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Corollary 10. If \( G \) is a group of prime order \( p \), then \( G \) is cyclic, hence \( G \cong {Z}_{p} \) . | Proof: Let \( x \in G, x \neq 1 \) . Thus \( \left| {\langle x\rangle }\right| > 1 \) and \( \left| {\langle x\rangle }\right| \) divides \( \left| G\right| \) . Since \( \left| G\right| \) is prime we must have \( \left| {\langle x\rangle }\right| = \left| G\right| \), hence \( G = \langle x\rangle \) is cyclic (with ... | No |
Theorem 11. (Cauchy’s Theorem) If \( G \) is a finite group and \( p \) is a prime dividing \( \left| G\right| \) , then \( G \) has an element of order \( p \) . | Proof: We shall give a proof of this in the next chapter and another elegant proof is outlined in Exercise 9. | No |
Proposition 13. If \( H \) and \( K \) are finite subgroups of a group then\n\n\[ \left| {HK}\right| = \frac{\left| H\right| \left| K\right| }{\left| H \cap K\right| } \] | Proof: Notice that \( {HK} \) is a union of left cosets of \( K \), namely,\n\n\[ {HK} = \mathop{\bigcup }\limits_{{h \in H}}{hK} \]\n\nSince each coset of \( K \) has \( \left| K\right| \) elements it suffices to find the number of distinct left cosets of the form \( {hK}, h \in H \) . But \( {h}_{1}K = {h}_{2}K \) fo... | Yes |
Proposition 14. If \( H \) and \( K \) are subgroups of a group, \( {HK} \) is a subgroup if and only if \( {HK} = {KH} \) . | Proof: Assume first that \( {HK} = {KH} \) and let \( a, b \in {HK} \) . We prove \( a{b}^{-1} \in {HK} \) so \( {HK} \) is a subgroup by the subgroup criterion. Let\n\n\[ a = {h}_{1}{k}_{1}\;\text{ and }\;b = {h}_{2}{k}_{2}, \]\n\nfor some \( {h}_{1},{h}_{2} \in H \) and \( {k}_{1},{k}_{2} \in K \) . Thus \( {b}^{-1} ... | Yes |
Corollary 15. If \( H \) and \( K \) are subgroups of \( G \) and \( H \leq {N}_{G}\left( K\right) \), then \( {HK} \) is a subgroup of \( G \) . In particular, if \( K \trianglelefteq G \) then \( {HK} \leq G \) for any \( H \leq G \) . | Proof: We prove \( {HK} = {KH} \) . Let \( h \in H, k \in K \) . By assumption, \( {hk}{h}^{-1} \in K \) ,\n\nhence\n\n\[ \n{hk} = \left( {{hk}{h}^{-1}}\right) h \in {KH}.\n\]\n\nThis proves \( {HK} \subseteq {KH} \) . Similarly, \( {kh} = h\left( {{h}^{-1}{kh}}\right) \in {HK} \), proving the reverse containment. The ... | Yes |
Corollary 17. Let \( \varphi : G \rightarrow H \) be a homomorphism of groups.\n\n(1) \( \varphi \) is injective if and only if \( \ker \varphi = 1 \) . | Proof: Exercise. | No |
Theorem 18. (The Second or Diamond Isomorphism Theorem) Let \( G \) be a group, let \( A \) and \( B \) be subgroups of \( G \) and assume \( A \leq {N}_{G}\left( B\right) \) . Then \( {AB} \) is a subgroup of \( G \) , \( B \trianglelefteq {AB}, A \cap B \trianglelefteq A \) and \( {AB}/B \cong A/A \cap B \) . | Proof: By Corollary 15, \( {AB} \) is a subgroup of \( G \) . Since \( A \leq {N}_{G}\left( B\right) \) by assumption and \( B \leq {N}_{G}\left( B\right) \) trivially, it follows that \( {AB} \leq {N}_{G}\left( B\right) \), i.e., \( B \) is a normal subgroup of the subgroup \( {AB} \) .\n\nSince \( B \) is normal in \... | Yes |
