Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Proposition 8. Let \( {l}_{0} \) be a squarefree positive integer, let \( {b}_{l} \) denote the \( q \) - expansion coefficient of \( {F}_{k/2} \), and set \( \lambda = \left( {k - 1}\right) /2 \) . Then\n\n\[ \mathop{\sum }\limits_{{{l}_{1} = 1}}^{\infty }{b}_{{l}_{0}{l}_{1}^{2}}{l}_{1}^{-s} = \frac{{b}_{{l}_{0}}}{1 -... | Proof. Let \( {l}_{1} = {p}_{1}^{{v}_{1}}\cdots {p}_{r}^{{v}_{r}} \) . Let \( {f}_{p}\left( s\right) \) denote the factor in the product corresponding to \( p \) (here \( {f}_{2}\left( s\right) = {\left( 1 - {2}^{k - 2 - s}\right) }^{-1} \) ). Then we must show that \( {b}_{{l}_{0}{l}_{1}^{2}} \) equals \( {b}_{{l}_{0}... | Yes |
The element \( {H}_{k/2} \in {M}_{k/2}\left( {{\widetilde{\Gamma }}_{0}\left( 4\right) }\right) \) given by \( {H}_{k/2} = \zeta \left( {1 - {2\lambda }}\right) \) \( \left( {{E}_{k/2} + \left( {1 + {i}^{k}}\right) {2}^{-k/2}{F}_{k/2}}\right) \) has q-expansion coefficients \( {c}_{l} \) which can be determined by the ... | Thus, there are two very different elegant properties satisfied by the \( q \) -expansion coefficients of \( {H}_{k/2} \) . First, the coefficients corresponding to discriminants of quadratic fields are the values of the \( L \) -function for the corresponding quadratic character at the fixed negative integer \( 1 - \l... | Yes |
Proposition 10. A complete set of double coset representatives of \( \Gamma \) in \( {\Delta }^{n} \) is \( \left\{ \left( \begin{matrix} {n}_{1} & 0 \\ 0 & {n}_{1}{n}_{0} \end{matrix}\right) \right\} \), where \( {n}_{0},{n}_{1} \) run through all positive integers such that \( n = {n}_{0}{n}_{1}^{2} \) . In particula... | Proof. Consider the abelian group \( {\mathbb{Z}}^{2} \) with standard basis generators \( {e}_{1} = \) \( \left( \begin{array}{l} 1 \\ 0 \end{array}\right) ,{e}_{2} = \left( \begin{array}{l} 0 \\ 1 \end{array}\right) \) . Any matrix \( \alpha \in {\Delta }^{n} \) gives a subgroup of index \( n \) in \( {\mathbb{Z}}^{2... | Yes |
Proposition 11. If g.c.d. \( \left( {n, N}\right) = 1 \), then a complete set of double coset representatives of \( {\Gamma }_{1}\left( N\right) \) in \( {\Delta }^{n} = {\Delta }^{n}\left( {N,\{ 1\} ,\mathbb{Z}}\right) \) is \( \left\{ {{\sigma }_{{n}_{1}}\left( \begin{matrix} {n}_{1} & 0 \\ 0 & {n}_{1}{n}_{0} \end{ma... | Proof. If \( \alpha \in {\Delta }^{n} \), we know by Proposition 10 that \( \alpha = {n}_{1}{\gamma }_{1}\left( \begin{matrix} 1 & 0 \\ 0 & {n}_{0} \end{matrix}\right) {\gamma }_{2} \), where \( {\gamma }_{1},{\gamma }_{2} \in \Gamma \) and \( {n}_{1} \) is the greatest common divisor of the entries of \( \alpha \) . W... | Yes |
Proposition 14. Let \( f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{e}^{2\pi inz} \in {M}_{k/2}\left( {{\widetilde{\Gamma }}_{0}\left( N\right) ,\chi }\right) \) be an eigenform for all of the Hecke operators \( {T}_{{p}^{2}} \) . Let \( {\lambda }_{p} \) be the corresponding eigenvalue, i.e., \(... | Proof. If \( {T}_{{p}^{2}}f = {\lambda }_{p}f \) with \( p/N \), then (3.5) gives for any \( {l}_{1} \) prime to \( p \) :\n\n\[ {\lambda }_{p}{a}_{{l}_{0}{l}_{1}^{2}} = {a}_{{l}_{0}{l}_{1}^{2}{p}^{2}} + {p}^{\lambda - 1}\chi \left( p\right) \left( \frac{{\left( -1\right) }^{\lambda }{l}_{0}}{p}\right) {a}_{{l}_{0}{l}_... | Yes |
If \( \dim V = 3 \) and \( \mathcal{G} \) is a finite subgroup of \( \mathcal{O}\left( V\right) \), then \( \mathcal{G} \) is one of the following:\n\n(a) \( {\mathcal{C}}_{3}^{n}, n \geq 1;{\mathcal{H}}_{3}^{n}, n \geq 2;\mathcal{T};\mathcal{W};\mathcal{I} \) ;\n\n(b) \( {\left( {\mathcal{C}}_{3}^{n}\right) }^{ * }, n... | Let us see geometrically why \( \mathcal{T} \) is a subgroup of \( \mathcal{W} \) . As shown in Figure 2.8, a regular tetrahedron may be inscribed in a cube. Moreover, this tetrahedron is invariant under the rotations in \( \mathcal{W} \) of order 3 about axes joining extreme opposite vertices, as well as rotations of ... | No |
The set \( F \) is a fundamental region for \( \mathcal{G} \) in \( V \) . | Since each \( {L}_{i} \) is open, \( F \) is open. If \( {T}_{i} \neq 1 \), then \( {T}_{i}F = {T}_{i}\left( {\cap {L}_{j}}\right) \), or\n\n\[ \n{T}_{i}F = {T}_{i}\left( \left\{ {x : d\left( {x,{x}_{0}}\right) < d\left( {x,{x}_{j}}\right) ,1 \leq j \leq N - 1}\right\} \right) \n\]\n\n\[ \n= \left\{ {{T}_{i}x : d\left(... | No |
If \( r \) is a root of \( \mathcal{G} \leq \mathcal{O}\left( V\right) \) and if \( T \in \mathcal{G} \), then \( {Tr} \) is also a root of \( \mathcal{G} \) . In fact, if \( {Tr} = x \), then \( {S}_{x} = T{S}_{r}{T}^{-1} \in \mathcal{G} \) . | Set \( \mathcal{P} = {r}^{ \bot } \) and \( {\mathcal{P}}^{\prime } = T\mathcal{P} \) . Then \( {\mathcal{P}}^{\prime } \) is a hyperplane, and\n\n\[{\mathcal{P}}^{\prime } = {\left( Tr\right) }^{ \bot } = {x}^{ \bot }\]\n\nsince \( T \in \mathcal{O}\left( V\right) \) . If \( y = {Tz} \in {\mathcal{P}}^{\prime } \), wi... | Yes |
If \( {r}_{i},{r}_{j} \in \Pi \), with \( i \neq j \), and \( {\lambda }_{i} \) and \( {\lambda }_{j} \) are positive real numbers, then the vector \( x = {\lambda }_{i}{r}_{i} - {\lambda }_{j}{r}_{j} \) is neither positive nor negative. | If \( x \) were positive, we could write\n\n\[ x = {\lambda }_{i}{r}_{i} - {\lambda }_{j}{r}_{j} = {\sum }_{k = 1}^{m}{u}_{k}{r}_{k} \]\n\nwith all \( {\mu }_{k} \geq 0 \) . If \( {\lambda }_{i} \leq {\mu }_{i} \), then\n\n\[ 0 = \left( {{\mu }_{i} - {\lambda }_{i}}\right) {r}_{i} + \left( {{\mu }_{j} + {\lambda }_{j}}... | Yes |
Proposition 4.1.6\n\nSuppose that \( {x}_{1},{x}_{2},\ldots ,{x}_{m} \in V \) are all on the same side of a hyperplane; i.e., \( \left( {{x}_{i}, x}\right) > 0,1 \leq i \leq m \), for some \( x \in V \) . If \( \left( {{x}_{i},{x}_{j}}\right) \leq 0 \) whenever \( i \neq j \) , then \( \left\{ {{x}_{1},\ldots ,{x}_{m}}... | Proof\n\nSuppose the contrary and relabel if necessary so that there is a dependence relation of the form\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i} = \mathop{\sum }\limits_{{i = k + 1}}^{m}{\mu }_{i}{x}_{i} \]\n\nwith all \( {\lambda }_{i} \geq 0 \), all \( {\mu }_{i} \geq 0 \), and some \( {\lambd... | Yes |
If \( \Pi \) is a \( t \) -base for \( \Delta \), then \( \Pi \) is a basis for \( V \) . | Since \( \mathcal{G} \) is effective \( \Delta \) spans \( V \), by Proposition 4.1.2. Since every \( r \in \Delta \) is a linear combination of roots in \( \Pi, V \) is spanned by \( \Pi \) . By Propositions 4.1.5 and 4.1.6, \( \Pi \) is linearly independent, so \( \Pi \) is a basis. | Yes |
