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Theorem 2.21 Any subgroup of an \( n \) -generated abelian group \( A \) is also \( n \) - generated. In particular, a subgroup of a cyclic group is cyclic. | Proof. Let \( H \leq A \) . The proof is by induction on \( n \) . If \( n = 1 \), then \( A = \langle a\rangle \) is cyclic. Let \( k \) be the smallest positive exponent for which \( {a}^{k} \in H \) . Then \( \left\langle {a}^{k}\right\rangle \leq H \) . However, if \( {a}^{m} \in H \), then \( m = {qk} + r \) where... | Yes |
Theorem 2.24 Let \( G \) be a group and let \( a \in G \) . The conjugation map \( {\gamma }_{a} : \operatorname{sub}\left( G\right) \rightarrow \operatorname{sub}\left( G\right) \) defined by \( {\gamma }_{a}H = {H}^{a} \) is an order isomorphism on \( \operatorname{sub}\left( G\right) \) . Hence, \( {\gamma }_{a} \) ... | Proof. It is clear that\n\n\[A \leq B\; \Leftrightarrow \;{A}^{a} \leq {B}^{a}\]\n\nand so \( {\gamma }_{a} \) is an order embedding of \( G \) into itself. But any subgroup \( A \) of \( G \) has the form \( A = {\gamma }_{a}\left( {{\gamma }_{{a}^{-1}}A}\right) \) and so \( {\gamma }_{a} \) is also surjective. \( ▱ \... | Yes |
Theorem 2.28 Let \( G \) be a finite group and let \( H \leq G \) have order \( o\left( G\right) /2 \) . Then any subgroup \( S \) of \( G \) is either a subgroup of \( H \) or else\n\n\[ \left| {S \cap H}\right| = \left| S\right| /2 \]\n\nIn words, \( S \) lies either completely in \( H \) or half-in and half-out of \... | Proof. If \( S \) is not contained in \( H \), then there is an \( a \in S \smallsetminus H \) and so\n\n\[ {SH} \supseteq H \sqcup {aH} \]\n\nBut the latter has size \( o\left( G\right) \) and so \( {SH} = G \) . Then\n\n\[ \left| H\right| \left| S\right| = \left| G\right| \left| {H \cap S}\right| \]\n\nimplies that \... | Yes |
Theorem 2.30 Let \( G \) be a group and let \( H \) and \( K \) be finite subgroups of \( G \) , with \( {HK} \leq G \) .\n\n1) If \( h \in H \) and \( k \in K \), then\n\n\[ o\left( {hk}\right) \mid o\left( H\right) o\left( K\right) \]\n\nIn particular, if \( \langle h\rangle \langle k\rangle \leq G \) is finite, then... | Proof. For part 1), Lagrange's theorem implies that\n\n\[ o\left( {hk}\right) \left| {o\left( {HK}\right) }\right| o\left( H\right) o\left( K\right) \]\n\nFor part 2), if \( h \in H \) then \( h = {ka} \) where \( k \in K \) and \( a \in N \) . Hence \( {k}^{-1}h = a \in N \) and so \( o\left( a\right) \mid o\left( K\r... | Yes |
Theorem 2.32 (Euler's theorem) Let \( n \) be a positive integer. If \( a \) is an integer relatively prime to \( n \), then\n\n\[ \n{a}^{\phi \left( n\right) } \equiv 1{\;\operatorname{mod}\;n} \n\]\n\nThis formula is called Euler's formula. | Proof. Since \( {\mathbb{Z}}_{n}^{ * } \) is a multiplicative group of order \( \phi \left( n\right) \), Lagrange’s theorem implies that Euler’s formula holds for \( a \in {\mathbb{Z}}_{n}^{ * } \) and since\n\n\[ \n{\left( a + kn\right) }^{\phi \left( n\right) } \equiv {a}^{\phi \left( n\right) } \equiv 1{\;\operatorn... | Yes |
Theorem 2.37 The subgroups of the dihedral group \( {D}_{2n} \) are of two types. For each \( d \mid n \), we have\n\n1) the cyclic subgroup \( \left\langle {\rho }^{n/d}\right\rangle \) of order \( d \) and\n\n2) for each \( 0 \leq k < n/d \), the dihedral subgroup\n\n\[ \n{S}_{d, k} = \left\langle {\sigma {\rho }^{k}... | \[ ▱ \] | No |
Theorem 3.1 Let \( G \) be a group.\n\n1) If \( H \leq G \), then\n\n\[ \left( {G : H}\right) = 1\; \Leftrightarrow \;G = H \] | Proof. We leave proof of part 1) and part 2) to the reader. | No |
Theorem 3.3 Let \( G \) be a finitely-generated group. If \( H \) is a subgroup of \( G \) of finite index, then \( H \) is also finitely generated. | Proof. Let \( T = \left\{ {{t}_{1},\ldots ,{t}_{m}}\right\} \) be a left transversal for \( G/H \), with \( {t}_{1} = 1 \) . Let \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) be a generating set for \( G \) and let\n\n\[ W = \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \cup \left\{ {{x}_{1}^{-1},\ldots ,{x}_{n}^{-1}}\ri... | Yes |
Theorem 3.8 Let \( H, K \leq G \). 3) The fact that \( {HK} \trianglelefteq G \) does not imply that either subgroup need be normal. | Proof. For part 3), let \( G = {S}_{4} \). Let \[ H = \{ \iota ,\left( {12}\right) \left( {34}\right) ,\left( {13}\right) \left( {24}\right) ,\left( {14}\right) \left( {23}\right) ,\left( {24}\right) ,\left( {13}\right) ,\left( {1234}\right) ,\left( {1432}\right) \} \] and \[ K = {S}_{3} = \{ \iota ,\left( {12}\right) ... | Yes |
Example 3.9 (The normal subgroups of \( {D}_{2n} \) ) We have seen that for \( d \mid n \), the subgroups of the dihedral group \( {D}_{2n} \) are\n\n1) the cyclic subgroup \( \left\langle {\rho }^{n/d}\right\rangle \) of order \( d \) ,\n\n2) for each \( 0 \leq k < n/d \), the dihedral subgroup\n\n\[ S = \sigma {\rho ... | Let\n\n\[ S = \left\langle {\sigma {\rho }^{k},{\rho }^{n/d}}\right\rangle \]\n\nbe a subgroup of type 2). Then since \( \left\langle {\rho }^{n/d}\right\rangle \trianglelefteq S \), it follows that \( S \) is normal if and only if the conjugates of the other generator \( \sigma {\rho }^{k} \) are in \( S \) . Conjugat... | Yes |
Theorem 3.10 Let \( G \) be a group.\n\n1) The subgroups \( \{ 1\} \) and \( G \) are normal in \( G \) .\n\n2) If \( \left\{ {{N}_{i} \mid i \in I}\right\} \) is a family of normal subgroups of \( G \), then\n\n\[ \mathop{\bigcap }\limits_{{i \in I}}{N}_{i}\;\text{ and }\;\mathop{\bigvee }\limits_{{i \in I}}{N}_{i} \]... | \( ▱ \) | No |
Theorem 3.12 If \( H, K \leq G \), then\n\n\[ \langle H, K\rangle = {H}^{K}K \] | Proof. Since \( {H}^{K} \) and \( K \) permute, it follows that \( {H}^{K}K \leq G \) . Since \( H \leq {H}^{K}K \) and \( K \leq {H}^{K}K \), it follows that \( \langle H, K\rangle \leq {H}^{K}K \) . The reverse inclusion is clear. \( ▱ \) | Yes |
Theorem 3.13 Let \( \mathcal{F} = \left\{ {{H}_{i} \mid i \in I}\right\} \) be a nonempty family of normal subgroups of a group \( G \) . Then the following are equivalent:\n\n1) \( \mathcal{F} \) is strongly disjoint\n\n2) If\n\n\[ \n{h}_{{i}_{1}}\cdots {h}_{{i}_{n}} = 1 \n\]\n\nwhere \( {h}_{{i}_{j}} \in {H}_{{i}_{j}... | Proof. If \( \mathcal{F} \) is strongly disjoint and \( {h}_{{i}_{1}}\cdots {h}_{{i}_{n}} = 1 \), where the factors are from different subgroups and \( n \geq 2 \), then\n\n\[ \n{h}_{{i}_{1}} = {\left( {h}_{{i}_{2}}\cdots {h}_{{i}_{n}}\right) }^{-1} \in {H}_{{i}_{1}} \cap {H}_{\left( {i}_{1}\right) } = \{ 1\} \n\]\n\na... | Yes |
