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Proposition 8.6. \( \operatorname{Im}\iota \cong A,\operatorname{Im}\kappa \cong B,\operatorname{Im}\iota \cap \operatorname{Im}\kappa = 1 \), and \( A \coprod B \) is generated by \( \operatorname{Im}\iota \cup \operatorname{Im}\kappa \) . | Proof. \( \operatorname{Im}\iota \cong A \) and \( \operatorname{Im}\kappa \cong B \) since \( \iota \) and \( \kappa \) are injective; \( \operatorname{Im}\iota \cap \operatorname{Im}\kappa = 1 \) since \( A \cap B = 1;A \coprod B \) is generated by \( \operatorname{Im}\iota \cup \operatorname{Im}\kappa \) since every... | Yes |
Proposition 8.7. Let \( A \) and \( B \) be groups. For every group \( G \) and homomorphisms \( \varphi : A \rightarrow G \) and \( \psi : B \rightarrow G \), there is a unique homomorphism \( \chi : A \coprod B \rightarrow G \) such that \( \chi \circ \iota = \varphi \) and \( \chi \circ \kappa = \psi \), where \( \i... | Proof. We may assume that \( A \cap B = 1 \), as readers will verify. Then it is convenient to combine \( \iota \) and \( \kappa \) into a single mapping \( \lambda : A \cup B \rightarrow A \coprod B \) , and to combine \( \varphi \) and \( \psi \) into a single mapping \( \omega : A \cup B \rightarrow G \) . Now, ever... | No |
Proposition 1.4. If \( m \) and \( n \) are relatively prime, then \( {C}_{mn} \cong {C}_{m} \times {C}_{n} \) . | Proof. Let \( {C}_{mn} = \langle c\rangle \) be cyclic of order \( {mn} \) . Then \( {c}^{n} \) has order \( m \) (since \( {\left( {c}^{m}\right) }^{k} = 1 \) if and only if \( {mn} \) divides \( {mk} \), if and only if \( n \) divides \( k \) ) and \( {c}^{m} \) has order \( n \) . The subgroups\n\n\[ A = \left\langl... | Yes |
Proposition 1.5. A cyclic group of order \( n \) has exactly \( \phi \left( n\right) \) elements of order \( n \) . | Proof. Let \( G = \langle a\rangle \) be cyclic of order \( n \) . Let \( 1 \leqq k \leqq n \) . The order of \( {a}^{k} \) divides \( n \), since \( {\left( {a}^{k}\right) }^{n} = {\left( {a}^{n}\right) }^{k} = 1 \) . We show that \( {a}^{k} \) has order \( n \) if and only if \( k \) and \( n \) are relatively prime.... | Yes |
Proposition 1.6. If \( m \) and \( n \) are relatively prime, then \( \phi \left( {mn}\right) = \phi \left( m\right) \phi \left( n\right) \) . | Proof. By 1.4 a cyclic group \( {C}_{mn} \) of order \( {mn} \) is, up to isomorphism, the direct product of a cyclic group \( {C}_{m} \) of order \( m \) and a cyclic group \( {C}_{n} \) of order \( n \) . In \( {C}_{m} \times {C}_{n},{\left( x, y\right) }^{k} = 1 \) if and only if \( {x}^{k} = 1 \) and \( {y}^{k} = 1... | Yes |
Corollary 1.7. \( \phi \left( n\right) = n\mathop{\prod }\limits_{{p\text{ prime,}p \mid n}}\left( {1 - 1/p}\right) \) . | Proof. This follows from 1.6 and \( \phi \left( {p}^{m}\right) = {p}^{m}\left( {1 - 1/p}\right) \), since \( n \) is a product of relatively prime powers of primes. | No |
Proposition 1.8. \( \mathop{\sum }\limits_{{d \mid n}}\phi \left( d\right) = n \) . | Proof. Let \( G = \langle c\rangle \) be a cyclic group of order \( n \) . By I.5.7, every divisor \( d \) of \( n \) is the order of a unique cyclic subgroup of \( G \), namely \( D = \left\{ {x \in G \mid {x}^{d} = 1}\right\} \) . Since \( D \) is cyclic of order \( d, G \) has exactly \( \phi \left( d\right) \) elem... | Yes |
Proposition 2.2. Every group of finite length is a direct sum of (finitely many) indecomposable subgroups. | Proof. Assume that there is a group \( G \) of finite length that is not a direct sum of indecomposable subgroups. Call a normal subgroup \( B \) of \( G \) bad when \( G = A \oplus B \) for some subgroup \( A \), but \( B \) is not a direct sum of indecomposable subgroups. For instance, \( G = 1 \oplus G \) is bad. Si... | Yes |
Lemma 2.4. If \( G \) has finite length, then a normal endomorphism of \( G \) is injective if and only if it is surjective, if and only if it is bijective. | Proof. Let \( \eta \in \operatorname{End}\left( G\right) \) be normal. For every \( n > 0,{\eta }^{n} \) is normal, so that \( \operatorname{Im}{\eta }^{n} \) and \( \operatorname{Ker}{\eta }^{n} \) are normal subgroups of \( G \) . The descending sequence\n\n\[ \operatorname{Im}\eta \supseteq \operatorname{Im}{\eta }^... | Yes |
Lemma 2.5. If \( G \) has finite length and \( \eta \) is a normal endomorphism of \( G \), then \( G = \operatorname{Im}{\eta }^{n} \oplus \operatorname{Ker}{\eta }^{n} \) for some \( n > 0 \) . | Proof. As in the proof of Lemma 2.4, the sequences\n\n\( \operatorname{Im}\eta \supseteq \cdots \supseteq \operatorname{Im}{\eta }^{n} \supseteq \cdots \) and \( \operatorname{Ker}\eta \subseteq \cdots \subseteq \operatorname{Ker}{\eta }^{n} \subseteq \cdots \)\n\ncannot be infinite, so that \( \operatorname{Im}{\eta }... | Yes |
Lemma 2.6. If \( G \) is an indecomposable group of finite length, then every normal endomorphism of \( G \) is either nilpotent or an automorphism. | Proof. By 2.5, either \( \operatorname{Im}{\eta }^{n} = 1 \) and \( \eta \) is nilpotent, or \( \operatorname{Ker}{\eta }^{n} = 1 \), and then \( \operatorname{Ker}\eta = 1 \) and \( \eta \) is bijective by 2.4. | Yes |
Lemma 2.8. Let \( {\eta }_{1},{\eta }_{2},\ldots ,{\eta }_{n} \) be normal endomorphisms of an indecomposable group \( G \) of finite length. If \( {\eta }_{i}x \) commutes with \( {\eta }_{j}y \) for every \( x, y \in G \) and every \( i \neq j \), and every \( {\eta }_{i} \) is nilpotent, then \( {\eta }_{1} \cdot {\... | Proof. We prove this when \( n = 2 \) ; the general case follows by induction on \( n \) . Assume that \( \eta ,\zeta \in \operatorname{End}\left( G\right) \) are normal and nilpotent, and that \( \alpha = \eta \cdot \zeta \) is defined but not nilpotent. Then \( \alpha \) is an automorphism, by 2.6. Let \( \varphi = \... | Yes |
Lemma 2.10. Let \( G = A \oplus B \) . Every normal subgroup of \( A \) is a normal subgroup of \( G \) . If \( \eta \) is a normal endomorphism of \( G \) and \( {\eta A} \subseteq A \), then the restriction \( {\eta }_{\mid A} \) of \( \eta \) to \( A \) is a normal endomorphism of \( A \) . | Proof. Let \( a \in A \) and \( b \in B \) . The inner automorphism \( x \mapsto {abx}{b}^{-1}{a}^{-1} \) of \( G \) has a restriction to \( A \), which is the inner automorphism \( x \mapsto {ax}{a}^{-1} \) of \( A \) , since \( b \) commutes with every element of \( A \) . Therefore every normal subgroup of \( A \) i... | Yes |