There is only one \( t \) -base for \( \Delta \) . | Suppose that \( {\Pi }_{1} \) and \( {\Pi }_{2} \) are \( t \) -bases. Since each root in \( {\Pi }_{1} \) is a nonnegative linear combination of elements of \( {\Pi }_{2} \), the change of basis matrix \( A \) from the basis \( {\Pi }_{2} \) to the basis \( {\Pi }_{1} \) has nonnegative entries. Likewise, the change o... | Yes |
If \( T \in \mathcal{G} \) and \( {T\Pi } = \Pi \), then \( T = 1 \) . | Suppose that \( T \neq 1 \) . By Theorem 4.1.12 we may write \( T \) as \( {S}_{{i}_{1}}{S}_{{i}_{2}}\cdots {S}_{{i}_{k}} \), a product of fundamental reflections. We may assume that \( T \) cannot be written as a product of fewer fundamental reflections, i.e., that \( k \) is minimal. Since \( T \neq 1, k \) is positi... | Yes |
If \( T \in \mathcal{G} \), then \( T\left( {\Delta }_{t}^{ + }\right) = {\Delta }_{T\left( t\right) }^{ + } \) ; consequently, \( T\left( {\Pi }_{t}\right) = {\Pi }_{T\left( t\right) } \) . | Since every root in \( T\left( {\Delta }_{t}^{ + }\right) \) is a nonnegative linear combination of roots in \( T\left( {\Pi }_{t}\right) \), the second statement follows from the first by Proposition 4.1.8. As for the first statement,\n\n\[ T\left( {\Delta }_{t}^{ + }\right) = T\{ r \in \Delta : \left( {t, r}\right) >... | Yes |
If \( T \in \mathcal{G} \) and \( T\left( {\Delta }^{ + }\right) = {\Delta }^{ + } \), then \( T = 1 \) . | By Proposition 4.2.2 we have\n\n\[ \n{\Delta }_{t}^{ + } = T\left( {\Delta }_{t}^{ + }\right) = {\Delta }_{T\left( t\right) }^{ + }\n\] \n\nso \( {\Pi }_{t} = {\Pi }_{T\left( t\right) } \) by Proposition 4.1.8. But then \( T{\Pi }_{t} = {\Pi }_{t} \) by Proposition 4.2.2, so \( T = 1 \) by Theorem 4.2.1. | Yes |
Suppose that \( \\left\\{ {{r}_{1},\\ldots ,{r}_{n}}\\right\\} \) is a basis for \( V \) with \( \\left\\{ {{r}_{i},{r}_{j}}\\right\\} \\leq 0 \) if \( i \\neq j \) , and let \( \\left\\{ {{s}_{1},\\ldots ,{s}_{n}}\\right\\} \) be the dual basis. Then \( \\left\\{ {{s}_{i},{s}_{j}}\\right\\} \\geq 0 \) for all \( i, j ... | Let \( A \) be the matrix whose \( {ij} \) th entry is \( \\left\\{ {{r}_{i},{r}_{j}}\\right\\} \) and \( B \) the matrix whose \( {ij} \) th entry is \( \\left\\{ {{s}_{i},{s}_{j}}\\right\\} \) . Then \( B = {A}^{-1} \) (see Exercise 4.15), and we must show that \( B \) has nonnegative entries. Since \( \\left\\{ {{r}... | Yes |
If \( {r}_{i},{r}_{j} \in \Pi \), then there is an integer \( {p}_{ij} \geq 1 \) such that\n\n\[ \left( {{r}_{i},{r}_{j}}\right) /\begin{Vmatrix}{r}_{i}\end{Vmatrix}\begin{Vmatrix}{r}_{j}\end{Vmatrix} = - \cos \left( {\pi /{p}_{ij}}\right) . \] | If \( i = j \), we may take \( {p}_{ij} = 1 \) . Assume then that \( i \neq j \), and denote by \( W \) the two-dimensional subspace of \( V \) spanned by \( {r}_{i} \) and \( {r}_{j} \) . Let \( \mathcal{H} \) be the subgroup of \( \mathcal{G} \) spanned by \( {S}_{i} \) and \( {S}_{j} \) . Since \( {S}_{i} \mid {W}^{... | Yes |
If \( G \) is a connected positive definite Coxeter graph, then \( G \) is one of the graphs \( {A}_{n},{B}_{n},{D}_{n},{H}_{2}^{n},{G}_{2},{I}_{3},{I}_{4},{F}_{4},{E}_{6},{E}_{7} \), or \( {E}_{8} \) . | Observe first that \( G \) can have no cycles as subgraphs since no \( {P}_{n} \) is positive definite. If \( {H}_{2}^{n} \) is a subgraph of \( G \) for any \( n \geq 7 \), then \( G = {H}_{2}^{n} \), for otherwise \( {U}_{3} \) would be a subgraph of \( G \) . Likewise, \( G = {G}_{2} \) if \( {G}_{2} \) is a subgrap... | Yes |
Every relation \( W = {S}_{{i}_{1}}\cdots {S}_{{i}_{k}} = 1 \) in a Coxeter group \( \mathcal{G} \) is a consequence of the relations \( {\left( {S}_{i}{S}_{j}\right) }^{{p}_{ij}} = 1 \), so \( \mathcal{G} \) has the presentation\n\n\[ \left\langle {{S}_{1},\ldots ,{S}_{n} \mid {\left( {S}_{i}{S}_{j}\right) }^{{p}_{ij}... | Suppose that \( u \) is the maximal length of partial words of \( W \) . Then we may write \( W \) as \( {W}_{1}{S}_{i}{S}_{j}{W}_{2} \), where \( l\left( {{W}_{1}{S}_{i}}\right) = u \) and every partial word of \( {W}_{1} \) has length less than \( u \) . Set \( p = {p}_{ij} \), let \( {W}^{\prime } = {W}_{1}{\left( {... | Yes |
Proposition 6.1.7\n\nSuppose that \( W \) is a subspace of \( {\mathcal{R}}^{n} \) such that \( {TW} \subseteq W \) for all \( T \in \mathcal{H} \) . Then either \( W = {\mathcal{R}}^{n} \) or \( W = 0 \) . | Proof\n\nSuppose that \( W \neq {\mathcal{R}}^{n}, W \neq 0 \), and set\n\n\[ \n{W}^{\prime } = \left\{ {x \in {\mathcal{R}}^{n} : C\left( {x, y}\right) = 0\text{ for all }y \in W}\right\} .\n\]\n\nThen \( {W}^{\prime } \) is the orthogonal complement of \( W \) with respect to the inner product \( C \), so \( {\mathca... | Yes |
The bilinear form \( B \) is a positive scalar multiple of the inner product \( C \) , so \( B \) is also an inner product on \( {\mathcal{R}}^{n} \) . | Since the matrix \( {P}^{-1} \) is positive definite, we may write \( {P}^{-1} = N{N}^{t} \) , where \( N \) is a nonsingular real matrix (see [1], p. 256). If \( T \in \mathcal{H} \) is represented by the matrix \( M \) with respect to the basis \( \left\{ {{e}_{1},\ldots ,{e}_{n}}\right\} \), then \( T \) is represen... | Yes |
Proposition 7.3.1\n\nSuppose \( \mathcal{G} \leq \mathcal{G}\mathcal{L}\left( V\right) \) is finite. Then\n\n(i) if \( f \in \mathfrak{P} \) then \( {Mf} \in \mathfrak{J} \), and\n\n(ii) \( {M}^{2} = M \), and \( M \) restricts to the identity map on \( \mathfrak{J} \) .\n\nThus \( M \) is a projection from \( \mathfra... | Proof\n\n(i) If \( S \in \mathcal{G} \) then \( {SMf} = {\left| \mathcal{G}\right| }^{-1}\sum \{ \left( {STf}\right) : T \in \mathcal{G}\} = {Mf} \), since \( {ST} \) ranges over all elements of \( \mathcal{G} \) as \( T \) does, so \( {Mf} \in \mathfrak{I} \) .\n\n(ii) If \( f \in \mathfrak{I} \) then \( {Mf} = {\left... | Yes |
Suppose \( {g}_{1},{g}_{2},\ldots ,{g}_{k} \in \mathfrak{I};{h}_{1},{h}_{2},\ldots ,{h}_{k} \) are homogeneous in \( \mathfrak{P} \) ; and \( {g}_{1}{h}_{1} + \cdots + {g}_{k}{h}_{k} = 0 \) . Then either \( {h}_{1} \in \mathfrak{a} \) or else \( {g}_{1} \) is in the ideal \( \mathfrak{b} \) in \( \mathfrak{I} \) genera... | Use induction on the degree of \( {h}_{1} \) . If \( {h}_{1} = 0 \) then \( {h}_{1} \in \mathfrak{a} \) ; if \( \deg {h}_{1} = 0 \) then \( {g}_{1} = - \left( {{h}_{2}/{h}_{1}}\right) {g}_{2} - \cdots - \left( {{h}_{k}/{h}_{1}}\right) {g}_{k} \), and hence \( {g}_{1} = M{g}_{1} = \) \( - \left( {M{h}_{2}/{h}_{1}}\right... | Yes |
\( {\sum }_{d = 0}^{\infty }\left( {\dim {\mathfrak{P}}_{d}}\right) {t}^{d} = {\left( 1 - t\right) }^{-n}. \) | Take \( T = 1 \) . | No |
The dimension of \( {\mathfrak{P}}_{d} \) is \( \left( \begin{matrix} n + d - 1 \\ d \end{matrix}\right) \), all \( d \) . | See Exercise 7.8. | No |
