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Lemma 2.22. Disjoint permutations \( \alpha ,\beta \in {S}_{n} \) commute. | Proof. It suffices to prove that if \( 1 \leq i \leq n \), then \( {\alpha \beta }\left( i\right) = {\beta \alpha }\left( i\right) \) . If \( \beta \) moves \( i \), say, \( \beta \left( i\right) = j \neq i \), then \( \beta \) also moves \( j \) [otherwise, \( \beta \left( j\right) = j \) and \( \beta \left( i\right) ... | Yes |
Lemma 2.23. Let \( X = \{ 1,2,\ldots, n\} \), let \( \alpha \in {S}_{X} = {S}_{n} \), and, if \( {i}_{1} \in X \), define \( {i}_{j} \) for all \( j \geq 1 \) by induction: \( {i}_{j + 1} = \alpha \left( {i}_{j}\right) \) . Write \( Y = \left\{ {{i}_{j} : j \geq 1}\right\} \), and let \( {Y}^{\prime } \) be the complem... | Proof. (i) Since \( X \) is finite, there is a smallest \( r > 1 \) with \( {i}_{1},\ldots ,{i}_{r} \) all distinct, but with \( {i}_{r + 1} = \alpha \left( {i}_{r}\right) \in \left\{ {{i}_{1},\ldots ,{i}_{r}}\right\} \) ; that is, \( \alpha \left( {i}_{r}\right) = {i}_{j} \) for \( 1 \leq j \leq r \) . If \( j > 1 \),... | Yes |
Every permutation \( \alpha \in {S}_{n} \) is either a cycle or a product of disjoint cycles. | The proof is by induction on the number \( k \geq 0 \) of points moved by \( \alpha \) . The base step \( k = 0 \) is true, for \( \alpha \) is now the identity, which is a 1-cycle.\n\nIf \( k > 0 \), let \( {i}_{1} \) be a point moved by \( \alpha \) . As in Lemma 2.23, define \( Y = \) \( \left\{ {{i}_{1},\ldots ,{i}... | Yes |
Theorem 2.26. Let \( \alpha \in {S}_{n} \) and let \( \alpha = {\beta }_{1}\cdots {\beta }_{t} \) be a complete factorization into disjoint cycles. This factorization is unique except for the order in which the cycles occur. | Proof. Let \( \alpha = {\gamma }_{1}\cdots {\gamma }_{s} \) be a second complete factorization of \( \alpha \) into disjoint cycles. Since every complete factorization of \( \alpha \) has exactly one 1-cycle for each \( i \) fixed by \( \alpha \), it suffices to prove, by induction on \( \ell \), the larger of \( t \) ... | Yes |
Example 2.30. | <table><thead><tr><th>Cycle Structure</th><th>Number</th></tr></thead><tr><td>(1)</td><td>1</td></tr><tr><td>(1 2)</td><td>10</td></tr><tr><td>(1 2 3)</td><td>20</td></tr><tr><td>(1 2 3 4)</td><td>30</td></tr><tr><td>(1 2 3 4 5)</td><td>24</td></tr><tr><td>(1 2)(3 4 5)</td><td>20</td></tr><tr><td>(1 2)(3 4)</td><td>\\(... | No |
Lemma 2.31. Let \( \alpha ,\gamma \in {S}_{n} \) . For all \( i \), if \( \gamma : i \rightarrow j \), then \( {\alpha \gamma }{\alpha }^{-1} : \alpha \left( i\right) \rightarrow \alpha \left( j\right) \) . | \[ {\alpha \gamma }{\alpha }^{-1}\left( {\alpha \left( i\right) }\right) = {\alpha \gamma }\left( i\right) = \alpha \left( j\right) . \] | Yes |
If \( \gamma ,\alpha \in {S}_{n} \), then \( {\alpha \gamma }{\alpha }^{-1} \) has the same cycle structure as \( \gamma \) . In more detail, if the complete factorization of \( \gamma \) is\n\n\[ \gamma = {\beta }_{1}{\beta }_{2}\cdots \left( {{ij}\ldots }\right) \cdots {\beta }_{t} \]\n\nthen \( {\alpha \gamma }{\alp... | Proof. If \( \gamma \) fixes \( i \), then Lemma 2.31 shows that \( \sigma \) fixes \( \alpha \left( i\right) \) . Assume that \( \gamma \) moves a symbol \( i \), say, \( \gamma \left( i\right) = j \), so that one of the cycles in the complete factorization of \( \gamma \) is\n\n\[ \left( \begin{array}{lll} i & j & \l... | Yes |
Proposition 2.33. If \( \gamma ,{\gamma }^{\prime } \in {S}_{n} \), then \( \gamma \) and \( {\gamma }^{\prime } \) have the same cycle structure if and only only if there exists \( \alpha \in {S}_{n} \) with \( {\gamma }^{\prime } = {\alpha \gamma }{\alpha }^{-1} \) . | Proof. Sufficiency has just been proved, in Proposition 2.32.\n\nConversely, assume that \( \gamma \) and \( {\gamma }^{\prime } \) have the same cycle structure; that is, \( \gamma = \) \( {\beta }_{1}\cdots {\beta }_{t} \) and \( {\gamma }^{\prime } = {\sigma }_{1}\cdots {\sigma }_{t} \) are complete factorizations w... | Yes |
If \[ \gamma = \left( \begin{array}{lll} 1 & 2 & 3 \end{array}\right) \left( \begin{array}{ll} 4 & 5 \end{array}\right) \left( 6\right) \;\text{ and }\;{\gamma }^{\prime } = \left( \begin{array}{lll} 2 & 5 & 6 \end{array}\right) \left( \begin{array}{ll} 3 & 1 \end{array}\right) \left( 4\right) ,\] then \( {\gamma }^{\p... | Note that there are other choices for \( \alpha \) as well. | Yes |
Proposition 2.35. If \( n \geq 2 \), then every \( \alpha \in {S}_{n} \) is a product of transpositions. | Proof. Of course, \( \left( 1\right) = \left( {12}\right) \left( {12}\right) \) is a product of transpositions, as is every transposition: \( \left( \begin{array}{ll} i & j \end{array}\right) = \left( \begin{array}{ll} i & j \end{array}\right) \left( \begin{array}{ll} 1 & 2 \end{array}\right) \left( \begin{array}{ll} 1... | Yes |
Lemma 2.37. If \( k,\ell \geq 0 \) and the letters \( a, b,{c}_{i},{d}_{j} \) are all distinct, then\n\n\[ \left( {ab}\right) \left( {a{c}_{1}\ldots {c}_{k}b{d}_{1}\ldots {d}_{\ell }}\right) = \left( {a{c}_{1}\ldots {c}_{k}}\right) \left( {b{d}_{1}\ldots {d}_{\ell }}\right) \]\n\nand\n\n\[ \left( \begin{array}{ll} a & ... | Proof. The left side of the first asserted equation sends\n\n\[ a \mapsto {c}_{1} \mapsto {c}_{1} \]\n\n\[ {c}_{i} \mapsto {c}_{i + 1} \mapsto {c}_{i + 1}\text{ if }i < k \]\n\n\[ {c}_{k} \mapsto b \mapsto a \]\n\n\[ b \mapsto {d}_{1} \mapsto {d}_{1} \]\n\n\[ {d}_{j} \mapsto {d}_{j + 1} \mapsto {d}_{j + 1}\text{if}j < ... | Yes |
Lemma 2.38. If \( \alpha ,\tau \in {S}_{n} \), where \( \tau \) is a transposition, then\n\n\[ \operatorname{sgn}\left( {\tau \alpha }\right) = - \operatorname{sgn}\left( \alpha \right) \] | Proof. Let \( \alpha = {\beta }_{1}\cdots {\beta }_{t} \) be a complete factorization of \( \alpha \) into disjoint cycles, and let \( \tau = \left( \begin{array}{ll} a & b \end{array}\right) \) . If \( a \) and \( b \) occur in the same \( \beta \), say, in \( {\beta }_{1} \), then \( {\beta }_{1} = \) \( \left( {a{c}... | Yes |
