G \) by \( A \), then every two splittings of \( E \) are conjugate. | Proof. Indeed, we have \( {H}^{1}\left( {G, A}\right) = 0 \), cf. cor.4.5. | No |
Theorem 4.16. There exists an extension of \( G \) by \( A \) corresponding to \( \psi \) is and only if \( c\left( \psi \right) = 0. \) | The construction of \( c\left( \psi \right) \) is as follows: select a map \( s : G \rightarrow \operatorname{Aut}\left( A\right) \) such that the composite map \( G\overset{s}{ \rightarrow }\operatorname{Aut}\left( A\right) \rightarrow \operatorname{Out}\left( A\right) \) is \( \psi \) . For every \( x, y \in G \), ch... | Yes |
Theorem 4.17. The action of \( {H}^{2}\left( {G, Z\left( A\right) }\right) \) on \( \operatorname{Ext}\left( {G, A,\psi }\right) \) is free; it is transitive if \( \operatorname{Ext}\left( {G, A,\psi }\right) \) is not empty. | Let us prove that the action is free. With the notation above, we have to show that, if \( {E}_{f} \simeq E \), then \( f \) is a coboundary. Let \( \varphi : {E}_{f} \rightarrow E \) be an isomorphism (of extensions). We may write \( \varphi \) as \( e \mapsto r\left( e\right) e \), where \( r \) is a map of \( E \) i... | Yes |
Corollary 4.21. Let \( S \) be a p-Sylow subgroup of a group \( G \), and let \( N = {N}_{G}\left( S\right) \) . The extension \( 1 \rightarrow S \rightarrow N \rightarrow N/S \rightarrow 1 \) splits. | Proof. This follows from the fact that \( \left| S\right| \) and \( \left| {N/S}\right| \) are relatively prime. | No |
Corollary 4.22. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a surjective homomorphism of finite groups. There exists a subgroup \( {G}_{1} \) of \( G \) such that \( \varphi \left( {G}_{1}\right) = {G}^{\prime } \) and such that the prime divisors of \( \left| {G}_{1}\right| \) and of \( \left| {G}^{\prime }\rig... | Proof. Use induction on \( \left| G\right| \) . It is enough to show that, if a prime number \( p \) does not divide \( \left| {G}^{\prime }\right| \), there is a subgroup \( H \) of \( G \), of order prime to \( p \), such that \( \varphi \left( H\right) = {G}^{\prime } \) . Let \( S \) be a \( p \) -Sylow of \( G \),... | Yes |
Proposition 4.23. The liftings of \( \varphi \) are in one-one correspondence with the splittings of \( {E}_{\varphi } \) . | Proof. This is clear on the set-theoretic level: a map \( \psi : G \rightarrow E \) such that \( \pi \circ \psi = \varphi \) corresponds to a map \( {\psi }_{0} : G \rightarrow {E}_{\varphi } \) such that the composite \( G \rightarrow {E}_{\varphi } \rightarrow G \) is the identity. Moreover, \( \psi \) is a homomorph... | Yes |
Theorem 4.25. Let \( G \) be a finite group of order prime to \( p \), and let \( \rho : G \rightarrow {\mathrm{{GL}}}_{N}\left( {\mathbf{F}}_{p}\right) \) be a homomorphism. There exists a homomorphism \( {\rho }_{0} : G \rightarrow {\mathrm{{GL}}}_{N}\left( {\mathbf{Z}}_{p}\right) \) which lifts \( \rho \) . | Proof. Let us first show that \( {\rho }_{1} = \rho \) can be lifted to \( {\rho }_{2} : G \rightarrow {\mathrm{{GL}}}_{N}\left( {\mathbf{Z}/{p}^{2}\mathbf{Z}}\right) \) . We have the exact sequence\n\n\[ 1 \rightarrow A \rightarrow {\mathrm{{GL}}}_{N}\left( {\mathbf{Z}/{p}^{2}\mathbf{Z}}\right) \rightarrow {\mathrm{{G... | Yes |