Theorem 7.4.4 (Molien's Theorem)\n\nIf \( \mathcal{G} \leq \mathcal{G}\mathcal{L}\left( V\right) \) is finite then its Molien series is \( \Phi \left( t\right) = \) \( {\left| \mathcal{G}\right| }^{-1}\sum \left\{ {\det {\left( 1 - tT\right) }^{-1} : T \in \mathcal{G}}\right\} \) | Proof\n\nApply Propositions 7.4.1 and 7.4.3:\n\n\[ \Phi \left( t\right) = \mathop{\sum }\limits_{{d = 0}}^{\infty }\left( {\dim {\mathfrak{I}}_{d}}\right) {t}^{d} = \mathop{\sum }\limits_{{d = 0}}^{\infty }{\left| \mathcal{G}\right| }^{-1}\sum \left\{ {{\operatorname{tr}}_{d}T : T \in \mathcal{G}}\right\} {t}^{d} \]\n\... | Yes |
Under the assumptions above the Molien series of \( \mathcal{G} \) is \( \Phi \left( t\right) = \) \( \mathop{\prod }\limits_{{i = 1}}^{n}{\left( 1 - {t}^{{d}_{i}}\right) }^{-1} \) | \( {\Pi }_{i = 1}^{n}{\left( 1 - {t}^{{d}_{i}}\right) }^{-1} = {\Pi }_{i = 1}^{n}\left( {1 + {t}^{{d}_{i}} + {t}^{2{d}_{i}} + \cdots }\right) = {\sum }_{d = 0}^{\infty }{\alpha }_{d}{t}^{d} \), where\n\n\[ \n{\alpha }_{d} = \left| \left\{ {\left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {\mathbb{Z}}_{ + }^{n} : {\sum }_{i}... | Yes |
The degrees \( {d}_{1},{d}_{2},\ldots ,{d}_{n} \) of the basic generators are uniquely determined by \( \mathcal{G} \) . | Suppose also that \( \mathfrak{I} = \mathcal{R}\left\lbrack {{f}_{1}^{\prime },\ldots ,{f}_{n}^{\prime }}\right\rbrack \), with \( \left\{ {f}_{i}^{\prime }\right\} \) algebraically independent and \( \deg {f}_{i}^{\prime } = {d}_{i}^{\prime } \), with \( {d}_{1}^{\prime } \leq {d}_{2}^{\prime } \leq \cdots \leq {d}_{n... | Yes |
Proposition 7.4.7\n\n\\( \\left| \\mathcal{G}\\right| = \\mathop{\\prod }\\limits_{{i = 1}}^{n}{d}_{i} \\) | Proof\n\nFor each \\( T \\neq 1 \\) in \\( \\mathcal{G} \\) define \\( {h}_{T}\\left( t\\right) = {\\left( 1 - t\\right) }^{n - 1}\\det {\\left( 1 - tT\\right) }^{-1} \\), using the factors \\( 1 - t \\) to cancel any factors of \\( \\det {\\left( 1 - tT\\right) }^{-1} \\) resulting from eigenvalues of 1 . Thus \\( {h}... | Yes |
Theorem 7.4.9 (Shephard and Todd)\n\nSuppose \( \mathcal{G} \leq \mathcal{G}\mathcal{L}\left( V\right) \) is finite and that \( \mathcal{G} \) has alrebraically independent homogeneous invariants \( {f}_{1},\ldots ,{f}_{n},\;n = \dim V \), such that \( \mathfrak{I} = \mathcal{R}\left\lbrack {{f}_{1},\ldots ,{f}_{n}}\ri... | Proof\n\nAs we observed above we may assume that \( \mathcal{G} \leq \mathcal{O}\left( V\right) \) . Let \( \mathcal{H} \) be the subgroup of \( \mathcal{G} \) generated by the reflections in \( \mathcal{G} \) . We may assume that \( \mathcal{G} \neq 1 \) and hence also that \( \mathcal{H} \neq 1 \) by the corollary ab... | Yes |
Proposition 1. Let \( f : A \rightarrow B \) .\n\n(1) The map \( f \) is injective if and only if \( f \) has a left inverse.\n\n(2) The map \( f \) is surjective if and only if \( f \) has a right inverse.\n\n(3) The map \( f \) is a bijection if and only if there exists \( g : B \rightarrow A \) such that \( f \circ ... | Proof: Exercise. | No |
(1) If \( \sim \) defines an equivalence relation on \( A \) then the set of equivalence classes of \( \sim \) form a partition of \( A \) . | Proof: Omitted. | No |
Theorem 3. The operations of addition and multiplication on \( \mathbb{Z}/n\mathbb{Z} \) defined above are both well defined, that is, they do not depend on the choices of representatives for the classes involved. More precisely, if \( {a}_{1},{a}_{2} \in \mathbb{Z} \) and \( {b}_{1},{b}_{2} \in \mathbb{Z} \) with \( \... | Proof: Suppose \( {a}_{1} \equiv {b}_{1}\left( {\;\operatorname{mod}\;n}\right) \), i.e., \( {a}_{1} - {b}_{1} \) is divisible by \( n \) . Then \( {a}_{1} = {b}_{1} + {sn} \) for some integer \( s \) . Similarly, \( {a}_{2} \equiv {b}_{2}{\;(\operatorname{mod}\;n)} \) means \( {a}_{2} = {b}_{2} + {tn} \) for some inte... | Yes |
Proposition 4. \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } = \{ \bar{a} \in \mathbb{Z}/n\mathbb{Z} \mid \left( {a, n}\right) = 1\} \) . | It is easy to see that if any representative of \( \widetilde{a} \) is relatively prime to \( n \) then all representatives are relatively prime to \( n \) so that the set on the right in the proposition is well defined.\n\nIf \( a \) is an integer relatively prime to \( n \) then the Euclidean Algorithm produces integ... | Yes |
If \( G \) is a group under the operation \( \star \), then (1) the identity of \( G \) is unique | If \( f \) and \( g \) are both identities, then by axiom (ii) of the definition of a group \( f \star g = f \) (take \( a = f \) and \( e = g \) ). By the same axiom \( f \star g = g \) (take \( a = g \) and \( e = f \) ). Thus \( f = g \), and the identity is unique. | Yes |
Proposition 2. Let \( G \) be a group and let \( a, b \in G \). The equations \( {ax} = b \) and \( {ya} = b \) have unique solutions for \( x, y \in G \). In particular, the left and right cancellation laws hold in \( G \), i.e.,\n\n(1) if \( {au} = {av} \), then \( u = v \), and\n\n(2) if \( {ub} = {vb} \), then \( u... | Proof: We can solve \( {ax} = b \) by multiplying both sides on the left by \( {a}^{-1} \) and simplifying to get \( x = {a}^{-1}b \). The uniqueness of \( x \) follows because \( {a}^{-1} \) is unique. Similarly, if \( {ya} = b, y = b{a}^{-1} \). If \( {au} = {av} \), multiply both sides on the left by \( {a}^{-1} \) ... | Yes |
Proposition 1. (The Subgroup Criterion) A subset \( H \) of a group \( G \) is a subgroup if and only if\n\n(1) \( H \neq \varnothing \), and\n\n(2) for all \( x, y \in H, x{y}^{-1} \in H \) . | Proof: If \( H \) is a subgroup of \( G \), then certainly (1) and (2) hold because \( H \) contains the identity of \( G \) and the inverse of each of its elements and because \( H \) is closed under multiplication.\n\nIt remains to show conversely that if \( H \) satisfies both (1) and (2), then \( H \leq G \) . Let ... | Yes |
Proposition 2. If \( H = \langle x\rangle \), then \( \left| H\right| = \left| x\right| \) (where if one side of this equality is infinite, so is the other). More specifically\n\n(1) if \( \left| H\right| = n < \infty \), then \( {x}^{n} = 1 \) and \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are all the distinct elements of... | Proof: Let \( \left| x\right| = n \) and first consider the case when \( n < \infty \) . The elements \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are distinct because if \( {x}^{a} = {x}^{b} \), with, say, \( 0 \leq a < b < n \), then \( {x}^{b - a} = {x}^{0} = 1 \), contrary to \( n \) being the smallest positive power of ... | Yes |