Theorem 2.39. For all \( \alpha ,\beta \in {S}_{n} \) ,\n\n\[ \operatorname{sgn}\left( {\alpha \beta }\right) = \operatorname{sgn}\left( \alpha \right) \operatorname{sgn}\left( \beta \right) \] | Proof. Assume that \( \alpha \in {S}_{n} \) is given and that \( \alpha \) has a factorization as a product of \( m \) transpositions: \( \alpha = {\tau }_{1}\cdots {\tau }_{m} \) . We prove, by induction on \( m \), that \( \operatorname{sgn}\left( {\alpha \beta }\right) = \) \( \operatorname{sgn}\left( \alpha \right)... | Yes |
Corollary 2.41. Let \( \alpha ,\beta \in {S}_{n} \) . If \( \alpha \) and \( \beta \) have the same parity, then \( {\alpha \beta } \) is even, while if \( \alpha \) and \( \beta \) have distinct parity, then \( {\alpha \beta } \) is odd. | Proof. If \( \operatorname{sgn}\left( \alpha \right) = {\left( -1\right) }^{q} \) and \( \operatorname{sgn}\left( \beta \right) = {\left( -1\right) }^{p} \), then Theorem 2.39 gives \( \operatorname{sgn}\left( {\alpha \beta }\right) = {\left( -1\right) }^{q + p} \), and the result follows. | Yes |
Lemma 2.43. If \( * \) is an associative operation on a set \( G \), then\n\n\[ \left( {a * b}\right) * \left( {c * d}\right) = \left\lbrack {a * \left( {b * c}\right) }\right\rbrack * d \]\n\nfor all \( a, b, c, d \in G \) | Proof. If we write \( g = a * b \), then \( \left( {a * b}\right) * \left( {c * d}\right) = g * \left( {c * d}\right) = \left( {g * c}\right) * d = \) \( \left\lbrack {\left( {a * b}\right) * c}\right\rbrack * d = \left\lbrack {a * \left( {b * c}\right) }\right\rbrack * d. \) | Yes |
Lemma 2.44. If \( G \) is a group and \( a \in G \) satisfies \( a * a = a \), then \( a = e \) . | Proof. There is \( {a}^{\prime } \in G \) with \( {a}^{\prime } * a = e \) . Multiplying both sides on the left by \( {a}^{\prime } \) gives \( {a}^{\prime } * \left( {a * a}\right) = {a}^{\prime } * a \) . The right side is \( e \), and the left side is \( {a}^{\prime } * \left( {a * a}\right) = \) \( \left( {{a}^{\pr... | Yes |
Proposition 2.45. Let \( G \) be a group with operation \( * \) and identity \( e \) .\n\n(i) \( a * {a}^{\prime } = e \) for all \( a \in G \) .\n\n(ii) \( a * e = a \) for all \( a \in G \) .\n\n(iii) If \( {e}_{0} \in G \) satisfies \( {e}_{0} * a = a \) for all \( a \in G \), then \( {e}_{0} = e \) .\n\n(iv) Let \(... | Proof.\n\n(i) We know that \( {a}^{\prime } * a = e \), and we now show that \( a * {a}^{\prime } = e \) . By Lemma 2.43,\n\n\[ \n\left( {a * {a}^{\prime }}\right) * \left( {a * {a}^{\prime }}\right) = \left\lbrack {a * \left( {{a}^{\prime } * a}\right) }\right\rbrack * {a}^{\prime } \n\]\n\n\[ \n= \left( {a * e}\right... | Yes |
Lemma 2.46. Let \( G \) be a group.\n\n(i) The cancellation laws hold: if \( a, b, x \in G \), and either \( x * a = x * b \) or \( a * x = b * x \), then \( a = b \) . | Proof.\n\n(i)\n\n\[ a = e * a = \left( {{x}^{-1} * x}\right) * a = {x}^{-1} * \left( {x * a}\right) \]\n\n\[ = {x}^{-1} * \left( {x * b}\right) = \left( {{x}^{-1} * x}\right) * b = e * b = b. \] | Yes |
(iv) The nonzero complex numbers \( {\mathbb{C}}^{ \times } \) form an abelian group under multiplication. It is easy to see that multiplication is an associative operation and that 1 is the identity. | Here is the simplest way to find inverses. If \( z = a + {ib} \in \mathbb{C} \), where \( a, b \in \mathbb{R} \), define its complex conjugate \( \bar{z} = a - {ib} \) . Note that \( z\bar{z} = {a}^{2} + {b}^{2} \), so that \( z \neq 0 \) if and only if \( z\bar{z} \neq 0 \) . If \( z \neq 0 \), then\n\n\[ \n{z}^{-1} =... | Yes |
Theorem 2.49 (Generalized Associativity). If \( n \geq 3 \), then every \( n \) -expression \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) in a group \( G \) needs no parentheses. | Proof. The proof is by (the second form of) induction. The base step \( n = 3 \) follows from associativity. For the inductive step, consider 2-expressions of \( G \) obtained from an \( n \) -expression \( \left( {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right) \) after two series of choices:\n\n\[ \left( {W, X}\right) = \lef... | Yes |
Corollary 2.50. If \( G \) is a group, if \( a \in G \), and if \( m, n \geq 1 \), then\n\n\[ \n{a}^{m + n} = {a}^{m}{a}^{n}\;\text{ and }\;{\left( {a}^{m}\right) }^{n} = {a}^{mn}.\n\] | Proof. Both \( {a}^{m + n} \) and \( {a}^{m}{a}^{n} \) arise from the expression having \( m + n \) factors each equal to \( a \) ; in the second instance, both \( {\left( {a}^{m}\right) }^{n} \) and \( {a}^{mn} \) arise from the expression having \( {mn} \) factors each equal to \( a \) . | Yes |
Proposition 2.51 (Laws of Exponents). Let \( G \) be a group, let \( a, b \in G \), and let \( m \) and \( n \) be (not necessarily positive) integers.\n\n(i) If \( a \) and \( b \) commute, then \( {\left( ab\right) }^{n} = {a}^{n}{b}^{n} \).\n\n(ii) \( {\left( {a}^{n}\right) }^{m} = {a}^{mn} \).\n\n(iii) \( {a}^{m}{a... | Proof. Exercises for the reader. - | No |
Lemma 2.53. Let \( G \) be a group and assume that \( a \in G \) has finite order \( k \) . If \( {a}^{n} = 1 \), then \( k \mid n \) . In fact, \( \left\{ {n \in \mathbb{Z} : {a}^{n} = 1}\right\} \) is the set of all the multiples of \( k \) . | Proof. It is easy to see that \( I = \left\{ {n \in \mathbb{Z} : {a}^{n} = 1}\right\} \subseteq \mathbb{Z} \) satisfies the hypotheses of Corollary 1.34.\n\n(i): \( 0 \in I \) because \( {a}^{0} = 1 \) .\n\n(ii): If \( n, m \in I \), then \( {a}^{n} = 1 \) and \( {a}^{m} = 1 \), so that \( {a}^{n - m} = {a}^{n}{a}^{-m}... | Yes |
Proposition 2.54. Let \( \alpha \in {S}_{n} \) .\n\n(i) If \( \alpha \) is an \( r \) -cycle, then \( \alpha \) has order \( r \) .\n\n(ii) If \( \alpha = {\beta }_{1}\cdots {\beta }_{t} \) is a product of disjoint \( {r}_{i} \) -cycles \( {\beta }_{i} \), then \( \alpha \) has order \( m = \operatorname{lcm}\left( {{r... | Proof.\n\n(i) This is Exercise 2.22(i) on page 120.\n\n(ii) Each \( {\beta }_{i} \) has order \( {r}_{i} \), by (i). Suppose that \( {\alpha }^{M} = \left( 1\right) \) . Since the \( {\beta }_{i} \) commute, \( \left( 1\right) = {\alpha }^{M} = {\left( {\beta }_{1}\cdots {\beta }_{t}\right) }^{M} = {\beta }_{1}^{M}\cdo... | No |