Proposition 3.1. In a (left) group action of a group \( G \) on a set \( X \), the action \( {\sigma }_{g} : x \mapsto g \cdot x \) of \( g \in G \) is a permutation of \( X \) ; moreover, \( g \mapsto {\sigma }_{g} \) is a homomorphism of \( G \) into the symmetric group \( {S}_{X} \) . | Proof. By definition, \( {\sigma }_{1} \) is the identity mapping on \( X \), and \( {\sigma }_{g} \circ {\sigma }_{h} = {\sigma }_{gh} \) for all \( g, h \in G \) . In particular, \( {\sigma }_{g} \circ {\sigma }_{{g}^{-1}} = {1}_{X} = {\sigma }_{{g}^{-1}} \circ {\sigma }_{g} \), so that \( {\sigma }_{g} \) and \( {\s... | Yes |
Corollary 3.2 (Cayley’s Theorem). Every group \( G \) is isomorphic to a subgroup of the symmetric group \( {S}_{G} \) . | Proof. Let \( G \) act on itself by left multiplication. The homomorphism \( \sigma \) : \( G \rightarrow {S}_{G} \) in 3.1 is injective: if \( {\sigma }_{g} = {1}_{G} \), then \( {gx} = x \) for all \( x \in G \) and \( g = 1 \) . Hence \( G \cong \operatorname{Im}\sigma \leqq {S}_{G} \) . \( ▱ \) | Yes |
Proposition 3.3. Let the group \( G \) act (on the left) on a set \( X \) . The relation\n\n\[ x \equiv y\\text{if and only if}y = g \\cdot x\\text{for some}g \\in G \]\n\nis an equivalence relation on \( X \) . | Proof. The relation \( \\equiv \) is reflexive since \( 1 \\cdot x = x \), symmetric since \( y = g \\cdot x \) implies \( x = {g}^{-1} \\cdot \\left( {g \\cdot x}\\right) = {g}^{-1} \\cdot y \), and transitive since \( y = g \\cdot x, z = h \\cdot y \) implies \( z = {hg} \\cdot x \) . \( ▱ \) | Yes |
Proposition 3.4. The order of the orbit of an element is equal to the index of its stabilizer. | Proof. Let \( G \) act on \( X \) . Let \( x \in X \) . The surjection \( \widehat{x} : g \mapsto g \cdot x \) of \( G \) onto the orbit of \( x \) induces a one-to-one correspondence between the elements of the orbit of \( x \) and the classes of the equivalence relation induced on \( G \) by \( \widehat{x} \) . The l... | Yes |
For every element \( g \) of a group \( G \), the mapping \( {\alpha }_{g} : x \mapsto \) \( {gx}{g}^{-1} \) is an automorphism of \( G \) ; moreover, \( g \mapsto {\alpha }_{g} \) is a homomorphism of \( G \) into \( \operatorname{Aut}\left( G\right) \) . | Proof. First, \( \left( {{gx}{g}^{-1}}\right) \left( {{gy}{g}^{-1}}\right) = {gxy}{g}^{-1} \) for all \( x, y \in G \), so that \( {\alpha }_{g} \) is a homomorphism. Also, \( {\alpha }_{1} \) is the identity mapping \( {1}_{G} \) on \( G \), and \( {\alpha }_{g} \circ {\alpha }_{h} = {\alpha }_{gh} \) for all \( g, h ... | Yes |
Proposition 3.6. \( Z\left( G\right) \) and all its subgroups are normal subgroups of \( G \) . | Proof. If \( z \in Z \), then \( {gz}{g}^{-1} = z \) for all \( g \in G \) . Hence \( {gH}{g}^{-1} = H \) for all \( H \leqq Z \) . \( ▱ \) | Yes |
Proposition 3.8 (The Class Equation). In a finite group \( G \) , \[ \left| G\right| = \sum \left| C\right| = \left| {Z\left( G\right) }\right| + \mathop{\sum }\limits_{{\left| C\right| > 1}}\left| C\right| . \] The first sum has one term for each conjugacy class \( C \) ; the second sum has one term for each nontrivia... | Proof. First, \( \left| G\right| = \sum \left| C\right| \), since the conjugacy classes constitute a partition of \( G \) . Now, the conjugacy class of \( x \) is trivial \( \left( {\left| C\right| = 1}\right) \) if and only if \( x \in Z\left( G\right) \) ; hence there are \( \left| {Z\left( G\right) }\right| \) trivi... | Yes |
Proposition 3.9. Every nontrivial p-group has a nontrivial center. | Proof. By 3.7, \( \left| C\right| \) divides \( \left| G\right| = {p}^{n} \) for every conjugacy class \( C \) . In particular, \( p \) divides \( \left| C\right| \) when \( \left| C\right| > 1 \) . In the class equation, \( p \) divides \( \left| G\right| \) and \( p \) divides \( \mathop{\sum }\limits_{{\left| C\righ... | Yes |
Proposition 3.10. Every group of order \( {p}^{2} \), where \( p \) is prime, is abelian. | Proof. Readers will delight in proving that \( G/Z\left( G\right) \) cyclic implies \( G \) abelian. If now \( \left| G\right| = {p}^{2} \), then \( \left| {Z\left( G\right) }\right| > 1 \), so that \( \left| {Z\left( G\right) }\right| = p \) or \( \left| {Z\left( G\right) }\right| = {p}^{2} \) . If \( \left| {Z\left( ... | No |
Proposition 4.1. Every permutation is a product of transpositions. | Proof. By induction on \( n \) . Proposition 4.1 is vacuous if \( n = 1 \) . Let \( n > 1 \) and \( \sigma \in {S}_{n} \) . If \( {\sigma n} = n \), then, by the induction hypothesis, the restriction of \( \sigma \) to \( \{ 1,2,\ldots, n - 1\} \) is a product of transpositions; therefore \( \sigma \) is a product of t... | Yes |
Proposition 4.2. If \( \sigma = {\tau }_{1}{\tau }_{2}\cdots {\tau }_{r} = {v}_{1}{v}_{2}\cdots {v}_{s} \) is a product of transpositions \( {\tau }_{1},{\tau }_{2},\ldots ,{\tau }_{r} \) and \( {v}_{1},{v}_{2},\ldots ,{v}_{s} \), then \( r \equiv s\left( {\;\operatorname{mod}\;2}\right) \) . | Proof. This proof uses the ring \( R \) of all polynomials with \( n \) indeterminates \( {X}_{1},\ldots ,{X}_{n} \), with integer (or real) coefficients. Let \( {S}_{n} \) act on \( R \) by\n\n\[ \sigma \cdot f\left( {{X}_{1},\ldots ,{X}_{n}}\right) = f\left( {{X}_{▟},{X}_{\blacksquare },\ldots ,{X}_{\sigma n}}\right)... | Yes |
Proposition 4.3. \( {A}_{n} \) is generated by all 3-cycles. | Proof. First, \( \left( {abc}\right) = \left( {ab}\right) \left( {cb}\right) \) for all distinct \( a, b, c \), so that 3-cycles are even and \( {A}_{n} \) contains all 3-cycles. Now we show that every even permutation is a product of 3-cycles. It is enough to show that every product \( \left( {ab}\right) \left( {cd}\r... | Yes |
Lemma 4.4. Disjoint permutations commute. | Proof. Let \( \sigma \) and \( \tau \) be disjoint. If \( x \) is not in the support of \( \sigma \) or \( \tau \), then \( {\sigma \tau x} = {x\tau \sigma x} \) . If \( x \) is in the support of \( \sigma \), then so is \( {\sigma x} \), since \( {\sigma x} \neq x \) implies \( {\sigma \sigma x} \neq {\sigma x} \) ; t... | Yes |
