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Proposition 9.13. Let \( G \) and \( H \) be groups. The set \( G \times H \) is a group under the operation \( \left( {{g}_{1},{h}_{1}}\right) \left( {{g}_{2},{h}_{2}}\right) = \left( {{g}_{1}{g}_{2},{h}_{1}{h}_{2}}\right) \) where \( {g}_{1},{g}_{2} \in G \) and \( {h}_{1},{h}_{2} \in H \) . | Proof. Clearly the binary operation defined above is closed. If \( {e}_{G} \) and \( {e}_{H} \) are the identities of the groups \( G \) and \( H \) respectively, then \( \left( {{e}_{G},{e}_{H}}\right) \) is the identity of \( G \times H \) . The inverse of \( \left( {g, h}\right) \in G \times H \) is \( \left( {{g}^{... | Yes |
Consider\n\n\[ \n{\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} = \{ \left( {0,0}\right) ,\left( {0,1}\right) ,\left( {1,0}\right) ,\left( {1,1}\right) \} .\n\]\n\nAlthough \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \) and \( {\mathbb{Z}}_{4} \) both contain four elements, they are not isomorphic. Every element \( \left( {a... | Every element \( \left( {a, b}\right) \) in \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \) has order 2, since \( \left( {a, b}\right) + \left( {a, b}\right) = \left( {0,0}\right) \) ; however, \( {\mathbb{Z}}_{4} \) is cyclic. | Yes |
Theorem 9.17. Let \( \\left( {g, h}\\right) \\in G \\times H \) . If \( g \) and \( h \) have finite orders \( r \) and \( s \) respectively, then the order of \( \\left( {g, h}\\right) \) in \( G \\times H \) is the least common multiple of \( r \) and \( s \) . | Proof. Suppose that \( m \) is the least common multiple of \( r \) and \( s \) and let \( n = \\left| \\left( {g, h}\\right) \\right| \) . Then\n\n\[ \n{\\left( g, h\\right) }^{m} = \\left( {{g}^{m},{h}^{m}}\\right) = \\left( {{e}_{G},{e}_{H}}\\right) \n\] \n\n\[ \n\\left( {{g}^{n},{h}^{n}}\\right) = {\\left( g, h\\ri... | Yes |
Let \( \left( {8,{56}}\right) \in {\mathbb{Z}}_{12} \times {\mathbb{Z}}_{60} \). Since \( \gcd \left( {8,{12}}\right) = 4 \), the order of 8 is \( {12}/4 = 3 \) in \( {\mathbb{Z}}_{12} \). Similarly, the order of 56 in \( {\mathbb{Z}}_{60} \) is 15. The least common multiple of 3 and 15 is 15; hence, \( \left( {8,{56}}... | Since \( \gcd \left( {8,{12}}\right) = 4 \), the order of 8 is \( {12}/4 = 3 \) in \( {\mathbb{Z}}_{12} \). Similarly, the order of 56 in \( {\mathbb{Z}}_{60} \) is 15. The least common multiple of 3 and 15 is 15; hence, \( \left( {8,{56}}\right) \) has order 15 in \( {\mathbb{Z}}_{12} \times {\mathbb{Z}}_{60} \). | Yes |
The group \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \) consists of the pairs\n\n\[ \left( {0,0}\right) ,\;\left( {0,1}\right) ,\;\left( {0,2}\right) ,\;\left( {1,0}\right) ,\;\left( {1,1}\right) ,\;\left( {1,2}\right) . \]\n\nIn this case, unlike that of \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \) and \( {\mathbb{... | It is easy to see that \( \left( {1,1}\right) \) is a generator for \( {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \) . | Yes |
Theorem 9.21. The group \( {\mathbb{Z}}_{m} \times {\mathbb{Z}}_{n} \) is isomorphic to \( {\mathbb{Z}}_{mn} \) if and only if \( \gcd \left( {m, n}\right) = 1 \) . | Proof. We will first show that if \( {\mathbb{Z}}_{m} \times {\mathbb{Z}}_{n} \cong {\mathbb{Z}}_{mn} \), then \( \gcd \left( {m, n}\right) = 1 \) . We will prove the contrapositive; that is, we will show that if \( \gcd \left( {m, n}\right) = d > 1 \), then \( {\mathbb{Z}}_{m} \times {\mathbb{Z}}_{n} \) cannot be cycl... | Yes |
Corollary 9.23. If\n\n\[ \nm = {p}_{1}^{{e}_{1}}\cdots {p}_{k}^{{e}_{k}} \n\]\n\nwhere the \( {p}_{i}s \) are distinct primes, then\n\n\[ \n{\mathbb{Z}}_{m} \cong {\mathbb{Z}}_{{p}_{1}^{{e}_{1}}} \times \cdots \times {\mathbb{Z}}_{{p}_{k}^{{e}_{k}}} \n\] | Proof. Since the greatest common divisor of \( {p}_{i}^{{e}_{i}} \) and \( {p}_{j}^{{e}_{j}} \) is 1 for \( i \neq j \), the proof follows from Corollary 9.22. | No |
The dihedral group \( {D}_{6} \) is an internal direct product of its two subgroups\n\n\[ H = \left\{ {\mathrm{{id}},{r}^{3}}\right\} \;\text{ and }\;K = \left\{ {\mathrm{{id}},{r}^{2},{r}^{4}, s,{r}^{2}s,{r}^{4}s}\right\} . \] | It can easily be shown that \( K \cong {S}_{3} \) ; consequently, \( {D}_{6} \cong {\mathbb{Z}}_{2} \times {S}_{3} \) . | No |
Not every group can be written as the internal direct product of two of its proper subgroups. | If the group \( {S}_{3} \) were an internal direct product of its proper subgroups \( H \) and \( K \), then one of the subgroups, say \( H \), would have to have order 3 . In this case \( H \) is the subgroup \( \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \) . The subgroup \( K \) must have order 2... | Yes |
Theorem 9.27. Let \( G \) be the internal direct product of subgroups \( H \) and \( K \). Then \( G \) is isomorphic to \( H \times K \). | Proof. Since \( G \) is an internal direct product, we can write any element \( g \in G \) as \( g = {hk} \) for some \( h \in H \) and some \( k \in K \). Define a map \( \phi : G \rightarrow H \times K \) by \( \phi \left( g\right) = \left( {h, k}\right) \). The first problem that we must face is to show that \( \phi... | No |
Let \( G \) be an abelian group. Every subgroup \( H \) of \( G \) is a normal subgroup. | Since \( {gh} = {hg} \) for all \( g \in G \) and \( h \in H \), it will always be the case that \( {gH} = {Hg} \). | Yes |
Let \( H \) be the subgroup of \( {S}_{3} \) consisting of elements (1) and (12). Since\n\n\[ \left( {123}\right) H = \{ \left( {123}\right) ,\left( {13}\right) \} \text{ and }H\left( {123}\right) = \{ \left( {123}\right) ,\left( {23}\right) \} \]\n\n\( H \) cannot be a normal subgroup of \( {S}_{3} \). | However, the subgroup \( N \), consisting of the permutations (1),(123), and (132), is normal since the cosets of \( N \) are\n\n\[ N = \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \]\n\n\[ \left( {12}\right) N = N\left( {12}\right) = \{ \left( {12}\right) ,\left( {13}\right) ,\left( {23}\right) \} .... | Yes |