Lemma 2.56. If \( \varphi \) is an isometry of the plane, then distinct points \( P, Q, R \) in \( {\mathbb{R}}^{2} \) are collinear if and only if \( \varphi \left( P\right) ,\varphi \left( Q\right) ,\varphi \left( R\right) \) are collinear. | Proof. Suppose that \( P, Q, R \) are collinear. Choose notation so that \( R \) is between \( P \) and \( Q \) ; hence, \( \parallel P - Q\parallel = \parallel P - R\parallel + \parallel R - Q\parallel \) . If \( \varphi \left( P\right) ,\varphi \left( Q\right) ,\varphi \left( R\right) \) are not collinear, then they ... | Yes |
Every isometry \( \varphi : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) is a bijection, and every isometry fixing 0 is a nonsingular linear transformation. | Proof. Let us first assume that \( \varphi \) fixes the origin: \( \varphi \left( 0\right) = 0 \) . By Proposition 2.57, \( \varphi \) is a linear transformation. Since \( \varphi \) is injective, \( P = \varphi \left( {e}_{1}\right), Q = \) \( \varphi \left( {e}_{2}\right) \) is a basis of \( {\mathbb{R}}^{2} \), wher... | Yes |
Proposition 2.59. Both \( \operatorname{Isom}\left( {\mathbb{R}}^{2}\right) \) and \( {O}_{2}\left( \mathbb{R}\right) \) are groups under composition. | Proof. We show that \( \operatorname{Isom}\left( {\mathbb{R}}^{2}\right) \) is a group. Clearly, \( {1}_{\mathbb{R}} \) is an isometry, so that \( {1}_{\mathbb{R}} \in \operatorname{Isom}\left( {\mathbb{R}}^{2}\right) \) . Let \( {\varphi }^{\prime } \) and \( \varphi \) be isometries. For all points \( P \) and \( Q \... | No |
Corollary 2.60. If \( O, P, Q \) are noncollinear points, and if \( \varphi \) and \( \psi \) are isometries of the plane such that \( \varphi \left( P\right) = \psi \left( P\right) \) and \( \varphi \left( Q\right) = \psi \left( Q\right) \), then \( \varphi = \psi \) . | Proof. Since \( O, P, Q \) are noncollinear points, the list \( P, Q \) is linearly independent in the vector space \( {\mathbb{R}}^{2} \) . Since \( \dim \left( {\mathbb{R}}^{2}\right) = 2 \), this is a basis [see Corollary 4.24(ii)], and any two linear transformations that agree on a basis are equal (see Corollary 4.... | Yes |
Proposition 2.64. Every subgroup \( H \leq G \) of a group \( G \) is itself a group. | Proof. Axiom (ii) (in the definition of subgroup) shows that \( H \) is closed under the operation of \( G \) ; that is, \( H \) has an operation (namely, the restriction of the operation \( * : G \times G \rightarrow G \) to \( H \times H \subseteq G \times G \) ). This operation is associative: since the equation \( ... | Yes |
Proposition 2.66. A subset \( H \) of a group \( G \) is a subgroup if and only if \( H \) is nonempty and, whenever \( x, y \in H \), then \( x{y}^{-1} \in H \) . | Proof. If \( H \) is a subgroup, then it is nonempty, for \( 1 \in H \) . If \( x, y \in H \), then \( {y}^{-1} \in H \), by part (iii) of the definition, and so \( x{y}^{-1} \in H \), by part (ii). Conversely, assume that \( H \) is a subset satisfying the new condition. Since \( H \) is nonempty, it contains some ele... | Yes |
Proposition 2.67. A nonempty subset \( H \) of a finite group \( G \) is a subgroup if and only if \( H \) is closed under the operation of \( G \) ; that is, if \( a, b \in H \), then \( {ab} \in H \) . In particular, a nonempty subset of \( {S}_{n} \) is a subgroup if and only if it is closed under composition. | Proof. Every subgroup is nonempty, by axiom (i) in the definition of subgroup, and it is closed, by axiom (ii). Conversely, assume that \( H \) is a nonempty subset of \( G \) closed under the operation on \( G \) ; thus, axiom (ii) holds. It follows that \( H \) contains all the powers of its elements. In particular, ... | Yes |
Proposition 2.69. If \( G = \langle a\rangle \) is a cyclic group of order \( n \), then \( {a}^{k} \) is a generator of \( G \) if and only if \( \gcd \left( {k, n}\right) = 1 \) . | Proof. If \( {a}^{k} \) is a generator, then \( a \in \left\langle {a}^{k}\right\rangle \), so there is \( s \) with \( a = {a}^{ks} \) . Hence, \( {a}^{{ks} - 1} = 1 \), so that Lemma 2.53 shows that \( n \mid \left( {{ks} - 1}\right) \) ; that is, there is an integer \( t \) with \( {ks} - 1 = {tn} \), or \( {sk} - {... | Yes |
Corollary 2.70. The number of generators of a cyclic group of order \( n \) is \( \phi \left( n\right) \) . | Proof. This follows at once from the Propositions 2.69 and 1.39. - | No |
Every subgroup \( S \) of a cyclic group \( G = \langle a\rangle \) is itself cyclic. In fact, \( {a}^{m} \) is a generator of \( S \), where \( m \) is the smallest positive integer with \( {a}^{m} \in S \) . | Proof. We may assume that \( S \) is nontrivial; that is, \( S \neq \{ 1\} \), for the proposition is obviously true when \( S = \{ 1\} \) . Let \( I = \left\{ {m \in \mathbb{Z} : {a}^{m} \in S}\right\} \) ; it is easy to check that \( I \) satisfies the conditions in Corollary 1.34. (i): \( 0 \in I \), for \( {a}^{0} ... | Yes |
Proposition 2.72. Let \( G \) be a finite group and let \( a \in G \) . Then the order of a is the number of elements in \( \langle a\rangle \) . | Proof. We will use the idea in Lemma 2.23. Since \( G \) is finite, there is an integer \( k \geq 1 \) with \( 1, a,{a}^{2},\ldots ,{a}^{k - 1} \) consisting of \( k \) distinct elements, while \( 1, a,{a}^{2},\ldots ,{a}^{k} \) has a repetition; hence \( {a}^{k} \in \left\{ {1, a,{a}^{2},\ldots ,{a}^{k - 1}}\right\} \... | Yes |
Proposition 2.73. A group \( G \) of order \( n \) is cyclic if and only if, for each divisor \( d \) of \( n \), there is at most one cyclic subgroup of order \( d \) . | Proof. Suppose that \( G = \langle a\rangle \) is a cyclic group of order \( n \) . We claim that \( \left\langle {a}^{n/d}\right\rangle \) has order \( d \) . Clearly, \( {\left( {a}^{n/d}\right) }^{d} = {a}^{n} = 1 \), and it suffices to show that \( d \) is the smallest such positive integer. If \( {\left( {a}^{n/d}... | Yes |
Proposition 2.74. The intersection \( \mathop{\bigcap }\limits_{{i \in I}}{H}_{i} \) of any family of subgroups of a group \( G \) is again a subgroup of \( G \) . In particular, if \( H \) and \( K \) are subgroups of \( G \), then \( H \cap K \) is a subgroup of \( G \) . | Proof. Let \( D = \mathop{\bigcap }\limits_{{i \in I}}{H}_{i} \) ; we prove that \( D \) is a subgroup by verifying each of the parts in the definition. Note first that \( D \neq \varnothing \) because \( 1 \in D \) since \( 1 \in {H}_{i} \) for all \( i \) . If \( x \in D \), then \( x \) got into \( D \) by being in ... | Yes |