Theorem 19. (The Third Isomorphism Theorem) Let \( G \) be a group and let \( H \) and \( K \) be normal subgroups of \( G \) with \( H \leq K \). Then \( K/H \trianglelefteq G/H \) and\n\n\[ \left( {G/H}\right) /\left( {K/H}\right) \cong G/K\text{.} \] | Proof: We leave as an easy exercise the verification that \( K/H \trianglelefteq G/H \). Define\n\n\[ \varphi : G/H \rightarrow G/K \]\n\n\[ \left( {gH}\right) \mapsto {gK}\text{.} \]\n\nTo show \( \varphi \) is well defined suppose \( {g}_{1}H = {g}_{2}H \). Then \( {g}_{1} = {g}_{2}h \), for some \( h \in H \). Becau... | No |
Theorem 20. (The Fourth or Lattice Isomorphism Theorem) Let \( G \) be a group and let \( N \) be a normal subgroup of \( G \) . Then there is a bijection from the set of subgroups \( A \) of \( G \) which contain \( N \) onto the set of subgroups \( \bar{A} = A/N \) of \( G/N \) . In particular, every subgroup of \( \... | Proof: The complete preimage of a subgroup in \( G/N \) is a subgroup of \( G \) by Exercise 1 of Section 1. The numerous details of the theorem to check are all completely straightforward. We therefore leave the proof of this theorem to the exercises. | No |
Proposition 21. If \( G \) is a finite abelian group and \( p \) is a prime dividing \( \left| G\right| \), then \( G \) contains an element of order \( p \) . | Proof: The proof proceeds by induction on \( \left| G\right| \), namely, we assume the result is valid for every group whose order is strictly smaller than the order of \( G \) and then prove the result valid for \( G \) (this is sometimes referred to as complete induction). Since \( \left| G\right| > 1 \), there is an... | Yes |
Theorem 22. (Jordan-Hölder) Let \( G \) be a finite group with \( G \neq 1 \) . Then\n\n(1) \( G \) has a composition series and\n\n(2) The composition factors in a composition series are unique, namely, if \( 1 = {N}_{0} \leq {N}_{1} \leq \cdots \leq {N}_{r} = G \) and \( 1 = {M}_{0} \leq {M}_{1} \leq \cdots \leq {M}_... | Proof: This is fairly straightforward. Since we shall not explicitly use this theorem to prove others in the text we outline the proof in a series of exercises at the end of this section. | No |
Proposition 23. The map \( \epsilon : {S}_{n} \rightarrow \{ \pm 1\} \) is a homomorphism (where \( \{ \pm 1\} \) is a multiplicative version of the cyclic group of order 2). | Proof: By definition,\n\n\[ \left( {\tau \sigma }\right) \left( \Delta \right) = \mathop{\prod }\limits_{{1 \leq i < j \leq n}}\left( {{x}_{{\tau \sigma }\left( i\right) } - {x}_{{\tau \sigma }\left( j\right) }}\right) \]\n\nSuppose that \( \sigma \left( \Delta \right) \) has exactly \( k \) factors of the form \( {x}_... | Yes |
Proposition 25. The permutation \( \sigma \) is odd if and only if the number of cycles of even length in its cycle decomposition is odd. | For example, \( \sigma = \left( {123456}\right) \left( {789}\right) \left( {1011}\right) \left( {12131415}\right) \left( {161718}\right) \) has 3 cycles of even length, so \( \epsilon \left( \sigma \right) = - 1 \) . On the other hand, \( \tau = \left( {1128104}\right) \left( {213}\right) \left( {5117}\right) \left( {6... | No |
Proposition 2. Let \( G \) be a group acting on the nonempty set \( A \) . The relation on \( A \) defined by\n\n\[ a \sim b\;\text{ if and only if }\;a = g \cdot b\text{ for some }g \in G \]\n\nis an equivalence relation. For each \( a \in A \), the number of elements in the equivalence class containing \( a \) is \( ... | Proof: We first prove \( \sim \) is an equivalence relation. By axiom 2 of an action, \( a = 1 \cdot a \) for all \( a \in A \), i.e., \( a \sim a \) and the relation is reflexive. If \( a \sim b \), then \( a = g \cdot b \) for some \( b \in G \) so that\n\n\[ {g}^{-1} \cdot a = {g}^{-1} \cdot \left( {g \cdot b}\right... | Yes |