Theorem 10.3. Let \( G \) be a group and \( N \) be a subgroup of \( G \) . Then the following statements are equivalent.\n\n1. The subgroup \( N \) is normal in \( G \) .\n\n2. For all \( g \in G,{gN}{g}^{-1} \subset N \) .\n\n3. For all \( g \in G,{gN}{g}^{-1} = N \) . | Proof. \( \;\left( 1\right) \Rightarrow \left( 2\right) \) . Since \( N \) is normal in \( G,{gN} = {Ng} \) for all \( g \in G \) . Hence, for a given \( g \in G \) and \( n \in N \), there exists an \( {n}^{\prime } \) in \( N \) such that \( {gn} = {n}^{\prime }g \) . Therefore, \( {gn}{g}^{-1} = {n}^{\prime } \in N ... | Yes |
Theorem 10.4. Let \( N \) be a normal subgroup of a group \( G \) . The cosets of \( N \) in \( G \) form a group \( G/N \) of order \( \left\lbrack {G : N}\right\rbrack \) . | Proof. The group operation on \( G/N \) is \( \left( {aN}\right) \left( {bN}\right) = {abN} \) . This operation must be shown to be well-defined; that is, group multiplication must be independent of the choice of coset representative. Let \( {aN} = {bN} \) and \( {cN} = {dN} \) . We must show that\n\n\[ \left( {aN}\rig... | Yes |
Consider the normal subgroup of \( {S}_{3}, N = \{ \left( 1\right) ,\left( {123}\right) ,\left( {132}\right) \} \) . The cosets of \( N \) in \( {S}_{3} \) are \( N \) and \( \left( {12}\right) N \) . The factor group \( {S}_{3}/N \) has the following multiplication table. | <table><thead><tr><th></th><th>\( N \)</th><th>\( \left( {12}\right) N \)</th></tr></thead><tr><td>\( N \)</td><td>\( N \)</td><td>\( \left( \begin{matrix} {12} \\ \end{matrix}\right) N \)</td></tr><tr><td>\( \left( {12}\right) N \)</td><td>\( \left( \begin{matrix} {12} \\ \end{matrix}\right) N \)</td><td>\( N \)</td><... | Yes |
Consider the normal subgroup \( 3\mathbb{Z} \) of \( \mathbb{Z} \). The cosets of \( 3\mathbb{Z} \) in \( \mathbb{Z} \) are\n\n\[ 0 + 3\mathbb{Z} = \{ \ldots , - 3,0,3,6,\ldots \}\]\n\n\[ 1 + 3\mathbb{Z} = \{ \ldots , - 2,1,4,7,\ldots \}\]\n\n\[ 2 + 3\mathbb{Z} = \{ \ldots , - 1,2,5,8,\ldots \}\] | The group \( \mathbb{Z}/3\mathbb{Z} \) is given by the multiplication table below.\n\n\[ \begin{matrix} + & 0 + 3\mathbb{Z} & 1 + 3\mathbb{Z} & 2 + 3\mathbb{Z} \\ 0 + 3\mathbb{Z} & 0 + 3\mathbb{Z} & 1 + 3\mathbb{Z} & 2 + 3\mathbb{Z} \\ 1 + 3\mathbb{Z} & 1 + 3\mathbb{Z} & 2 + 3\mathbb{Z} & 0 + 3\mathbb{Z} \\ 2 + 3\mathb... | Yes |
Lemma 10.8. The alternating group \( {A}_{n} \) is generated by 3-cycles for \( n \geq 3 \) . | Proof. To show that the 3-cycles generate \( {A}_{n} \), we need only show that any pair of transpositions can be written as the product of 3-cycles. Since \( \left( {ab}\right) = \left( {ba}\right) \), every pair of transpositions must be one of the following:\n\n\[ \left( {ab}\right) \left( {ab}\right) = \mathrm{{id}... | Yes |
Lemma 10.9. Let \( N \) be a normal subgroup of \( {A}_{n} \), where \( n \geq 3 \) . If \( N \) contains a 3-cycle, then \( N = {A}_{n} \) . | Proof. We will first show that \( {A}_{n} \) is generated by 3-cycles of the specific form \( \left( {ijk}\right) \) , where \( i \) and \( j \) are fixed in \( \{ 1,2,\ldots, n\} \) and we let \( k \) vary. Every 3-cycle is the product of 3-cycles of this form, since\n\n\[ \left( {iaj}\right) = {\left( ija\right) }^{2... | Yes |
Theorem 10.11. The alternating group, \( {A}_{n} \), is simple for \( n \geq 5 \) . | Proof. Let \( N \) be a normal subgroup of \( {A}_{n} \) . By Lemma 10.10, \( N \) contains a 3-cycle. By Lemma 10.9, \( N = {A}_{n} \) ; therefore, \( {A}_{n} \) contains no proper nontrivial normal subgroups for \( n \geq 5 \) . | Yes |
Let \( G \) be a group and \( g \in G \) . Define a map \( \phi : \mathbb{Z} \rightarrow G \) by \( \phi \left( n\right) = {g}^{n} \) . Then \( \phi \) is a group homomorphism, since | \( \phi \left( {m + n}\right) = {g}^{m + n} = {g}^{m}{g}^{n} = \phi \left( m\right) \phi \left( n\right) . | Yes |
Recall that the circle group \( \mathbb{T} \) consists of all complex numbers \( z \) such that \( \left| z\right| = 1 \) . We can define a homomorphism \( \phi \) from the additive group of real numbers \( \mathbb{R} \) to \( \mathbb{T} \) by \( \phi : \theta \mapsto \cos \theta + i\sin \theta \) . | Indeed, \[ \phi \left( {\alpha + \beta }\right) = \cos \left( {\alpha + \beta }\right) + i\sin \left( {\alpha + \beta }\right) \] \[ = \left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta }\right) + i\left( {\sin \alpha \cos \beta + \cos \alpha \sin \beta }\right) \] \[ = \left( {\cos \alpha + i\sin \alpha }\right) ... | Yes |
Proposition 11.4. Let \( \phi : {G}_{1} \rightarrow {G}_{2} \) be a homomorphism of groups. Then\n\n1. If \( e \) is the identity of \( {G}_{1} \), then \( \phi \left( e\right) \) is the identity of \( {G}_{2} \) ;\n\n2. For any element \( g \in {G}_{1},\phi \left( {g}^{-1}\right) = {\left\lbrack \phi \left( g\right) \... | Proof. (1) Suppose that \( e \) and \( {e}^{\prime } \) are the identities of \( {G}_{1} \) and \( {G}_{2} \), respectively; then\n\n\[ \n{e}^{\prime }\phi \left( e\right) = \phi \left( e\right) = \phi \left( {ee}\right) = \phi \left( e\right) \phi \left( e\right) .\n\]\n\nBy cancellation, \( \phi \left( e\right) = {e}... | Yes |
Let us examine the homomorphism \( \phi : G{L}_{2}\left( \mathbb{R}\right) \rightarrow {\mathbb{R}}^{ * } \) defined by \( A \mapsto \) \( \det \left( A\right) \). | Since 1 is the identity of \( {\mathbb{R}}^{ * } \), the kernel of this homomorphism is all \( 2 \times 2 \) matrices having determinant one. That is, \( \ker \phi = S{L}_{2}\left( \mathbb{R}\right) \). | Yes |
The kernel of the group homomorphism \( \phi : \mathbb{R} \rightarrow {\mathbb{C}}^{ * } \) defined by \( \phi \left( \theta \right) = \) \( \cos \theta + i\sin \theta \) is \( \{ {2\pi n} : n \in \mathbb{Z}\} \). | Notice that \( \ker \phi \cong \mathbb{Z} \). | No |
Suppose that we wish to determine all possible homomorphisms \( \phi \) from \( {\mathbb{Z}}_{7} \) to \( {\mathbb{Z}}_{12} \) | Since the kernel of \( \phi \) must be a subgroup of \( {\mathbb{Z}}_{7} \), there are only two possible kernels, \( \{ 0\} \) and all of \( {\mathbb{Z}}_{7} \). The image of a subgroup of \( {\mathbb{Z}}_{7} \) must be a subgroup of \( {\mathbb{Z}}_{12} \). Hence, there is no injective homomorphism; otherwise, \( {\ma... | Yes |