If \( X \) is a subset of a group \( G \), then there is a subgroup \( \langle X\rangle \) of \( G \) containing \( X \) that is smallest in the sense that \( \langle X\rangle \leq H \) for every subgroup \( H \) of \( G \) which contains \( X \) . | First of all, note that there exist subgroups of \( G \) which contain \( X \) ; for example, \( G \) itself contains \( X \) . Define \( \langle X\rangle = \mathop{\bigcap }\limits_{{X \subseteq H}}H \), the intersection of all the subgroups \( H \) of \( G \) which contain \( X \) . By Proposition 2.74, \( \langle X\... | Yes |
Proposition 2.77. If \( X \) is a nonempty subset of a group \( G \), then \( \langle X\rangle \) is the set of all the words on \( X \) . | Proof. We begin by showing that \( W \), the set of all words on \( X \), is a subgroup of \( G \) . By definition, \( 1 \in W \) . If \( w,{w}^{\prime } \in W \), then \( w = {x}_{1}^{{e}_{1}}{x}_{2}^{{e}_{2}}\cdots {x}_{n}^{{e}_{n}} \) and \( {w}^{\prime } = {y}_{1}^{{f}_{1}}{y}_{2}^{{f}_{2}}\cdots {y}_{m}^{{f}_{m}} ... | Yes |
Proposition 2.78. Let \( a \) and \( b \) be integers and let \( A = \langle a\rangle \) and \( B = \langle b\rangle \) be the cyclic subgroups of \( \mathbb{Z} \) they generate.\n\n(i) If \( A + B \) is defined to be \( \{ a + b : a \in A \) and \( b \in B\} \), then \( A + B = \langle d\rangle \) , where \( d = \gcd ... | Proof.\n\n(i) It is straightforward to check that \( A + B \) is a subgroup of \( \mathbb{Z} \) (in fact, \( A + B \) is precisely the set of all the linear combinations of \( a \) and \( b \) ). By Proposition 2.71, the subgroup \( A + B \) is cyclic: \( A + B = \langle d\rangle \), where \( d \) can be chosen to be t... | Yes |
Lemma 2.80. Let \( H \) be a subgroup of a group \( G \), and let \( a, b \in G \). (i) \( {aH} = {bH} \) if and only if \( {b}^{-1}a \in H \). In particular, \( {aH} = H \) if and only if \( a \in H \). | (i) This is a special case of Lemma 2.19, for cosets are equivalence classes. The second statement follows because \( H = {1H} \). | No |
Theorem 2.81 (Lagrange’s Theorem). If \( H \) is a subgroup of a finite group \( G \), then \( \left| H\right| \) is a divisor of \( \left| G\right| \) . | Proof. Let \( \left\{ {{a}_{1}H,{a}_{2}H,\ldots ,{a}_{t}H}\right\} \) be the family of all the distinct cosets of \( H \) in \( G \) . Then\n\n\[ G = {a}_{1}H \cup {a}_{2}H \cup \cdots \cup {a}_{t}H \]\n\nbecause each \( g \in G \) lies in the coset \( {gH} \), and \( {gH} = {a}_{i}H \) for some \( i \) . Moreover, Lem... | Yes |
Corollary 2.82. If \( H \) is a subgroup of a finite group \( G \), then\n\n\[ \left\lbrack {G : H}\right\rbrack = \left| G\right| /\left| H\right| . \] | Proof. This follows at once from Lagrange's theorem. - | Yes |
Corollary 2.83. If \( G \) is a finite group and \( a \in G \), then the order of \( a \) is a divisor of \( \left| G\right| \) . | Proof. By Proposition 2.72, the order of the element \( a \) is equal to the order of the subgroup \( H = \langle a\rangle \) . | No |
Corollary 2.84. If a finite group \( G \) has order \( m \), then \( {a}^{m} = 1 \) for all \( a \in G \) . | Proof. By Corollary 2.83, \( a \) has order \( d \), where \( d \mid m \) ; that is, \( m = {dk} \) for some integer \( k \) . Thus, \( {a}^{m} = {a}^{dk} = {\left( {a}^{d}\right) }^{k} = 1 \) . | Yes |
Corollary 2.85. If \( p \) is a prime, then every group \( G \) of order \( p \) is cyclic. | Proof. Choose \( a \in G \) with \( a \neq 1 \), and let \( H = \langle a\rangle \) be the cyclic subgroup generated by \( a \) . By Lagrange’s theorem, \( \left| H\right| \) is a divisor of \( \left| G\right| = p \) . Since \( p \) is a prime and \( \left| H\right| > 1 \), it follows that \( \left| H\right| = p = \lef... | Yes |
Lemma 2.87. Let \( f : G \rightarrow H \) be a homomorphism.\n\n(i) \( f\left( 1\right) = 1 \) ;\n\n(ii) \( f\left( {x}^{-1}\right) = f{\left( x\right) }^{-1} \) ;\n\n(iii) \( f\left( {x}^{n}\right) = f{\left( x\right) }^{n} \) for all \( n \in \mathbb{Z} \) . | Proof.\n\n(i) Applying \( f \) to the equation \( 1 \cdot 1 = 1 \) in \( G \) gives the equation \( f\left( 1\right) f\left( 1\right) = f\left( 1\right) \) in \( H \), and multiplying both sides by \( f{\left( 1\right) }^{-1} \) gives \( f\left( 1\right) = 1 \) .\n\n(ii) Apply \( f \) to the equation \( {x}^{-1}x = 1 \... | Yes |
We show that any two finite cyclic groups \( G \) and \( H \) of the same order \( m \) are isomorphic. It will then follow from Corollary 2.85 that any two groups of prime order \( p \) are isomorphic. | Suppose that \( G = \langle x\rangle \) and \( H = \langle y\rangle \) . Define \( f : G \rightarrow H \) by \( f\left( {x}^{i}\right) = {y}^{i} \) for \( 0 \leq i < m \) . Now \( G = \left\{ {1, x,{x}^{2},\ldots ,{x}^{m - 1}}\right\} \) and \( H = \left\{ {1, y,{y}^{2},\ldots ,{y}^{m - 1}}\right\} \) , and so it follo... | Yes |
Proposition 2.91. Let \( f : G \rightarrow H \) be a homomorphism.\n\n(i) \( \ker f \) is a subgroup of \( G \) and \( \operatorname{im}f \) is a subgroup of \( H \) .\n\n(ii) If \( x \in \ker f \) and if \( a \in G \), then \( {ax}{a}^{-1} \in \ker f \) .\n\n(iii) \( f \) is an injection if and only if \( \ker f = \{ ... | Proof.\n\n(i) Lemma 2.87 shows that \( 1 \in \ker f \), for \( f\left( 1\right) = 1 \) . Next, if \( x, y \in \ker f \), then \( f\left( x\right) = 1 = f\left( y\right) \) ; hence, \( f\left( {xy}\right) = f\left( x\right) f\left( y\right) = 1 \cdot 1 = 1 \), and so \( {xy} \in \ker f \) . Finally, if \( x \in \ker f \... | Yes |
Proposition 2.97. The alternating group \( {A}_{4} \) is a group of order 12 having no subgroup of order 6. | Proof. First of all, \( \left| {A}_{4}\right| = {12} \), by Exercise 2.29 on page 121. If \( {A}_{4} \) contains a subgroup \( H \) of order 6, then \( H \) has index 2, and so \( {\alpha }^{2} \in H \) for every \( \alpha \in {A}_{4} \) , by Proposition 2.95(i). If \( \alpha \) is a 3-cycle, however, then \( \alpha \)... | No |