Theorem 3.15 Let \( \mathcal{F} = \left\{ {{H}_{i} \mid i \in I}\right\} \) be a nonempty family of normal subgroups of a group \( G \) . For any \( K \trianglelefteq G \), there is a \( J \subseteq I \) that is maximal with respect to the property that the family\n\n\[ \n{\mathcal{F}}_{J} = \left\{ {{H}_{j} \mid j \in... | Proof. Write\n\n\[ \n\mathcal{I} = \left\{ {J \subseteq I \mid {\mathcal{F}}_{J}}\right. \text{is strongly disjoint}\} \n\]\n\nThen \( \mathcal{I} \) is nonempty and the union of any chain in \( \mathcal{I} \) is in \( \mathcal{I} \) . Hence, Zorn’s lemma implies that \( \mathcal{I} \) has a maximal member. \( ▱ \) | Yes |
Theorem 3.17 Let \( H \) be a subgroup of \( G \) of index 2.\n\n1) \( H \trianglelefteq G \).\n\n2) If \( a \in G \), then \( {a}^{2} \in H \).\n\n3) If \( G \) is finite, then any subgroup \( S \) of \( G \) is either a subgroup of \( H \) or else\n\n\[ \left| {S \cap H}\right| = \left| S\right| /2 \]\n\nIn words, \(... | Proof. For part 1), if \( a \notin H \), then \( \{ H,{aH}\} \) and \( \{ H,{Ha}\} \) are both partitions of \( G \) and so \( {aH} = {Ha} \) for all \( a \in G \). Hence, \( H \trianglelefteq G \). For part 2), if \( a \notin H \) then \( {aH} \neq H \) has order 2 in \( G/H \) and so \( {a}^{2}H = {\left( aH\right) }... | Yes |
The alternating group \( {A}_{4} \) has order \( 4!/2 = {12} \), but has no subgroups of order 6. | For if \( H \leq {A}_{4} \) has order 6, then \( H \) has index 2 and so \( {\sigma }^{2} \in H \) for all \( \sigma \in {A}_{4} \) . But there are eight 3-cycles \( \sigma \) in \( {A}_{4} \) and each one is a square:\n\n\[ \sigma = {\sigma }^{4} = {\left( {\sigma }^{2}\right) }^{2} \in H \]\n\nSince these 8 elements ... | Yes |
Theorem 3.21 A finite group \( G \) is a p-group if and only if the order of \( G \) is a power of \( p \) . | Proof. If \( o\left( G\right) = {p}^{k} \), then Lagrange’s theorem implies that every element of \( G \) is a \( p \) -element and so \( G \) is a \( p \) -group. Conversely, if \( G \) is a finite \( p \) -group but \( q \mid o\left( G\right) \) where \( q \neq p \) is prime, then Cauchy’s theorem implies that \( G \... | Yes |
Theorem 3.24 Let \( G \) be a group. The normalizer of \( H \leq G \) is\n\n\[ \n{N}_{G}\left( H\right) = \left\{ {a \in G \mid {H}^{a} = H}\right\} \n\] | Will see in the chapter on free groups (Theorem 12.21) that it is possible to have \( {H}^{a} \subset H \), in which case \( a \notin {N}_{G}\left( H\right) \) . However, since\n\n\[ \n{H}^{a} = H\; \Leftrightarrow \;{H}^{a} \subseteq H\;\text{ and }\;{H}^{{a}^{-1}} \subseteq H \n\]\n\nif a subset \( S \subseteq G \) i... | No |
Theorem 3.26 Let \( H \) and \( K \) be essentially disjoint subgroups of a group \( G \). 1) Then \( H \) and \( K \) commute elementwise if and only if each subgroup is contained in the normalizer of the other, that is, \[ H \leq {N}_{G}\left( K\right) \text{and} K \leq {N}_{G}\left( H\right) \] In particular, if \( ... | 2) If \( G = H \bullet K \), then \( H \) and \( K \) commute elementwise if and only if \( H, K \trianglelefteq G \). \( ▱ \) | No |
Theorem 3.30 (R. Dedekind, c. 1880) Let \( G \) be a group. Then for any subgroup \( H \leq G \) , \[ H \trianglelefteq G\text{ and }G/H\text{ is abelian }\; \Leftrightarrow \;{G}^{\prime } \leq H \] | Proof. If \( H \trianglelefteq G \) and \( G/H \) is abelian, then in \( G/H \), we have \[ \left\lbrack {a, b}\right\rbrack H = \left\lbrack {{aH},{bH}}\right\rbrack = H \] and so \( \left\lbrack {a, b}\right\rbrack \in H \), whence \( {G}^{\prime } \leq H \) . Conversely, if \( {G}^{\prime } \leq H \), then \( H \) i... | Yes |
For the symmetric group \( {S}_{n} \) on \( n \geq 3 \) symbols, we have\n\n\[ \n{S}_{n}^{\prime } = {A}_{n} = \mathcal{C}\left( {S}_{n}\right) \n\] | Since all commutators are even permutations, we have \( {S}_{n}^{\prime } \leq {A}_{n} \) .\n\nNow we make a few observations. First the product of any finite number of disjoint commutators is a commutator, since if \( {\alpha }_{i}{\beta }_{i} \) and \( {\alpha }_{j}{\beta }_{j} \) are disjoint, then\n\n\[ \n\left\lbr... | Yes |
Example 3.33 (Cassidy [5]) Let \( G \) be the set of all matrices of the form\n\n\[ m\left( {f, g, h}\right) \mathrel{\text{:=}} \left\lbrack \begin{matrix} 1 & f\left( x\right) & h\left( {x, y}\right) \\ 0 & 1 & g\left( y\right) \\ 0 & 0 & 1 \end{matrix}\right\rbrack \]\n\nwhere \( f\left( x\right), g\left( x\right) \... | A straightforward calculation shows that\n\n\[ m\left( {{f}_{1},{g}_{1},{h}_{1}}\right) m\left( {{f}_{2},{g}_{2},{h}_{2}}\right) = m\left( {{f}_{1} + {f}_{2},{g}_{1} + {g}_{2},{h}_{1} + {h}_{2} + {f}_{1}{g}_{2}}\right) \]\n\nand\n\n\[ m{\left( f, g, h\right) }^{-1} = m\left( {-f, - g, - h + {fg}}\right) \]\n\nfrom whic... | Yes |
Theorem 3.34 (Speigel [31],1976) If a group \( G \) contains a normal abelian subgroup \( A \) whose quotient \( G/A \) is cyclic, then \( {G}^{\prime } = \mathcal{C}\left( G\right) \) . | Proof. If we can find a normal subgroup \( B \trianglelefteq G \) for which \( B \subseteq \mathcal{C}\left( G\right) \) and \( G/B \) is abelian, then\n\n\[ \n{G}^{\prime } \leq B \subseteq \mathcal{C}\left( G\right) \subseteq {G}^{\prime } \n\]\n\nwhence \( {G}^{\prime } = \mathcal{C}\left( G\right) \) . To this end,... | Yes |
Theorem 3.36 (MacDonald [22],1986) Let \( G \) be a group with \( Z = Z\left( G\right) \) . If\n\n\[{\left( G : Z\right) }^{2} < o\left( {G}^{\prime }\right)\]\n\nthen \( {G}^{\prime } \neq \mathcal{C}\left( G\right) \) . | Proof. Since \( z, w \in Z \) implies that\n\n\[ \left\lbrack {{az},{bw}}\right\rbrack = \left\lbrack {a, b}\right\rbrack \]\n\nit follows that the number of distinct commutators is at most the number of commutators of \( G/Z \) and that is certainly at most \( {\left( G : Z\right) }^{2}.▱ \) | No |
Theorem 3.37 Let \( G \) be a group and let \( a, b, c \in G \). 1) \[ \left\lbrack {b, a}\right\rbrack = {\left\lbrack {a}^{-1}, b\right\rbrack }^{a} = {\left\lbrack a,{b}^{-1}\right\rbrack }^{b} \] \[ \left\lbrack {a, b}\right\rbrack = {\left\lbrack {a}^{-1},{b}^{-1}\right\rbrack }^{ba} \] | Proof. The proofs of part 1) and part 2) are straightforward calculations. | No |
5) H normalizes \( \left\lbrack {H, K}\right\rbrack \) and so\n\n\[ \left\lbrack {H, K}\right\rbrack \trianglelefteq \langle H, K\rangle \] | Proof. We prove only part 5). Theorem 3.37 implies that if \( {h}_{1} \in H \), then\n\n\[ {\left\lbrack h, k\right\rbrack }^{{h}_{1}} = \left\lbrack {{h}_{1}h, k}\right\rbrack {\left\lbrack {h}_{1}, k\right\rbrack }^{-1} \in \left\lbrack {H, K}\right\rbrack \]\n\nHence, \( H \) normalizes \( \left\lbrack {H, K}\right\... | Yes |