Theorem 11.10 First Isomorphism Theorem. If \( \psi : G \rightarrow H \) is a group homomorphism with \( K = \ker \psi \), then \( K \) is normal in \( G \) . Let \( \phi : G \rightarrow G/K \) be the canonical homomorphism. Then there exists a unique isomorphism \( \eta : G/K \rightarrow \psi \left( G\right) \) such t... | Proof. We already know that \( K \) is normal in \( G \) . Define \( \eta : G/K \rightarrow \psi \left( G\right) \) by \( \eta \left( {gK}\right) = \) \( \psi \left( g\right) \) . We first show that \( \eta \) is a well-defined map. If \( {g}_{1}K = {g}_{2}K \), then for some \( k \in K \) , \( {g}_{1}k = {g}_{2} \) ; ... | Yes |
Let \( G \) be a cyclic group with generator \( g \) . Define a map \( \phi : \mathbb{Z} \rightarrow G \) by \( n \mapsto {g}^{n} \) . This map is a surjective homomorphism since | \[ \phi \left( {m + n}\right) = {g}^{m + n} = {g}^{m}{g}^{n} = \phi \left( m\right) \phi \left( n\right) . \] Clearly \( \phi \) is onto. If \( \left| g\right| = m \), then \( {g}^{m} = e \) . Hence, \( \ker \phi = m\mathbb{Z} \) and \( \mathbb{Z}/\ker \phi = \mathbb{Z}/m\mathbb{Z} \cong G \) . On the other hand, if th... | Yes |
Theorem 11.12 Second Isomorphism Theorem. Let \( H \) be a subgroup of a group \( G \) (not necessarily normal in \( G \) ) and \( N \) a normal subgroup of \( G \) . Then \( {HN} \) is a subgroup of \( G, H \cap N \) is a normal subgroup of \( H \), and\n\n\[ H/H \cap N \cong {HN}/N\text{.} \] | Proof. We will first show that \( {HN} = \{ {hn} : h \in H, n \in N\} \) is a subgroup of \( G \) . Suppose that \( {h}_{1}{n}_{1},{h}_{2}{n}_{2} \in {HN} \) . Since \( N \) is normal, \( {\left( {h}_{2}\right) }^{-1}{n}_{1}{h}_{2} \in N \) . So\n\n\[ \left( {{h}_{1}{n}_{1}}\right) \left( {{h}_{2}{n}_{2}}\right) = {h}_... | Yes |
Theorem 11.13 Correspondence Theorem. Let \( N \) be a normal subgroup of a group \( G \). Then \( H \mapsto H/N \) is a one-to-one correspondence between the set of subgroups \( H \) containing \( N \) and the set of subgroups of \( G/N \). Furthermore, the normal subgroups of \( G \) containing \( N \) correspond to ... | Proof. Let \( H \) be a subgroup of \( G \) containing \( N \). Since \( N \) is normal in \( H, H/N \) makes sense. Let \( {aN} \) and \( {bN} \) be elements of \( H/N \). Then \( \left( {aN}\right) \left( {{b}^{-1}N}\right) = a{b}^{-1}N \in H/N \); hence, \( H/N \) is a subgroup of \( G/N \). Let \( S \) be a subgrou... | Yes |
Example 11.15. By the Third Isomorphism Theorem,\n\n\[ \n\mathbb{Z}/m\mathbb{Z} \cong \left( {\mathbb{Z}/{mn}\mathbb{Z}}\right) /\left( {m\mathbb{Z}/{mn}\mathbb{Z}}\right) .\n\]\n | Since \( \left| {\mathbb{Z}/{mn}\mathbb{Z}}\right| = {mn} \) and \( \left| {\mathbb{Z}/m\mathbb{Z}}\right| = m \), we have \( \left| {m\mathbb{Z}/{mn}\mathbb{Z}}\right| = n \) . | Yes |
If we let \( T : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) be the map given by\n\n\[ T\left( {{x}_{1},{x}_{2}}\right) = \left( {2{x}_{1} + 5{x}_{2}, - 4{x}_{1} + 3{x}_{2}}\right) ,\]\nthe axioms that \( T \) must satisfy to be a linear transformation are easily verified. | The column vectors \( T{\mathbf{e}}_{1} = {\left( 2, - 4\right) }^{\mathrm{t}} \) and \( T{\mathbf{e}}_{2} = {\left( 5,3\right) }^{\mathrm{t}} \) tell us that \( T \) is given by the matrix\n\n\[ A = \left( \begin{matrix} 2 & 5 \\ - 4 & 3 \end{matrix}\right) \] | Yes |
If \( A \) is the matrix \n\n\[ \left( \begin{array}{ll} 2 & 1 \\ 5 & 3 \end{array}\right) \]\n\nthen the inverse of \( A \) is \n\n\[ {A}^{-1} = \left( \begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix}\right) \] | We are guaranteed that \( {A}^{-1} \) exists, since \( \det \left( A\right) = 2 \cdot 3 - 5 \cdot 1 = 1 \) is nonzero. | Yes |
Given a \( 2 \times 2 \) matrix\n\n\[ A = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \]\n\nthe determinant of \( A \) is \( {ad} - {bc} \) . The group \( G{L}_{2}\left( \mathbb{R}\right) \) consists of those matrices in which \( {ad} - {bc} \neq 0 \) . The inverse of \( A \) is\n\n\[ {A}^{-1} = \frac{1}... | If \( A \) is in \( S{L}_{2}\left( \mathbb{R}\right) \), then\n\n\[ {A}^{-1} = \left( \begin{matrix} d & - b \\ - c & a \end{matrix}\right) \] | No |
Example 12.5. The following matrices are orthogonal: | \[ \left( \begin{matrix} 3/5 & - 4/5 \\ 4/5 & 3/5 \end{matrix}\right) ,\;\left( \begin{matrix} 1/2 & - \sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{matrix}\right) ,\;\left( \begin{matrix} - 1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 1/\sqrt{6} & - 2/\sqrt{6} & 1/\sqrt{6} \\ 1/\sqrt{3} & 1/\sqrt{3} & 1/\sqrt{3} \end{matrix}\right) \] | Yes |
The vector \( \mathbf{x} = {\left( 3,4\right) }^{\mathrm{t}} \) has length \( \sqrt{{3}^{2} + {4}^{2}} = 5 \) . We can also see that the orthogonal matrix \[ A = \left( \begin{matrix} 3/5 & - 4/5 \\ 4/5 & 3/5 \end{matrix}\right) \] preserves the length of this vector. The vector \( A\mathbf{x} = {\left( -7/5,{24}/5\rig... | Since \( \det \left( {A{A}^{\mathrm{t}}}\right) = \det \left( I\right) = 1 \) and \( \det \left( A\right) = \det \left( {A}^{\mathrm{t}}\right) \), the determinant of any orthogonal matrix is either 1 or -1 . Consider the column vectors \[ {\mathbf{a}}_{j} = \left( \begin{matrix} {a}_{1j} \\ {a}_{2j} \\ \vdots \\ {a}_{... | Yes |
Theorem 12.8. Let \( A \) be an \( n \times n \) matrix. The following statements are equivalent.\n\n1. The columns of the matrix \( A \) form an orthonormal set.\n\n2. \( {A}^{-1} = {A}^{t} \) .\n\n3. For vectors \( \mathbf{x} \) and \( \mathbf{y},\langle A\mathbf{x}, A\mathbf{y}\rangle = \langle \mathbf{x},\mathbf{y}... | Proof. We have already shown (1) and (2) to be equivalent.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) \) .\n\n\[ \langle A\mathbf{x}, A\mathbf{y}\rangle = {\left( A\mathbf{x}\right) }^{\mathrm{t}}A\mathbf{y} \]\n\n\[ = {\mathbf{x}}^{\mathrm{t}}{A}^{\mathrm{t}}A\mathbf{y} \]\n\n\[ = {\mathbf{x}}^{\mathrm{t}}\math... | Yes |
Let us examine the orthogonal group on \( {\mathbb{R}}^{2} \) a bit more closely. An element \( T \in O\left( 2\right) \) is determined by its action on \( {\mathbf{e}}_{1} = {\left( 1,0\right) }^{\mathrm{t}} \) and \( {\mathbf{e}}_{2} = {\left( 0,1\right) }^{\mathrm{t}} \) . If \( T\left( {\mathbf{e}}_{1}\right) = {\l... | \[ A = \left( \begin{matrix} a & - b \\ b & a \end{matrix}\right) = \left( \begin{matrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{matrix}\right) \] where \( 0 \leq \theta < {2\pi } \) . A matrix \( T \) in \( O\left( 2\right) \) either reflects or rotates a vector in \( {\mathbb{R}}^{2} \) (Figure... | No |