Proposition 2.98. \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) in \( {\mathbb{I}}_{m} \) if and only if \( a \equiv b{\;\operatorname{mod}\;m} \) . | Proof. If \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \), then \( a \in \left\lbrack a\right\rbrack \), by reflexivity, and so \( a \in \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) . Therefore, \( a \equiv b{\;\operatorname{mod}\;m} \) .\n\nConversely, if \( c \in \left\lbrack a\right\r... | Yes |
Proposition 2.99. Let \( m \geq 2 \) be given.\n\n(i) If \( a \in \mathbb{Z} \), then \( \left\lbrack a\right\rbrack = \left\lbrack r\right\rbrack \) for some \( r \) with \( 0 \leq r < m \) . | Proof.\n\n(i) For each \( a \in \mathbb{Z} \), the division algorithm gives \( a = {qm} + r \), where \( 0 \leq r < m \) ; hence \( a - r = {qm} \) and \( a \equiv r{\;\operatorname{mod}\;m} \) . Therefore \( \left\lbrack a\right\rbrack = \left\lbrack r\right\rbrack \), where \( r \) is the remainder after dividing \( ... | Yes |
Lemma 2.100. If \( m \geq 2 \), then the function \( \alpha : {\mathbb{I}}_{m} \times {\mathbb{I}}_{m} \rightarrow {\mathbb{I}}_{m} \), given by \[ \alpha \left( {\left\lbrack a\right\rbrack ,\left\lbrack b\right\rbrack }\right) = \left\lbrack {a + b}\right\rbrack \] is an operation on \( {\mathbb{I}}_{m} \) . | Proof. The operation appears to depend on choosing names \( \left\lbrack a\right\rbrack \) and \( \left\lbrack b\right\rbrack \) ; what if we had chosen names \( \left\lbrack {a}^{\prime }\right\rbrack \) and \( \left\lbrack {b}^{\prime }\right\rbrack \) ? To see that \( \alpha \) is a (well-defined) function, we must ... | Yes |
Proposition 2.101. \( {\mathbb{I}}_{m} \) is an additive cyclic group of order \( m \) with generator [1]. | Proof.\n\nIn this proof only, we shall write \( \boxplus \) for addition of congruence classes:\n\n\[ \alpha \left( {\left\lbrack a\right\rbrack ,\left\lbrack b\right\rbrack }\right) = \left\lbrack a\right\rbrack \boxplus \left\lbrack b\right\rbrack = \left\lbrack {a + b}\right\rbrack . \]\n\nAssociativity of the opera... | Yes |
Corollary 2.102. Every cyclic group of order \( m \geq 2 \) is isomorphic to \( {\mathbb{I}}_{m} \) . | Proof. We have already seen, in Example 2.88, that any two finite cyclic groups of the same order are isomorphic. | No |
Proposition 2.103. The function \( \mu : {\mathbb{I}}_{m} \times {\mathbb{I}}_{m} \rightarrow {\mathbb{I}}_{m} \), given by\n\n\[ \mu \left( {\left\lbrack a\right\rbrack ,\left\lbrack b\right\rbrack }\right) = \left\lbrack {ab}\right\rbrack \]\n\nis an operation on \( {\mathbb{I}}_{m} \). This operation is associative ... | Proof. The operation appears to depend on choosing names \( \left\lbrack a\right\rbrack \) and \( \left\lbrack b\right\rbrack \) ; what if we had chosen names \( \left\lbrack {a}^{\prime }\right\rbrack \) and \( \left\lbrack {b}^{\prime }\right\rbrack \) ? To see that \( \mu \) is a (well-defined) function, we must sho... | Yes |
Corollary 2.105 (Fermat). If \( p \) is a prime and \( a \in \mathbb{Z} \), then\n\n\[ \n{a}^{p} \equiv a{\;\operatorname{mod}\;p} \n\] | Proof. By Proposition 2.98, it suffices to show that \( \left\lbrack {a}^{p}\right\rbrack = \left\lbrack a\right\rbrack \) in \( {\mathbb{I}}_{p} \) . If \( \left\lbrack a\right\rbrack = \) [0], then Proposition 2.103 gives \( \left\lbrack {a}^{p}\right\rbrack = {\left\lbrack a\right\rbrack }^{p} = {\left\lbrack 0\righ... | Yes |
\[ U\left( {I}_{m}\right) = \left\{ {\left\lbrack r\right\rbrack \in {\mathbb{I}}_{m} : \left( {r, m}\right) = 1}\right\} . \] | Let \( E = \left\{ {\left\lbrack r\right\rbrack \in {\mathbb{I}}_{m} : \left( {r, m}\right) = 1}\right\} \) . If \( \left\lbrack r\right\rbrack \in E \), then \( \left( {r, m}\right) = 1 \), so there are integers \( s \) and \( t \) with \( {sr} + {tm} = 1 \) . Hence, \( {sr} \equiv 1{\;\operatorname{mod}\;m} \) . Ther... | Yes |
Theorem 2.107 (Euler). If \( \left( {r, m}\right) = 1 \), then\n\n\[{r}^{\phi \left( m\right) } \equiv 1{\;\operatorname{mod}\;m}.\] | Proof. If \( G \) is a finite group of order \( n \), then Corollary 2.84 to Lagrange’s theorem gives \( {x}^{n} = 1 \) for all \( x \in G \) . Here, if \( \left\lbrack r\right\rbrack \in U\left( {I}_{m}\right) \), then \( {\left\lbrack r\right\rbrack }^{\phi \left( m\right) } = \left\lbrack 1\right\rbrack \) , by Lemm... | Yes |
Theorem 2.109 (Wilson's Theorem). An integer \( p \) is a prime if and only if\n\n\[ \left( {p - 1}\right) ! \equiv - 1{\;\operatorname{mod}\;p}. \]\n | Proof. Assume that \( p \) is a prime. If \( {a}_{1},{a}_{2},\ldots ,{a}_{n} \) is a list of all the elements of a finite abelian group \( G \), then the product \( {a}_{1}{a}_{2}\ldots {a}_{n} \) is the same as the product of all elements \( a \) with \( {a}^{2} = 1 \), for any other element cancels against its invers... | Yes |
Lemma 2.110. If \( K \) is a normal subgroup of a group \( G \), then\n\n\[ \n{bK} = {Kb} \n\]\n\nfor every \( b \in G \) . | Proof. Let \( {bk} \in {bK} \) . Since \( K \) is normal, \( {bk}{b}^{-1} \in K \), say \( {bk}{b}^{-1} = {k}^{\prime } \in K \), so that \( {bk} = \left( {{bk}{b}^{-1}}\right) b = {k}^{\prime }b \in {Kb} \), and so \( {bK} \subseteq {Kb} \) . For the reverse inclusion, let \( {kb} \in {Kb} \) . Since \( K \) is normal... | Yes |
Theorem 2.111. Let \( G/K \) denote the family of all the cosets of a subgroup \( K \) of \( G \) . If \( K \) is a normal subgroup, then\n\n\[ \n{aKbK} = {abK} \n\]\n\nfor all \( a, b \in G \), and \( G/K \) is a group under this operation. | Proof. The product of two cosets \( \left( {aK}\right) \left( {bK}\right) \) can also be viewed as the product of 4 elements in \( \mathcal{S}\left( G\right) \) . Hence, associativity in \( \mathcal{S}\left( G\right) \) gives generalized associativity:\n\n\[ \n\left( {aK}\right) \left( {bK}\right) = a\left( {Kb}\right)... | Yes |
Every normal subgroup \( K \vartriangleleft G \) is the kernel of some homomorphism. | Proof. Define the natural map \( \pi : G \rightarrow G/K \) by \( \pi \left( a\right) = {aK} \) . With this notation, the formula \( {aKbK} = {abK} \) can be rewritten as \( \pi \left( a\right) \pi \left( b\right) = \pi \left( {ab}\right) \) ; thus, \( \pi \) is a (surjective) homomorphism. Since \( K \) is the identit... | Yes |