Proposition 4.5. Every permutation is a product of pairwise disjoint cycles, and this decomposition is unique up to the order of the terms. | Proof. Given \( \sigma \in {S}_{n} \), let \( \mathbb{Z} \) act on \( X = \{ 1,2,\ldots, n\} \) by \( m \cdot x = {\sigma }^{m}x \) . This is a group action since \( {\sigma }^{0} = 1 \) and \( {\sigma }^{\ell }{\sigma }^{m} = {\sigma }^{\ell + m} \) . It partitions \( X \) into orbits. We see that \( {\sigma x} = x \)... | Yes |
Lemma 4.6. If \( \gamma = \left( {{a}_{1}{a}_{2}\ldots {a}_{k}}\right) \) is a \( k \) -cycle, then so is \( {\sigma \gamma }{\sigma }^{-1} = \) \( \left( {\sigma {a}_{1}\sigma {a}_{2}\ldots \sigma {a}_{k}}\right) \) . | Proof. If \( x \neq \sigma {a}_{1},\sigma {a}_{2},\ldots ,\sigma {a}_{k} \), then \( {\sigma }^{-1}x \neq {a}_{1},{a}_{2},\ldots ,{a}_{k},\gamma {\sigma }^{-1}x = \) \( {\sigma }^{-1}x \), and \( {\sigma \gamma }{\sigma }^{-1}x = x \) . But if \( x = \sigma {a}_{i} \), where \( i < k \), then \( {\sigma \gamma }{\sigma... | Yes |
Proposition 4.7. Two permutations are conjugate if and only if they have the same cycle structure. | Proof. Let \( \sigma \) be a product of disjoint cycles \( \sigma = {\gamma }_{1}{\gamma }_{2}\cdots {\gamma }_{r} \) . Each \( {\gamma }_{i} \) is a \( {k}_{i} \) -cycle for some \( {k}_{i} \geqq 2 \) ; by 4.4 we may assume that \( {k}_{1} \geqq {k}_{2} \geqq \cdots \geqq {k}_{r} \), and then the cycle structure of \(... | Yes |
Theorem 5.1 (First Sylow Theorem). Let \( G \) be a finite group and let \( p \) be a prime number. If \( {p}^{k} \) divides the order of \( G \), then \( G \) has a subgroup of order \( {p}^{k} \) . | Proof. First we prove a particular case: if \( G \) is abelian and \( p \) divides \( \left| G\right| \) , then \( G \) has a subgroup of order \( p \) . Readers will easily derive this statement from Theorem 1.2 but may prefer a direct proof. If \( \left| G\right| = p \), then \( G \) itself serves. Otherwise, \( \lef... | Yes |
Proposition 5.5. If a Sylow p-subgroup of a finite group \( G \) is normal in \( G \) , then it is the largest \( p \) -subgroup of \( G \) and the only Sylow \( p \) -subgroup of \( G \). | Proof. Let the Sylow \( p \) -subgroup \( S \) be normal in \( G \) . If \( T \) is a \( p \) -subgroup of \( G \), then \( {ST} \leqq G \) and \( \left| {ST}\right| = \left| S\right| \left| T\right| /\left| {S \cap T}\right| \geqq \left| S\right| \), by I.5.9. Hence \( \left| {ST}\right| = \left| S\right| \) , by the ... | Yes |
Theorem 5.7 (Third Sylow Theorem). Let \( p \) be a prime number. All Sylow p-subgroups of a finite group are conjugate. | Proof. Let \( S \) be a Sylow \( p \) -subgroup. A conjugate of a Sylow \( p \) -subgroup is a Sylow \( p \) -subgroup; therefore \( S \) acts on the set \( \mathcal{S} \) of all Sylow \( p \) -subgroups by inner automorphisms. Under this action, \( \{ S\} \) is an orbit, since \( {aS}{a}^{-1} = S \) for all \( a \in S... | Yes |
Corollary 5.8. A Sylow p-subgroup is normal if and only if it is the only Sylow p-subgroup. | The use of Theorems 5.6 and 5.7 may be shown by an example. Let \( G \) be a group of order 15 . The divisors of 15 are \( 1,3,5 \), and 15 ; its prime divisors are 3 and 5 . Since 1 is the only divisor of 15 that is congruent to 1 (mod 3), \( G \) has only one Sylow 3-subgroup \( S \) ; since 1 is the only divisor of ... | Yes |
Proposition 5.9. In a finite group, every p-subgroup is contained in a Sylow p-subgroup. | Proof. As above, a \( p \) -subgroup \( H \) of a finite group \( G \) acts by inner auto-morphisms on the set \( \mathcal{S} \) of all Sylow \( p \) -subgroups. Since \( \left| \mathcal{S}\right| \equiv 1\left( {\;\operatorname{mod}\;p}\right) \) there is at least one trivial orbit \( \{ S\} \) . Then \( {hS}{h}^{-1} ... | Yes |
Proposition 5.10. In a finite group, a subgroup that contains the normalizer of a Sylow p-subgroup is its own normalizer. | Proof. Let \( S \) be a Sylow \( p \) -subgroup of a finite group \( G \), and let \( H \) be a subgroup of \( G \) that contains \( {N}_{G}\left( S\right) \) . Let \( a \in {N}_{G}\left( H\right) \) . Then \( {aH}{a}^{-1} = H \) , so that \( S \) and \( {aS}{a}^{-1} \) are Sylow \( p \) -subgroups of \( H \) . By 5.7,... | Yes |
Proposition 5.11. A p-subgroup of a finite group that is not a Sylow p-subgroup is not its own normalizer. | Proof. Let \( H \) be a \( p \) -subgroup of a finite group \( G \) . If \( H \) is not a Sylow \( p \) -subgroup, then \( p \) divides \( \left\lbrack {G : H}\right\rbrack \) . Now, \( H \leqq {N}_{G}\left( H\right) \), and \( \left\lbrack {G : {N}_{G}\left( H\right) }\right\rbrack \) divides \( \left\lbrack {G : H}\r... | Yes |
Proposition 6.2. If \( p > q \) are primes, and \( q \) does not divide \( p - 1 \), then every group of order \( {pq} \) is cyclic. | Proof. By the Sylow theorems, a group \( G \) of order \( {pq} \) has a Sylow \( p \) -subgroup \( P \) of order \( p \) and a Sylow \( q \) -subgroup \( Q \) of order \( q \), both of which are cyclic. Among the divisors \( 1, p, q,{pq} \) of \( {pq} \), only 1 is congruent to 1 (mod \( p \) ), since \( q < p \), and ... | Yes |
Proposition 6.3. A nonabelian group of order 8 is isomorphic to either \( {D}_{4} \) or \( Q \) . | Proof. Let \( G \) be a nonabelian group of order 8 . No element of \( G \) has order 8, since \( G \) is not cyclic, and the elements of \( G \) cannot all have order 1 or 2 : if \( x = {x}^{-1} \) for all \( x \in G \), then \( {xy} = {\left( xy\right) }^{-1} = {y}^{-1}{x}^{-1} = {yx} \) for all \( x, y \in G \) . Th... | Yes |
Proposition 6.4. A nonabelian group of order 12 is isomorphic to either \( {D}_{4} \) or \( T \) or \( {A}_{4} \) . | Proof. A nonabelian group \( G \) of order 12 has a subgroup \( P \) of order 3 . Then \( G \) acts by left multiplication on the set of all four left cosets of \( P : g \cdot {xP} = {gxP} \) . By 3.1, this group action induces a homomorphism of \( G \) into \( {S}_{4} \), whose kernel \( K \) is a normal subgroup of \... | Yes |