Proposition 3. Let \( G \) be an arbitrary group, \( x \in G \) and let \( m, n \in \mathbb{Z} \) . If \( {x}^{n} = 1 \) and \( {x}^{m} = 1 \), then \( {x}^{d} = 1 \), where \( d = \left( {m, n}\right) \) . In particular, if \( {x}^{m} = 1 \) for some \( m \in \mathbb{Z} \), then \( \left| x\right| \) divides \( m \) . | Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers \( r \) and \( s \) such that \( d = {mr} + {ns} \), where \( d \) is the g.c.d. of \( m \) and \( n \) . Thus\n\n\[ \n{x}^{d} = {x}^{{mr} + {ns}} = {\left( {x}^{m}\right) }^{r}{\left( {x}^{n}\right) }^{s} = {1}^{r}{1}^{s} = 1.\n\]\n\nThis pro... | Yes |
Theorem 4. Any two cyclic groups of the same order are isomorphic. More specifically, (1) if \( n \in {\mathbb{Z}}^{ + } \) and \( \langle x\rangle \) and \( \langle y\rangle \) are both cyclic groups of order \( n \), then the map\n\n\[ \n\varphi : \langle x\rangle \rightarrow \langle y\rangle \n\]\n\n\[ \n{x}^{k} \ma... | Proof: Suppose \( \langle x\rangle \) and \( \langle y\rangle \) are both cyclic groups of order \( n \) . Let \( \varphi : \langle x\rangle \rightarrow \langle y\rangle \) be defined by \( \varphi \left( {x}^{k}\right) = {y}^{k} \) ; we must first prove \( \varphi \) is well defined, that is,\n\n\[ \n\text{if}{x}^{r} ... | Yes |
Proposition 5. Let \( G \) be a group, let \( x \in G \) and let \( a \in \mathbb{Z} - \{ 0\} \) . (1) If \( \left| x\right| = \infty \), then \( \left| {x}^{a}\right| = \infty \) . | Proof: (1) By way of contradiction assume \( \left| x\right| = \infty \) but \( \left| {x}^{a}\right| = m < \infty \) . By definition of order \[ 1 = {\left( {x}^{a}\right) }^{m} = {x}^{am}. \] Also, \[ {x}^{-{am}} = {\left( {x}^{am}\right) }^{-1} = {1}^{-1} = 1. \] Now one of \( a{\;m} \) or \( - a{\;m} \) is positive... | Yes |
Assume \( \left| x\right| = n < \infty \) . Then \( H = \left\langle {x}^{a}\right\rangle \) if and only if \( \left( {a, n}\right) = 1 \) . In particular, the number of generators of \( H \) is \( \varphi \left( n\right) \) (where \( \varphi \) is Euler’s \( \varphi \) -function). | Proof: We leave (1) as an exercise. In (2) if \( \left| x\right| = n < \infty \), Proposition 2 says \( {x}^{a} \) generates a subgroup of \( H \) of order \( \left| {x}^{a}\right| \) . This subgroup equals all of \( H \) if and only if \( \left| {x}^{a}\right| = \left| x\right| \) . By Proposition 5,\n\n\[ \left| {x}^... | No |
Theorem 7. Let \( H = \langle x\rangle \) be a cyclic group.\n\n(1) Every subgroup of \( H \) is cyclic. More precisely, if \( K \leq H \), then either \( K = \{ 1\} \) or \( K = \left\langle {x}^{d}\right\rangle \), where \( d \) is the smallest positive integer such that \( {x}^{d} \in K \) . | Proof: (1) Let \( K \leq H \) . If \( K = \{ 1\} \), the proposition is true for this subgroup, so we assume \( K \neq \{ 1\} \) . Thus there exists some \( a \neq 0 \) such that \( {x}^{a} \in K \) . If \( a < 0 \) then since \( K \) is a group also \( {x}^{-a} = {\left( {x}^{a}\right) }^{-1} \in K \) . Hence \( K \) ... | Yes |
Proposition 8. If \( \mathcal{A} \) is any nonempty collection of subgroups of \( G \), then the intersection of all members of \( \mathcal{A} \) is also a subgroup of \( G \) . | Proof: This is an easy application of the subgroup criterion (see also Exercise 10, Section 1). Let\n\n\[ K = \mathop{\bigcap }\limits_{{H \in \mathcal{A}}}H \]\n\nSince each \( H \in \mathcal{A} \) is a subgroup, \( 1 \in H \), so \( 1 \in K \), that is, \( K \neq \varnothing \) . If \( a, b \in K \) , then \( a, b \i... | No |
Proposition 9. \( \bar{A} = \langle A\rangle \) . | Proof: We first prove \( \bar{A} \) is a subgroup. Note that \( \bar{A} \neq \varnothing \) (even if \( A = \varnothing \) ). If \( a, b \in \bar{A} \) with \( a = {a}_{1}^{{\epsilon }_{1}}{a}_{2}^{{\epsilon }_{2}}\ldots {a}_{n}^{{\epsilon }_{n}} \) and \( b = {b}_{1}^{{\delta }_{1}}{b}_{2}^{{\delta }_{2}}\ldots {b}_{m... | Yes |
Proposition 1. Let \( G \) and \( H \) be groups and let \( \varphi : G \rightarrow H \) be a homomorphism.\n\n(1) \( \varphi \left( {1}_{G}\right) = {1}_{H} \), where \( {1}_{G} \) and \( {1}_{H} \) are the identities of \( G \) and \( H \), respectively.\n\n(2) \( \varphi \left( {g}^{-1}\right) = \varphi {\left( g\ri... | Proof: (1) Since \( \varphi \left( {1}_{G}\right) = \varphi \left( {{1}_{G}{1}_{G}}\right) = \varphi \left( {1}_{G}\right) \varphi \left( {1}_{G}\right) \), the cancellation laws show that (1) holds.\n\n(2) \( \varphi \left( {1}_{G}\right) = \varphi \left( {g{g}^{-1}}\right) = \varphi \left( g\right) \varphi \left( {g}... | Yes |
Proposition 2. Let \( \varphi : G \rightarrow H \) be a homomorphism of groups with kernel \( K \) . Let\n\n\( X \in G/K \) be the fiber above \( a \), i.e., \( X = {\varphi }^{-1}\left( a\right) \) . Then\n\n(1) For any \( u \in X,\;X = \{ {uk} \mid k \in K\} \)\n\n(2) For any \( u \in X,\;X = \{ {ku} \mid k \in K\} \... | Proof: We prove (1) and leave the proof of (2) as an exercise. Let \( u \in X \) so, by definition of \( X,\varphi \left( u\right) = a \) . Let\n\n\[ \n{uK} = \{ {uk} \mid k \in K\} .\n\]\n\nWe first prove \( {uK} \subseteq X \) . For any \( k \in K \), \n\n\[ \n\varphi \left( {uk}\right) = \varphi \left( u\right) \var... | No |
Proposition 4. Let \( N \) be any subgroup of the group \( G \) . The set of left cosets of \( N \) in \( G \) form a partition of \( G \) . Furthermore, for all \( u, v \in G,{uN} = {vN} \) if and only if \( {v}^{-1}u \in N \) and in particular, \( {uN} = {vN} \) if and only if \( u \) and \( v \) are representatives ... | Proof: First of all note that since \( N \) is a subgroup of \( G,1 \in N \) . Thus \( g = g \cdot 1 \in {gN} \) for all \( g \in G \), i.e.,\n\n\[ G = \mathop{\bigcup }\limits_{{g \in G}}{gN} \]\n\nTo show that distinct left cosets have empty intersection, suppose \( {uN} \cap {vN} \neq \varnothing \) . We show \( {uN... | Yes |
Proposition 5. Let \( G \) be a group and let \( N \) be a subgroup of \( G \) .\n\n(1) The operation on the set of left cosets of \( N \) in \( G \) described by\n\n\[ \n{uN} \cdot {vN} = \left( {uv}\right) N \n\]\n\nis well defined if and only if \( {gn}{g}^{-1} \in N \) for all \( g \in G \) and all \( n \in N \) . | Proof: (1) Assume first that this operation is well defined, that is, for all \( u, v \in G \), \n\n\[ \n\text{if}u,{u}_{1} \in {uN}\text{and}v,{v}_{1} \in {vN}\;\text{then}\;{uvN} = {u}_{1}{v}_{1}N\text{.} \n\]\n\nLet \( g \) be an arbitrary element of \( G \) and let \( n \) be an arbitrary element of \( N \) . Letti... | Yes |