Lemma 5.5. If \( A \) and \( B \) are two subgroups of a group \( G \), the following properties are equivalent :\n\n(1) \( {AB} = {BA} \) .\n\n(2) \( {AB} \) is a subgroup of \( G \) . | Proof of \( \left( 1\right) \Rightarrow \left( 2\right) \) . If \( {AB} = {BA} \), we have \( {ABAB} = {AABB} = {AB} \) and \( {\left( AB\right) }^{-1} = \) \( {BA} = {AB} \) ; thus, \( {AB} \) is a subgroup of \( G \) .\n\nProof of \( \left( 2\right) \Rightarrow \left( 1\right) \) . If \( {AB} \) is a subgroup of \( G... | Yes |
Lemma 5.6. Let \( {A}_{1},\ldots ,{A}_{n} \) be subgroups of a group \( G \) that are pairwise permutable. Then \( {A}_{1}\cdots {A}_{n} \) is a subgroup of \( G \) . | Proof. The proof is by induction on \( n \) . Lemma 5.5 gives the case \( n = 2 \) . According to the induction hypothesis, \( {A}_{1}\cdots {A}_{n - 1} \) is a subgroup. It is permutable with \( {A}_{n} \) since \( {A}_{1}\cdots {A}_{n - 1}.{A}_{n} = {A}_{1}\cdots {A}_{n}.{A}_{n - 1} = \cdots \) (iterating \( \left( {... | Yes |
Proposition 5.7. Let \( A \) and \( B \) be two subgroups of a group \( G \) . The following properties are equivalent :\n\n(1) \( {AB} = G \) .\n\n\( \left( {1}^{\prime }\right) A \) acts transitively on \( G/B \) .\n\n(2) \( {BA} = G \) .\n\n\( \left( {2}^{\prime }\right) B \) acts transitively on \( G/A \) .\n\n(3) ... | Proof. The equivalences \( \left( 1\right) \Leftrightarrow \left( {1}^{\prime }\right) ,\left( 2\right) \Leftrightarrow \left( {2}^{\prime }\right) \), and \( \left( 3\right) \Leftrightarrow \left( {3}^{\prime }\right) \) are clear.\n\n(1) \( \Leftrightarrow \) (2). If \( {AB} = G \), then \( {AB} \) is a subgroup of \... | Yes |
Corollary 5.8. If the indices of \( A \) and \( B \) in \( G \) are finite and relatively prime, then the properties (1) to \( \left( {4}^{\prime }\right) \) of prop. 5.7 hold. | Proof. Indeed, \( \left( {G : A \cap B}\right) \) is divisible by \( \left( {G : A}\right) \) and \( \left( {G : B}\right) \), and hence by their product; this shows that property \( \left( {4}^{\prime }\right) \) holds. | No |
Theorem 5.9. If \( G \) is solvable, \( G \) has a permutable family of Sylow subgroups. | Proof. The proof is by induction on \( \left| G\right| \) . The case \( G = 1 \) is trivial. Assume \( G \neq 1 \) . By cor.3.3, there exist a prime \( {p}_{0} \) and a nontrivial normal \( {p}_{0} \) -subgroup \( A \) of \( G \) . By the induction hypothesis, the group \( G/A \) has a permutable family \( \left\{ {H}_... | Yes |
Proposition 5.10. Let \( G \) be a finite solvable group. Let \( H \) be a subgroup of \( G \) and let \( P \) be a \( \pi \) -Sylow subgroup of \( H \) . There exists a \( \pi \) -Sylow subgroup \( S \) of \( G \) such that \( P = S \cap H \) . | Proof. This follows from part (2) of th.5.1 applied to \( P \) . | No |