Theorem 3. Let \( G \) be a group, let \( H \) be a subgroup of \( G \) and let \( G \) act by left multiplication on the set \( A \) of left cosets of \( H \) in \( G \) . Let \( {\pi }_{H} \) be the associated permutation representation afforded by this action. Then\n\n(1) \( G \) acts transitively on \( A \)\n\n(2) ... | Proof: To see that \( G \) acts transitively on \( A \), let \( {aH} \) and \( {bH} \) be any two elements of \( A \), and let \( g = b{a}^{-1} \) . Then \( g \cdot {aH} = \left( {b{a}^{-1}}\right) {aH} = {bH} \), and so the two arbitrary elements \( {aH} \) and \( {bH} \) of \( A \) lie in the same orbit, which proves... | Yes |
Corollary 4. (Cayley's Theorem) Every group is isomorphic to a subgroup of some symmetric group. If \( G \) is a group of order \( n \), then \( G \) is isomorphic to a subgroup of \( {S}_{n} \) . | Proof: Let \( H = 1 \) and apply the preceding theorem to obtain a homomorphism of \( G \) into \( {S}_{G} \) (here we are identifying the cosets of the identity subgroup with the elements of \( G \) ). Since the kernel of this homomorphism is contained in \( H = 1, G \) is isomorphic to its image in \( {S}_{G} \) . | No |
Corollary 5. If \( G \) is a finite group of order \( n \) and \( p \) is the smallest prime dividing \( \left| G\right| \) , then any subgroup of index \( p \) is normal. | Proof: Suppose \( H \leq G \) and \( \left| {G : H}\right| = p \) . Let \( {\pi }_{H} \) be the permutation representation afforded by multiplication on the set of left cosets of \( H \) in \( G \), let \( K = \ker {\pi }_{H} \) and let \( \left| {H : K}\right| = k \) . Then \( \left| {G : K}\right| = \left| {G : H}\ri... | Yes |
Proposition 6. The number of conjugates of a subset \( S \) in a group \( G \) is the index of the normalizer of \( S,\left| {G : {N}_{G}\left( S\right) }\right| \) . In particular, the number of conjugates of an element \( s \) of \( G \) is the index of the centralizer of \( s,\left| {G : {C}_{G}\left( s\right) }\rig... | Proof: The second assertion of the proposition follows from the observation that \( {N}_{G}\left( {\{ s\} }\right) = {C}_{G}\left( s\right) \) . | No |
Theorem 7. (The Class Equation) Let \( G \) be a finite group and let \( {g}_{1},{g}_{2},\ldots ,{g}_{r} \) be representatives of the distinct conjugacy classes of \( G \) not contained in the center \( Z\left( G\right) \) of \( G \) . Then\n\n\[ \left| G\right| = \left| {Z\left( G\right) }\right| + \mathop{\sum }\limi... | Proof: As noted in Example 2 above the element \( \{ x\} \) is a conjugacy class of size 1 if and only if \( x \in Z\left( G\right) \), since then \( {gx}{g}^{-1} = x \) for all \( g \in G \) . Let \( Z\left( G\right) = \left\{ {1,{z}_{2},\ldots ,{z}_{m}}\right\} \) , let \( {\mathcal{K}}_{1},{\mathcal{K}}_{2},\ldots ,... | Yes |