Theorem 6. Let \( N \) be a subgroup of the group \( G \) . The following are equivalent:\n\n(1) \( N \trianglelefteq G \)\n\n(2) \( {N}_{G}\left( N\right) = G \) (recall \( {N}_{G}\left( N\right) \) is the normalizer in \( G \) of \( N \) )\n\n(3) \( {gN} = {Ng} \) for all \( g \in G \)\n\n(4) the operation on left co... | Proof: We have already done the hard equivalences; the others are left as exercises. | No |
Proposition 7. A subgroup \( N \) of the group \( G \) is normal if and only if it is the kernel of some homomorphism. | Proof: If \( N \) is the kernel of the homomorphism \( \varphi \), then Proposition 2 shows that the left cosets of \( N \) are the same as the right cosets of \( N \) (and both are the fibers of the\n\nmap \( \varphi \) ). By (3) of Theorem \( 6, N \) is then a normal subgroup. (Another direct proof of this from the d... | No |
Theorem 8. (Lagrange’s Theorem) If \( G \) is a finite group and \( H \) is a subgroup of \( G \) , then the order of \( H \) divides the order of \( G \) (i.e., \( \left| H\right| \left| \right| G \mid \) ) and the number of left cosets of \( H \) in \( G \) equals \( \frac{\left| G\right| }{\left| H\right| } \) . | Proof: Let \( \left| H\right| = n \) and let the number of left cosets of \( H \) in \( G \) equal \( k \) . By\n\n# Proposition 4 the set of left cosets of \( H \) in \( G \) partition \( G \) . By definition of a left coset\n\nthe map:\n\n\[ H \rightarrow {gH}\;\text{ defined by }\;h \mapsto {gh} \]\n\nis a surjectio... | Yes |
Corollary 9. If \( G \) is a finite group and \( x \in G \), then the order of \( x \) divides the order of \( G \) . In particular \( {x}^{\left| G\right| } = 1 \) for all \( x \) in \( G \) . | Proof: By Proposition 2.2, \( \left| x\right| = \left| {\langle x\rangle }\right| \) . The first part of the corollary follows from Lagrange’s Theorem applied to \( H = \langle x\rangle \) . The second statement is clear since now \( \left| G\right| \) is a multiple of the order of \( x \) . | Yes |
Corollary 10. If \( G \) is a group of prime order \( p \), then \( G \) is cyclic, hence \( G \cong {Z}_{p} \) . | Proof: Let \( x \in G, x \neq 1 \) . Thus \( \left| {\langle x\rangle }\right| > 1 \) and \( \left| {\langle x\rangle }\right| \) divides \( \left| G\right| \) . Since \( \left| G\right| \) is prime we must have \( \left| {\langle x\rangle }\right| = \left| G\right| \), hence \( G = \langle x\rangle \) is cyclic (with ... | No |
Theorem 11. (Cauchy’s Theorem) If \( G \) is a finite group and \( p \) is a prime dividing \( \left| G\right| \) , then \( G \) has an element of order \( p \) . | Proof: We shall give a proof of this in the next chapter and another elegant proof is outlined in Exercise 9. | No |
Proposition 13. If \( H \) and \( K \) are finite subgroups of a group then\n\n\[ \left| {HK}\right| = \frac{\left| H\right| \left| K\right| }{\left| H \cap K\right| } \] | Proof: Notice that \( {HK} \) is a union of left cosets of \( K \), namely,\n\n\[ {HK} = \mathop{\bigcup }\limits_{{h \in H}}{hK} \]\n\nSince each coset of \( K \) has \( \left| K\right| \) elements it suffices to find the number of distinct left cosets of the form \( {hK}, h \in H \) . But \( {h}_{1}K = {h}_{2}K \) fo... | Yes |
Proposition 14. If \( H \) and \( K \) are subgroups of a group, \( {HK} \) is a subgroup if and only if \( {HK} = {KH} \) . | Proof: Assume first that \( {HK} = {KH} \) and let \( a, b \in {HK} \) . We prove \( a{b}^{-1} \in {HK} \) so \( {HK} \) is a subgroup by the subgroup criterion. Let\n\n\[ a = {h}_{1}{k}_{1}\;\text{ and }\;b = {h}_{2}{k}_{2}, \]\n\nfor some \( {h}_{1},{h}_{2} \in H \) and \( {k}_{1},{k}_{2} \in K \) . Thus \( {b}^{-1} ... | Yes |
Corollary 15. If \( H \) and \( K \) are subgroups of \( G \) and \( H \leq {N}_{G}\left( K\right) \), then \( {HK} \) is a subgroup of \( G \) . In particular, if \( K \trianglelefteq G \) then \( {HK} \leq G \) for any \( H \leq G \) . | Proof: We prove \( {HK} = {KH} \) . Let \( h \in H, k \in K \) . By assumption, \( {hk}{h}^{-1} \in K \) , hence\n\n\[ \n{hk} = \left( {{hk}{h}^{-1}}\right) h \in {KH}.\n\]\n\nThis proves \( {HK} \subseteq {KH} \) . Similarly, \( {kh} = h\left( {{h}^{-1}{kh}}\right) \in {HK} \), proving the reverse containment. The cor... | Yes |
Corollary 17. Let \( \varphi : G \rightarrow H \) be a homomorphism of groups.\n\n(1) \( \varphi \) is injective if and only if \( \ker \varphi = 1 \) . | Proof: Exercise. | No |
Theorem 18. (The Second or Diamond Isomorphism Theorem) Let \( G \) be a group, let \( A \) and \( B \) be subgroups of \( G \) and assume \( A \leq {N}_{G}\left( B\right) \) . Then \( {AB} \) is a subgroup of \( G \) , \( B \trianglelefteq {AB}, A \cap B \trianglelefteq A \) and \( {AB}/B \cong A/A \cap B \) . | Proof: By Corollary 15, \( {AB} \) is a subgroup of \( G \) . Since \( A \leq {N}_{G}\left( B\right) \) by assumption and \( B \leq {N}_{G}\left( B\right) \) trivially, it follows that \( {AB} \leq {N}_{G}\left( B\right) \), i.e., \( B \) is a normal subgroup of the subgroup \( {AB} \) . Since \( B \) is normal in \( {... | Yes |
Theorem 19. (The Third Isomorphism Theorem) Let \( G \) be a group and let \( H \) and \( K \) be normal subgroups of \( G \) with \( H \leq K \). Then \( K/H \trianglelefteq G/H \) and\n\n\[ \left( {G/H}\right) /\left( {K/H}\right) \cong G/K\text{.} \] | Proof: We leave as an easy exercise the verification that \( K/H \trianglelefteq G/H \). Define\n\n\[ \varphi : G/H \rightarrow G/K \]\n\n\[ \left( {gH}\right) \mapsto {gK}\text{.} \]\n\nTo show \( \varphi \) is well defined suppose \( {g}_{1}H = {g}_{2}H \). Then \( {g}_{1} = {g}_{2}h \), for some \( h \in H \). Becau... | No |
Theorem 20. (The Fourth or Lattice Isomorphism Theorem) Let \( G \) be a group and let \( N \) be a normal subgroup of \( G \) . Then there is a bijection from the set of subgroups \( A \) of \( G \) which contain \( N \) onto the set of subgroups \( \bar{A} = A/N \) of \( G/N \) . In particular, every subgroup of \( \... | Proof: The complete preimage of a subgroup in \( G/N \) is a subgroup of \( G \) by Exercise 1 of Section 1. The numerous details of the theorem to check are all completely straightforward. We therefore leave the proof of this theorem to the exercises. | No |
Proposition 21. If \( G \) is a finite abelian group and \( p \) is a prime dividing \( \left| G\right| \), then \( G \) contains an element of order \( p \) . | Proof: The proof proceeds by induction on \( \left| G\right| \), namely, we assume the result is valid for every group whose order is strictly smaller than the order of \( G \) and then prove the result valid for \( G \) (this is sometimes referred to as complete induction). Since \( \left| G\right| > 1 \), there is an... | Yes |