Theorem 3.41 Let \( A, H, K \leq G \). 1) \[ \left\lbrack {A,{HK}}\right\rbrack = \left\lbrack {A, H}\right\rbrack {\left\lbrack A, K\right\rbrack }^{H} \] where \[ {\left\lbrack A, K\right\rbrack }^{H} = \left\langle {{\left\lbrack a, k\right\rbrack }^{h} \mid a \in A, k \in K, h \in H}\right\rangle \] | Proof. For part 1), if \( a \in A, h \in H \) and \( k \in K \), then \[ \left\lbrack {a,{hk}}\right\rbrack = \left\lbrack {a, h}\right\rbrack {\left\lbrack a, k\right\rbrack }^{h} \in \left\lbrack {A, H}\right\rbrack {\left\lbrack A, K\right\rbrack }^{H} \] and so \( \left\lbrack {A,{HK}}\right\rbrack \leq \left\lbrac... | Yes |
Theorem 3.42 Let \( G \) be a group. The following are equivalent:\n\n1) (Associativity) For all \( a, b, c \in G \),\n\n\[ \left\lbrack {\left\lbrack {a, b}\right\rbrack, c}\right\rbrack = \left\lbrack {a,\left\lbrack {b, c}\right\rbrack }\right\rbrack \]\n\n2) (Distributivity) For all \( a, b, c \in G \),\n\n\[ \left... | Proof. Since\n\n\[ \left\lbrack {a,{bc}}\right\rbrack = \left\lbrack {a, b}\right\rbrack {\left\lbrack a, c\right\rbrack }^{b} \]\n\nit follows that 2) holds if and only if \( {\left\lbrack a, c\right\rbrack }^{b} = \left\lbrack {a, c}\right\rbrack \) for all \( a, b, c \in G \), which holds if and only if \( \left\lbr... | Yes |
Theorem 3.43 (Hall-Witt Identity) Let \( a, b, c \in G \) . Then\n\n\[{\left\lbrack a,{b}^{-1}, c\right\rbrack }^{b}{\left\lbrack b,{c}^{-1}, a\right\rbrack }^{c}{\left\lbrack c,{a}^{-1}, b\right\rbrack }^{a} = 1\] | Proof. For the first factor we have\n\n\[{\left\lbrack a,{b}^{-1}, c\right\rbrack }^{b} = b\left\lbrack {a,\left\lbrack {{b}^{-1}, c}\right\rbrack }\right\rbrack {b}^{-1}\]\n\n\[= {ba}\left\lbrack {{b}^{-1}, c}\right\rbrack {a}^{-1}{\left\lbrack {b}^{-1}, c\right\rbrack }^{-1}{b}^{-1}\]\n\n\[= {ba}{b}^{-1}{cb}{c}^{-1}{... | Yes |
Corollary 3.44 (Three subgroups lemma) Let \( G \) be a group. Let \( H, K \) and \( L \) be subgroups of \( G \). Then if any two of the commutators \( \left\lbrack {H, K, L}\right\rbrack ,\left\lbrack {L, H, K}\right\rbrack \) or \( \left\lbrack {K, L, H}\right\rbrack \) are contained in a normal subgroup \( N \) of ... | Proof. We assume that \( \left\lbrack {H, K, L}\right\rbrack \leq N \) and \( \left\lbrack {L, H, K}\right\rbrack \leq N \) and use the Hall-Witt Identity. Since \( \left\lbrack {h,{k}^{-1},\ell }\right\rbrack \in \left\lbrack {H, K, L}\right\rbrack \) and \( \left\lbrack {\ell ,{h}^{-1}, k}\right\rbrack \in \left\lbra... | Yes |
Example 4.2 Let \( {D}_{6} = \langle \rho ,\sigma \rangle \) be the dihedral group, where \( o\left( \rho \right) = 3 \) and \( o\left( \sigma \right) = 2 \) . Let \( {S}_{3} \) be the symmetric group of order 6 . The map \( f : {D}_{6} \rightarrow {S}_{3} \) defined by\n\n\[ f\left( {{\sigma }^{i}{\rho }^{k}}\right) =... | This tells us that the group of symmetries of the triangle is the group of permutations of the vertices. \( ▱ \) | No |
Theorem 4.3 Let \( {\sigma }_{1},\ldots ,{\sigma }_{n} \in \operatorname{End}\left( G\right) \) have the property that the images \( \operatorname{im}\left( {\sigma }_{i}\right) \) commute elementwise. If each \( {\sigma }_{i} \) is nilpotent, then so is the sum\n\n\[ \tau = {\sigma }_{1} + \cdots + {\sigma }_{n} \] | Proof. For any \( m > 0 \) ,\n\n\[ {\tau }^{nm} = \mathop{\sum }\limits_{{{j}_{1} + \cdots + {j}_{n} = {nm}}}\left( \begin{matrix} {nm} \\ {j}_{1},\ldots ,{j}_{n} \end{matrix}\right) {\sigma }_{1}^{{j}_{1}}\cdots {\sigma }_{n}^{{j}_{n}} \]\n\nBut if each \( {\sigma }_{i} \) is nilpotent, then there is a positive intege... | Yes |
Theorem 4.6 (Universal pairs) Let \( G \) be a group and let \( K \trianglelefteq G \). 1) (Existence) The pair \[ \left( {G/K,{\pi }_{K} : G \rightarrow G/K}\right) \] where \( {\pi }_{K} \) is the canonical projection modulo \( K \) is universal in \( \mathcal{F}\left( {G;K}\right) \). The mediating morphism for \( \... | Proof. For part 1), a mediating morphism for \( \sigma : G \rightarrow H \), if it exists, must satisfy \[ \tau \left( {gK}\right) = {\sigma g} \] for all \( g \in G \) and so must be unique. But the condition \( K \subseteq \ker \left( \sigma \right) \) implies that \( \tau \) is a well defined map and it is easy to s... | Yes |
1) (First isomorphism theorem) Every group homomorphism \( \\sigma : G \\rightarrow H \) induces an embedding \( \\bar{\sigma } : G/\\ker \\left( \\sigma \\right) \\hookrightarrow H \) of \( G/\\ker \\left( \\sigma \\right) \) defined by\n\n\[ \n\\bar{\sigma }\\left( {g\\ker \\left( \\sigma \\right) }\\right) = {\\sigm... | For part 1), since \( \\left( {G/K,{\\pi }_{K}}\\right) \) is universal, there is a unique homomorphism \( \\tau : G/K \\rightarrow \\operatorname{im}\\left( \\sigma \\right) \) for which \( \\tau \\circ {\\pi }_{K} = \\sigma \) . The rest follows from the fact that \( \\operatorname{im}\\left( \\tau \\right) = \\opera... | Yes |
Theorem 4.8 Let \( {G}_{1} \) and \( {G}_{2} \) be groups and let \( {H}_{i} \trianglelefteq {G}_{i} \) for \( i = 1,2 \) . Then\n\n\[ \frac{{G}_{1} \boxtimes {G}_{2}}{{H}_{1} \boxtimes {H}_{2}} \approx \frac{{G}_{1}}{{H}_{1}} \boxtimes \frac{{G}_{2}}{{H}_{2}} \] | Proof. Let \( \tau : {G}_{1} \boxtimes {G}_{2} \rightarrow \left( {{G}_{1}/{H}_{1}}\right) \boxtimes \left( {{G}_{2}/{H}_{2}}\right) \) be defined by\n\n\[ \tau \left( {{a}_{1},{a}_{2}}\right) = \left( {{a}_{1}{H}_{1},{a}_{2}{H}_{2}}\right) \]\n\nWe leave it to the reader to show that \( \tau \) is an epimorphism with ... | No |
Theorem 4.9 (Correspondence theorem) Let \( G \) be a group, let \( N \trianglelefteq G \) and let \( \pi : G \rightarrow G/N \) be the natural projection. Referring to Figure 4.3, let \( \bar{\pi } : \operatorname{sub}\left( {N;G}\right) \rightarrow \operatorname{sub}\left( {G/N}\right) \) be the map defined by \[ \ba... | Proof. For the surjectivity of \( \bar{\pi } \), any subgroup of \( G/N \) has the form \[ S = \{ {xN} \mid x \in X\} \] for some index set \( X \) . The set \( H = \bigcup \left( {xN}\right) \) is a subgroup of \( G \) since \( x, y \in H \) implies that \( {xN},{yN} \in S \) and so \( {x}^{-1}N \) and \( {xyN} \) are... | No |
Theorem 4.10 Let \( G \) be a group with \( A \leq B \leq G \) . Let \( \mathcal{P} \) be an isomorphism-invariant property of groups.\n\n2) (Normal lifting) Normality is preserved by normal lifting, that is, for \( N \trianglelefteq G \)\n\n\[ A \trianglelefteq B\; \Rightarrow \;{AN} \trianglelefteq {BN} \] | Proof. For part 2), to see that \( {AN} \trianglelefteq {BN} \), note that \( B \) normalizes both \( A \) and \( N \) and so \( B \) normalizes \( {AN} \) . Also, \( N \leq {AN} \) implies that \( N \) normalizes \( {AN} \) and so \( {BN} \) normalizes \( {AN} \), that is, \( {AN} \trianglelefteq {BN} \) . Since \( {B... | Yes |