Proposition 7.4. Every finite group has a composition series. | Proof. In a group \( G \) of order \( n \), every strictly ascending normal series \( \mathcal{A} : 1 = \) \( {A}_{0} \lneqq {A}_{1} \lneqq {A}_{2} \lneqq \cdots \lneqq {A}_{m} = G \) has length \( m \leqq n \) . Hence \( G \) has a strictly ascending normal series of maximal length, i.e., a composition series. | No |
Theorem 7.5 (Jordan-Hölder). Any two composition series of a group are equivalent. | Proof. Let \( \mathcal{A} : 1 = {A}_{0} \lneqq {A}_{1} \lneqq {A}_{2} \lneqq \cdots \lneqq {A}_{m} = G \) and \( \mathcal{B} : 1 = \) \( {B}_{0} \lneqq {B}_{1} \lneqq {B}_{2} \lneqq \cdots \lneqq {B}_{n} = G \) be two composition series of a group \( G \) . By Schreier’s theorem (7.1), \( \mathcal{A} \) and \( \mathcal... | Yes |
Proposition 7.6. A finite abelian group is simple if and only if it is a cyclic group of prime order. | This follows from, say, Theorem 1.2. | No |
Proposition 7.7. \( {A}_{n} \) is simple for all \( n \geqq 5 \) . | Proof. The simplicity of \( {A}_{5} \) is proved by counting the elements of its conjugacy classes, which readers will verify consist of:\n\n12 5-cycles;\n\n12 more 5-cycles;\n\n20 3-cycles;\n\n15 products of two disjoint transpositions; and\n\n1 identity element.\n\nA normal subgroup of \( {A}_{5} \) is the union of \... | Yes |
Proposition 8.2. For a linear transformation \( T \in {GL}\left( V\right) \) the following are equivalent: (i) \( T : V \rightarrow V \) is elementary; (ii) \( \det T = 1 \) and\n\n\[ F\left( T\right) = \{ x \in V \mid {Tx} = x\} = \operatorname{Ker}\left( {T - 1}\right) \]\n\nhas dimension \( \dim V - 1 \) ; (iii) \( ... | Proof. Let \( T \) be elementary, so that there is a basis \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) of \( V \) such that \( T{b}_{1} = {b}_{1} + {b}_{2} \) and \( T{b}_{i} = {b}_{i} \) for all \( i \geqq 2 \) . The matrix of \( T \) in that basis is triangular, with 1 ’s on the main diagonal; hence det \( T = 1 \) . Also ... | Yes |
Proposition 8.3. For a linear transformation \( T \in {GL}\left( V\right) \) the following are equivalent: (i) \( T \) is in the center of \( {GL}\left( V\right) \) ; (ii) \( T \) commutes with every elementary transformation; (iii) \( {Tx} \in {Kx} \) for all \( x \in V \) ; (iv) \( T = \lambda {1}_{V} \) for some \( ... | Proof. Let \( T \in {GL}\left( V\right) \) . We see that (i) implies (ii).\n\nAssume (ii). If \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) is a basis of \( V \), then there is an elementary transformation \( E \) such that \( E{b}_{1} = {b}_{1} + {b}_{2} \) and \( E{b}_{i} = {b}_{i} \) for all \( i \geqq 2 \) . Then \( \opera... | Yes |
Proposition 8.4. For every vector space \( V \) of finite dimension \( n \geqq 2 \) over a field \( K \), the group \( {SL}\left( V\right) \) is generated by all elementary transformations. | Proof. We use matrices. As long as rows are not permuted or multiplied by scalars, Gauss-Jordan reduction is equivalent to left multiplication by elementary matrices. Therefore, when \( M \) is an invertible matrix, there are elementary matrices \( {E}_{1},\ldots ,{E}_{n} \) such that \( {E}_{1}\cdots {E}_{n}M \) is di... | Yes |
Proposition 8.5. Elementary transformations constitute a conjugacy class of \( {GL}\left( V\right) \) ; if \( \dim V \geqq 3 \), they constitute a conjugacy class of \( {SL}\left( V\right) \) . | Proof. Let \( E \) be an elementary transformation and let \( {TE}{T}^{-1} \) be a conjugate of \( E \) in \( {GL}\left( V\right) \) . There is a basis \( {b}_{1},{b}_{2},\ldots ,{b}_{n} \) of \( V \) such that \( E{b}_{1} = {b}_{1} + {b}_{2} \) and \( E{b}_{i} = {b}_{i} \) for all \( i \geqq 2 \) . Then \( T{b}_{1}, T... | Yes |
Proposition 8.8. If \( K \) has \( q \) elements, then \( \left| {{GL}\left( {n, K}\right) }\right| = \mathop{\prod }\limits_{{0 \leq i < n}}\left( {{q}^{n} - {q}^{i}}\right) \) ; \( \left| {{SL}\left( {n, K}\right) }\right| = \left| {{GL}\left( {n, K}\right) }\right| /\left( {q - 1}\right) \) ; and \( \left| {{PSL}\le... | Proof. An \( n \times n \) matrix \( M \) is invertible if and only if its columns constitute a basis of the vector space \( {K}^{n} \) . If \( \left| K\right| = q \), there are \( {q}^{n} \) possible columns and \( {q}^{n} - 1 \) ways to choose the first column of \( M \), which must not be the zero column; there are ... | Yes |
Proposition 9.1. \( {G}^{\prime } \) is a normal subgroup of \( G \) ; in fact, \( {G}^{\prime } \) is the smallest normal subgroup \( N \) of \( G \) such that \( G/N \) is abelian. | Proof. The inverse of a commutator \( {xy}{x}^{-1}{y}^{-1} \) is a commutator, and a conjugate of a commutator is again a commutator:\n\n\[ \n{axy}{x}^{-1}{y}^{-1}{a}^{-1} = {ax}{a}^{-1}{ay}{a}^{-1}{\left( ax{a}^{-1}\right) }^{-1}{\left( ay{a}^{-1}\right) }^{-1}. \n\]\n\nHence every \( x \in {G}^{\prime } \) is a produ... | Yes |
Proposition 9.2. A group \( G \) is solvable if and only if \( {G}^{\left( r\right) } = 1 \) for some \( r \geqq 0 \) . | Proof. If \( {G}^{\left( r\right) } = 1 \), then \( 1 = {G}^{\left( r\right) } \trianglelefteq {G}^{\left( r - 1\right) } \trianglelefteq \cdots \trianglelefteq {G}^{\prime } \trianglelefteq G \) is a normal series whose factors are abelian, by 9.1. Conversely, assume that \( G \) has a normal series \( 1 = {A}_{0} \le... | Yes |
Proposition 9.6. Every finite p-group is solvable. | Proof. That a group \( G \) of order \( {p}^{n} \) is solvable is proved by induction on \( n \) . If \( n \leqq 2 \), then \( G \) is abelian, hence solvable, by 3.10. In general, \( G \) has a subgroup \( N \) of order \( {p}^{n - 1} \), which is normal by 5.12; then \( N \) and \( G/N \) are solvable, by the inducti... | Yes |
Proposition 9.7. Every group of order \( {p}^{n}q \) (where \( p \) and \( q \) are primes) is solvable. | Proof. We may assume that \( p \neq q \) . The proof is by induction on \( n \) .\n\nLet \( S \) be a Sylow \( p \) -subgroup of \( G \) . If \( S \leqq G \), then \( G/S \) is cyclic, since \( \left| {G/S}\right| = q \) is prime, \( S \) is solvable by 9.6, and \( G \) is solvable by 9.5 .\n\nNow assume that \( S \) i... | Yes |