Proposition 5.11. Let \( G \) be a finite solvable group. Let \( H \) be a normal subgroup of \( G \) and let \( S \) be a \( \pi \) -Sylow subgroup of \( G \) . Then :\n\n(1) \( S \cap H \) is a \( \pi \) -Sylow subgroup of \( H \) .\n\n(2) The image \( {S}^{\prime } \) of \( S \) in \( G/H \) is a \( \pi \) -Sylow su... | Proof.\n\n(1) By prop.5.10 there exists a \( p \) -Sylow \( {S}^{\prime } \) of \( G \) such that \( {S}^{\prime } \cap H \) is a \( \pi \) -Sylow subgroup of \( H \) . Since \( S \) and \( {S}^{\prime } \) are conjugate, and \( H \) is normal, \( {S}^{\prime } \cap H \) and \( S \cap H \) are \( G \) -conjugate, hence... | Yes |
Theorem 5.13. If \( G \) has a p-complement for every prime \( p \), then \( G \) is solvable. | Proof.\n\nUse induction on \( \left| G\right| \) . By Burnside’s theorem (th.5.4), we may assume that \( \left| G\right| \) has at least three distinct prime factors, \( {p}_{1},{p}_{2},{p}_{3} \) . For \( i = 1,2,3 \), let \( {H}_{i} \) be a \( {p}_{i} \) -complement of \( G \) . Then the indices \( \left( {G : {H}_{i... | Yes |
Lemma 6.1. Let \( H \) be a subgroup of a finite group \( G \) . Let \( {H}^{\mathrm{{cl}}} \) be the union of the conjugates of \( H \) . Then :\na) \( \left| G\right| - \left| {H}^{\mathrm{{cl}}}\right| \geq n - 1 \), where \( n = \left( {G : H}\right) \).\nb) There is equality in a) if and only if \( H \cap {gH}{g}^... | Proof. Let \( X = G/H \) ; we have \( \left| X\right| = n \) ; if \( x \in X \), denote by \( {G}_{x} \) the stabilizer of \( x \) . We\n\nhave\n\n\[ \n{H}^{\mathrm{{cl}}} = \{ 1\} \cup \mathop{\bigcup }\limits_{{x \in X}}\left( {{G}_{x}-\{ 1\} }\right) .\n\]\n\nSince \( {G}_{x} \) is a conjugate of \( H \), this gives... | Yes |
Theorem 6.2 (Jordan [61]). Let \( X \) be a finite set on which a group \( G \) acts transitively. Assume \( \left| X\right| \geq 2 \) . Then there exists an element of \( G \) which fixes no element of \( X \) . | Proof. By replacing \( G \) by its image in \( {\mathcal{S}}_{X} \) we may assume that \( G \) is finite. Let \( H \) be the stabilizer of a point of \( X \) . Part a) of lemma 6.1 shows that \( \left| G\right| - \left| {H}^{\mathrm{{cl}}}\right| > 0 \), hence that there exists an element of \( G \) which is not conjug... | Yes |
Theorem 6.6. Let \( p \) be a prime number. The following properties are equivalent :\n\n(1) There exists an element of \( G \), of order a power of \( p \), which is not conjugate to any element of \( H \) .\n\n\( \left( {1}^{\prime }\right) \) There exists a cyclic subgroup \( C \) of \( G \) such that the order of e... | Note. The equivalence of (1) and \( \left( {1}^{\prime }\right) \) is easy. That of \( \left( {1}^{\prime }\right) \) and \( \left( {1}^{\prime \prime }\right) \) is a consequence of Chebotarev’s density theorem, cf. [65]. That of \( \left( {1}^{\prime \prime }\right) \),(2) and (3) uses standard properties of number f... | No |
Theorem 6.7 (Frobenius [19], p.199). Let \( \left( {G, H}\right) \) be a Frobenius pair. Then the identity, together with the elements of \( G \) which do not belong to any conjugate of \( H \), make up a normal subgroup \( N \) of \( G \) such that \( N \cap H = 1 \) and \( G = {NH} \) . | Proof. The main point is to show that the set \( N = \{ 1\} \cup \left( {G - {H}^{\mathrm{{cl}}}}\right) \) is a subgroup of \( G \) ; the proof, which uses character theory, will be given in chap. 8 (th.8.54). The other statements are easy:\n\nThe group \( N \) is obviously invariant under conjugation; it is also clea... | No |