Theorem 4.15 Let \( G \) be an elementary abelian group.\n\n1) Every nonidentity element of \( G \) has prime order \( p \).\n\n2) Writing the group product additively, \( G \) is a vector space over \( {\mathbb{Z}}_{p} \), where if \( \alpha \in {\mathbb{Z}}_{p} \) and \( a \in G \), then \( {\alpha a} \) is the sum o... | Proof. We use additive notation for \( G \) . For part 1), if \( o\left( a\right) = {mn} \) where \( m > 1 \) , then \( o\left( {na}\right) = m \) and so \( {mn} = m \), that is, \( n = 1 \) . Thus, \( o\left( a\right) = p \) is prime for all nonidentity \( a \in G \) . For part 2), let \( \oplus \) and \( \odot \) den... | No |
Theorem 4.16 The following are equivalent for an abelian group \( G \) :\n\n1) \( G \) is elementary abelian\n\n2) \( G \) is characteristically simple and has at least one nonidentity torsion element. | Proof. If \( G \) is an elementary abelian group, then the automorphisms of \( G \) are the linear automorphisms of \( G \) . It follows that if \( S \leq G \) is nontrivial and proper, then for any nonzero \( a \in S \) and nonzero \( x \in G \smallsetminus S \), there is an automorphism sending \( a \) to \( x \) and... | Yes |
Theorem 4.17 Let \( H \) be a proper subgroup of a group \( G \) . Then there are distinct group homomorphisms \( \sigma ,\tau : G \rightarrow K \) into some group \( K \) that agree on \( H \) . | Proof. For \( x \notin G/H \), we construct two distinct group actions of \( G \) on the set\n\n\[ X = G/H \cup \{ x\} \]\n\nthat agree on \( H \) . Let \( \sigma : G \rightarrow {S}_{X} \) be left translation on \( G/H \) that also leaves \( x \) fixed, that is, for any \( a \in G \), \n\n\[ {\sigma }_{a}\left( {bH}\r... | Yes |
Theorem 4.19 Let \( H \leq G \) . The largest subgroup of \( H \) that is normal in \( G \) is called the normal interior or core of \( H \), which we denote by \( {H}^{ \circ } \) . The core of \( H \) is the kernel | \[ {H}^{ \circ } = \mathop{\bigcap }\limits_{{a \in G}}{H}^{a} \] of the action of \( G \) on \( G/H \) by left translation. \( ▱ \) | Yes |
Theorem 4.20 Let \( G \) be a group and let \( H \leq G \) have finite index. Then\n\n\[ G/{H}^{ \circ } \hookrightarrow {S}_{G/H} \]\n\nand so\n\n\[ \left( {G : {H}^{ \circ }}\right) \mid \left( {G : H}\right) ! \]\n\nIn particular, \( \left( {G : {H}^{ \circ }}\right) \) is also finite and\n\n\[ \left( {H : {H}^{ \ci... | Proof. For part 1), first note that \( p \) is a prime. If \( q \) is a prime dividing \( \left( {H : {H}^{ \circ }}\right) \) , then Cauchy’s theorem implies that there is an \( h \in H \) for which \( q = o\left( {h{H}^{ \circ }}\right) \mid o\left( h\right) \), whence \( q \geq p \), a contradiction to the fact that... | No |
If \( o\left( G\right) = {p}^{n}u \) with \( p \) prime and \( p > u \), then we will prove in a later chapter that \( G \) has a subgroup \( H \) of order \( {p}^{n} \) (a Sylow \( p \) -subgroup of \( G \) ). | Since \( {p}^{n} \) and \( \left( {u - 1}\right) \) ! are relatively prime, it follows that \( H \trianglelefteq G.▱ | No |
Example 4.25 Let \( G \) be the infinite dihedral group,\n\n\[ G = \\left\\{ {{\\rho }^{i},\\sigma {\\rho }^{i} \\mid i \\in \\mathbb{Z}}}\\right\\} \]\n\nwhere \( o\\left( \\sigma \\right) = 2, o\\left( \\rho \\right) = \\infty \) and \( {\\rho \\sigma } = \\sigma {\\rho }^{-1} \) . If\n\n\[ {H}_{i} = \\left\\langle {... | \[ ▱ \] | No |
Theorem 4.31 Let \( G \) be a group and let \( H, K \trianglelefteq \trianglelefteq G \) . The following are equivalent:\n\n1) \( \langle H, K\rangle \trianglelefteq \trianglelefteq G \)\n\n2) \( {H}^{K} \trianglelefteq \trianglelefteq G \)\n\n3) \( \left\lbrack {H, K}\right\rbrack \trianglelefteq \trianglelefteq G \) ... | Proof. Recall from Theorem 3.40 that\n\n\[ \left\lbrack {H, K}\right\rbrack \trianglelefteq H\left\lbrack {H, K}\right\rbrack = {H}^{K} \trianglelefteq {H}^{K}K = \langle H, K\rangle \]\n\nThus,1) \( \Rightarrow 2) \Rightarrow 3) \) . On the other hand, if 3) holds, then since \( H \trianglelefteq \trianglelefteq G \) ... | Yes |
Theorem 4.36 The following are equivalent for a finite group \( G \): 1) Every subgroup of \( G \) is subnormal 2) \( G \) has the normalizer condition. | If \( G \) has the normalizer condition, then any maximal subgroup \( M \) must be normal in \( G \), for we have \( M < {N}_{G}\left( M\right) \) and the maximality of \( M \) implies that \( {N}_{G}\left( M\right) = G \) . Thus, with respect to the following conditions on a finite group \( G \): 1) Every subgroup of ... | No |
Theorem 4.37 Let \( G \) be a group and let \( \mathcal{S} \subseteq \operatorname{sub}\left( G\right) \).\n\n1) G has the maximal condition on \( \mathcal{S} \) if and only if \( G \) has the \( {ACC} \) on \( \mathcal{S} \).\n\n2) \( G \) has the minimal condition on \( \mathcal{S} \) if and only if \( G \) has the D... | Proof. Suppose \( G \) satisfies the maximal condition on \( \mathcal{S} \) and that\n\n\[ {H}_{1} \leq {H}_{2} \leq \cdots \]\n\n is an ascending sequence of members of \( \mathcal{S} \). Then the subgroups \( {H}_{k} \) have a maximal member \( {H}_{n} \), which implies that \( {H}_{n + k} \leq {H}_{n} \) for all \( ... | Yes |
Theorem 4.38 Let \( G \) be a group and let \( \mathcal{S} \subseteq \operatorname{sub}\left( G\right) \) be closed under arbitrary intersections and closed under unions of ascending sequences. Then \( G \) has the ACC on \( \mathcal{S} \) if and only if every \( H \in \mathcal{S} \) is finitely \( \mathcal{S} \) -gene... | Proof. Suppose \( G \) satisfies the ACC on \( \mathcal{S} \) and let \( H \in \mathcal{S} \) . If \( H \) is not finitely \( \mathcal{S} \) - generated, then for any \( {h}_{1} \in H \), we have\n\n\[ \n{\left\langle {h}_{1}\right\rangle }_{\mathcal{S}} < H \n\]\n\nHence, there is an \( {h}_{2} \in H \smallsetminus {\... | Yes |
\[ \mathcal{F}\left( {N;G}\right) \in \mathrm{{ACC}}\; \Leftrightarrow \;\mathcal{F}\left( {G/N}\right) \in \mathrm{{ACC}} \] | Proof. Part 1) follows from the correspondence theorem. | No |