Theorem 8. If \( p \) is a prime and \( P \) is a group of prime power order \( {p}^{\alpha } \) for some \( \alpha \geq 1 \) , then \( P \) has a nontrivial center: \( Z\left( P\right) \neq 1 \) . | Proof: By the class equation\n\n\[ \left| P\right| = \left| {Z\left( P\right) }\right| + \mathop{\sum }\limits_{{i = 1}}^{r}\left| {P : {C}_{P}\left( {g}_{i}\right) }\right| \]\n\nwhere \( {g}_{1},\ldots ,{g}_{r} \) are representatives of the distinct non-central conjugacy classes. By definition, \( {C}_{P}\left( {g}_{... | Yes |
Corollary 9. If \( \\left| P\\right| = {p}^{2} \) for some prime \( p \), then \( P \) is abelian. More precisely, \( P \) is isomorphic to either \( {Z}_{{p}^{2}} \) or \( {Z}_{p} \\times {Z}_{p} \) . | Proof: Since \( Z\\left( P\\right) \\neq 1 \) by the theorem, it follows that \( P/Z\\left( P\\right) \) is cyclic. By Exercise 36, Section 3.1, \( P \) is abelian. If \( P \) has an element of order \( {p}^{2} \), then \( P \) is cyclic. Assume therefore that every nonidentity element of \( P \) has order \( p \) . Le... | Yes |
Proposition 10. Let \( \sigma ,\tau \) be elements of the symmetric group \( {S}_{n} \) and suppose \( \sigma \) has cycle decomposition\n\n\[ \left( {{a}_{1}{a}_{2}\ldots {a}_{{k}_{1}}}\right) \left( {{b}_{1}{b}_{2}\ldots {b}_{{k}_{2}}}\right) \ldots \]\n\nThen \( {\tau \sigma }{\tau }^{-1} \) has cycle decomposition\... | Proof: Observe that if \( \sigma \left( i\right) = j \), then\n\n\[ {\tau \sigma }{\tau }^{-1}\left( {\tau \left( i\right) }\right) = \tau \left( j\right) \]\n\nThus, if the ordered pair \( i, j \) appears in the cycle decomposition of \( \sigma \), then the ordered pair \( \tau \left( i\right) ,\tau \left( j\right) \)... | Yes |
Proposition 11. Two elements of \( {S}_{n} \) are conjugate in \( {S}_{n} \) if and only if they have the same cycle type. The number of conjugacy classes of \( {S}_{n} \) equals the number of partitions of \( n \) . | Proof: By Proposition 10, conjugate permutations have the same cycle type. Conversely, suppose the permutations \( {\sigma }_{1} \) and \( {\sigma }_{2} \) have the same cycle type. Order the cycles in nondecreasing length, including 1-cycles (if several cycles of \( {\sigma }_{1} \) and \( {\sigma }_{2} \) have the sa... | Yes |
Corollary 14. If \( K \) is any subgroup of the group \( G \) and \( g \in G \), then \( K \cong {gK}{g}^{-1} \) . Conjugate elements and conjugate subgroups have the same order. | Proof: Letting \( G = H \) in the proposition shows that conjugation by \( g \in G \) is an automorphism of \( G \), from which the corollary follows. | No |
For any subgroup \( H \) of a group \( G \), the quotient group \( {N}_{G}\left( H\right) /{C}_{G}\left( H\right) \) is isomorphic to a subgroup of \( \operatorname{Aut}\left( H\right) \) . In particular, \( G/Z\left( G\right) \) is isomorphic to a subgroup of \( \operatorname{Aut}\left( G\right) \) . | Proof: Since \( H \) is a normal subgroup of the group \( {N}_{G}\left( H\right) \), Proposition 13 (applied with \( {N}_{G}\left( H\right) \) playing the role of \( G \) ) implies the first assertion. The second assertion is the special case when \( H = G \), in which case \( {N}_{G}\left( G\right) = G \) and \( {C}_{... | Yes |
Proposition 16. The automorphism group of the cyclic group of order \( n \) is isomorphic to \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \), an abelian group of order \( \varphi \left( n\right) \) (where \( \varphi \) is Euler’s function). | Proof: Let \( x \) be a generator of the cyclic group \( {Z}_{n} \) . If \( \psi \in \operatorname{Aut}\left( {Z}_{n}\right) \), then \( \psi \left( x\right) = {x}^{a} \) for some \( a \in \mathbb{Z} \) and the integer \( a \) uniquely determines \( \psi \) . Denote this automorphism by \( {\psi }_{a} \) . As usual, si... | Yes |