Lemma 9.8. Let \( M \) be the intersection of two distinct Sylow p-subgroups of a group \( G \) . If \( M \) has the greatest possible number of elements, then \( H = {N}_{G}\left( M\right) \) has more than one Sylow p-subgroup; \( M \) is the intersection of all the Sylow p-subgroups of \( H \) ; and every Sylow p-sub... | Proof. We have \( M \subsetneqq S \) for some Sylow \( p \) -subgroup \( S \) of \( G \) . By 5.11, \( M \subsetneqq {N}_{S}\left( M\right) = H \cap S \) . Now, \( {N}_{S}\left( M\right) \subseteq S \) is a \( p \) -subgroup of \( H \) and is by 5.9 contained in a Sylow \( p \) -subgroup \( P \) of \( H \), which is in... | Yes |
Lemma 9.9. Every nontrivial finite solvable group contains a nontrivial abelian normal p-subgroup for some prime \( p \) . | Proof. Let \( G \) be be a finite solvable group. There is a smallest integer \( r > 0 \) such that \( {G}^{\left( r\right) } = 1 \) . Then \( A = {G}^{\left( r - 1\right) } \) is a nontrivial abelian normal subgroup of \( G \) . Some prime \( p \) divides \( \left| A\right| > 1 \) ; let \( N \) be the set of all eleme... | Yes |
Theorem 9.10. Let \( m \) and \( n \) be relatively prime. Every solvable group of order mn contains a subgroup of order \( m \) . | Proof. Let \( G \) be solvable of order \( {mn} \) . If \( m \) is a power of a prime, then 9.10 follows from the first Sylow theorem. Otherwise, we proceed by induction on \( \left| G\right| \) . By 9.9, \( G \) contains contains a nontrivial abelian normal subgroup \( N \) of order \( {p}^{k} > 1 \) for some prime \(... | Yes |
Lemma 9.11. Let \( m \) and \( n \) be relatively prime and let \( G \) be a group of order \( {mn} \) with an abelian normal subgroup of order \( n \) . All subgroups of \( G \) of order \( m \) are conjugate. | Proof. Let \( \left| G\right| = {mn} \) and let \( N \leqq G \), with \( \left| N\right| = n \) and \( N \) abelian. Let\n\n\( A \) and \( B \) be subgroups of \( G \) of order \( m \) . Since \( m \) and \( n \) are relatively prime we have \( A \cap N = B \cap N = 1 \) ; hence \( {AN} = {BN} = G \) . Therefore every ... | Yes |
Theorem 9.12. In a solvable group of order \( {mn} \), where \( m \) and \( n \) are relatively prime, all subgroups of order \( m \) are conjugate. | Proof. Let \( G \) be solvable of order \( {mn} \). If \( m \) is a power of a prime, then 9.12 follows from the third Sylow theorem. Otherwise, we proceed by induction on \( \left| G\right| \). By 9.9, \( G \) contains an abelian normal subgroup \( N \) of order \( {p}^{k} > 1 \) for some prime \( p \), and \( {p}^{k}... | Yes |
Theorem 9.13. In a solvable group of order \( {mn} \), where \( m \) and \( n \) are relatively prime, every subgroup whose order divides \( m \) is contained in a subgroup of order \( m \) . | Proof. Let \( G \) be solvable of order \( {mn} \) . If \( m \) is a power of a prime, then 9.13 follows from 5.9. Otherwise, we proceed by induction on \( \left| G\right| \) . By 9.9, \( G \) contains an abelian normal subgroup \( N \) of order \( {p}^{k} > 1 \) for some prime \( p \), and \( {p}^{k} \) divides \( m \... | Yes |
Proposition 10.2. Every group \( G \) has unique normal subgroups \( {Z}_{k}\left( G\right) \) such that \( {Z}_{0}\left( G\right) = 1 \) and \( {Z}_{k + 1}\left( G\right) /{Z}_{k}\left( G\right) = Z\left( {G/{Z}_{k}\left( G\right) }\right) \) for all \( k \geqq 0 \) . | Proof. First, \( {Z}_{0}\left( G\right) = 1 \) is normal in \( G \) . If \( {Z}_{k} \leqq G \), then \( Z\left( {G/{Z}_{k}\left( G\right) }\right) \) is a normal subgroup of \( G/{Z}_{k}\left( G\right) \) ; by I.4.9, there is a unique normal subgroup \( {Z}_{k + 1}\left( G\right) \supseteq {Z}_{k}\left( G\right) \) of ... | Yes |
A group \( G \) is nilpotent if and only if \( {G}^{r} = 1 \) for some \( r \geqq 0 \) , if and only if \( {Z}_{r}\left( G\right) = G \) for some \( r \geqq 0 \) . | If \( {G}^{r} = 1 \) for some \( r \geqq 0 \), or if \( {Z}_{r}\left( G\right) = G \) for some \( r \geqq 0 \), then truncating the descending central series, or the ascending central series, yields a central normal series, and \( G \) is nilpotent.\n\nConversely, assume that \( G \) has a central normal series \( 1 = ... | Yes |
Proposition 10.8. Every finite p-group is nilpotent. | Proof. That a group \( G \) of order \( {p}^{n} \) is nilpotent is proved by induction on \( n \) . If \( n \leqq 2 \), then \( G \) is abelian, hence nilpotent, by 3.10. If \( n > 2 \), then \( G \) has a nontrivial center, by 3.9; then \( G/Z\left( G\right) \) is nilpotent, by the induction hypothesis, and \( G \) is... | Yes |
Proposition 10.9. A finite group is nilpotent if and only if all its Sylow subgroups are normal, if and only if it is isomorphic to a direct product of p-groups (for various primes \( p \) ). | Proof. The ascending central series of any group \( G \) has the following property: if \( {Z}_{k} \subseteq H \leqq G \), then \( {Z}_{k + 1} \subseteq {N}_{G}\left( H\right) \) . Indeed, let \( x \in {Z}_{k + 1} \) and \( y \in H \) . Since \( {Z}_{k}x \in {Z}_{k + 1}/{Z}_{k} \subseteq C\left( {G/{Z}_{k}}\right) \) w... | Yes |
Proposition 11.1. Let the group \( B \) act on a group \( A \) by automorphisms. The mapping \( \varphi : b \mapsto \varphi \left( b\right) \) defined by \( \varphi \left( b\right) : a \mapsto {}^{b}a \) is a homomorphism of \( B \) into the group \( \operatorname{Aut}\left( A\right) \) of all automorphisms of \( A \) ... | Proof. By definition, every \( \varphi \left( b\right) \) is an automorphism of \( A \) ; moreover, \( \varphi \left( 1\right) \) is the identity mapping on \( A \), and \( \varphi \left( b\right) \circ \varphi \left( {b}^{\prime }\right) = \varphi \left( {b{b}^{\prime }}\right) \) for all \( b,{b}^{\prime } \in B \) .... | Yes |
Lemma 12.1. Let \( G\overset{\kappa }{ \rightarrow }E\overset{\rho }{ \rightarrow }Q \) be a group extension and let \( p \) be a cross-section of \( E \) . Every element of \( E \) can be written in the form \( \kappa \left( x\right) {p}_{a} \) for some unique \( x \in G \) and \( a \in Q \) (then \( a = \rho \left( {... | Lemma 12.1 provides a bijection \( \left( {x, a}\right) \mapsto \kappa \left( x\right) {p}_{a} \) of \( G \times Q \) onto \( E \) . Now we put every product \( \kappa \left( x\right) {p}_{a}\kappa \left( y\right) {p}_{b} \) in the form \( \kappa \left( z\right) {p}_{c} \) . We start with the simpler products \( {p}_{a... | Yes |