Theorem 22. (Jordan-Hölder) Let \( G \) be a finite group with \( G \neq 1 \) . Then\n\n(1) \( G \) has a composition series and\n\n(2) The composition factors in a composition series are unique, namely, if \( 1 = {N}_{0} \leq {N}_{1} \leq \cdots \leq {N}_{r} = G \) and \( 1 = {M}_{0} \leq {M}_{1} \leq \cdots \leq {M}_... | Proof: This is fairly straightforward. Since we shall not explicitly use this theorem to prove others in the text we outline the proof in a series of exercises at the end of this section. | No |
Proposition 23. The map \( \epsilon : {S}_{n} \rightarrow \{ \pm 1\} \) is a homomorphism (where \( \{ \pm 1\} \) is a multiplicative version of the cyclic group of order 2). | Proof: By definition,\n\n\[ \left( {\tau \sigma }\right) \left( \Delta \right) = \mathop{\prod }\limits_{{1 \leq i < j \leq n}}\left( {{x}_{{\tau \sigma }\left( i\right) } - {x}_{{\tau \sigma }\left( j\right) }}\right) \]\n\nSuppose that \( \sigma \left( \Delta \right) \) has exactly \( k \) factors of the form \( {x}_... | Yes |
Proposition 24. Transpositions are all odd permutations and \( \epsilon \) is a surjective homomorphism. | Moreover, since \( \epsilon \) is a homomorphism and every \( \sigma \in {S}_{n} \) is a product of transpositions, say \( \sigma = {\tau }_{1}{\tau }_{2}\cdots {\tau }_{k} \), then \( \epsilon \left( \sigma \right) = \epsilon \left( {\tau }_{1}\right) \cdots \epsilon \left( {\tau }_{k}\right) \) ; since \( \epsilon \l... | Yes |
Proposition 25. The permutation \( \sigma \) is odd if and only if the number of cycles of even length in its cycle decomposition is odd. | For example, \( \sigma = \left( {123456}\right) \left( {789}\right) \left( {1011}\right) \left( {12131415}\right) \left( {161718}\right) \) has 3 cycles of even length, so \( \epsilon \left( \sigma \right) = - 1 \) . On the other hand, \( \tau = \left( {1128104}\right) \left( {213}\right) \left( {5117}\right) \left( {6... | No |
Proposition 2. Let \( G \) be a group acting on the nonempty set \( A \). The relation on \( A \) defined by\n\n\[ a \sim b\;\text{ if and only if }\;a = g \cdot b\text{ for some }g \in G \]\n\n is an equivalence relation. For each \( a \in A \), the number of elements in the equivalence class containing \( a \) is \( ... | Proof: We first prove \( \sim \) is an equivalence relation. By axiom 2 of an action, \( a = 1 \cdot a \) for all \( a \in A \), i.e., \( a \sim a \) and the relation is reflexive. If \( a \sim b \), then \( a = g \cdot b \) for some \( b \in G \) so that\n\n\[ {g}^{-1} \cdot a = {g}^{-1} \cdot \left( {g \cdot b}\right... | Yes |
Theorem 3. Let \( G \) be a group, let \( H \) be a subgroup of \( G \) and let \( G \) act by left multiplication on the set \( A \) of left cosets of \( H \) in \( G \) . Let \( {\pi }_{H} \) be the associated permutation representation afforded by this action. Then\n\n(1) \( G \) acts transitively on \( A \)\n\n(2) ... | Proof: To see that \( G \) acts transitively on \( A \), let \( {aH} \) and \( {bH} \) be any two elements of \( A \), and let \( g = b{a}^{-1} \) . Then \( g \cdot {aH} = \left( {b{a}^{-1}}\right) {aH} = {bH} \), and so the two arbitrary elements \( {aH} \) and \( {bH} \) of \( A \) lie in the same orbit, which proves... | Yes |
Corollary 4. (Cayley's Theorem) Every group is isomorphic to a subgroup of some symmetric group. If \( G \) is a group of order \( n \), then \( G \) is isomorphic to a subgroup of \( {S}_{n} \) . | Proof: Let \( H = 1 \) and apply the preceding theorem to obtain a homomorphism of \( G \) into \( {S}_{G} \) (here we are identifying the cosets of the identity subgroup with the elements of \( G \) ). Since the kernel of this homomorphism is contained in \( H = 1, G \) is isomorphic to its image in \( {S}_{G} \) . | No |
Corollary 5. If \( G \) is a finite group of order \( n \) and \( p \) is the smallest prime dividing \( \left| G\right| \) , then any subgroup of index \( p \) is normal. | Proof: Suppose \( H \leq G \) and \( \left| {G : H}\right| = p \) . Let \( {\pi }_{H} \) be the permutation representation afforded by multiplication on the set of left cosets of \( H \) in \( G \), let \( K = \ker {\pi }_{H} \) and let \( \left| {H : K}\right| = k \) . Then \( \left| {G : K}\right| = \left| {G : H}\ri... | Yes |
Proposition 6. The number of conjugates of a subset \( S \) in a group \( G \) is the index of the normalizer of \( S,\left| {G : {N}_{G}\left( S\right) }\right| \) . In particular, the number of conjugates of an element \( s \) of \( G \) is the index of the centralizer of \( s,\left| {G : {C}_{G}\left( s\right) }\rig... | Proof: The second assertion of the proposition follows from the observation that \( {N}_{G}\left( {\{ s\} }\right) = {C}_{G}\left( s\right) \) . | No |
Theorem 7. (The Class Equation) Let \( G \) be a finite group and let \( {g}_{1},{g}_{2},\ldots ,{g}_{r} \) be representatives of the distinct conjugacy classes of \( G \) not contained in the center \( Z\left( G\right) \) of \( G \) . Then\n\n\[ \left| G\right| = \left| {Z\left( G\right) }\right| + \mathop{\sum }\limi... | Proof: As noted in Example 2 above the element \( \{ x\} \) is a conjugacy class of size 1 if and only if \( x \in Z\left( G\right) \), since then \( {gx}{g}^{-1} = x \) for all \( g \in G \) . Let \( Z\left( G\right) = \left\{ {1,{z}_{2},\ldots ,{z}_{m}}\right\} \) , let \( {\mathcal{K}}_{1},{\mathcal{K}}_{2},\ldots ,... | Yes |
Theorem 8. If \( p \) is a prime and \( P \) is a group of prime power order \( {p}^{\alpha } \) for some \( \alpha \geq 1 \) , then \( P \) has a nontrivial center: \( Z\left( P\right) \neq 1 \) . | Proof: By the class equation\n\n\[ \left| P\right| = \left| {Z\left( P\right) }\right| + \mathop{\sum }\limits_{{i = 1}}^{r}\left| {P : {C}_{P}\left( {g}_{i}\right) }\right| \]\n\nwhere \( {g}_{1},\ldots ,{g}_{r} \) are representatives of the distinct non-central conjugacy classes. By definition, \( {C}_{P}\left( {g}_{... | Yes |
Corollary 9. If \( \\left| P\\right| = {p}^{2} \) for some prime \( p \), then \( P \) is abelian. More precisely, \( P \) is isomorphic to either \( {Z}_{{p}^{2}} \) or \( {Z}_{p} \\times {Z}_{p} \) . | Proof: Since \( Z\\left( P\\right) \\neq 1 \) by the theorem, it follows that \( P/Z\\left( P\\right) \) is cyclic. By Exercise 36, Section 3.1, \( P \) is abelian. If \( P \) has an element of order \( {p}^{2} \), then \( P \) is cyclic. Assume therefore that every nonidentity element of \( P \) has order \( p \) . Le... | Yes |