Lemma 19. Let \( P \in {\operatorname{Syl}}_{p}\left( G\right) \) . If \( Q \) is any \( p \) -subgroup of \( G \), then \( Q \cap {N}_{G}\left( P\right) = Q \cap P \) . | Proof: Let \( H = {N}_{G}\left( P\right) \cap Q \) . Since \( P \leq {N}_{G}\left( P\right) \) it is clear that \( P \cap Q \leq H \), so we must prove the reverse inclusion. Since by definition \( H \leq Q \), this is equivalent to showing \( H \leq P \) . We do this by demonstrating that \( {PH} \) is a \( p \) -subg... | Yes |
Corollary 20. Let \( P \) be a Sylow \( p \) -subgroup of \( G \) . Then the following are equivalent:\n\n(1) \( P \) is the unique Sylow \( p \) -subgroup of \( G \), i.e., \( {n}_{p} = 1 \)\n\n(2) \( P \) is normal in \( G \)\n\n(3) \( P \) is characteristic in \( G \)\n\n(4) All subgroups generated by elements of \(... | Proof: If (1) holds, then \( {gP}{g}^{-1} = P \) for all \( g \in G \) since \( {gP}{g}^{-1} \in {Sy}{l}_{p}\left( G\right) \), i.e., \( P \) is normal in \( G \) . Hence (1) implies (2). Conversely, if \( P \trianglelefteq G \) and \( Q \in {\operatorname{Syl}}_{p}\left( G\right) \), then by Sylow’s Theorem there exis... | Yes |
Proposition 21. If \( \left| G\right| = {60} \) and \( G \) has more than one Sylow 5-subgroup, then \( G \) is simple. | Proof: Suppose by way of contradiction that \( \left| G\right| = {60} \) and \( {n}_{5} > 1 \) but that there exists \( H \) a normal subgroup of \( G \) with \( H \neq 1 \) or \( G \) . By Sylow’s Theorem the only possibility for \( {n}_{5} \) is 6 . Let \( P \in {\operatorname{Syl}}_{5}\left( G\right) \), so that \( ... | Yes |
Proposition 23. If \( G \) is a simple group of order 60, then \( G \cong {A}_{5} \) . | Proof: Let \( G \) be a simple group of order 60, so \( {n}_{2} = 3,5 \) or 15 . Let \( P \in {\operatorname{Syl}}_{2}\left( G\right) \) and let \( N = {N}_{G}\left( P\right) \), so \( \left| {G : N}\right| = {n}_{2} \) . First observe that \( G \) has no proper subgroup \( H \) of index less that 5, as follows: if \( ... | Yes |
Proposition 1. If \( {G}_{1},\ldots ,{G}_{n} \) are groups, their direct product is a group of order \( \left| {G}_{1}\right| \left| {G}_{2}\right| \cdots \left| {G}_{n}\right| \) (if any \( {G}_{i} \) is infinite, so is the direct product). | Proof: Let \( G = {G}_{1} \times {G}_{2} \times \cdots \times {G}_{n} \) . The proof that the group axioms hold for \( G \) is straightforward since each axiom is a consequence of the fact that the same axiom holds in each factor, \( {G}_{i} \), and the operation on \( G \) is defined componentwise. For example, the as... | Yes |
Proposition 2. Let \( {G}_{1},{G}_{2},\ldots ,{G}_{n} \) be groups and let \( G = {G}_{1} \times \cdots \times {G}_{n} \) be their direct product.\n\n(1) For each fixed \( i \) the set of elements of \( G \) which have the identity of \( {G}_{j} \) in the \( {j}^{\text{th }} \) position for all \( j \neq i \) and arbit... | Proof: (1) Since the operation in \( G \) is defined componentwise, it follows easily from the subgroup criterion that \( \left\{ {\left( {1,1,\ldots ,1,{g}_{i},1,\ldots ,1}\right) \mid {g}_{i} \in {G}_{i}}\right\} \) is a subgroup of \( G \) . Furthermore, the map \( {g}_{i} \mapsto \left( {1,1,\ldots ,1,{g}_{i},1,\ld... | Yes |