Lemma 12.2. Relative to any cross-section, \( Q \) acts on \( G \) by automorphisms; in particular, \( {}^{a}\left( {xy}\right) = {}^{a}x{}^{a}y \) and \( {}^{a}1 = 1 \) for all \( x, y \in G \) and \( a \in Q \) . Moreover, the cross-section \( p \) can be chosen so that \( {p}_{1} = 1 \), and then\n\n\[ \n{}^{1}x = x... | This is straightforward. The automorphism \( x \mapsto {}^{a}x \) is induced on \( G \), via \( \kappa \), by the inner automorphism \( x \mapsto {p}_{a}x{p}_{a}^{-1} \), which has a restriction to \( \operatorname{Im}\kappa \) ; \( \left( N\right) \) is the normalization condition. | No |
Lemma 12.3. If \( \left( N\right) \) holds, then \( \left( M\right) \) is associative if and only if\n\n\[ \n{}^{a}\left( {{}^{b}x}\right) {s}_{a, b} = {s}_{a, b}{}^{ab}x\text{ and }{s}_{a, b}{s}_{{ab}, c} = {}^{a}{s}_{b, c}{s}_{a,{bc}}, \]\n\n(A)\n\nfor all \( x \in G \) and \( a, b, c \in Q \) . | Proof. By \( \left( M\right) \) and 12.2,\n\n\[ \n\left( {\left( {x, a}\right) \left( {y, b}\right) }\right) \left( {z, c}\right) = \left( {{x}^{a}y{s}_{a, b},{ab}}\right) \left( {z, c}\right) \]\n\n\[ = \left( {{x}^{a}y{s}_{a, b}{}^{ab}z{s}_{{ab}, c},\left( {ab}\right) c}\right) , \]\n\n\[ \left( {x, a}\right) \left( ... | Yes |
Theorem 12.4 (Schreier [1926]). Let \( G \) and \( Q \) be groups, and let \( s : Q \times \) \( Q \rightarrow G \) and \( \varphi : Q \rightarrow \operatorname{Aut}\left( G\right) \) be mappings such that \( \left( N\right) \) and \( \left( A\right) \) hold. Then \( E\left( {s,\varphi }\right) = G \times Q \) with mul... | Proof. If \( s \) and \( \varphi \) satisfy \( \left( N\right) \) and \( \left( A\right) \), then \( \left( M\right) \) is associative by 12.3, and \( \left( {1,1}\right) \) is an identity element of \( E\left( {s,\varphi }\right) \) . Moreover, every element \( \left( {y, b}\right) \) of \( E\left( {s,\varphi }\right)... | Yes |
Proposition 12.5. \( E\left( {s,\varphi }\right) \) and \( E\left( {t,\psi }\right) \) are equivalent if and only if there exists a mapping \( u : a \mapsto {u}_{a} \) of \( Q \) into \( G \) such that\n\n\[ \n{u}_{1} = 1,{}_{\varphi }^{a}x = {u}_{a}{}_{\psi }^{a}x{u}_{a}^{-1},\text{ and }{s}_{a, b} = {u}_{a}{}_{\psi }... | Proof. Let \( \theta : E\left( {s,\varphi }\right) \rightarrow E\left( {t,\psi }\right) \) be an equivalence of group extensions:\n\n\n\nWe have \( \theta \left( {x,1}\right) = \left( {x,1}\right) \) and \( \theta \l... | Yes |
Proposition 12.6. For a group extension \( G\overset{\kappa }{ \rightarrow }E\overset{\rho }{ \rightarrow }Q \) the following conditions are equivalent:\n\n(1) There exists a homomorphism \( \mu : Q \rightarrow E \) such that \( \rho \circ \mu = {1}_{Q} \) .\n\n(2) There is a cross-section of \( E \) relative to which ... | Proof. (1) implies (2). If (1) holds, then \( {p}_{a} = \mu \left( a\right) \) is a cross-section of \( E \) , relative to which \( {s}_{a, b} = 1 \) for all \( a, b \), since \( \mu \left( a\right) \mu \left( b\right) = \mu \left( {ab}\right) \).\n\n(2) implies (3). If \( {s}_{a, b} = 1 \) for all \( a, b \), then \( ... | Yes |
Theorem 12.9 (Hölder). A group \( G \) is an extension of a cyclic group of order \( m \) by a cyclic group of order \( n \) if and only if \( G \) is generated by two elements \( a \) and \( b \) such that \( a \) has order \( m,{b}^{n} = {a}^{t},{b}^{i} \notin \langle a\rangle \) when \( 0 < i < n \), and \( {ba}{b}^... | Proof. First let \( G = \langle a, b\rangle \), where \( a \) has order \( m,{b}^{n} = {a}^{t},{b}^{i} \notin \langle a\rangle \) when \( 0 < i < n \), and \( {ba}{b}^{-1} = {a}^{r} \), where \( {r}^{n} \equiv 1 \) and \( {rt} \equiv 1\left( {\;\operatorname{mod}\;m}\right) \) . Then \( A = \langle a\rangle \) is cycli... | Yes |
Theorem 12.10 (Schur). If \( m \) and \( n \) are relatively prime, then a group of order mn that contains an abelian normal subgroup of order \( n \) also contains a subgroup of order \( m \) . | Proof. Let \( G \) be a group of order \( {mn} \) with an abelian normal subgroup \( N \) of order \( n \) . Then \( G \) is an extension of \( N \) by a group \( Q = G/N \) of order \( m \) . Let \( s \) be any factor set of this extension. For every \( a \in Q \) let \( {t}_{a} = \mathop{\prod }\limits_{{c \in Q}}{s}... | Yes |
Proposition 1.1. The set \( \operatorname{End}\left( A\right) \) of all endomorphisms of an abelian group \( A \) is a ring with identity. | In the additive notation for \( A \), addition on \( \operatorname{End}\left( A\right) \) is pointwise \( (\left( {\eta + \zeta }\right) x = \) \( {\eta x} + {\zeta x}) \) ; multiplication on \( \operatorname{End}\left( A\right) \) is composition \( \left( {\left( {\eta \zeta }\right) x = \eta \left( {\zeta x}\right) }... | No |
Proposition 1.5 (Binomial Theorem). In a commutative ring \( R \) , \[ {\left( x + y\right) }^{n} = \mathop{\sum }\limits_{{0 \leqq i \leqq n}}\left( \begin{matrix} n \\ i \end{matrix}\right) {x}^{i}{y}^{n - i}\text{, where }\left( \begin{matrix} n \\ i \end{matrix}\right) = \frac{n!}{i!\left( {n - i}\right) !}. \] | In fact,1.5 works in every ring, as long as \( {xy} = {yx} \) . | No |
Proposition 1.7. For every ring \( R \), the set \( {R}^{1} = R \times \mathbb{Z} \), with operations\n\n\[ \left( {x, m}\right) + \left( {y, n}\right) = \left( {x + y, m + n}\right) ,\;\left( {x, m}\right) \left( {y, n}\right) = \left( {{xy} + {nx} + {my},{mn}}\right) ,\]\n\nis a ring with identity. Moreover, \( \iota... | The proof is straightforward but no fun, and left to our poor, abused readers. | No |
Proposition 1.8. Every homomorphism \( \varphi \) of \( R \) into a ring \( S \) with identity factors uniquely through \( \iota : R \rightarrow {R}^{1} \) (there is a homomorphism \( \psi : {R}^{1} \rightarrow S \) of rings with identity, unique such that \( \varphi = \psi \circ \iota \) ). | Proof. In \( {R}^{1} \), the identity element is \( \left( {0,1}\right) \) and \( \left( {x, n}\right) = \left( {x,0}\right) + n\left( {0,1}\right) \) . If now \( \psi \left( {0,1}\right) = 1 \) and \( \psi \circ \iota = \varphi \), then necessarily \( \psi \left( {x, n}\right) = \varphi \left( x\right) + {n1} \in S \)... | Yes |