Theorem 2.114 (First Isomorphism Theorem). If \( f : G \rightarrow H \) is a homomorphism, then\n\n\[ \ker f \vartriangleleft G\;\text{ and }\;G/\ker f \cong \operatorname{im}f. \]\n\nIn more detail, if \( \ker f = K \), then the function \( \varphi : G/K \rightarrow \operatorname{im}f \leq H \), given by \( \varphi : ... | Proof. We have already seen, in Proposition 2.91(ii), that \( K = \ker f \) is a normal subgroup of \( G \) . Now \( \varphi \) is well-defined: if \( {aK} = {bK} \), then \( a = {bk} \) for some \( k \in K \), and so \( f\left( a\right) = f\left( {bk}\right) = f\left( b\right) f\left( k\right) = f\left( b\right) \), b... | Yes |
Proposition 2.118 (Product Formula). If \( H \) and \( K \) are subgroups of a finite group \( G \), then\n\n\[ \left| {HK}\right| \left| {H \cap K}\right| = \left| H\right| \left| K\right| \]\n\nwhere \( {HK} = \{ {hk} : h \in H \) and \( k \in K\} \) . | Proof. Define a function \( f : H \times K \rightarrow {HK} \) by \( f : \left( {h, k}\right) \mapsto {hk} \) . Clearly, \( f \) is a surjection. It suffices to show, for every \( x \in {HK} \), that \( \left| {{f}^{-1}\left( x\right) }\right| = \left| {H \cap K}\right| \) , where \( {f}^{-1}\left( x\right) = \{ \left(... | Yes |
Theorem 2.119 (Second Isomorphism Theorem). If \( H \) and \( K \) are subgroups of a group \( G \) with \( H \vartriangleleft G \), then \( {HK} \) is a subgroup, \( H \cap K \vartriangleleft K \), and \[ K/\left( {H \cap K}\right) \cong {HK}/H. \] | Proof. We begin by showing first that \( {HK}/H \) makes sense and then describing its elements. Since \( H \vartriangleleft G \), Proposition 2.117 shows that \( {HK} \) is a subgroup. Normality of \( H \) in \( {HK} \) follows from a more general fact: if \( H \leq S \leq G \) and if \( H \) is normal in \( G \), the... | Yes |
Theorem 2.120 (Third Isomorphism Theorem). If \( H \) and \( K \) are normal subgroups of a group \( G \) with \( K \leq H \), then \( H/K \vartriangleleft G/K \) and\n\n\[ \left( {G/K}\right) /\left( {H/K}\right) \cong G/H \] | Proof. Define \( f : G/K \rightarrow G/H \) by \( f : {aK} \mapsto {aH} \) . Note that \( f \) is a (well-defined) function, for if \( {a}^{\prime } \in G \) and \( {a}^{\prime }K = {aK} \), then \( {a}^{-1}{a}^{\prime } \in K \leq H \), and so \( {aH} = {a}^{\prime }H \) . It is easy to see that \( f \) is a surjectiv... | Yes |
Proposition 2.121 (Correspondence Theorem). If \( G \) is a group and \( K \vartriangleleft G \) , then \( S \mapsto S/K \) is a bijection \( \operatorname{Sub}\left( {G;K}\right) \rightarrow \operatorname{Sub}\left( {G/K}\right) \) . Denoting \( S/K \) by \( {S}^{ * } \) , we have\n\n(i) \( T \leq S \leq G \) in \( \m... | Proof. Let \( \Phi : \operatorname{Sub}\left( {G;K}\right) \rightarrow \operatorname{Sub}\left( {G/K}\right) \) denote the function \( \Phi : S \mapsto S/K \) (it is routine to check that if \( S \) is subgroup of \( G \) containing \( K \), then \( S/K \) is a subgroup of \( G/K \) ).\n\nTo see that \( \Phi \) is inje... | Yes |
If \( G \) is a finite abelian group, then \( G \) has a subgroup of order \( d \) for every divisor \( d \) of \( \left| G\right| \) . In particular, if \( p \) is a prime divisor of \( \left| G\right| \) , then \( G \) contains an element of order \( p \) . | We begin by proving, by induction on \( n = \left| G\right| \), that for every prime divisor \( p \) of \( \left| G\right| \), there is an element of order \( p \) in \( G \) . The base step \( n = 1 \) is true, for there are no prime divisors of 1 . For the inductive step, choose \( a \in G \) of order \( k > 1 \) . I... | No |
The four-group \( \mathbf{V} \) is isomorphic to \( {\mathbb{I}}_{2} \times {\mathbb{I}}_{2} \). | The reader may check that the function \( f : \mathbf{V} \rightarrow {\mathbb{I}}_{2} \times {\mathbb{I}}_{2} \), defined by\n\n\[ f : \left( 1\right) \mapsto \left( {\left\lbrack 0\right\rbrack ,\left\lbrack 0\right\rbrack }\right) \]\n\n\[ f : \left( {12}\right) \left( {34}\right) \mapsto \left( {\left\lbrack 1\right... | Yes |
Proposition 2.124. Let \( G \) and \( {G}^{\prime } \) be groups, and let \( K \vartriangleleft G \) and \( {K}^{\prime } \vartriangleleft {G}^{\prime } \) be normal subgroups. Then \( K \times {K}^{\prime } \) is a normal subgroup of \( G \times {G}^{\prime } \), and there is an isomorphism\n\n\[ \left( {G \times {G}^... | Proof. Let \( \pi : G \rightarrow G/K \) and \( {\pi }^{\prime } : {G}^{\prime } \rightarrow {G}^{\prime }/{K}^{\prime } \) be the natural maps. The reader may check that \( f : G \times {G}^{\prime } \rightarrow \left( {G/K}\right) \times \left( {{G}^{\prime }/{K}^{\prime }}\right) \), given by\n\n\[ f : \left( {g,{g}... | No |
Proposition 2.125. If \( G \) is a group containing normal subgroups \( H \) and \( K \) with \( H \cap K = \{ 1\} \) and \( {HK} = G \), then \( G \cong H \times K \) . | Proof. We show first that if \( g \in G \), then the factorization \( g = {hk} \), where \( h \in H \) and \( k \in K \), is unique. If \( {hk} = {h}^{\prime }{k}^{\prime } \), then \( {h}^{\prime - 1}h = {k}^{\prime }{k}^{-1} \in H \cap K = \{ 1\} \) . Therefore, \( {h}^{\prime } = h \) and \( {k}^{\prime } = k \) . W... | No |
Theorem 2.126. If \( m \) and \( n \) are relatively prime, then\n\n\[{\mathbb{I}}_{mn} \cong {\mathbb{I}}_{m} \times {\mathbb{I}}_{n}\] | Proof. If \( a \in \mathbb{Z} \), denote its congruence class in \( {\mathbb{I}}_{m} \) by \( {\left\lbrack a\right\rbrack }_{m} \) . The reader can show that the function \( f : \mathbb{Z} \rightarrow {\mathbb{I}}_{m} \times {\mathbb{I}}_{n} \), given by \( a \mapsto \left( {{\left\lbrack a\right\rbrack }_{m},{\left\l... | No |
Proposition 2.127. Let \( G \) be a group, and let \( a, b \in G \) be commuting elements of orders \( m \) and \( n \), respectively. If \( \left( {m, n}\right) = 1 \), then ab has order \( {mn} \) . | Proof. Since \( a \) and \( b \) commute, we have \( {\left( ab\right) }^{r} = {a}^{r}{b}^{r} \) for all \( r \), so that \( {\left( ab\right) }^{mn} = \) \( {a}^{mn}{b}^{mn} = 1 \) . It suffices to prove that if \( {\left( ab\right) }^{k} = 1 \), then \( {mn} \mid k \) . If \( 1 = {\left( ab\right) }^{k} = \) \( {a}^{... | Yes |