Proposition 10. Let \( \sigma ,\tau \) be elements of the symmetric group \( {S}_{n} \) and suppose \( \sigma \) has cycle decomposition\n\n\[ \left( {{a}_{1}{a}_{2}\ldots {a}_{{k}_{1}}}\right) \left( {{b}_{1}{b}_{2}\ldots {b}_{{k}_{2}}}\right) \ldots \]\n\nThen \( {\tau \sigma }{\tau }^{-1} \) has cycle decomposition\... | Proof: Observe that if \( \sigma \left( i\right) = j \), then\n\n\[ {\tau \sigma }{\tau }^{-1}\left( {\tau \left( i\right) }\right) = \tau \left( j\right) \]\n\nThus, if the ordered pair \( i, j \) appears in the cycle decomposition of \( \sigma \), then the ordered pair \( \tau \left( i\right) ,\tau \left( j\right) \)... | Yes |
Proposition 11. Two elements of \( {S}_{n} \) are conjugate in \( {S}_{n} \) if and only if they have the same cycle type. The number of conjugacy classes of \( {S}_{n} \) equals the number of partitions of \( n \) . | Proof: By Proposition 10, conjugate permutations have the same cycle type. Conversely, suppose the permutations \( {\sigma }_{1} \) and \( {\sigma }_{2} \) have the same cycle type. Order the cycles in nondecreasing length, including 1-cycles (if several cycles of \( {\sigma }_{1} \) and \( {\sigma }_{2} \) have the sa... | Yes |
Corollary 14. If \( K \) is any subgroup of the group \( G \) and \( g \in G \), then \( K \cong {gK}{g}^{-1} \) . Conjugate elements and conjugate subgroups have the same order. | Proof: Letting \( G = H \) in the proposition shows that conjugation by \( g \in G \) is an automorphism of \( G \), from which the corollary follows. | No |
For any subgroup \( H \) of a group \( G \), the quotient group \( {N}_{G}\left( H\right) /{C}_{G}\left( H\right) \) is isomorphic to a subgroup of \( \operatorname{Aut}\left( H\right) \) . In particular, \( G/Z\left( G\right) \) is isomorphic to a subgroup of \( \operatorname{Aut}\left( G\right) \) . | Proof: Since \( H \) is a normal subgroup of the group \( {N}_{G}\left( H\right) \), Proposition 13 (applied with \( {N}_{G}\left( H\right) \) playing the role of \( G \) ) implies the first assertion. The second assertion is the special case when \( H = G \), in which case \( {N}_{G}\left( G\right) = G \) and \( {C}_{... | Yes |
Proposition 16. The automorphism group of the cyclic group of order \( n \) is isomorphic to \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \), an abelian group of order \( \varphi \left( n\right) \) (where \( \varphi \) is Euler’s function). | Proof: Let \( x \) be a generator of the cyclic group \( {Z}_{n} \) . If \( \psi \in \operatorname{Aut}\left( {Z}_{n}\right) \), then \( \psi \left( x\right) = {x}^{a} \) for some \( a \in \mathbb{Z} \) and the integer \( a \) uniquely determines \( \psi \) . Denote this automorphism by \( {\psi }_{a} \) . As usual, si... | Yes |
Lemma 19. Let \( P \in {Sy}{l}_{p}\left( G\right) \) . If \( Q \) is any \( p \) -subgroup of \( G \), then \( Q \cap {N}_{G}\left( P\right) = Q \cap P \) . | Proof: Let \( H = {N}_{G}\left( P\right) \cap Q \) . Since \( P \leq {N}_{G}\left( P\right) \) it is clear that \( P \cap Q \leq H \), so we must prove the reverse inclusion. Since by definition \( H \leq Q \), this is equivalent to showing \( H \leq P \) . We do this by demonstrating that \( {PH} \) is a \( p \) -subg... | Yes |
Corollary 20. Let \( P \) be a Sylow \( p \) -subgroup of \( G \) . Then the following are equivalent:\n\n(1) \( P \) is the unique Sylow \( p \) -subgroup of \( G \), i.e., \( {n}_{p} = 1 \)\n\n(2) \( P \) is normal in \( G \)\n\n(3) \( P \) is characteristic in \( G \)\n\n(4) All subgroups generated by elements of \(... | Proof: If (1) holds, then \( {gP}{g}^{-1} = P \) for all \( g \in G \) since \( {gP}{g}^{-1} \in {Sy}{l}_{p}\left( G\right) \), i.e., \( P \) is normal in \( G \) . Hence (1) implies (2). Conversely, if \( P \trianglelefteq G \) and \( Q \in {\operatorname{Syl}}_{p}\left( G\right) \), then by Sylow’s Theorem there exis... | Yes |
Proposition 21. If \( \left| G\right| = {60} \) and \( G \) has more than one Sylow 5-subgroup, then \( G \) is simple. | Proof: Suppose by way of contradiction that \( \left| G\right| = {60} \) and \( {n}_{5} > 1 \) but that there exists \( H \) a normal subgroup of \( G \) with \( H \neq 1 \) or \( G \) . By Sylow’s Theorem the only possibility for \( {n}_{5} \) is 6 . Let \( P \in {\operatorname{Syl}}_{5}\left( G\right) \), so that \( ... | Yes |
Corollary 22. \( {A}_{5} \) is simple. | Proof: The subgroups \( \langle \;\left( {1\;2\;3\;4\;5}\right) \;\rangle \;\mathrm{{and}}\;\langle \;\left( {1\;3\;2\;4\;5}\right) \;\rangle \;\mathrm{{are}}\;\mathrm{{distinct}}\;\mathrm{{Sylow}}\;5 \) -subgroups of \( {A}_{5} \) so the result follows immediately from the proposition. | No |
Proposition 23. If \( G \) is a simple group of order 60, then \( G \cong {A}_{5} \) . | Proof: Let \( G \) be a simple group of order 60, so \( {n}_{2} = 3,5 \) or 15 . Let \( P \in {\operatorname{Syl}}_{2}\left( G\right) \) and let \( N = {N}_{G}\left( P\right) \), so \( \left| {G : N}\right| = {n}_{2} \) . First observe that \( G \) has no proper subgroup \( H \) of index less that 5, as follows: if \( ... | Yes |
Proposition 1. If \( {G}_{1},\ldots ,{G}_{n} \) are groups, their direct product is a group of order \( \left| {G}_{1}\right| \left| {G}_{2}\right| \cdots \left| {G}_{n}\right| \) (if any \( {G}_{i} \) is infinite, so is the direct product). | Proof: Let \( G = {G}_{1} \times {G}_{2} \times \cdots \times {G}_{n} \) . The proof that the group axioms hold for \( G \) is straightforward since each axiom is a consequence of the fact that the same axiom holds in each factor, \( {G}_{i} \), and the operation on \( G \) is defined componentwise. For example, the as... | Yes |
For each fixed \( i \) the set of elements of \( G \) which have the identity of \( {G}_{j} \) in the \( {j}^{\text{th }} \) position for all \( j \neq i \) and arbitrary elements of \( {G}_{i} \) in position \( i \) is a subgroup of \( G \) isomorphic to \( {G}_{i} \): | Since the operation in \( G \) is defined componentwise, it follows easily from the subgroup criterion that \( \left\{ {\left( {1,1,\ldots ,1,{g}_{i},1,\ldots ,1}\right) \mid {g}_{i} \in {G}_{i}}\right\} \) is a subgroup of \( G \) . Furthermore, the map \( {g}_{i} \mapsto \left( {1,1,\ldots ,1,{g}_{i},1,\ldots ,1}\rig... | Yes |
Theorem 3. (Fundamental Theorem of Finitely Generated Abelian Groups) Let \( G \) be a finitely generated abelian group. Then (1)\n\n\[ G \cong {\mathbb{Z}}^{r} \times {Z}_{{n}_{1}} \times {Z}_{{n}_{2}} \times \cdots \times {Z}_{{n}_{s}} \]\n\nfor some integers \( r,{n}_{1},{n}_{2},\ldots ,{n}_{s} \) satisfying the fol... | Proof: We shall derive this theorem in Section 12.1 as a consequence of a more general classification theorem. For finite groups we shall give an alternate proof at the end of Section 6.1. | No |