Theorem 3. (Fundamental Theorem of Finitely Generated Abelian Groups) Let \( G \) be a finitely generated abelian group. Then\n\n(1)\n\n\[ G \cong {\mathbb{Z}}^{r} \times {Z}_{{n}_{1}} \times {Z}_{{n}_{2}} \times \cdots \times {Z}_{{n}_{s}}, \]\n\nfor some integers \( r,{n}_{1},{n}_{2},\ldots ,{n}_{s} \) satisfying the... | Proof: We shall derive this theorem in Section 12.1 as a consequence of a more general classification theorem. For finite groups we shall give an alternate proof at the end of Section 6.1. | No |
Proposition 6. Let \( m, n \in {\mathbb{Z}}^{ + } \). (1) \( {Z}_{m} \times {Z}_{n} \cong {Z}_{mn} \) if and only if \( \left( {m, n}\right) = 1 \) . | Proof: Since (2) is an easy exercise using (1) and induction on \( k \), we concentrate on proving (1). Let \( {Z}_{m} = \langle x\rangle ,{Z}_{n} = \langle y\rangle \) and let \( l = \) l.c.m. \( \left( {m, n}\right) \) . Note that \( l = {mn} \) if and only if \( \left( {m, n}\right) = 1 \) . Let \( {x}^{a}{y}^{b} \)... | No |
Proposition 7. Let \( G \) be a group, let \( x, y \in G \) and let \( H \leq G \) . Then\n\n(1) \( {xy} = {yx}\left\lbrack {x, y}\right\rbrack \) (in particular, \( {xy} = {yx} \) if and only if \( \left\lbrack {x, y}\right\rbrack = 1 \) ). | Proof: (1) This is immediate from the definition of \( \left\lbrack {x, y}\right\rbrack \) . | No |
Proposition 8. Let \( H \) and \( K \) be subgroups of the group \( G \) . The number of distinct ways of writing each element of the set \( {HK} \) in the form \( {hk} \), for some \( h \in H \) and \( k \in K \) is \( \left| {H \cap K}\right| \) . In particular, if \( H \cap K = 1 \), then each element of \( {HK} \) ... | Proof: Exercise. | No |
Theorem 9. Suppose \( G \) is a group with subgroups \( H \) and \( K \) such that\n\n(1) \( H \) and \( K \) are normal in \( G \), and\n\n(2) \( H \cap K = 1 \).\n\nThen \( {HK} \cong H \times K \). | Proof: Observe that by hypothesis (1), H \( K \) is a subgroup of \( G \) (see Corollary 3.15). Let \( h \in H \) and let \( k \in K \) . Since \( H \trianglelefteq G,{k}^{-1}{hk} \in H \), so that \( {h}^{-1}\left( {{k}^{-1}{hk}}\right) \in H \) . Similarly, \( \left( {{h}^{-1}{k}^{-1}h}\right) k \in K \) . Since \( H... | Yes |
Theorem 10. Let \( H \) and \( K \) be groups and let \( \varphi \) be a homomorphism from \( K \) into Aut \( \left( H\right) \) . Let-denote the (left) action of \( K \) on \( H \) determined by \( \varphi \) . Let \( G \) be the set of ordered pairs \( \left( {h, k}\right) \) with \( h \in H \) and \( k \in K \) and... | Proof: It is straightforward to check that \( G \) is a group under this multiplication using the fact that \( \cdot \) is an action of \( K \) on \( H \) . For example, the associative law is verified as follows:\n\n\[ \left( {\left( {a, x}\right) \left( {b, y}\right) }\right) \left( {c, z}\right) = \left( {{ax} \cdot... | No |