Proposition 2.129. If \( G \) is a finite abelian group having a unique subgroup of order \( p \) for every prime divisor \( p \) of \( \left| G\right| \), then \( G \) is cyclic. | Proof. Choose \( a \in G \) of largest order, say, \( n \) . If \( p \) is a prime divisor of \( \left| G\right| \), let \( C = {C}_{p} \) be the unique subgroup of \( G \) having order \( p \) ; the subgroup \( C \) must be cyclic, say \( C = \langle c\rangle \) . We show that \( p \mid n \) by showing that \( c \in \... | Yes |
Theorem 2.130 (Cayley). Every group \( G \) is (isomorphic to) a subgroup of the symmetric group \( {S}_{G} \) . In particular, if \( \left| G\right| = n \), then \( G \) is isomorphic to a subgroup of \( {S}_{n} \) . | Proof. For each \( a \in G \), define \ | No |
Theorem 2.131 (Representation on Cosets). Let \( G \) be a group, and let \( H \) be a subgroup of \( G \) having finite index \( n \) . Then there exists a homomorphism \( \varphi : G \rightarrow {S}_{n} \) with \( \ker \varphi \leq H \) . | Proof. Even though \( H \) may not be a normal subgroup, we still denote the family of all the cosets of \( H \) in \( G \) by \( G/H \).\n\nFor each \( a \in G \), define \ | No |
Proposition 2.132. Every group \( G \) of order 4 is isomorphic to either \( {\mathbb{I}}_{4} \) or the four-group \( \mathbf{V} \) . Moreover, \( {\mathbb{I}}_{4} \) and \( \mathbf{V} \) are not isomorphic. | Proof. By Lagrange’s theorem, every element in \( G \), other than 1, has order either 2 or 4 . If there is an element of order 4, then \( G \) is cyclic. Otherwise, \( {x}^{2} = 1 \) for all \( x \in G \), so that Exercise 2.38 on page 143 shows that \( G \) is abelian.\n\nIf distinct elements \( x \) and \( y \) in \... | No |
Proposition 2.133. If \( G \) is a group of order 6, then \( G \) is isomorphic to either \( {\mathbb{I}}_{6} \) or \( {S}_{3}.{}^{20} \) Moreover, \( {\mathbb{I}}_{6} \) and \( {S}_{3} \) are not isomorphic. | Proof. By Lagrange's theorem, the only possible orders of nonidentity elements are 2,3, and 6 . Of course, \( G \cong {\mathbb{I}}_{6} \) if \( G \) has an element of order 6 . Now Exercise 2.40 on page 144 shows that \( G \) must contain an element of order 2, say, \( t \) . Let \( T = \langle t\rangle \) .\n\nSince \... | No |
Proposition 2.134. If \( \alpha : G \times X \rightarrow X \) is an action of a group \( G \) on a set \( X \) , then \( g \mapsto {\alpha }_{g} \) defines a homomorphism \( G \rightarrow {S}_{X} \) . Conversely, if \( B : G \rightarrow {S}_{X} \) is a homomorphism, then \( \beta : G \times X \rightarrow X \), defined ... | Proof. If \( \alpha : G \times X \rightarrow X \) is an action, then we claim that each \( {\alpha }_{g} \) is a permutation of \( X \) . Indeed, its inverse is \( {\alpha }_{{g}^{-1}} \), because \( {\alpha }_{g}{\alpha }_{{g}^{-1}} = {\alpha }_{g{g}^{-1}} = {\alpha }_{1} = {1}_{X} \) . It follows that \( A : G \right... | Yes |
Let \( X = \{ 1,2,\ldots, n\} \), let \( \sigma \in {S}_{n} \), and regard the cyclic group \( G = \langle \sigma \rangle \) as acting on \( X \) . If \( i \in X \), then\n\n\[ \mathcal{O}\left( i\right) = \left\{ {{\sigma }^{k}\left( i\right) : k \in \mathbb{Z}}\right\} \] | Let \( \sigma = {\beta }_{1}\cdots {\beta }_{t\left( \sigma \right) } \) be the complete factorization of \( \sigma \), and let \( i = {i}_{0} \) be moved by \( \sigma \) . If the cycle involving \( {i}_{0} \) is \( {\beta }_{j} = \left( \begin{array}{llll} {i}_{0} & {i}_{1} & \ldots & {i}_{r - 1} \end{array}\right) \)... | Yes |
Proposition 2.140. If \( G \) acts on a set \( X \), then \( X \) is the disjoint union of the orbits. If \( X \) is finite, then\n\n\[ \left| X\right| = \mathop{\sum }\limits_{i}\left| {\mathcal{O}\left( {x}_{i}\right) }\right| \]\n\nwhere one \( {x}_{i} \) is chosen from each orbit. | Proof. This follows from Proposition 2.20, for the orbits form a partition of \( X \) .\n\nThe count given in the second statement is correct: since the orbits are disjoint, no element in \( X \) is counted twice. \( \; \bullet \) | Yes |
Theorem 2.141. If \( G \) acts on a set \( X \) and \( x \in X \), then\n\n\[ \left| {\mathcal{O}\left( x\right) }\right| = \left\lbrack {G : {G}_{x}}\right\rbrack \] | Proof. Let \( G/{G}_{x} \) denote the family of all the cosets of \( {G}_{x} \) in \( G \) . We will exhibit a bijection \( \varphi : \mathcal{O}\left( x\right) \rightarrow G/{G}_{x} \) ; this will give the result, since \( \left| {G/{G}_{x}}\right| = \) \( \left\lbrack {G : {G}_{x}}\right\rbrack \), by Corollary 2.82 ... | Yes |
Corollary 2.142. If a finite group \( G \) acts on a set \( X \), then the number of elements in any orbit is a divisor of \( \left| G\right| \) . | Proof. This follows at once from Theorem 2.141 and Lagrange's theorem. - | Yes |
If \( x \) lies in a finite group \( G \), then the number of conjugates of \( x \) is the index of its centralizer: | Proof. As in Example 2.138, the orbit of \( x \) is its conjugacy class \( {x}^{G} \), and the stabilizer \( {G}_{x} \) is the centralizer \( {C}_{G}\left( x\right) \) . | Yes |
If \( \alpha \in {S}_{n} \), then the number of permutations in \( {S}_{n} \) having the same cycle structure as \( \alpha \) is a divisor of \( n \) !. | This follows at once from Corollary 2.143 once one recalls Proposition 2.33 which says that two permutations in \( {S}_{n} \) are conjugate in \( {S}_{n} \) if and only if they have the same cycle structure. | Yes |
Theorem 2.145 (Cauchy). If \( G \) is a finite group whose order is divisible by a prime \( p \), then \( G \) contains an element of order \( p \) . | Proof. We prove the theorem by induction on \( \\left| G\\right| \) ; the base step \( \\left| G\\right| = 1 \) is vacuously true, for there are no prime divisors of 1 . If \( x \\in G \), then the number of conjugates of \( x \) is \( \\left| {x}^{G}\\right| = \\left\\lbrack {G : {C}_{G}\\left( x\\right) }\\right\\rbr... | Yes |