Theorem 5.3 Let \( H \leq G \) . The following are equivalent for a subgroup \( K \leq G \) . 1) \( K \) is a complement of \( H \) in \( G \) . 2) \( K \) is a left transversal for \( H \) in \( G \) . 3) \( K \) is a right transversal for \( H \) in \( G \) . | Proof. Suppose that \( K \) is a complement of \( H \) in \( G \) . If \( {k}_{1}H = {k}_{2}H \), then \( {k}_{2}^{-1}{k}_{1} \in K \cap H = \{ 1\} \) and so \( {k}_{1} = {k}_{2} \) . Also, since \( G = {KH} \), every \( a \in G \) has the form \( a = {kh} \in {kH} \) for some \( k \in K \) . Thus, the cosets \( {kH} \... | Yes |
Theorem 5.4 Let \( \left\{ {{H}_{i} \mid i \in I}\right\} \) be a family of groups.\n\n1) The pair\n\n\[ \mathcal{P} = \left( {P,{\left\{ {\rho }_{i} : P \rightarrow {H}_{i}\right\} }_{i \in I}}\right) \]\nwhere \( P \) is the direct product of the family \( \left\{ {H}_{i}\right\} \) and \( {\rho }_{i} \) is the ith p... | Proof. For part 1), for any pair \( \mathcal{G} \in \mathcal{F} \), there must exist a unique mediating morphism \( \tau : G \rightarrow P \) satisfying\n\n\[ {\rho }_{i} \circ \tau = {f}_{i} \]\n\nBut this is equivalent to\n\n\[ {\rho }_{i}\left( {\tau a}\right) = {f}_{i}\left( a\right) \]\n\nfor all \( a \in G \) and... | No |
Theorem 5.5 Let \( \left\{ {{H}_{i} \mid i \in I}\right\} \) be a family of groups.\n\n1) The pair\n\n\[ \mathcal{S} = \left( {S,{\left\{ {\kappa }_{i} : {H}_{i} \rightarrow S\right\} }_{i \in I}}\right) \]\n\nwhere \( S \) is the direct sum of the family \( \left\{ {H}_{i}\right\} \) and \( {\kappa }_{i} \) is the ith... | Proof. For part 1), for any pair \( \left( {K,{\left\{ {k}_{i} : {H}_{i} \rightarrow K\right\} }_{i \in I}}\right) \) in \( \mathcal{F} \), there must be a unique mediating morphism \( \tau : S \rightarrow G \) satisfying\n\n\[ \tau \circ {\kappa }_{i} = {f}_{i} \]\n\nBut this specifies how \( \tau \) is defined on any... | No |
Theorem 5.6 Any finite group \( G \) is cancellable in direct sums. | Proof. It is sufficient to show that\n\n\[ A \ltimes G = B \ltimes H,\;G \approx H\; \Rightarrow \;A \approx B \]\n\nand we prove this by induction on \( o\left( G\right) \) . If \( o\left( G\right) = 1 \), the result is clear. Assume that any group of order less than \( o\left( G\right) \) is cancellable and let\n\n\[... | Yes |
Theorem 5.7 Let \( A \) be a finite abelian group. If \( u \in A \) has maximum order among all elements of \( A \), then \( M = \langle u\rangle \) has a direct complement, that is, there is a subgroup \( V \) for which\n\n\[ A = M \ltimes V \] | Proof. The proof is by induction on the order of \( A \) . If \( o\left( A\right) = 1 \), the result is clear. Assume the result holds for all abelian groups of order less than that of \( A \) . We may also assume that \( M < A \) . If we find a subgroup \( X \leq A \) for which \( X \cap M = \{ 0\} \), since then \( o... | Yes |
Theorem 5.10 Let \( G \) be a group. If\n\n\[ G = A \ltimes {A}^{\prime } = B \ltimes {B}^{\prime },\;A \subseteq B \]\n\n then we can replace either \( {A}^{\prime } \) or \( {B}^{\prime } \) by another subgroup to get an equation in good order. Specifically, the following are in good order:\n\n\[ G = A \ltimes \left\... | Proof. The first equation follows directly from Dedekind's law. For the second equation,\n\n\[ B \cap \left( {{A}^{\prime } \cap A{B}^{\prime }}\right) = {A}^{\prime } \cap \left( {B \cap A{B}^{\prime }}\right) \]\n\n\[ = {A}^{\prime } \cap A\left( {B \cap {B}^{\prime }}\right) \]\n\n\[ = {A}^{\prime } \cap {A}^{\prime... | Yes |
Theorem 5.11 A group \( G \) has the \( {ACC} \) on \( \mathcal{{DS}}\left( G\right) \) if and only if it has the \( {DCC} \) on \( \mathcal{{DS}}\left( G\right) \) . | Proof. Suppose that \( G \) has the DCC on direct summands and let\n\n\[ \n{D}_{1} \leq {D}_{2} \leq \cdots \n\]\n\nbe an ascending sequence in \( \mathcal{{DS}}\left( G\right) \) . Theorem 5.10 implies that there is a descending sequence\n\n\[ \n{E}_{1} \geq {E}_{2} \geq \cdots \n\]\n\nwhere \( G = {D}_{i} \boxtimes {... | No |
Theorem 5.12 (Remak) If a group \( G \) has either (and therefore both) chain condition on direct summands, then \( G \) has a Remak decomposition\n\n\[ G = {R}_{1} \ltimes \cdots \ltimes {R}_{n} \] | Proof. Let \( {R}_{1} \) be a minimal direct summand of \( G \) . If \( {R}_{1} = G \), then \( G \) is indecomposable and we are done. Otherwise,\n\n\[ G = {R}_{1} \ltimes {E}_{1} \]\n\nwhere \( {E}_{1} \neq \{ 1\} \) also has BCC on direct summands. Let \( {R}_{2} \) be a minimal direct summand of \( {E}_{1} \), whic... | Yes |
Theorem 5.14 Suppose that \( G \) is the join of a family \( \mathcal{F} = \left\{ {{H}_{i} \mid i \in I}\right\} \) of minimal normal subgroups. For any \( K \trianglelefteq G \), there is a \( J \subseteq I \) for which\n\n\[ G = K \ltimes \left( {\underset{j \in J}{ \ltimes }{H}_{j}}\right) \]\n\nand so\n\n\[ \opera... | Proof. Let \( J \subseteq I \) be maximal with respect to the fact that the direct sum\n\n\[ {M}_{J} = K \ltimes \left( {\underset{j \in J}{ \ltimes }{N}_{j}}\right) \]\n\nexists. If \( i \notin J \), then \( {M}_{J} \cap {N}_{i} \trianglelefteq G \) implies that \( {M}_{J} \cap {N}_{i} = \{ 1\} \) or \( {N}_{i} \leq {... | Yes |
Theorem 5.15 (aD-groups: Kertész [21], 1952) A group \( G \) is an aD-group if and only if it is the direct product of cyclic groups of prime order. | Proof. If \( G \) is a direct sum of cyclic subgroups \( {C}_{i} \) of prime order, then since each \( {C}_{i} \) is minimal normal, Theorem 5.14 implies that\n\n\[ \operatorname{sub}\left( G\right) = \operatorname{nor}\left( G\right) = \mathcal{D}\mathcal{S}\left( G\right) \]\n\nand so \( G \) is an aD-group.\n\nFor t... | Yes |
Theorem 5.17 (aNC-groups: Weigold [34], 1960) A group is an aNC-group if and only if it is an aD-group, that is, if and only if it is a direct sum of cyclic subgroups of prime order. | Proof. An aD-group is an aNC-group. For the converse, we show that an aNC-group \( G \) has trivial commutator subgroup and so is abelian. If \( A \leq G \) is abelian, then \( G = N \rtimes A \) for some \( N \trianglelefteq G \) and so\n\n\[{G}^{\prime } = \left\lbrack {{NA},{NA}}\right\rbrack \leq N\]\n\nHence, \( {... | No |
The dihedral group \( {D}_{2n} \) is a nontrivial semidirect product: | \[ {D}_{2n} = \langle \rho \rangle \rtimes \langle \sigma \rangle \] | Yes |
Theorem 5.23 Let \( G \) and \( {G}_{1} \) be groups, let \( H \leq G \) and let \( \sigma : H \rightarrow {G}_{1} \) be an epimorphism.\n\n1) Given any normal complement \( K \) of \( H \) modulo \( \ker \left( \sigma \right) \), \n\n\[ \nG = K \rtimes H\left\lbrack {{\;\operatorname{mod}\;\ker }\left( \sigma \right) ... | Proof. For part 2), the Dedekind law implies that \n\n\[ \nH \cap K\ker \left( \sigma \right) = \ker \left( \sigma \right) \left( {H \cap K}\right) = \ker \left( \sigma \right) \n\] \n\nand so \( K\ker \left( \sigma \right) \) is a normal complement of \( H \) modulo \( \ker \left( \sigma \right) .▱ | No |