Lemma 12.13. An isometry \( f \) that fixes the origin in \( {\mathbb{R}}^{2} \) is a linear transformation. In particular, \( f \) is given by an element in \( O\left( 2\right) \) . | Proof. Let \( f \) be an isometry in \( {\mathbb{R}}^{2} \) fixing the origin. We will first show that \( f \) preserves inner products. Since \( f\left( 0\right) = 0,\parallel f\left( \mathbf{x}\right) \parallel = \parallel \mathbf{x}\parallel \) ; therefore,\n\n\[ \parallel \mathbf{x}{\parallel }^{2} - 2\langle f\lef... | Yes |
Not all groups are finitely generated. Consider the rational numbers \( \mathbb{Q} \) under the operation of addition. Suppose that \( \mathbb{Q} \) is finitely generated with generators \( {p}_{1}/{q}_{1},\ldots ,{p}_{n}/{q}_{n} \), where each \( {p}_{i}/{q}_{i} \) is a fraction expressed in its lowest terms. | Let \( p \) be some prime that does not divide any of the denominators \( {q}_{1},\ldots ,{q}_{n} \). We claim that \( 1/p \) cannot be in the subgroup of \( \mathbb{Q} \) that is generated by \( {p}_{1}/{q}_{1},\ldots ,{p}_{n}/{q}_{n} \), since \( p \) does not divide the denominator of any element in this subgroup. T... | Yes |
Proposition 13.3. Let \( H \) be the subgroup of a group \( G \) that is generated by \( \left\{ {{g}_{i} \in G : i \in I}\right\} \) . Then \( h \in H \) exactly when it is a product of the form\n\n\[ h = {g}_{{i}_{1}}^{{\alpha }_{1}}\cdots {g}_{{i}_{n}}^{{\alpha }_{n}} \]\n\nwhere the \( {g}_{{i}_{k}}s \) are not nec... | Proof. Let \( K \) be the set of all products of the form \( {g}_{{i}_{1}}^{{\alpha }_{1}}\cdots {g}_{{i}_{n}}^{{\alpha }_{n}} \), where the \( {g}_{{i}_{k}}\mathrm{\;s} \) are not necessarily distinct. Certainly \( K \) is a subset of \( H \) . We need only show that \( K \) is a subgroup of \( G \) . If this is the c... | Yes |
Suppose that we wish to classify all abelian groups of order \( {540} = {2}^{2} \cdot {3}^{3} \cdot 5 \) . | The Fundamental Theorem of Finite Abelian Groups tells us that we have the following six possibilities.\n\n\[ \text{-}{\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2} \times {\mathbb{Z}}_{3} \times {\mathbb{Z}}_{3} \times {\mathbb{Z}}_{3} \times {\mathbb{Z}}_{5}\text{;} \]\n\n\[ \text{-}{\mathbb{Z}}_{2} \times {\mathbb{Z}}_{2}... | Yes |
Lemma 13.6. Let \( G \) be a finite abelian group of order \( n \) . If \( p \) is a prime that divides \( n \) , then \( G \) contains an element of order \( p \) . | Proof. We will prove this lemma by induction. If \( n = 1 \), then there is nothing to show. Now suppose that the order of \( G \) is \( n \) the lemma is true for all groups of order \( k \), where \( k < n \) . Furthermore, let \( p \) be a prime that divides \( n \) .\n\nIf \( G \) has no proper nontrivial subgroups... | Yes |
Lemma 13.7. A finite abelian group is a p-group if and only if its order is a power of \( p \) . | Proof. If \( \left| G\right| = {p}^{n} \) then by Lagrange’s theorem, then the order of any \( g \in G \) must divide \( {p}^{n} \), and therefore must be a power of \( p \) . Conversely, if \( \left| G\right| \) is not a power of \( p \), then it has some other prime divisor \( q \), so by Lemma 13.6, \( G \) has an e... | Yes |
Lemma 13.8. Let \( G \) be a finite abelian group of order \( n = {p}_{1}^{{\alpha }_{1}}\cdots {p}_{k}^{{\alpha }_{k}} \), where where \( {p}_{1},\ldots ,{p}_{k} \) are distinct primes and \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{k} \) are positive integers. Then \( G \) is the internal direct product of subg... | Proof. Since \( G \) is an abelian group, we are guaranteed that \( {G}_{i} \) is a subgroup of \( G \) for \( i = 1,\ldots, n \) . Since the identity has order \( {p}_{i}^{0} = 1 \), we know that \( 1 \in {G}_{i} \) . If \( g \in {G}_{i} \) has order \( {p}_{i}^{r} \), then \( {g}^{-1} \) must also have order \( {p}_{... | Yes |
Lemma 13.9. Let \( G \) be a finite abelian p-group and suppose that \( g \in G \) has maximal order. Then \( G \) is isomorphic to \( \langle g\rangle \times H \) for some subgroup \( H \) of \( G \) . | Proof. By Lemma 13.7, we may assume that the order of \( G \) is \( {p}^{n} \) . We shall induct on \( n \) . If \( n = 1 \), then \( G \) is cyclic of order \( p \) and must be generated by \( g \) . Suppose now that the statement of the lemma holds for all integers \( k \) with \( 1 \leq k < n \) and let \( g \) be o... | Yes |
Any series of subgroups of an abelian group is a normal series. | Consider the following series of groups:\n\n\[ \mathbb{Z} \supset 9\mathbb{Z} \supset {45}\mathbb{Z} \supset {180}\mathbb{Z} \supset \{ 0\} \]\n\n\[ {\mathbb{Z}}_{24} \supset \langle 2\rangle \supset \langle 6\rangle \supset \langle {12}\rangle \supset \{ 0\} . \] | No |
A subnormal series need not be a normal series. Consider the following subnormal series of the group \( {D}_{4} \) :\n\n\[ \n{D}_{4} \supset \{ \left( 1\right) ,\left( {12}\right) \left( {34}\right) ,\left( {13}\right) \left( {24}\right) ,\left( {14}\right) \left( {23}\right) \} \supset \{ \left( 1\right) ,\left( {12}\... | The subgroup \( \{ \left( 1\right) ,\left( {12}\right) \left( {34}\right) \} \) is not normal in \( {D}_{4} \) ; consequently, this series is not a normal series. | Yes |
The two normal series\n\n\[ \n{\\mathbb{Z}}_{60} \\supset \\langle 3\\rangle \\supset \\langle {15}\\rangle \\supset \\{ 0\\}\n\]\n\n\[ \n{\\mathbb{Z}}_{60} \\supset \\langle 4\\rangle \\supset \\langle {20}\\rangle \\supset \\{ 0\\}\n\]\n\nof the group \( {\\mathbb{Z}}_{60} \) are isomorphic | since\n\n\[ \n{\\mathbb{Z}}_{60}/\\langle 3\\rangle \\cong \\langle {20}\\rangle /\\{ 0\\} \\cong {\\mathbb{Z}}_{3}\n\]\n\n\[ \n\\langle 3\\rangle /\\langle {15}\\rangle \\cong \\langle 4\\rangle /\\langle {20}\\rangle \\cong {\\mathbb{Z}}_{5}\n\]\n\n\[ \n\\langle {15}\\rangle /\\{ 0\\} \\cong {\\mathbb{Z}}_{60}/\\lang... | Yes |