Proposition 2.1. Every intersection of ideals of a ring \( R \) is an ideal of \( R \) . | By 2.1 there is for every subset \( S \) of \( R \) a smallest ideal of \( R \) that contains \( S \) , namely the intersection of all the ideals of \( R \) that contain \( S \) . | No |
Proposition 2.2. In a ring \( R \) [with identity], the ideal \( \left( S\right) \) generated by a subset \( S \) is the set of all finite sums of elements of the form \( {xsy} \), with \( s \in S \) and \( x, y \in R \) . If \( R \) is commutative, then \( \left( S\right) \) is the set of all finite linear combination... | Proof. An ideal that contains \( S \) must also contain all elements of the form \( {xsy} \) with \( s \in S \) and \( x, y \in R \), and all finite sums of such elements. We show that the set \( I \) of all such sums is an ideal of \( R \) . First, \( I \) contains the empty sum \( 0;I \) is closed under sums by defin... | Yes |
Proposition 2.3. In a commutative ring \( R \) [with identity], the principal ideal generated by \( a \in R \) is the set \( \left( a\right) = {Ra} \) of all multiples of \( a \) . | This follows from Proposition 2.2: by distributivity, a linear combination \( {x}_{1}a + \) \( \cdots + {x}_{n}a \) of copies of \( a \) is a multiple \( \left( {{x}_{1} + \cdots + {x}_{n}}\right) a \) of \( a \) . | Yes |
Proposition 2.6. In a ring \( R \) [with identity], every proper ideal is contained in a maximal ideal. | Proof. An ideal that contains the identity element must contain all its multiples and is not proper. Hence an ideal is proper if and only if it does not contain the identity element. Therefore the union of a nonempty chain of proper ideals, which is an ideal by 2.5 , is a proper ideal.\n\nGiven an ideal \( I \neq R \) ... | Yes |
Let \( I \) be an ideal of a ring \( R \) . The cosets of \( I \) in the abelian group \( \left( {R, + }\right) \) constitute a ring \( R/I \) . In \( R/I \), the sum of two cosets is their sum as subsets, so that \( \left( {x + I}\right) + \left( {y + I}\right) = \left( {x + y}\right) + I \) ; the product of two coset... | Proof. \( R/I \) is already an abelian group, by I.4.7. If \( x + I, y + I \in R/I \), then the product \( \left( {x + I}\right) \left( {y + I}\right) \) of subsets is contained in the single coset \( {xy} + I \), since \( \left( {x + i}\right) \left( {y + j}\right) = {xy} + {xj} + {iy} + {ij} \in {xy} + I \) for all \... | Yes |
Proposition 3.4. Let \( I \) be an ideal of a ring \( R \) . Every subring of \( R/I \) is the quotient \( S/I \) of a unique subring \( S \) of \( R \) that contains \( I \) . Every ideal of \( R/I \) is the quotient \( J/I \) of a unique ideal \( J \) of \( R \) that contains \( I \) . | This follows from I.4.9. Theorem I.5.1 also extends to quotient rings: | No |
Theorem 3.5 (Factorization Theorem). Let \( I \) be an ideal of a ring \( R \) . Every homomorphism of rings \( \varphi : R \rightarrow S \) whose kernel contains I factors uniquely through the canonical projection \( \pi : R \rightarrow R/I \) (there exists a homomorphism \( \psi : R/I \rightarrow S \) unique such tha... | Proof. By I.5.1 there is a homomorphism of abelian groups \( \psi \) of \( \left( {R/I, + }\right) \) into \( \left( {S, + }\right) \) unique such that \( \varphi = \psi \circ \pi \) ; equivalently, \( \psi \left( {x + I}\right) = \varphi \left( x\right) \) for all \( x \in R \) . Now, \( \psi \) is a homomorphism of r... | Yes |
Theorem 3.6 (Homomorphism Theorem). If \( \varphi : R \rightarrow S \) is a homomorphism of rings, then\n\n\[ R/\operatorname{Ker}\varphi \cong \operatorname{Im}\varphi \]\nin fact, there is an isomorphism \( \theta : R/\operatorname{Ker}f \rightarrow \operatorname{Im}f \) unique such that \( \varphi = \iota \circ \the... | Proof. By I.5.2 there is an isomorphism of abelian groups \( \theta : \left( {R/\operatorname{Ker}f, + }\right) \) \( \rightarrow \left( {\operatorname{Im}f, + }\right) \) unique such that \( \varphi = \iota \circ \theta \circ \pi \) ; equivalently, \( \theta \left( {x + \operatorname{Ker}\varphi }\right) = \varphi \le... | Yes |
Proposition 3.7. Let \( R \) be a ring [with identity]. There is a unique homomorphism of rings of \( \mathbb{Z} \) into \( R \) . Its image is the smallest subring of \( R \) ; it consists of all integer multiples of the identity element of \( R \), and is isomorphic either to \( \mathbb{Z} \) or to \( {\mathbb{Z}}_{n... | Proof. If \( \varphi : \mathbb{Z} \rightarrow R \) is a homomorphism of rings [with identity], then \( \varphi \left( 1\right) = 1 \) and \( \varphi \left( n\right) = \varphi \left( {n1}\right) = {n1} \in R \) for all \( n \in \mathbb{Z} \) . Hence \( \varphi \) is unique. Conversely, we saw that the mapping \( \varphi... | Yes |
Proposition 4.1. Let \( n > 0 \) . The ring \( {\mathbb{Z}}_{n} \) is a domain if and only if \( n \) is prime, and then \( {\mathbb{Z}}_{n} \) is a field. | Proof. If \( n > 0 \) is not prime, then either \( n = 1 \), in which case \( {\mathbb{Z}}_{n} = 0 \), or \( n = {xy} \) for some \( 1 < x, y < n \), in which case \( \bar{x}\bar{y} = \overline{0} \) in \( {\mathbb{Z}}_{n} \) and \( {\mathbb{Z}}_{n} \) has a zero divisor. In either case \( {\mathbb{Z}}_{n} \) is not a ... | Yes |
Proposition 4.2. In a domain, \( {xy} = {xz} \) implies \( y = z \), when \( x \neq 0 \) . | Proof. If \( x\left( {y - z}\right) = 0 \) and \( x \neq 0 \), then \( y - z = 0 \) : otherwise, \( x \) would be a zero divisor. \( ▱ \) | Yes |
Proposition 4.3. The characteristic of a domain is either 0 or a prime number. | Proof. The smallest subring of a domain has no zero divisors; by 3.7, 4.1, it is isomorphic either to \( \mathbb{Z} \), or to \( {\mathbb{Z}}_{p} \) for some prime \( p \) . \( ▱ \) | Yes |
Proposition 4.4. In a commutative ring \( R \) of prime characteristic \( p,{\left( x + y\right) }^{p} = \) \( {x}^{p} + {y}^{p} \) and \( {\left( x - y\right) }^{p} = {x}^{p} - {y}^{p} \), for all \( x, y \in R \) . | Proof. By the binomial theorem, \( {\left( x + y\right) }^{p} = \mathop{\sum }\limits_{{0 \leqq i \leqq p}}\left( \begin{array}{l} p \\ i \end{array}\right) {x}^{i}{y}^{p - i} \), where \( \left( \begin{matrix} p \\ i \end{matrix}\right) = \frac{p!}{i!\left( {p - i}\right) !} \) . If \( 0 < i < p \), then \( p \) divid... | Yes |