Theorem 5.25 Let \( H \) and \( K \) be indecomposable groups. Let \( \alpha : K \rightarrow H \) and \( \beta : H \rightarrow K \) be homomorphisms for which \( {\alpha \beta } \in \operatorname{Aut}\left( H\right) \) and \( \operatorname{im}\left( \beta \right) \trianglelefteq K \) . Then \( \alpha \) and \( \beta \)... | Proof. The map\n\n\[{\beta }_{L} = {\left( \alpha \beta \right) }^{-1} \circ \alpha : K \rightarrow H\]\n\nis a left inverse of \( \beta \) and so \( \beta \) is injective and\n\n\[H = \ker \left( {\beta }_{L}\right) \ltimes \operatorname{im}\left( \beta \right)\]\n\nBut since \( {\beta }_{L} \) is not the zero map, \(... | Yes |
Theorem 5.26 Let \( N \) and \( H \) be groups and let \( \theta : H \rightarrow \operatorname{Aut}\left( N\right) \) be a homomorphism. We denote \( \theta \left( h\right) \) by \( {\theta }_{h} \) . Let \( N{ \rtimes }_{\theta }H \) be the cartesian product \( N \times H \) together with the binary operation defined ... | Proof. To see that multiplication is associative, we have\n\n\[ \left\lbrack {\left( {a, x}\right) \left( {b, y}\right) }\right\rbrack \left( {c, z}\right) = \left( {a{\theta }_{x}\left( b\right) ,{xy}}\right) \left( {c, z}\right) \]\n\n\[ = \left( {a{\theta }_{x}\left( b\right) {\theta }_{xy}\left( c\right) ,{xyz}}\ri... | Yes |
Given a group \( G \), there is one rather obvious way to create an external semidirect product \( G{ \rtimes }_{\theta }H \), namely, by taking \( H = \operatorname{Aut}\left( G\right) \) and \( \theta : H \rightarrow H \) to be the identity. The group product in this case is | \[ \left( {a\sigma }\right) \left( {b\tau }\right) = {a\sigma }\left( b\right) {\sigma \tau } \] The group \( G{ \rtimes }_{\theta }\operatorname{Aut}\left( G\right) \) is called the holomorph of \( G \). | Yes |
Recall that if \( C \) is an infinite cyclic group, then \( \operatorname{Aut}\left( C\right) = \{ \iota ,\tau \} \) where \( \tau : a \mapsto {a}^{-1} \) and if \( C \) is cyclic of order \( n \), then \( \operatorname{Aut}\left( C\right) \approx {\mathbb{Z}}_{n}^{ * } \) . Let \( {C}_{\infty }\left( a\right) \) and \... | If \( {\theta }_{b} = \iota \) , then \( \theta \) is the zero map and \( {C}_{\infty }\left( a\right) { \rtimes }_{\theta }{C}_{\infty }\left( b\right) \) is direct. If \( {\theta }_{b} = \tau \), then the commutativity rule in the group \( {C}_{\infty }\left( a\right) { \rtimes }_{\theta }{C}_{\infty }\left( b\right)... | Yes |
To define a semidirect product \( {C}_{3}\left( a\right) \rtimes {}_{\theta }{C}_{4}\left( b\right) \), we must specify a homomorphism\n\n\[ \theta : {C}_{4}\left( b\right) \rightarrow \operatorname{Aut}\left( {C}_{3}\right) = \left\{ {\iota ,{\sigma }_{2}}\right\} \]\n\nwhere \( {\sigma }_{2}\left( a\right) = {a}^{2} ... | \[ T = \langle a, b\rangle ,\;o\left( a\right) = 3,\;o\left( b\right) = 4,\;{ba} = {a}^{2}b \] | Yes |
Example 5.30 Let us examine the possibilities for an external semidirect product of the form\n\n\[ \n{C}_{{p}^{m}}\left( a\right) { \rtimes }_{\theta }{C}_{{p}^{n}}\left( b\right) \n\]\n\nwhere \( p \) is prime. The automorphisms of \( {C}_{{p}^{m}}\left( a\right) \) are the \( k \) th power maps \( {\sigma }_{k} : a \... | As an example, for \( m = 1 \), Fermat’s theorem implies that (5.31) is equivalent to\n\n\[ \nk \equiv 1{\;\operatorname{mod}\;p} \n\]\n\nand since \( 1 \leq k < p \), it follows that \( k = 1 \) . Hence, the only semidirect product of the form\n\n\[ \n{C}_{p}\left( a\right) { \rtimes }_{\theta }{C}_{{p}^{n}}\left( b\r... | Yes |
Let \( D \) be a group and let \( G = {D}^{3} \mathrel{\text{:=}} D \boxtimes D \boxtimes D \). Then each permutation \( \sigma \in {S}_{3} \) defines an automorphism \( {\theta }_{\sigma } \) of \( G \) by permuting the coordinates in \( G \). For example,\n\n\[ \n{\theta }_{\left( {13}\right) }\left( {x, y, z}\right)... | Moreover, the map \( \theta : {S}_{3} \rightarrow \operatorname{Aut}\left( G\right) \) sending \( \sigma \) to \( {\theta }_{\sigma } \) is a homomorphism, since \( {\theta }_{\sigma }{\theta }_{\tau } = {\theta }_{\sigma \tau } \). Thus, the semidirect product\n\n\[ \nG{ \rtimes }_{\theta }{S}_{3} = {D}^{3}{ \rtimes }... | Yes |
Theorem 5.34 Let \( W = D{u}_{\Omega }Q \) be a wreath product and suppose that \( Q \) acts faithfully on \( \Omega \) . Suppose also that \( D \) acts faithfully on \( \Lambda \) . Then the map \( \sigma : W \rightarrow {S}_{\Lambda \times \Omega } \) defined by\n\n\[ \sigma \left( {f, q}\right) = {\left( f, q\right)... | It is convenient to drop the \( * \) notation and to think of elements of a wreath product \( D\wr Q \) as permutations of \( \Lambda \times \Omega \) . | No |
A permutation matrix \( P \) is an \( n \times n \) matrix with entries from the set \( \{ 0,1\} \) with the property that each row contains exactly one 1 and each column contains exactly one 1. Multiplication of a matrix \( A \) on the left by a permutation matrix \( P \) permutes the rows of \( A \) . Similarly, mult... | For \( P \in {\mathcal{P}}_{n} \), let \( {P}_{i} \) denote the \( i \) th row of \( P \) and let \( {P}^{\left( j\right) } \) denote the \( j \) th column. The rows of \( P \) also define a permutation \( {\pi }_{P} \) of \( \Omega = \{ 1,\ldots, n\} \), in particular, \( {\pi }_{P}\left( i\right) \) is the column num... | Yes |
Theorem 6.1 For a permutation group \( {S}_{X} \), the following hold.\n\n1) Disjoint cycles in \( {S}_{X} \) commute.\n\n2) Every permutation in \( {S}_{X} \) is a product of disjoint cycles, the product being unique except for the order of factors. A representation of \( \sigma \) as a product of disjoint cycles (wit... | Proof. Part 1) we leave to the reader. For part 2), let \( \sigma \in {S}_{X} \) and define an equivalence relation on \( X \) by \( x \equiv y \) if \( y = {\sigma }^{k}x \) for some integer \( k \) . For \( x \in X \) , the equivalence class containing \( x \) is\n\n\[ \left\lbrack x\right\rbrack = \left\{ {{\sigma }... | No |
Theorem 6.5 Let \( \sigma \in {S}_{n} \). 1) Exactly one of the following holds: a) \( \sigma \) can be written as a product of an even number of transpositions, in which case we say that \( \sigma \) is \( \operatorname{even} \) (or has even parity). b) \( \sigma \) can be written as a product of an odd number of tran... | Proof. For part 1), if \( \sigma \) can be written as a product of an even number of transpositions and an odd number of transpositions, say \[ \sigma = {\rho }_{1}\cdots {\rho }_{2v} = {\tau }_{1}\cdots {\tau }_{{2u} + 1} \] then the identity can be written as a product of an odd number of transpositions \[ \iota = {\... | Yes |