The group \( {\mathbb{Z}}_{60} \) has a composition series | \[ {\mathbb{Z}}_{60} \supset \langle 3\rangle \supset \langle {15}\rangle \supset \langle {30}\rangle \supset \{ 0\} \] with factor groups \[ {\mathbb{Z}}_{60}/\langle 3\rangle \cong {\mathbb{Z}}_{3} \] \[ \langle 3\rangle /\langle {15}\rangle \cong {\mathbb{Z}}_{5} \] \[ \langle {15}\rangle /\langle {30}\rangle \cong ... | Yes |
The group \( {S}_{4} \) is solvable since | \[ {S}_{4} \supset {A}_{4} \supset \{ \left( 1\right) ,\left( {12}\right) \left( {34}\right) ,\left( {13}\right) \left( {24}\right) ,\left( {14}\right) \left( {23}\right) \} \supset \{ \left( 1\right) \} \] has abelian factor groups; | Yes |
Let \( G = {D}_{4} \) be the symmetry group of a square. If \( X = \{ 1,2,3,4\} \) is the set of vertices of the square, then we can consider \( {D}_{4} \) to consist of the following permutations:\n\n\[ \n\{ \left( 1\right) ,\left( {13}\right) ,\left( {24}\right) ,\left( {1432}\right) ,\left( {1234}\right) ,\left( {12... | It is easy to see that the axioms of a group action are satisfied. | No |
If we let \( X = G \), then every group \( G \) acts on itself by the left regular representation; that is, \( \left( {g, x}\right) \mapsto {\lambda }_{g}\left( x\right) = {gx} \), where \( {\lambda }_{g} \) is left multiplication: | \[ e \cdot x = {\lambda }_{e}x = {ex} = x \] \[ \left( {gh}\right) \cdot x = {\lambda }_{gh}x = {\lambda }_{g}{\lambda }_{h}x = {\lambda }_{g}\left( {hx}\right) = g \cdot \left( {h \cdot x}\right) . \] | Yes |
Let \( G \) be a group and suppose that \( X = G \) . If \( H \) is a subgroup of \( G \), then \( G \) is an \( H \) -set under conjugation; that is, we can define an action of \( H \) on \( G \ | via\n\n\[ \left( {h, g}\right) \mapsto {hg}{h}^{-1} \]\n\nfor \( h \in H \) and \( g \in G \) . Clearly, the first axiom for a group action holds. Observing that\n\n\[ \left( {{h}_{1}{h}_{2}, g}\right) = {h}_{1}{h}_{2}g{\left( {h}_{1}{h}_{2}\right) }^{-1} \]\n\n\[ = {h}_{1}\left( {{h}_{2}g{h}_{2}^{-1}}\right) {h}_{1}^{... | Yes |
Let \( H \) be a subgroup of \( G \) and \( {\mathcal{L}}_{H} \) the set of left cosets of \( H \). The set \( {\mathcal{L}}_{H} \) is a \( G \)-set under the action \[ \left( {g,{xH}}\right) \mapsto {gxH}. \] | Again, it is easy to see that the first axiom is true. Since \( \left( {g{g}^{\prime }}\right) {xH} = g\left( {{g}^{\prime }{xH}}\right) \), the second axiom is also true. | No |
Proposition 14.6. Let \( X \) be a \( G \) -set. Then \( G \) -equivalence is an equivalence relation on \( X \) . | Proof. The relation \( \sim \) is reflexive since \( {ex} = x \) . Suppose that \( x \sim y \) for \( x, y \in X \) . Then there exists a \( g \) such that \( {gx} = y \) . In this case \( {g}^{-1}y = x \) ; hence, \( y \sim x \) . To show that the relation is transitive, suppose that \( x \sim y \) and \( y \sim z \) ... | No |
Let \( X = \{ 1,2,3,4,5,6\} \) and suppose that \( G \) is the permutation group given by the permutations \[ \{ \left( 1\right) ,\left( {12}\right) \left( {3456}\right) ,\left( {35}\right) \left( {46}\right) ,\left( {12}\right) \left( {3654}\right) \} . \] | Then the fixed point sets of \( X \) under the action of \( G \) are \[ {X}_{\left( 1\right) } = X \] \[ {X}_{\left( {35}\right) \left( {46}\right) } = \{ 1,2\} \] \[ {X}_{\left( {12}\right) \left( {3456}\right) } = {X}_{\left( {12}\right) \left( {3654}\right) } = \varnothing , \] and the stabilizer subgroups are \[ {G... | Yes |
Proposition 14.10. Let \( G \) be a group acting on a set \( X \) and \( x \in X \) . The stabilizer group of \( x,{G}_{x} \), is a subgroup of \( G \) . | Proof. Clearly, \( e \in {G}_{x} \) since the identity fixes every element in the set \( X \) . Let \( g, h \in {G}_{x} \) . Then \( {gx} = x \) and \( {hx} = x \) . So \( \left( {gh}\right) x = g\left( {hx}\right) = {gx} = x \) ; hence, the product of two elements in \( {G}_{x} \) is also in \( {G}_{x} \) . Finally, i... | Yes |
Theorem 14.11. Let \( G \) be a finite group and \( X \) a finite \( G \) -set. If \( x \in X \), then \( \left| {\mathcal{O}}_{x}\right| = \lbrack G \) : \( \left. {G}_{x}\right\rbrack \) . | Proof. We know that \( \left| G\right| /\left| {G}_{x}\right| \) is the number of left cosets of \( {G}_{x} \) in \( G \) by Lagrange’s Theorem (Theorem 6.10). We will define a bijective map \( \phi \) between the orbit \( {\mathcal{O}}_{x} \) of \( X \) and the set of left cosets \( {\mathcal{L}}_{{G}_{x}} \) of \( {G... | Yes |
For \( {S}_{n} \) it takes a bit of work to find the conjugacy classes. We begin with cycles. Suppose that \( \sigma = \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) is a cycle and let \( \tau \in {S}_{n} \) . By Theorem 6.16, | \[ {\tau \sigma }{\tau }^{-1} = \left( {\tau \left( {a}_{1}\right) ,\ldots ,\tau \left( {a}_{k}\right) }\right) . \] Consequently, any two cycles of the same length are conjugate. Now let \( \sigma = {\sigma }_{1}{\sigma }_{2}\cdots {\sigma }_{r} \) be a cycle decomposition, where the length of each cycle \( {\sigma }_... | Yes |
Theorem 14.15. Let \( G \) be a group of order \( {p}^{n} \) where \( p \) is prime. Then \( G \) has a nontrivial center. | Proof. We apply the class equation\n\n\[ \left| G\right| = \left| {Z\left( G\right) }\right| + {n}_{1} + \cdots + {n}_{k} \]\n\nSince each \( {n}_{i} > 1 \) and \( {n}_{i}\left| \right| G \mid \), it follows that \( p \) must divide each \( {n}_{i} \) . Also, \( p\left| \right| G \mid \) ; hence, \( p \) must divide \(... | Yes |
Corollary 14.16. Let \( G \) be a group of order \( {p}^{2} \) where \( p \) is prime. Then \( G \) is abelian. | Proof. By Theorem 14.15, \( \left| {Z\left( G\right) }\right| = p \) or \( {p}^{2} \) . If \( \left| {Z\left( G\right) }\right| = {p}^{2} \), then we are done. Suppose that \( \left| {Z\left( G\right) }\right| = p \) . Then \( Z\left( G\right) \) and \( G/Z\left( G\right) \) both have order \( p \) and must both be cyc... | Yes |
Lemma 14.18. Let \( X \) be a \( G \) -set and suppose that \( x \sim y \) . Then \( {G}_{x} \) is isomorphic to \( {G}_{y} \) . In particular, \( \left| {G}_{x}\right| = \left| {G}_{y}\right| \). | Proof. Let \( G \) act on \( X \) by \( \left( {g, x}\right) \mapsto g \cdot x \) . Since \( x \sim y \), there exists a \( g \in G \) such that \( g \cdot x = y \) . Let \( a \in {G}_{x} \) . Since\n\n\[ \n{ga}{g}^{-1} \cdot y = {ga} \cdot {g}^{-1}y = {ga} \cdot x = g \cdot x = y, \n\]\n\nwe can define a map \( \phi :... | Yes |