Proposition 4.5. If \( \mathfrak{a} \) is an ideal of a commutative ring \( R \) [with identity], then \( R/\mathfrak{a} \) is a domain if and only if \( \mathfrak{a} \) is a prime ideal. | The proof is an exercise. | No |
Proposition 4.6. If \( \mathfrak{a} \) is an ideal of a commutative ring \( R \) [with identity], then \( R/\mathfrak{a} \) is a field if and only if \( \mathfrak{a} \) is a maximal ideal. | Proof. A field \( F \) has no proper ideal \( \mathfrak{c} \neq 0 \) : indeed, if \( x \in \mathfrak{c}, x \neq 0 \), then \( 1 \in \mathfrak{c} \) since \( x \) is a unit, and \( \mathfrak{c} = F \) . Conversely, let \( R \neq 0 \) be a commutative ring with no proper ideal \( \mathfrak{c} \neq 0 \) . For every \( x \... | Yes |
Proposition 4.10. For every domain \( R, Q\left( R\right) \) is a field and \( \iota : x \mapsto x/1 \) is an injective homomorphism. | Then \( R \) is isomorphic to the subring \( \operatorname{Im}\iota \) of the field \( Q\left( R\right) \) . It is common practice to identify \( x \in R \) and \( \iota \left( x\right) = x/1 \in Q\left( R\right) \) ; then \( \iota \) is an inclusion homomorphism and \( R \) is a subring of \( Q\left( R\right) \) . | No |
Proposition 4.11. Let \( R \) be a domain. Every injective homomorphism \( \varphi \) of \( R \) into a field \( F \) factors uniquely through \( \iota : R \mapsto Q\left( R\right) : \varphi = \psi \circ \iota \) for some unique homomorphism \( \psi : Q\left( R\right) \rightarrow F \), namely \( \psi \left( {x/y}\right... | Proof. Every homomorphism \( \psi \) of fields preserves inverses: if \( x \neq 0 \), then \( \psi \left( x\right) \psi \left( {x}^{-1}\right) = \psi \left( 1\right) = 1 \), so that \( \psi \left( x\right) \neq 0 \) and \( \psi \left( {x}^{-1}\right) = \psi {\left( x\right) }^{-1} \) . In \( Q\left( R\right) ,{\left( x... | Yes |
Proposition 4.12. Let \( R \) be a subring of a field \( K \) . The identity on \( R \) extends to an isomorphism \( K \cong Q\left( R\right) \) if and only if every element of \( K \) can be written in the form \( a{b}^{-1} \) for some \( a, b \in R, b \neq 0 \) . | Proof. This condition is necessary since every element of \( Q\left( R\right) \) can be written in the form \( a/b = a{b}^{-1} \) for some \( a, b \in R, b \neq 0 \) . Conversely, by 4.11, the inclusion homomorphism \( R \rightarrow K \) extends to a homomorphism \( \theta : Q\left( R\right) \rightarrow \) \( K \), whi... | Yes |
Proposition 5.1. For every ring \( R \) [with identity], \( R\left\lbrack X\right\rbrack \), with the operations above, is a ring. | Proof. For each \( A \in R\left\lbrack X\right\rbrack \) there exists some \( m \geqq 0 \) such that \( {a}_{k} = 0 \) for all \( k > m \) : otherwise, \( \left\{ {k \geqq 0 \mid {a}_{k} \neq 0}\right\} \) is not finite. If \( {a}_{k} = 0 \) for all \( k > m \) , and \( {b}_{k} = 0 \) for all \( k > n \), then \( {a}_{... | No |
Proposition 5.2. \( A = \mathop{\sum }\limits_{{n \geqq 0}}{a}_{n}{X}^{n} \), for every \( A \in R\left\lbrack X\right\rbrack \) ; if \( {a}_{i} = 0 \) for all \( i > n \), then \( A = {a}_{0} + {a}_{1}X + \cdots + {a}_{n}{X}^{n} \) . | Proof. The infinite sum \( \mathop{\sum }\limits_{{n \geqq 0}}{a}_{n}{X}^{n} \) exists since \( {a}_{n}{X}^{n} = 0 \) for almost all \( n \) . Its coefficients are found as follows. By induction on \( k,{X}^{k} \) has coefficients \( {a}_{n} = 0 \) if \( n \neq k,{a}_{k} = 1 \) . Then \( r{X}^{k} \) has coefficients \(... | Yes |
Proposition 5.5. Let \( B \in R\left\lbrack X\right\rbrack \) be a nonzero polynomial whose leading coefficient is a unit of \( R \) . For every polynomial \( A \in R\left\lbrack X\right\rbrack \) there exist polynomials \( Q, S \in R\left\lbrack X\right\rbrack \) such that \( A = {BQ} + S \) and \( \deg S < \deg B \) ... | Proof. First we assume that \( B \) is monic and prove existence by induction on \( \deg A \) . Let \( \deg B = n \) . If \( \deg A < n \), then \( Q = 0 \) and \( S = A \) serve. Now let \( \deg A = m \geqq n \) . Then \( B{a}_{m}{X}^{m - n} \) has degree \( m \) and leading coefficient \( {a}_{m} \) . Hence \( A - B{... | Yes |
Proposition 5.6. If \( R \) is commutative, then evaluation at \( r \in R \) is a homomorphism of \( R\left\lbrack X\right\rbrack \) into \( R \) . More generally, if \( R \) is a subring of \( S \) and \( s \in S \) commutes with every element of \( R \), then evaluation at \( s \) is a homomorphism of \( R\left\lbrac... | Proof. For all \( A, B \in R\left\lbrack X\right\rbrack ,\left( {A + B}\right) \left( s\right) = A\left( s\right) + B\left( s\right) \) and\n\n\[ A\left( s\right) B\left( s\right) = \left( {\mathop{\sum }\limits_{i}{a}_{i}{s}^{i}}\right) \left( {\mathop{\sum }\limits_{j}{b}_{j}{s}^{j}}\right) = \mathop{\sum }\limits_{{... | Yes |
Proposition 5.7. Let \( r \in R \) and \( A \in R\left\lbrack X\right\rbrack \) . If \( R \) is commutative, then \( A \) is a multiple of \( X - r \) if and only if \( A\left( r\right) = 0 \) . | Proof. By polynomial division, \( A = \left( {X - r}\right) B + S \), where \( B \) and \( S \) are unique with \( \deg S < 1 \) . Then \( S \) is constant. Evaluating at \( r \) yields \( S = A\left( r\right) \), by 5.6. Hence \( X - r \) divides \( A \) if and only if \( A\left( r\right) = 0.▱ \) | Yes |
Proposition 5.9. If \( R \) is commutative, then a root \( r \in R \) of a polynomial \( A \in R\left\lbrack X\right\rbrack \) is simple if and only if \( {A}^{\prime }\left( r\right) \neq 0 \) . | Proof. If \( r \) has multiplicity \( m \), then \( A = {\left( X - r\right) }^{m}B \), where \( B\left( r\right) \neq 0 \) : otherwise \( {\left( X - r\right) }^{m + 1} \) divides \( A \) by 5.7. If \( m = 1 \), then \( {A}^{\prime } = B + \left( {X - r}\right) {B}^{\prime } \) and \( {A}^{\prime }\left( r\right) = B\... | Yes |
Proposition 5.10. Every homomorphism of rings \( \varphi : R \rightarrow S \) induces a homomorphism of rings \( A \mapsto {}^{\varphi }A \) of \( R\left\lbrack X\right\rbrack \) into \( S\left\lbrack X\right\rbrack \), namely, | \[ \varphi \left( {{a}_{0} + {a}_{1}X + \cdots + {a}_{n}{X}^{n}}\right) = \varphi \left( {a}_{0}\right) + \varphi \left( {a}_{1}\right) X + \cdots + \varphi \left( {a}_{n}\right) {X}^{n}. \] | Yes |
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