Proposition 6.9. Let \( \sigma \) be an automorphism of order \( p \) (not necessarily prime) of a finite group \( N \) . Suppose that 1 is the only fixed point of \( \sigma \) . Then :\n\n(1) The map \( x \mapsto {x}^{-1}\sigma \left( x\right) \) of \( N \) into \( N \) is bijective.\n\n(2) If \( x \) and \( \sigma \l... | Proof.\n\n(1) Since \( N \) is finite, it is enough to show that the map \( x \mapsto {x}^{-1}\sigma \left( x\right) \) is injective. Suppose that \( {x}^{-1}\sigma \left( x\right) = {y}^{-1}\sigma \left( y\right) \) with \( x, y \in N \) . Thus, \( y{x}^{-1} = \sigma \left( {y{x}^{-1}}\right) \) . Therefore, the eleme... | Yes |
Corollary 6.10. If \( \ell \) is a prime number, there exists an \( \ell \) -Sylow subgroup of \( N \) which is stable under the action of \( \sigma \) . | Proof. Let \( S \) be an \( \ell \) -Sylow subgroup of \( N \) . The group \( \sigma \left( S\right) \) is also an \( \ell \) -Sylow subgroup of \( N \), hence there exists \( a \in N \) such that \( {aS}{a}^{-1} = \sigma \left( S\right) \) . According to prop.6.9 (1), we can write \( {a}^{-1} \) in the form \( {a}^{-1... | Yes |
If \( a \in N \), the automorphism \( {\sigma }_{a} = {\operatorname{int}}_{a} \circ \sigma : x \mapsto {a\sigma }\left( x\right) {a}^{-1} \) is conjugate to \( \sigma \) in \( \operatorname{Aut}\left( G\right) \) ; in particular, it is of order \( p \) and it has no fixed point \( \neq 1 \) . | Proof. According to prop.6.9(1), there exists \( b \in G \) such that \( a = {b}^{-1}\sigma \left( b\right) \) ; then \( {\sigma }_{a}\left( x\right) = {b}^{-1}\sigma \left( {{bx}{b}^{-1}}\right) b \), hence \( b{\sigma }_{a}\left( x\right) {b}^{-1} = \sigma \left( {{bx}{b}^{-1}}\right) \), i.e., \( {\operatorname{int}... | Yes |
Theorem 6.12. Every Frobenius kernel is nilpotent. | For a proof, see [78] and Huppert [25], Kap.V, Haupsatz 8.14. | No |
Theorem 6.13. Let \( H \) be a finite group. The following properties are equivalent :\n\n(1) \( H \) has property \( \mathcal{F} \) .\n\n(2) There exist a field \( K \) and a linear representation \( {}^{4}\rho : H \rightarrow {\mathbf{{GL}}}_{n}\left( K\right) \), with \( n \geq 1 \), which is almost free, i.e., such... | Proof of theorem 6.13. First part.\n\nIf \( K \) is a field, we shall say that it has \ | No |
Corollary 6.14. If \( H \) has property \( \mathcal{F} \), its abelian subgroups are cyclic. | Proof. This means that an abelian group having property \( \mathcal{F} \) is cyclic, which is clear from (4), since an irreducible representation of an abelian group is one-dimensional, cf. chap.8, cor.8.14. | No |
Theorem 7.1. The map \( \operatorname{Ver} : G \rightarrow {H}^{\text{ab }} \) defined above is a homomorphism and it does not depend on the choice of the section \( \varphi \) . | Proof. Let us show first that Ver does not depend on the choice of \( \varphi \) . Let \( {\varphi }^{\prime } \) be another section; we have \( {\varphi }^{\prime }\left( x\right) = \varphi \left( x\right) \theta \left( x\right) \), where \( \theta \) is a map of \( X \) into \( H \) . If \( g \in G \), we have\n\n\[ ... | Yes |