Theorem 14.19 Burnside. Let \( G \) be a finite group acting on a set \( X \) and let \( k \) denote the number of orbits of \( X \) . Then\n\n\[ k = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{g \in G}}\left| {X}_{g}\right| \] | Proof. We look at all the fixed points \( x \) of all the elements in \( g \in G \) ; that is, we look at all \( g \) ’s and all \( x \) ’s such that \( {gx} = x \) . If viewed in terms of fixed point sets, the number of all \( g \) ’s fixing \( x \) ’s is\n\n\[ \mathop{\sum }\limits_{{g \in G}}\left| {X}_{g}\right| \]... | Yes |
Let \( X = \{ 1,2,3,4,5\} \) and suppose that \( G \) is the permutation group \( G = \{ \left( 1\right) ,\left( {13}\right) ,\left( {13}\right) \left( {25}\right) ,\left( {25}\right) \} \) . The orbits of \( X \) are \( \{ 1,3\} ,\{ 2,5\} \), and \( \{ 4\} \) . The fixed point sets are | \[ \n{X}_{\left( 1\right) } = X \n\] \n\[ \n{X}_{\left( {13}\right) } = \{ 2,4,5\} \n\] \n\[ \n{X}_{\left( {13}\right) \left( {25}\right) } = \{ 4\} \n\] \n\[ \n{X}_{\left( {25}\right) } = \{ 1,3,4\} \n\] \nBurnside's Theorem says that \n\[ \nk = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{g \in G}}\left| {X}_{g}... | Yes |
Proposition 14.21. Let \( G \) be a permutation group of \( X \) and \( \widetilde{X} \) the set of functions from \( X \) to \( Y \) . Then there exists a permutation group \( \widetilde{G} \) acting on \( \widetilde{X} \), where \( \widetilde{\sigma } \in \widetilde{G} \) is defined by \( \widetilde{\sigma }\left( f\... | Proof. Let \( \sigma \in G \) and \( f \in \widetilde{X} \) . Clearly, \( f \circ \sigma \) is also in \( \widetilde{X} \) . Suppose that \( g \) is another function from \( X \) to \( Y \) such that \( \widetilde{\sigma }\left( f\right) = \widetilde{\sigma }\left( g\right) \) . Then for each \( x \in X \) ,\n\n\[ f\le... | Yes |
Let \( X = \{ 1,2,\ldots ,7\} \) and suppose that \( Y = \{ A, B, C\} \) . If \( g \) is the permutation of \( X \) given by \( \left( {13}\right) \left( {245}\right) = \left( {13}\right) \left( {245}\right) \left( 6\right) \left( 7\right) \), then \( n = 4 \) . Any \( f \in {\widetilde{X}}_{g} \) must have the same va... | There are \( \left| Y\right| = 3 \) such choices for any value, so \( \left| {\widetilde{X}}_{g}\right| = {3}^{4} = {81} \). | Yes |
Suppose that we wish to color the vertices of a square using four different colors. By Proposition 14.21, we can immediately decide that there are | \[ \frac{1}{8}\left( {{4}^{4} + {4}^{1} + {4}^{2} + {4}^{1} + {4}^{2} + {4}^{2} + {4}^{3} + {4}^{3}}\right) = {55} \] | Yes |
Theorem 15.1 Cauchy. Let \( G \) be a finite group and \( p \) a prime such that \( p \) divides the order of \( G \) . Then \( G \) contains a subgroup of order \( p \) . | Proof. We will use induction on the order of \( G \) . If \( \left| G\right| = p \), then clearly \( G \) itself is the required subgroup. We now assume that every group of order \( k \), where \( p \leq k < n \) and \( p \) divides \( k \), has an element of order \( p \) . Assume that \( \left| G\right| = n \) and \(... | Yes |
Theorem 15.4 First Sylow Theorem. Let \( G \) be a finite group and \( p \) a prime such that \( {p}^{r} \) divides \( \left| G\right| \) . Then \( G \) contains a subgroup of order \( {p}^{r} \) . | Proof. We induct on the order of \( G \) once again. If \( \left| G\right| = p \), then we are done. Now suppose that the order of \( G \) is \( n \) with \( n > p \) and that the theorem is true for all groups of order less than \( n \), where \( p \) divides \( n \) . We shall apply the class equation once again:\n\n... | Yes |
Lemma 15.5. Let \( P \) be a Sylow p-subgroup of a finite group \( G \) and let \( x \) have as its order a power of \( p \) . If \( {x}^{-1}{Px} = P \), then \( x \in P \) . | Proof. Certainly \( x \in N\left( P\right) \), and the cyclic subgroup, \( \langle {xP}\rangle \subset N\left( P\right) /P \), has as its order a power of \( p \) . By the Correspondence Theorem there exists a subgroup \( H \) of \( N\left( P\right) \) containing \( P \) such that \( H/P = \langle {xP}\rangle \) . Sinc... | Yes |
Lemma 15.6. Let \( H \) and \( K \) be subgroups of \( G \) . The number of distinct \( H \) -conjugates of \( K \) is \( \left\lbrack {H : N\left( K\right) \cap H}\right\rbrack \) . | Proof. We define a bijection between the conjugacy classes of \( K \) and the right cosets of \( N\left( K\right) \cap H \) by \( {h}^{-1}{Kh} \mapsto \left( {N\left( K\right) \cap H}\right) h \) . To show that this map is a bijection, let \( {h}_{1},{h}_{2} \in H \) and suppose that \( \left( {N\left( K\right) \cap H}... | Yes |
Theorem 15.7 Second Sylow Theorem. Let \( G \) be a finite group and \( p \) a prime dividing \( \left| G\right| \) . Then all Sylow p-subgroups of \( G \) are conjugate. That is, if \( {P}_{1} \) and \( {P}_{2} \) are two Sylow p-subgroups, there exists a \( g \in G \) such that \( g{P}_{1}{g}^{-1} = {P}_{2} \) . | Proof. Let \( P \) be a Sylow \( p \) -subgroup of \( G \) and suppose that \( \left| G\right| = {p}^{r}m \) with \( \left| P\right| = {p}^{r} \) . Let\n\n\[ \mathcal{S} = \left\{ {P = {P}_{1},{P}_{2},\ldots ,{P}_{k}}\right\} \]\n\nconsist of the distinct conjugates of \( P \) in \( G \) . By Lemma 15.6, \( k = \left\l... | Yes |
Theorem 15.8 Third Sylow Theorem. Let \( G \) be a finite group and let \( p \) be a prime dividing the order of \( G \) . Then the number of Sylow p-subgroups is congruent to \( 1\left( {\;\operatorname{mod}\;p}\right) \) and divides \( \left| G\right| \) . | Proof. Let \( P \) be a Sylow \( p \) -subgroup acting on the set of Sylow \( p \) -subgroups,\n\n\[ \mathcal{S} = \left\{ {P = {P}_{1},{P}_{2},\ldots ,{P}_{k}}\right\} \]\n\nby conjugation. From the proof of the Second Sylow Theorem, the only \( P \) -conjugate of \( P \) is itself and the order of the other \( P \) -... | Yes |