Proposition 11. Let \( H \) and \( K \) be groups and let \( \varphi : K \rightarrow \operatorname{Aut}\left( H\right) \) be a homomorphism. Then the following are equivalent:\n\n(1) the identity (set) map between \( H \rtimes K \) and \( H \times K \) is a group homomorphism (hence an isomorphism)\n\n(2) \( \varphi \)... | Proof: \( \left( 1\right) \Rightarrow \left( 2\right) \) By definition of the group operation in \( H \rtimes K \)\n\n\[ \left( {{h}_{1},{k}_{1}}\right) \left( {{h}_{2},{k}_{2}}\right) = \left( {{h}_{1}{k}_{1} \cdot {h}_{2},{k}_{1}{k}_{2}}\right) \]\n\nfor all \( {h}_{1},{h}_{2} \in H \) and \( {k}_{1},{k}_{2} \in K \)... | Yes |
Theorem 12. Suppose \( G \) is a group with subgroups \( H \) and \( K \) such that\n\n(1) \( H \trianglelefteq G \), and\n\n(2) \( H \cap K = 1 \) .\n\nLet \( \varphi : K \rightarrow \operatorname{Aut}\left( H\right) \) be the homomorphism defined by mapping \( k \in K \) to the automorphism of left conjugation by \( ... | Proof: Note that since \( H \trianglelefteq G,{HK} \) is a subgroup of \( G \) . By Proposition 8 every element of \( {HK} \) can be written uniquely in the form \( {hk} \), for some \( h \in H \) and \( k \in K \) . Thus the map \( {hk} \mapsto \left( {h, k}\right) \) is a set bijection from \( {HK} \) onto \( H \rtim... | Yes |
Let \( p \) be a prime and let \( P \) be a group of order \( {p}^{a}, a \geq 1 \) . Then\n\n(1) The center of \( P \) is nontrivial: \( Z\left( P\right) \neq 1 \) .\n\n(2) If \( H \) is a nontrivial normal subgroup of \( P \) then \( H \) intersects the center non-trivially: \( H \cap Z\left( P\right) \neq 1 \) . In p... | These results rely ultimately on the class equation and it may be useful for the reader to review Section 4.3.\n\nPart 1 is Theorem 8 of Chapter 4 and is also the special case of part 2 when \( H = P \) . We therefore begin by proving (2); we shall not quote Theorem 8 of Chapter 4 although the argument that follows is ... | Yes |
Proposition 2. Let \( p \) be a prime and let \( P \) be a group of order \( {p}^{a} \) . Then \( P \) is nilpotent of nilpotence class at most \( a - 1 \) . | Proof: For each \( i \geq 0, P/{Z}_{i}\left( P\right) \) is a \( p \) -group, so\n\n\[ \text{if}\left| {P/{Z}_{i}\left( P\right) }\right| > 1\text{then}Z\left( {P/{Z}_{i}\left( P\right) }\right) \neq 1 \]\n\nby Theorem 1(1). Thus if \( {Z}_{i}\left( P\right) \neq G \) then \( \left| {{Z}_{i + 1}\left( P\right) }\right|... | Yes |
Theorem 3. Let \( G \) be a finite group, let \( {p}_{1},{p}_{2},\ldots ,{p}_{s} \) be the distinct primes dividing its order and let \( {P}_{i} \in {\operatorname{Syl}}_{{p}_{i}}\left( G\right) ,1 \leq i \leq s \) . Then the following are equivalent:\n\n(1) \( G \) is nilpotent\n\n(2) if \( H < G \) then \( H < {N}_{G... | Proof: The proof that (1) implies (2) is the same argument as for \( p \) -groups - the only fact we needed was if \( G \) is nilpotent then so is \( G/Z\left( G\right) \) - so the details are omitted (cf. the exercises).\n\nTo show that (2) implies (3) let \( P = {P}_{i} \) for some \( i \) and let \( N = {N}_{G}\left... | No |
Proposition 5. If \( G \) is a finite group such that for all positive integers \( n \) dividing its order, \( G \) contains at most \( n \) elements \( x \) satisfying \( {x}^{n} = 1 \), then \( G \) is cyclic. | Proof: Let \( \left| G\right| = {p}_{1}^{{\alpha }_{1}}\cdots {p}_{s}^{{\alpha }_{s}} \) and let \( {P}_{i} \) be a Sylow \( {p}_{i} \) -subgroup of \( G \) for \( i = 1,2,\ldots, s \) . Since \( {p}_{i}^{{\alpha }_{i}}\left| \right| G| \) and the \( {p}_{i}^{{\alpha }_{i}} \) elements of \( {P}_{i} \) are solutions of... | Yes |