Theorem 7.2. Let \( n = \left( {G : H}\right) \) and let \( i : {H}^{\text{ab }} \rightarrow {G}^{\text{ab }} \) be the map deduced from the injection \( H \rightarrow G \) . The composite map \( {G}^{\text{ab }}\overset{\text{ Ver }}{ \rightarrow }{H}^{\text{ab }}\overset{i}{ \rightarrow }{G}^{\text{ab }} \) is \( g \... | Proof. Let us keep the notation of the proof of the theorem above. If we make the product (in the abelian group \( {G}^{\mathrm{{ab}}} \) ) of the equations \( {g\varphi }\left( x\right) = \varphi \left( {gx}\right) {h}_{g, x}^{\varphi } \), we find \( {g}^{n}\prod \varphi \left( x\right) = \prod \varphi \left( {gx}\ri... | Yes |
Theorem 7.5. We have :\n\n\[ \operatorname{Ver}\left( g\right) = \mathop{\prod }\limits_{\alpha }{h}_{\alpha } = \mathop{\prod }\limits_{\alpha }{z}_{\alpha }^{-1}{g}^{{f}_{\alpha }}{z}_{\alpha }{\;\operatorname{mod}\;D}\left( H\right) . \] | Proof. Let \( \varphi : G/H \rightarrow G \) be the section defined by \( \varphi \left( {{g}^{i}{x}_{\alpha }}\right) = {g}^{i}{z}_{\alpha } \) for \( i = 0,1,\ldots ,{f}_{\alpha } - 1 \) . We have \( \varphi \left( {gx}\right) = {g\varphi }\left( x\right) \) for every \( x \in X \), except when \( x \) is equal to \(... | Yes |
Proposition 7.6. Let \( \varphi : H \rightarrow A \) be a homomorphism of \( H \) into an abelian group \( A \) . Suppose that \( \varphi \left( h\right) = \varphi \left( {h}^{\prime }\right) \) whenever \( h,{h}^{\prime } \in H \) are conjugate in \( G \) . Then\n\n\[ \varphi \left( {\operatorname{Ver}\left( h\right) ... | Proof. Indeed, we have \( \varphi \left( {\operatorname{Ver}\left( h\right) }\right) = \prod \varphi \left( {{z}_{\alpha }^{-1}{h}^{{f}_{\alpha }}{z}_{\alpha }}\right) \) . On the other hand, the elements \( {z}_{\alpha }^{-1}{h}^{{f}_{\alpha }}{z}_{\alpha } \) and \( {h}^{{f}_{\alpha }} \) are conjugate in \( G \), he... | Yes |
Corollary 7.7. Let \( C \) be a subgroup of \( H \) having the following property :\n\n(*) If \( c \in C \) and \( h \in H \) are \( G \) -conjugate, they are equal.\n\nThen \( \operatorname{Ver}\left( c\right) = {c}^{n}{\;\operatorname{mod}\;D}\left( H\right) \) for every \( c \in C \) . | Proof. Apply formula (7.3) to \( g = c \) ; this gives \( \operatorname{Ver}\left( c\right) = \prod {h}_{\alpha } \), with \( {h}_{\alpha } = {z}_{\alpha }^{-1}{c}^{{f}_{\alpha }}{z}_{\alpha } \) . The elements \( {h}_{\alpha } \in H \) and \( {c}^{{f}_{\alpha }} \in C \) are \( G \) -conjugate, hence equal because of ... | Yes |
Assume that, if two elements of \( H \) are conjugate in \( G \), they are equal. Then :\n\n(1) \( \operatorname{Ver}\left( h\right) = {h}^{n} \) for every \( h \in H \) .\n\n(2) Assume that \( \left| H\right| \) and \( \left( {G : H}\right) \) are relatively prime. Let \( N = \operatorname{Ker}\left( {\operatorname{Ve... | Proof. Note that the hypothesis on \( H \) implies that \( H \) is abelian, so that the transfer Ver : \( G \rightarrow {H}^{\text{ab }} \) does map \( G \) into \( H \) . Assertion (1) then follows from cor.7.7; under the assumption of (2), it implies that the composite map \( H \rightarrow G\overset{\text{ Ver }}{ \r... | Yes |
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