Using the Sylow Theorems, we can determine that \( {A}_{5} \) has subgroups of orders \( 2,3,4 \), and 5 . The Sylow \( p \) -subgroups of \( {A}_{5} \) have orders 3,4, and 5 . The Third Sylow Theorem tells us exactly how many Sylow \( p \) -subgroups \( {A}_{5} \) has. Since the number of Sylow 5-subgroups must divid... | All Sylow 5-subgroups are conjugate. If there were only a single Sylow 5-subgroup, it would be conjugate to itself; that is, it would be a normal subgroup of \( {A}_{5} \) . Since \( {A}_{5} \) has no normal subgroups, this is impossible; hence, we have determined that there are exactly six distinct Sylow 5-subgroups o... | Yes |
Theorem 15.10. If \( p \) and \( q \) are distinct primes with \( p < q \), then every group \( G \) of order \( {pq} \) has a single subgroup of order \( q \) and this subgroup is normal in \( G \) . Hence, \( G \) cannot be simple. Furthermore, if \( q ≢ 1\\left( {\\operatorname{mod}\\;p}\\right) \), then \( G \) is ... | Proof. We know that \( G \) contains a subgroup \( H \) of order \( q \) . The number of conjugates of \( H \) divides \( {pq} \) and is equal to \( 1 + {kq} \) for \( k = 0,1,\\ldots \) . However, \( 1 + q \) is already too large to divide the order of the group; hence, \( H \) can only be conjugate to itself. That is... | Yes |
Every group of order 15 is cyclic. | This is true because \( {15} = 5 \cdot 3 \) and \( 5 ≢ 1\left( {\;\operatorname{mod}\;3}\right) \). | No |
Let us classify all of the groups of order \( {99} = {3}^{2} \cdot {11} \) up to isomorphism. First we will show that every group \( G \) of order 99 is abelian. | By the Third Sylow Theorem, there are \( 1 + {3k} \) Sylow 3-subgroups, each of order 9, for some \( k = 0,1,2,\ldots \) Also, \( 1 + {3k} \) must divide 11; hence, there can only be a single normal Sylow 3-subgroup \( H \) in \( G \) . Similarly, there are \( 1 + {11k} \) Sylow 11-subgroups and \( 1 + {11k} \) must di... | Yes |
We will now show that every group of order \( 5 \cdot 7 \cdot {47} = {1645} \) is abelian, and cyclic by Corollary 9.21. | By the Third Sylow Theorem, \( G \) has only one subgroup \( {H}_{1} \) of order 47. So \( G/{H}_{1} \) has order 35 and must be abelian by Theorem 15.10. Hence, the commutator subgroup of \( G \) is contained in \( H \) which tells us that \( \left| {G}^{\prime }\right| \) is either 1 or 47 . If \( \left| {G}^{\prime ... | Yes |
Let us show that no group \( G \) of order 20 can be simple. | By the Third Sylow Theorem, \( G \) contains one or more Sylow 5-subgroups. The number of such subgroups is congruent to \( 1\left( {\;\operatorname{mod}\;5}\right) \) and must also divide 20 . The only possible such number is 1 . Since there is only a single Sylow 5-subgroup and all Sylow 5-subgroups are conjugate, th... | Yes |
No group of order \( 56 = 2^3 \cdot 7 \) is simple. | We have seen that if we can show that there is only one Sylow \( p \)-subgroup for some prime \( p \) dividing 56, then this must be a normal subgroup and we are done. By the Third Sylow Theorem, there are either one or eight Sylow 7-subgroups. If there is only a single Sylow 7-subgroup, then it must be normal.\n\nOn t... | Yes |
Lemma 15.18. Let \( H \) and \( K \) be finite subgroups of a group \( G \) . Then\n\n\[ \left| {HK}\right| = \frac{\left| H\right| \cdot \left| K\right| }{\left| H \cap K\right| } \] | Proof. Recall that\n\n\[ {HK} = \{ {hk} : h \in H, k \in K\} . \]\n\nCertainly, \( \left| {HK}\right| \leq \left| H\right| \cdot \left| K\right| \) since some element in \( {HK} \) could be written as the product of different elements in \( H \) and \( K \) . It is quite possible that \( {h}_{1}{k}_{1} = {h}_{2}{k}_{2}... | Yes |
To demonstrate that a group \( G \) of order 48 is not simple, we will show that \( G \) contains either a normal subgroup of order 8 or a normal subgroup of order 16. | By the Third Sylow Theorem, \( G \) has either one or three Sylow 2-subgroups of order 16 . If there is only one subgroup, then it must be a normal subgroup.\n\nSuppose that the other case is true, and two of the three Sylow 2-subgroups are \( H \) and \( K \) . We claim that \( \left| {H \cap K}\right| = 8 \) . If \( ... | Yes |
For an example of a noncommutative division ring, let\n\n\[ 1 = \left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) ,\;\mathbf{i} = \left( \begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}\right) ,\;\mathbf{j} = \left( \begin{matrix} 0 & i \\ i & 0 \end{matrix}\right) ,\;\mathbf{k} = \left( \begin{matrix} i & 0 \\ 0 ... | Let \( \mathbb{H} \) consist of elements of the form \( a + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} \), where \( a, b, c, d \) are real numbers. Equivalently, \( \mathbb{H} \) can be considered to be the set of all \( 2 \times 2 \) matrices of the form\n\n\[ \left( \begin{matrix} \alpha & \beta \\ - \bar{\beta } & \bar... | Yes |
Proposition 16.8. Let \( R \) be a ring with \( a, b \in R \) . Then\n\n1. \( {a0} = {0a} = 0 \) ;\n\n2. \( a\left( {-b}\right) = \left( {-a}\right) b = - {ab} \) ;\n\n3. \( \left( {-a}\right) \left( {-b}\right) = {ab} \) . | Proof. To prove (1), observe that\n\n\[ \n{a0} = a\left( {0 + 0}\right) = {a0} + {a0} \n\]\n\nhence, \( {a0} = 0 \) . Similarly, \( {0a} = 0 \) . For (2), we have \( {ab} + a\left( {-b}\right) = a\left( {b - b}\right) = {a0} = 0 \) ; consequently, \( - {ab} = a\left( {-b}\right) \) . Similarly, \( - {ab} = \left( {-a}\... | Yes |
Let \( R = {\mathbb{M}}_{2}\left( \mathbb{R}\right) \) be the ring of \( 2 \times 2 \) matrices with entries in \( \mathbb{R} \). If \( T \) is the set of upper triangular matrices in \( R \); i.e., \( T = \left\{ {\left( \begin{array}{ll} a & b \\ 0 & c \end{array}\right) : a, b, c \in \mathbb{R}}\right\} \), then \( ... | If \( A = \left( \begin{array}{ll} a & b \\ 0 & c \end{array}\right) \) and \( B = \left( \begin{matrix} {a}^{\prime } & {b}^{\prime } \\ 0 & {c}^{\prime } \end{matrix}\right) \) are in \( T \), then clearly \( A - B \) is also in \( T \). Also, \( {AB} = \left( \begin{matrix} a{a}^{\prime } & a{b}^{\prime } + b{c}^{\p... | Yes |
If \( {i}^{2} = - 1 \), then the set \( \mathbb{Z}\left\lbrack i\right\rbrack = \{ m + {ni} : m, n \in \mathbb{Z}\} \) forms a ring known as the Gaussian integers. | It is easily seen that the Gaussian integers are a subring of the complex numbers since they are closed under addition and multiplication. Let \( \alpha = a + {bi} \) be a unit in \( \mathbb{Z}\left\lbrack i\right\rbrack \) . Then \( \bar{\alpha } = a - {bi} \) is also a unit since if \( {\alpha \beta } = 1 \), then \(... | Yes |