(1) \( {Z}_{m} \times {Z}_{n} \cong {Z}_{mn} \) if and only if \( \left( {m, n}\right) = 1 \) . | Proof: Since (2) is an easy exercise using (1) and induction on \( k \), we concentrate on proving (1). Let \( {Z}_{m} = \langle x\rangle ,{Z}_{n} = \langle y\rangle \) and let \( l = \) 1.c.m. \( \left( {m, n}\right) \) . Note that \( l = {mn} \) if and only if \( \left( {m, n}\right) = 1 \) . Let \( {x}^{a}{y}^{b} \)... | Yes |
Proposition 7. Let \( G \) be a group, let \( x, y \in G \) and let \( H \leq G \) . Then\n\n(1) \( {xy} = {yx}\left\lbrack {x, y}\right\rbrack \) (in particular, \( {xy} = {yx} \) if and only if \( \left\lbrack {x, y}\right\rbrack = 1 \) ). | Proof: (1) This is immediate from the definition of \( \left\lbrack {x, y}\right\rbrack \) . | No |
Proposition 8. Let \( H \) and \( K \) be subgroups of the group \( G \). The number of distinct ways of writing each element of the set \( {HK} \) in the form \( {hk} \), for some \( h \in H \) and \( k \in K \) is \( \left| {H \cap K}\right| \). In particular, if \( H \cap K = 1 \), then each element of \( {HK} \) ca... | Proof: Exercise. | No |
Theorem 9. Suppose \( G \) is a group with subgroups \( H \) and \( K \) such that\n\n(1) \( H \) and \( K \) are normal in \( G \), and\n\n(2) \( H \cap K = 1 \).\n\nThen \( {HK} \cong H \times K \) . | Proof: Observe that by hypothesis (1), \( {HK} \) is a subgroup of \( G \) (see Corollary 3.15). Let \( h \in H \) and let \( k \in K \) . Since \( H \trianglelefteq G,{k}^{-1}{hk} \in H \), so that \( {h}^{-1}\left( {{k}^{-1}{hk}}\right) \in H \) . Similarly, \( \left( {{h}^{-1}{k}^{-1}h}\right) k \in K \) . Since \( ... | Yes |
Theorem 10. Let \( H \) and \( K \) be groups and let \( \varphi \) be a homomorphism from \( K \) into Aut \( \left( H\right) \) . Let \( \cdot \) denote the (left) action of \( K \) on \( H \) determined by \( \varphi \) . Let \( G \) be the set of ordered pairs \( \left( {h, k}\right) \) with \( h \in H \) and \( k ... | Proof: It is straightforward to check that \( G \) is a group under this multiplication using the fact that \( \cdot \) is an action of \( K \) on \( H \) . For example, the associative law is verified as follows:\n\n\[ \left( {\left( {a, x}\right) \left( {b, y}\right) }\right) \left( {c, z}\right) = \left( {{ax} \cdot... | No |
Proposition 11. Let \( H \) and \( K \) be groups and let \( \varphi : K \rightarrow \operatorname{Aut}\left( H\right) \) be a homomorphism. Then the following are equivalent:\n\n(1) the identity (set) map between \( H \rtimes K \) and \( H \times K \) is a group homomorphism (hence an isomorphism)\n\n(2) \( \varphi \)... | Proof: \( \left( 1\right) \Rightarrow \left( 2\right) \) By definition of the group operation in \( H \rtimes K \)\n\n\[ \left( {{h}_{1},{k}_{1}}\right) \left( {{h}_{2},{k}_{2}}\right) = \left( {{h}_{1}{k}_{1} \cdot {h}_{2},{k}_{1}{k}_{2}}\right) \]\n\nfor all \( {h}_{1},{h}_{2} \in H \) and \( {k}_{1},{k}_{2} \in K \)... | Yes |
Theorem 12. Suppose \( G \) is a group with subgroups \( H \) and \( K \) such that\n\n(1) \( H \trianglelefteq G \), and\n\n(2) \( H \cap K = 1 \) .\n\nLet \( \varphi : K \rightarrow \operatorname{Aut}\left( H\right) \) be the homomorphism defined by mapping \( k \in K \) to the automorphism of left conjugation by \( ... | Proof: Note that since \( H \trianglelefteq G,{HK} \) is a subgroup of \( G \) . By Proposition 8 every element of \( {HK} \) can be written uniquely in the form \( {hk} \), for some \( h \in H \) and \( k \in K \) . Thus the map \( {hk} \mapsto \left( {h, k}\right) \) is a set bijection from \( {HK} \) onto \( H \rtim... | Yes |
Let \( p \) be a prime and let \( P \) be a group of order \( {p}^{a}, a \geq 1 \) . Then\n\n(1) The center of \( P \) is nontrivial: \( Z\left( P\right) \neq 1 \) .\n\n(2) If \( H \) is a nontrivial normal subgroup of \( P \) then \( H \) intersects the center non-trivially: \( H \cap Z\left( P\right) \neq 1 \) . In p... | These results rely ultimately on the class equation and it may be useful for the reader to review Section 4.3.\n\nPart 1 is Theorem 8 of Chapter 4 and is also the special case of part 2 when \( H = P \) . We therefore begin by proving (2); we shall not quote Theorem 8 of Chapter 4 although the argument that follows is ... | Yes |
Proposition 2. Let \( p \) be a prime and let \( P \) be a group of order \( {p}^{a} \) . Then \( P \) is nilpotent of nilpotence class at most \( a - 1 \) . | Proof: For each \( i \geq 0, P/{Z}_{i}\left( P\right) \) is a \( p \) -group, so\n\n\[ \text{if}\left| {P/{Z}_{i}\left( P\right) }\right| > 1\text{then}Z\left( {P/{Z}_{i}\left( P\right) }\right) \neq 1 \]\n\nby Theorem 1(1). Thus if \( {Z}_{i}\left( P\right) \neq G \) then \( \left| {{Z}_{i + 1}\left( P\right) }\right|... | Yes |
Theorem 3. Let \( G \) be a finite group, let \( {p}_{1},{p}_{2},\ldots ,{p}_{s} \) be the distinct primes dividing its order and let \( {P}_{i} \in {\operatorname{Syl}}_{{p}_{i}}\left( G\right) ,1 \leq i \leq s \) . Then the following are equivalent:\n\n(1) \( G \) is nilpotent\n\n(2) if \( H < G \) then \( H < {N}_{G... | Proof: The proof that (1) implies (2) is the same argument as for \( p \) -groups - the only fact we needed was if \( G \) is nilpotent then so is \( G/Z\left( G\right) \) - so the details are omitted (cf. the exercises).\n\nTo show that (2) implies (3) let \( P = {P}_{i} \) for some \( i \) and let \( N = {N}_{G}\left... | No |
Proposition 5. If \( G \) is a finite group such that for all positive integers \( n \) dividing its order, \( G \) contains at most \( n \) elements \( x \) satisfying \( {x}^{n} = 1 \), then \( G \) is cyclic. | Proof: Let \( \left| G\right| = {p}_{1}^{{\alpha }_{1}}\cdots {p}_{s}^{{\alpha }_{s}} \) and let \( {P}_{i} \) be a Sylow \( {p}_{i} \) -subgroup of \( G \) for \( i = 1,2,\ldots, s \) . Since \( {p}_{i}^{{\alpha }_{i}}\left| \right| G| \) and the \( {p}_{i}^{{\alpha }_{i}} \) elements of \( {P}_{i} \) are solutions of... | Yes |
Proposition 6. (Frattini’s Argument) Let \( G \) be a finite group, let \( H \) be a normal subgroup of \( G \) and let \( P \) be a Sylow \( p \) -subgroup of \( H \) . Then \( G = H{N}_{G}\left( P\right) \) and \( \left| {G : H}\right| \) divides \( \left| {{N}_{G}\left( P\right) }\right| \) . | Proof: By Corollary 3.15, \( H{N}_{G}\left( P\right) \) is a subgroup of \( G \) and \( H{N}_{G}\left( P\right) = {N}_{G}\left( P\right) H \) since \( H \) is a normal subgroup of \( G \) . Let \( g \in G \) . Since \( {P}^{g} \leq {H}^{g} = H \), both \( P \) and \( {P}^{g} \) are Sylow \( p \) -subgroups of \( H \) .... | Yes |
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