Theorem 6.6 The following sets generate \( {S}_{n} \) . | Proof. For part 2) we have\n\n\[ \left( {ab}\right) = {\left( 1b\right) }^{\left( 1a\right) } \]\n\nand so the subgroup generated by the transpositions of 1 contains all transpositions. For part 3), if \( A \) is the subgroup generated by the adjacent transpositions, then \( \left( {12}\right) \in A \) and if \( \left(... | Yes |
Theorem 6.7 For \( n \geq 3,{A}_{n} \) is generated by the 3-cycles. | Proof. Any even permutation is a product of pairs of distinct transpositions. For \ | No |
Theorem 6.10 If \( n \neq 4 \), then \( {S}_{n} \) has no normal subgroups other than \( \{ 1\} ,{A}_{n} \) and \( {S}_{n} \) . | Proof. This is easily checked for \( n \leq 2 \) so assume that \( n \geq 3 \) . If \( N \trianglelefteq {S}_{n} \), then \( N \cap {A}_{n} \trianglelefteq {A}_{n} \) . Hence, \( N \cap {A}_{n} = \{ \iota \} \) or \( N \cap {A}_{n} = {A}_{n} \) . But if \( N \cap {A}_{n} = \{ \iota \} \) , then \( N = \{ \iota ,\sigma ... | Yes |
Theorem 7.1 The kernel of an action \( \lambda : G \rightarrow {S}_{X} \) is the intersection of the stabilizers of all elements of \( X \) , | \[ \ker \left( \lambda \right) = \mathop{\bigcap }\limits_{{x \in X}}\operatorname{stab}\left( x\right) \] | Yes |
Theorem 7.7 If a p-group \( G \) acts on a set \( X \), then\n\n\[ \left| X\right| \equiv \left| {{\operatorname{Fix}}_{X}\left( G\right) }\right| {\;\operatorname{mod}\;p} \] | Note that a finite group \( G \) has the center-intersection property if and only if every nontrivial normal subgroup of \( G \) contains a central subgroup of prime order. Any finite \( p \) -group \( G \) has the center-intersection property, for if \( N \trianglelefteq G \) is nontrivial, then \( G \) acts on the el... | No |
Theorem 7.8 A finite p-group \( G \) has the center-intersection property. | 1) \( Z\left( G\right) \) is nontrivial.\n\n2) \( G \) is simple if and only if \( o\left( G\right) = p.▱ \) | No |
Corollary 7.9 If \( o\left( G\right) = {p}^{2} \), then \( G \) is abelian. In fact, \( G \) is either cyclic or is the direct product of two cyclic subgroups of order \( p \) . | Proof. We must have \( \left| {Z\left( G\right) }\right| = p \) or \( {p}^{2} \) . But if \( \left| {Z\left( G\right) }\right| = p \), then \( G/Z\left( G\right) \) is cyclic and so \( G \) is abelian, which is a contradiction. Hence, \( \left| {Z\left( G\right) }\right| = {p}^{2} \) and \( G \) is abelian. If \( G \) ... | Yes |
Theorem 7.10 Let \( G \) be a finite p-group. Then every \( H < G \) has a p-cover \( K \) and if \( H \vartriangleleft G \), then \( K \) can be chosen so that \( H \vartriangleleft K \) is central in \( G \) . | Proof. The proof is by induction on \( o\left( G\right) \) . The theorem is true if \( o\left( G\right) = p \) . Assume \( o\left( G\right) > p \) and that the theorem is true for all groups smaller than \( G \) . Let \( H < G \) and let \( N \) be central in \( G \) of order \( p \) . If \( H = \{ 1\} \), then \( N \)... | Yes |
Theorem 7.13 Let \( p \) be a prime. Let \( G \) be a group of order \( {p}^{n} \), with Frattini subgroup \( \Phi \left( G\right) \) of order \( {p}^{m} \). 1) \( \Phi \left( G\right) \) is the smallest normal subgroup of \( G \) for which \( G/\Phi \left( G\right) \) is an elementary abelian group. Moreover, \( G/\Ph... | Proof. Part 1) has been proved. For part 2), since \( \left\{ {{G}^{\prime }{a}^{p} \mid a \in G}\right\} \) is a subgroup of \( G/{G}^{\prime } \), it follows that \( {G}^{\prime }{G}^{p} \) is a normal subgroup of \( G \) . In fact, \( G/{G}^{\prime }{G}^{p} \) is elementary abelian of exponent \( p \) and so part 1)... | Yes |
Theorem 7.14 Let \( G \) be a nontrivial p-group of order \( {p}^{n} \). 1) The number of maximal subgroups of \( G \) is \[ \left| {{\operatorname{sub}}_{n - 1}\left( G\right) }}\right| = \left| {{\operatorname{nor}}_{n - 1}\left( G\right) }}\right| = \frac{{p}^{n - m} - 1}{p - 1} \equiv 1{\;\operatorname{mod}\;p} \] ... | Proof. For part 1), if \( o\left( {\Phi \left( G\right) }}\right) = {p}^{m} \), then Theorem 7.13 implies that \( A = G/\Phi \left( G\right) \) is a vector space over \( {\mathbb{Z}}_{p} \) of dimension \( n - m \). In general, if \( V \) is a vector space over \( {\mathbb{Z}}_{p} \) of dimension \( k \), then the numb... | No |
Theorem 7.15 Let \( G \) be a finite p-group and let \( a \in G \) have order \( o\left( a\right) = {p}^{m} \) . Let \( b \in G \) and suppose that\n\n\[ \n{a}^{b} = {a}^{\alpha }\n\]\n\nfor some integer \( \alpha ≢ 1{\;\operatorname{mod}\;{p}^{m}} \) . Let \( {a}^{\langle b\rangle } \) be the set of conjugates of a by... | Proof. The conjugates of \( a \) by \( \langle b\rangle \) are\n\n\[ \n{b}^{k}a{b}^{-k} = {a}^{{\alpha }^{k}}\n\]\n\nfor \( k = 1,\ldots, r \), where \( r = \left| {a}^{\langle b\rangle }\right| \) . In fact, \( r \) is the smallest positive integer for which \( {b}^{r} \) commutes with \( a \) . Also, \( {a}^{\langle ... | Yes |
Theorem 7.16 Let \( p > 2 \) be prime and let \( o\left( G\right) = {p}^{n} > 1 \). 1) If \( G \) is noncyclic and if \( a \in G \) is an element of maximum order, then \( G \) has an element of order \( p \) that is not contained in \( \langle a\rangle \). 2) If \( G \) has a unique subgroup of order \( {p}^{s} \) for... | Proof. We have already seen that part 2) follows from part 1). To prove part 1), assume that \( G \) is noncyclic. Let \( A = \langle a\rangle \), where \( a \in G \) has maximum order \( {p}^{m} \). If \( m = 1 \), then all nonidentity elements of \( G \) have order \( p \) and so we may assume that \( 2 \leq m < n \)... | Yes |
Theorem 7.18 Let \( p > 2 \) be a prime. Let \( G \) be a nonabelian group of order \( {p}^{n} \) with an element \( a \) of order \( {p}^{n - 1} \) . Then\n\n\[ G = \langle a\rangle \rtimes \langle b\rangle \]\n\nwhere \( o\left( b\right) = p \) and\n\n\[ {ba}{b}^{-1} = {a}^{1 + {p}^{n - 2}} \] | To see that such a group exists, recall from Example 5.30 that there is a semidirect product\n\n\[ {C}_{{p}^{n - 1}}\left( a\right) { \rtimes }_{\theta }{C}_{p}\left( b\right) \]\n\nwhere\n\n\[ {\theta }_{b}\left( a\right) = {a}^{1 + {p}^{n - 2}} \] | Yes |