The set \( \mathbb{Q}\left( \sqrt{2}\right) = \{ a + b\sqrt{2} : a, b \in \mathbb{Q}\} \) is a field. | The inverse of an element \( a + b\sqrt{2} \) in \( \mathbb{Q}\left( \sqrt{2}\right) \) is\n\n\[ \frac{a}{{a}^{2} - 2{b}^{2}} + \frac{-b}{{a}^{2} - 2{b}^{2}}\sqrt{2} \] | Yes |
Proposition 16.15 Cancellation Law. Let \( D \) be a commutative ring with identity. Then \( D \) is an integral domain if and only if for all nonzero elements \( a \in D \) with \( {ab} = {ac} \) , we have \( b = c \) . | Proof. Let \( D \) be an integral domain. Then \( D \) has no zero divisors. Let \( {ab} = {ac} \) with \( a \neq 0 \) . Then \( a\left( {b - c}\right) = 0 \) . Hence, \( b - c = 0 \) and \( b = c \) .\n\nConversely, let us suppose that cancellation is possible in \( D \) . That is, suppose that \( {ab} = {ac} \) impli... | Yes |
Theorem 16.16. Every finite integral domain is a field. | Proof. Let \( D \) be a finite integral domain and \( {D}^{ * } \) be the set of nonzero elements of \( D \) . We must show that every element in \( {D}^{ * } \) has an inverse. For each \( a \in {D}^{ * } \) we can define a map \( {\lambda }_{a} : {D}^{ * } \rightarrow {D}^{ * } \) by \( {\lambda }_{a}\left( d\right) ... | Yes |
For every prime \( p,{\mathbb{Z}}_{p} \) is a field of characteristic \( p \) | By Proposition 3.4, every nonzero element in \( {\mathbb{Z}}_{p} \) has an inverse; hence, \( {\mathbb{Z}}_{p} \) is a field. If \( a \) is any nonzero element in the field, then \( {pa} = 0 \), since the order of any nonzero element in the abelian group \( {\mathbb{Z}}_{p} \) is \( p \). | Yes |
Lemma 16.18. Let \( R \) be a ring with identity. If 1 has order \( n \), then the characteristic of \( R \) is \( n \). | Proof. If 1 has order \( n \), then \( n \) is the least positive integer such that \( {n1} = 0 \) . Thus, for all \( r \in R \) ,\n\n\[ \n{nr} = n\left( {1r}\right) = \left( {n1}\right) r = {0r} = 0.\n\]\n\nOn the other hand, if no positive \( n \) exists such that \( {n1} = 0 \), then the characteristic of \( R \) is... | Yes |
Theorem 16.19. The characteristic of an integral domain is either prime or zero. | Proof. Let \( D \) be an integral domain and suppose that the characteristic of \( D \) is \( n \) with \( n \neq 0 \) . If \( n \) is not prime, then \( n = {ab} \), where \( 1 < a < n \) and \( 1 < b < n \) . By Lemma 16.18, we need only consider the case \( {n1} = 0 \) . Since \( 0 = {n1} = \left( {ab}\right) 1 = \l... | Yes |
For any integer \( n \) we can define a ring homomorphism \( \phi : \mathbb{Z} \rightarrow {\mathbb{Z}}_{n} \) by \( a \mapsto a\left( {\;\operatorname{mod}\;n}\right) \) . This is indeed a ring homomorphism, since | \n\n\[
\phi \left( {a + b}\right) = \left( {a + b}\right) \;\left( {\;\operatorname{mod}\;n}\right)
\]
\n\n\[
= a\;\left( {\;\operatorname{mod}\;n}\right) + b\;\left( {\;\operatorname{mod}\;n}\right)
\]
\n\n\[
= \phi \left( a\right) + \phi \left( b\right)
\]
\n\nand
\n\n\[
\phi \left( {ab}\right) = {ab}\;\left( {\;\ope... | Yes |
Let \( C\left\lbrack {a, b}\right\rbrack \) be the ring of continuous real-valued functions on an interval \( \left\lbrack {a, b}\right\rbrack \) as in Example 16.5. For a fixed \( \alpha \in \left\lbrack {a, b}\right\rbrack \), we can define a ring homomorphism \( {\phi }_{\alpha } : C\left\lbrack {a, b}\right\rbrack ... | This is a ring homomorphism since\n\n\[ \n{\phi }_{\alpha }\left( {f + g}\right) = \left( {f + g}\right) \left( \alpha \right) = f\left( \alpha \right) + g\left( \alpha \right) = {\phi }_{\alpha }\left( f\right) + {\phi }_{\alpha }\left( g\right) \n\]\n\n\[ \n{\phi }_{\alpha }\left( {fg}\right) = \left( {fg}\right) \le... | Yes |
Every ring \( R \) has at least two ideals, \( \{ 0\} \) and \( R \) . These ideals are called the trivial ideals. | Let \( R \) be a ring with identity and suppose that \( I \) is an ideal in \( R \) such that 1 is in \( I \) . Since for any \( r \in R,{r1} = r \in I \) by the definition of an ideal, \( I = R \). | No |
If \( a \) is any element in a commutative ring \( R \) with identity, then the set\n\n\[ \langle a\rangle = \{ {ar} : r \in R\} \]\n\n is an ideal in \( R \) . | Certainly, \( \langle a\rangle \) is nonempty since both \( 0 = {a0} \) and \( a = {a1} \) are in \( \langle a\rangle \) . The sum of two elements in \( \langle a\rangle \) is again in \( \langle a\rangle \) since \( {ar} + a{r}^{\prime } = a\left( {r + {r}^{\prime }}\right) \) . The inverse of \( {ar} \) is \( - {ar} ... | Yes |
Theorem 16.25. Every ideal in the ring of integers \( \mathbb{Z} \) is a principal ideal. | Proof. The zero ideal \( \{ 0\} \) is a principal ideal since \( \langle 0\rangle = \{ 0\} \) . If \( I \) is any nonzero ideal in \( \mathbb{Z} \), then \( I \) must contain some positive integer \( m \) . There exists a least positive integer \( n \) in \( I \) by the Principle of Well-Ordering. Now let \( a \) be an... | Yes |
The set \( n\mathbb{Z} \) is ideal in the ring of integers. | If \( {na} \) is in \( n\mathbb{Z} \) and \( b \) is in \( \mathbb{Z} \) , then \( {nab} \) is in \( n\mathbb{Z} \) as required. | No |
Proposition 16.27. The kernel of any ring homomorphism \( \phi : R \rightarrow S \) is an ideal in \( R \) . | Proof. We know from group theory that \( \ker \phi \) is an additive subgroup of \( R \) . Suppose that \( r \in R \) and \( a \in \ker \phi \) . Then we must show that \( {ar} \) and \( {ra} \) are in \( \ker \phi \) . However,\n\n\[ \phi \left( {ar}\right) = \phi \left( a\right) \phi \left( r\right) = {0\phi }\left( ... | Yes |
Theorem 16.29. Let \( I \) be an ideal of \( R \) . The factor group \( R/I \) is a ring with multiplication defined by\n\n\[ \left( {r + I}\right) \left( {s + I}\right) = {rs} + I. \] | Proof. We already know that \( R/I \) is an abelian group under addition. Let \( r + I \) and \( s + I \) be in \( R/I \) . We must show that the product \( \left( {r + I}\right) \left( {s + I}\right) = {rs} + I \) is independent of the choice of coset; that is, if \( {r}^{\prime } \in r + I \) and \( {s}^{\prime } \in... | No |
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