Theorem 2.146. If \( p \) is a prime and \( G \) is a \( p \) -group with more than one element, then \( Z\left( G\right) \neq \{ 1\} \) . | Proof. Consider the class equation\n\n\[ \left| G\right| = \left| {Z\left( G\right) }\right| + \mathop{\sum }\limits_{i}\left\lbrack {G : {C}_{G}\left( {x}_{i}\right) }\right\rbrack \]\n\nEach \( {C}_{G}\left( {x}_{i}\right) \) is a proper subgroup of \( G \), for \( {x}_{i} \notin Z\left( G\right) \) . Since \( G \) i... | Yes |
Corollary 2.147. If \( p \) is a prime, then every group \( G \) of order \( {p}^{2} \) is abelian. | Proof. If \( G \) is not abelian, then its center \( Z\left( G\right) \) is a proper subgroup, so that \( \left| {Z\left( G\right) }\right| = 1 \) or \( p \), by Lagrange’s theorem. But Theorem 2.146 says that \( Z\left( G\right) \neq \) \( \{ 1\} \), and so \( \left| {Z\left( G\right) }\right| = p \) . The center is a... | Yes |
Proposition 2.150. If \( G \) is a group of order \( \left| G\right| = {p}^{e} \), then \( G \) has a normal subgroup of order \( {p}^{k} \) for every \( k \leq e \) . | Proof. We prove the result by induction on \( e \geq 0 \) . The base step is obviously true, and so we proceed to the inductive step. By Theorem 2.146, the center of \( G \) is nontrivial: \( Z\left( G\right) \neq \{ 1\} \) . If \( Z\left( G\right) = G \), then \( G \) is abelian, and we have already proved the result ... | Yes |
Proposition 2.151. An abelian group \( G \) is simple if and only if it is finite and of prime order. | Proof. If \( G \) is finite of prime order \( p \), then \( G \) has no subgroups \( H \) other than \( \{ 1\} \) and \( G \), otherwise Lagrange’s theorem would show that \( \left| H\right| \) is a divisor of \( p \) . Therefore, \( G \) is simple.\n\nConversely, assume that \( G \) is simple. Since \( G \) is abelian... | Yes |
Lemma 2.152. All 3-cycles are conjugate in \( {A}_{5} \) . | Proof. Let \( G = {S}_{5},\alpha = \left( \begin{array}{lll} 1 & 2 & 3 \end{array}\right) \), and \( H = {A}_{5} \) . We know that \( \left| {\alpha }^{{S}_{5}}\right| = {20} \) , for there are 20 3-cycles in \( {S}_{5} \) (as we saw in Example 2.30). Therefore, \( {20} = \) \( \left| {S}_{5}\right| /\left| {{C}_{{S}_{... | Yes |
Lemma 2.153. Every element in \( {A}_{5} \) is a 3-cycle or a product of 3-cycles. | Proof. If \( \alpha \in {A}_{5} \), then \( \alpha \) is a product of an even number of transpositions: \( \alpha = {\tau }_{1}{\tau }_{2}\cdots {\tau }_{{2n} - 1}{\tau }_{2n} \) . As the transpositions may be grouped in pairs \( {\tau }_{{2i} - 1}{\tau }_{2i} \), it suffices to consider products \( \tau {\tau }^{\prim... | Yes |
Theorem 2.154. \( {A}_{5} \) is a simple group. | Proof. We shall show that if \( H \) is a normal subgroup of \( {A}_{5} \) and \( H \neq \{ \left( 1\right) \} \), then \( H = {A}_{5} \) . Now if \( H \) contains a 3-cycle, then normality forces \( H \) to contain all its conjugates. By Lemma 2.152, \( H \) contains every 3-cycle, and by Lemma 2.153, \( H = {A}_{5} \... | Yes |
Lemma 2.155. \( {A}_{6} \) is a simple group. | Proof. Let \( H \neq \{ \left( 1\right) \} \) be a normal subgroup of \( {A}_{6} \) ; we must show that \( H = {A}_{6} \) . Assume that there is some \( \alpha \in H \) with \( \alpha \neq \left( 1\right) \) which fixes some \( i \), where \( 1 \leq i \leq 6 \) . Define\n\n\[ F = \left\{ {\sigma \in {A}_{6} : \sigma \l... | Yes |
Theorem 2.156. \( {A}_{n} \) is a simple group for all \( n \geq 5 \) . | Proof. If \( H \) is a nontrivial normal subgroup of \( {A}_{n} \) [that is, \( H \neq \left( 1\right) \) ], then we must show that \( H = {A}_{n} \) ; by Exercise 2.116 on page 205, it suffices to prove that \( H \) contains a 3-cycle. If \( \beta \in H \) is nontrivial, then there exists some \( i \) that \( \beta \)... | No |
Theorem 2.158 (Burnside’s Lemma). \( {}^{23} \) Let \( G \) act on a finite set \( X \) . If \( N \) is the number of orbits, then\n\n\[ N = \frac{1}{\\left| G\\right| }\\mathop{\\sum }\\limits_{{\\tau \\in G}}F\\left( \\tau \\right) \]\n\nwhere \( F\\left( \\tau \\right) \) is the number of \( x \\in X \) fixed by \( ... | Proof. List the elements of \( X \) as follows: choose \( {x}_{1} \\in X \), and then list all the elements in the orbit \( \\mathcal{O}\\left( {x}_{1}\\right) \) ; say, \( \\mathcal{O}\\left( {x}_{1}\\right) = \\left\\{ {{x}_{1},{x}_{2},\\ldots ,{x}_{r}}\\right\\} \) ; then choose \( {x}_{r + 1} \\notin \\mathcal{O}\\... | No |
Theorem 2.160. Let \( \mathcal{C} \) be a set of \( q \) colors, and let \( \tau \in {S}_{n} \) .\n\n(i) If \( F\left( \tau \right) \) is the number of \( x \in {\mathcal{C}}^{n} \) fixed by \( \tau \), and if \( t\left( \tau \right) \) is the number of cycles in the complete factorization of \( \tau \), then\n\n\[ F\l... | Proof.\n\n(i) Let \( \tau \in {S}_{n} \) and let \( \tau = {\beta }_{1}\cdots {\beta }_{t} \) be a complete factorization, where each \( {\beta }_{j} \) is an \( {r}_{j} \) -cycle. If \( {i}_{1},\ldots ,{i}_{{r}_{j}} \) are the symbols moved by \( {\beta }_{j} \), then \( {i}_{k + 1} = {\tau }^{k}{i}_{1} \) for \( k < ... | Yes |
Proposition 2.162. If \( \left| X\right| = n \) and \( G \) is a subgroup of \( {S}_{n} \), then the number of \( \left( {q, G}\right) \) -colorings of \( X \) is \( {P}_{G}\left( {q,\ldots, q}\right) \), where \( {P}_{G}\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is the cycle index. | Proof. By Theorem 2.160, the number of \( \left( {q, G}\right) \) -colorings of \( X \) is\n\n\[ \n\frac{1}{\left| G\right| }\mathop{\sum }\limits_{{\tau \in G}}{q}^{t\left( \tau \right) }\n\]\n\nwhere \( t\left( \tau \right) \) is the number of cycles in the complete factorization of \( \tau \) . On the other hand,\n\... | Yes |
Theorem 2.163 (Pólya). Let \( G \leq {S}_{X} \), where \( \left| X\right| = n \), let \( \left| \mathcal{C}\right| = q \), and, for each \( i \geq 1 \), define \( {\sigma }_{i} = {c}_{1}^{i} + \cdots + {c}_{q}^{i} \) . Then the number of \( \left( {q, G}\right) \) -colorings of \( X \) having \( {f}_{r} \) elements of ... | Proofs of P 'olya's theorem can be found in combinatorics books (for example, see Biggs, Discrete Mathematics). | No |
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