Theorem 8.1 Let \( G \) be a finite group and let \( S \in {\operatorname{Syl}}_{p}\left( G\right) \). 1) \( S \) is the set of all \( p \) -elements of \( {N}_{G}\left( S\right) \). 2) Any p-element \( a \in G \smallsetminus S \) moves \( S \) by conjugation, that is, \( {S}^{a} \neq S \). 3) \( S \) is the only Sylow... | If \( S \in {\operatorname{Syl}}_{p}\left( G\right) \), then \[ {S}^{a} \leq {N}_{G}{\left( S\right) }^{a} \] for any \( a \in G \). Hence, if \( a \) normalizes \( {N}_{G}\left( S\right) \), then \[ {S}^{a} \leq {N}_{G}\left( S\right) \] and since \( {S}^{a} \) is also a Sylow \( p \) -subgroup of \( {N}_{G}\left( S\r... | No |
Theorem 8.7 Let \( G \) be a finite group and let \( S \in {\operatorname{Syl}}_{p}\left( G\right) \) . If\n\n\[ S \leq {N}_{G}\left( S\right) \leq H \leq G \]\n\nthen \( H \) is self-normalizing. In particular, if \( H < G \), then \( H \) is not normal in \( G \) . | Proof. Conjugating by any \( a \in {N}_{G}\left( H\right) \) gives\n\n\[ {S}^{a} \leq {N}_{G}{\left( S\right) }^{a} \leq H \leq G \]\n\nand so both \( S \) and \( {S}^{a} \) are Sylow \( p \) -subgroups of \( H \) . It follows that \( S \) and \( {S}^{a} \) are conjugate in \( H \) . Hence, there is an \( h \in H \) fo... | Yes |
Theorem 8.10 Let \( G \) be a group in which every Sylow subgroup is normal. Let the Sylow subgroups of \( G \) be \( \left\{ {{Y}_{p} \mid p \in \mathcal{P}}\right\} \). Then\n\n\[ \n{G}_{\text{tor }} = \underset{p \in \mathcal{P}}{ \ltimes }{Y}_{p}\n\]\n\nand so \( {G}_{\text{tor }} \leq G \). Thus, the product of tw... | Proof. Since the Sylow \( p \) -subgroups are normal and pairwise essentially disjoint, they commute elementwise. In particular, if \( {a}_{1},\ldots ,{a}_{n} \) come from distinct Sylow subgroups, then\n\n\[ \no\left( {{a}_{1}\cdots {a}_{n}}\right) = o\left( {a}_{1}\right) \cdots o\left( {a}_{n}\right)\n\]\n\nand so t... | Yes |
Theorem 8.14 (Frattini [12],1885) If \( G \) is a finite group, then the Frattini subgroup \( \Phi \left( G\right) \) has the property that all of its Sylow subgroups are normal in \( G \) . | Proof. If \( S \in {\operatorname{Syl}}_{p}\left( \Phi \right) \), then \( S \leq \Phi \trianglelefteq G \) and the Frattini argument shows that\n\n\[ G = \Phi {N}_{G}\left( S\right) \]\n\nBut if \( {N}_{G}\left( S\right) < G \), then there is a maximal subgroup \( M \) of \( G \) for which \( {N}_{G}\left( S\right) \l... | Yes |
Example 8.15 If \( o\left( G\right) = {9982} = 2 \cdot 7 \cdot {23} \cdot {31} \), then routine calculation shows that the \( {n}_{7} \) argument holds: | \[ 1 + {7k} \mid o\left( G\right) \; \Rightarrow \;k = 0 \] and so \( {Y}_{7} \vartriangleleft G \) and \( G \) is not simple. \( ▱ \) | Yes |
Example 8.17 Let \( o\left( G\right) = {30} = 2 \cdot 3 \cdot 5 \) . Then based on the fact that \( {n}_{p} = 1 + {kp} \mid o\left( G\right) \), we can conclude only that \( {n}_{3} \in \{ 1,{10}\} \) and \( {n}_{5} \in \{ 1,6\} \) . However, if \( {n}_{3} = {10} \) and \( {n}_{5} = 6 \), then \( G \) contains at least... | \( ▱ \) | No |
If \( o\left( G\right) = 3 \cdot {5}^{2} = {75} \), then \( {Y}_{5} \vartriangleleft G \) and \( G = {Y}_{3} \rtimes {Y}_{5} \) | Also, \( 1 + {3k} \mid {25} \) holds only for \( k = 0 \) or \( k = 8 \) and so \( {n}_{3} = 1 \) or \( {n}_{3} = {25} \) . Note that if \( {n}_{3} = 1 \), then \( G = {Y}_{3} \ltimes {Y}_{5} \) is abelian. \( ▱ \) | No |
Theorem 8.19 Let \( o\left( G\right) = {pq} \), with \( p < q \) primes. Then\n\n\[ G = {C}_{q}\left( b\right) \rtimes {C}_{p}\left( a\right) \]\n\nwhere\n\n\[ {b}^{a} = {b}^{k} \]\n\nfor some \( 1 \leq k < q \) and \( {k}^{p} \equiv 1{\;\operatorname{mod}\;q} \) . Moreover, \( G \) is cyclic if and only if \( p \nmid ... | Proof. We have seen that\n\n\[ G = {Y}_{q} \rtimes {Y}_{p} = {C}_{q}\left( b\right) \rtimes {C}_{p}\left( a\right) \]\n\nThus, \( {ab}{a}^{-1} = {b}^{k} \) for some \( 1 \leq k < q \) and repeated conjugation by \( a \) gives\n\n\[ b = {a}^{p}b{a}^{-p} = {b}^{{k}^{p}} \]\n\nwhich implies that \( {k}^{p} \equiv 1{\;\ope... | Yes |
Example 8.20 Let us return to the case \( o\left( G\right) = {30} = 2 \cdot 3 \cdot 5 \) . We saw in Example 8.17 that one of \( {Y}_{3} \) or \( {Y}_{5} \) must be normal in \( G \) . It follows that \( {Y}_{3}{Y}_{5} \trianglelefteq G \) has order 15 and so is cyclic. Hence, \( {Y}_{3},{Y}_{5} \sqsubseteq {Y}_{3}{Y}_... | ▱ | No |
If \( o\left( G\right) = {108} = {3}^{3} \cdot 4 \), then \( 1 + {3k} \mid 4 \) and so \( k = 0 \) or \( k = 1 \) . Thus, this case is not amenable to the \( {n}_{p} \) -argument. However, if \( k = 1 \) then | Theorem 8.21 implies that\n\n\[ K = \mathop{\bigcap }\limits_{{Y \in {\operatorname{Syl}}_{3}\left( G\right) }}Y \]\n\nis a nontrivial proper normal subgroup of \( G \) of order 9 . Thus, \( G \) is not simple. | Yes |
If \( o\left( G\right) = p\left( {p + 1}\right) \), where \( p \) is prime. Then \( {n}_{p} = 1 \) or \( {n}_{p} = p + 1 \). | While the previous theorem does not apply, if \( {n}_{p} = p + 1 \), then\n\n\[ K = \mathop{\bigcap }\limits_{{Y \in {\operatorname{Syl}}_{p}\left( G\right) }}Y = \{ 1\} \]\n\nHence, \( G \hookrightarrow {S}_{p + 1} \) . As an example, if \( o\left( G\right) = {12} = 3 \cdot 4 \), then either \( {Y}_{3} \vartrianglelef... | No |
Example 8.25 Let \( o\left( G\right) = {6201} = {3}^{2} \cdot {13} \cdot {53} \) . Then \( {n}_{3} = 1 + {3k} \mid {13} \cdot {53} \) , which implies that \( {n}_{3} \in \{ 1,{13}\} \) . If \( {n}_{3} = 1 \), then \( {Y}_{3} \vartriangleleft G \) . If \( {n}_{3} = {13} < {53} \), then Theorem 8.24 implies that \( {N}_{... | Thus \( G \) is not simple. \( ▱ \) | Yes |
If \( o\left( G\right) = {3675} = 3 \cdot {5}^{2} \cdot {7}^{2} \), then it is easy to see that \( {n}_{7} \in \{ 1,{15}\} \) . If \( {n}_{7} = {15} \), then \( o\left( {{N}_{G}\left( {Y}_{7}\right) }\right) = 5 \cdot {7}^{2} \) . | Let \( P \) be a Sylow 5- subgroup of \( {N}_{G}\left( {Y}_{7}\right) \) . The number of such subgroups is \( 1 + {5k} \mid {7}^{2} \) and so \( P \vartriangleleft {N}_{G}\left( {Y}_{7}\right) \) . Hence, \[ 5 \cdot {7}^{2} \mid o\left( {{N}_{G}\left( P\right) }\right) \] Also, \( P \) has index 5 in \( {P}^{ * } \in {... | Yes |
If \( o\left( G\right) = {1785} = 3 \cdot 5 \cdot 7 \cdot {17} \), then a routine calculation gives\n\n\[ \n{n}_{3} \in \{ 1,7,{85},{595}\} \;\text{ and }\;{n}_{17} \in \{ 1,{35}\} \n\] | But\n\n\[ \n3 < {17},\;3/\left( {{17} - 1}\right) ,\;3//{n}_{17}\; \Rightarrow \;{n}_{3}\left| {\;\frac{o\left( G\right) }{3 \cdot {17}} = {35}}\right. \n\]\n\nand so \( {n}_{3} = 1 \) or \( {n}_{3} = 7 \) . Hence, Theorem 8.24 now implies that one of \( {Y}_{3} \) or \( N{\left( {Y}_{3}\right) }^{ \circ } \) is a prop... | Yes |
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