Proposition 6. (Frattini’s Argument) Let \( G \) be a finite group, let \( H \) be a normal subgroup of \( G \) and let \( P \) be a Sylow \( p \) -subgroup of \( H \) . Then \( G = H{N}_{G}\left( P\right) \) and \( \left| {G : H}\right| \) divides \( \left| {{N}_{G}\left( P\right) }\right| \) . | Proof: By Corollary 3.15, \( H{N}_{G}\left( P\right) \) is a subgroup of \( G \) and \( H{N}_{G}\left( P\right) = {N}_{G}\left( P\right) H \) since \( H \) is a normal subgroup of \( G \) . Let \( g \in G \) . Since \( {P}^{g} \leq {H}^{g} = H \), both \( P \) and \( {P}^{g} \) are Sylow \( p \) -subgroups of \( H \) .... | Yes |
Proposition 7. A finite group is nilpotent if and only if every maximal subgroup is normal. | Proof: Let \( G \) be a finite nilpotent group and let \( M \) be a maximal subgroup of \( G \) . As in the proof of Theorem 1, since \( M < {N}_{G}\left( M\right) \) (by Theorem 3(2)) maximality of \( M \) forces \( {N}_{G}\left( M\right) = G \), i.e., \( M \trianglelefteq G \) . Conversely, assume every maximal subgr... | Yes |
Theorem 8. A group \( G \) is nilpotent if and only if \( {G}^{n} = 1 \) for some \( n \geq 0 \) . More precisely, \( G \) is nilpotent of class \( c \) if and only if \( c \) is the smallest nonnegative integer such that \( {G}^{c} = 1 \) . If \( G \) is nilpotent of class \( c \) then\n\n\[ \n{Z}_{i}\left( G\right) \... | Proof: This is proved by a straightforward induction on the length of either the upper or lower central series. | No |
Theorem 9. A group \( G \) is solvable if and only if \( {G}^{\left( n\right) } = 1 \) for some \( n \geq 0 \) . | Proof: Assume first that \( G \) is solvable and so possesses a series\n\n\[ 1 = {H}_{0} \trianglelefteq {H}_{1} \trianglelefteq \cdots \trianglelefteq {H}_{s} = G \]\n\nsuch that each factor \( {H}_{i + 1}/{H}_{i} \) is abelian. We prove by induction that \( {G}^{\left( i\right) } \leq {H}_{s - i} \) . This is true fo... | Yes |
Proposition 10. Let \( G \) and \( K \) be groups, let \( H \) be a subgroup of \( G \) and let \( \varphi : G \rightarrow K \) be a surjective homomorphism.\n\n(1) \( {H}^{\left( i\right) } \leq {G}^{\left( i\right) } \) for all \( i \geq 0 \) . In particular, if \( G \) is solvable, then so is \( H \), i.e., subgroup... | Proof: Part 1 follows from the observation that since \( H \leq G \), by definition of commutator subgroups, \( \left\lbrack {H, H}\right\rbrack \leq \left\lbrack {G, G}\right\rbrack \), i.e., \( {H}^{\left( 1\right) } \leq {G}^{\left( 1\right) } \) . Then, by induction,\n\n\[ \n{H}^{\left( i\right) } \leq {G}^{\left( ... | Yes |
Theorem 11. Let \( G \) be a finite group.\n\n(1) (Burnside) If \( \left| G\right| = {p}^{a}{q}^{b} \) for some primes \( p \) and \( q \), then \( G \) is solvable. | We shall prove Burnside’s Theorem in Chapter 19 and deduce Philip Hall’s generalization of it. | No |
Lemma 13. In a finite group \( G \) if \( {n}_{p} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), then there are distinct Sylow \( p \) -subgroups \( P \) and \( R \) of \( G \) such that \( P \cap R \) is of index \( p \) in both \( P \) and \( R \) (hence is normal in each). | Proof: The argument is an easy refinement of the proof of the congruence part of Sylow’s Theorem (cf. the exercises at the end of Section 4.5). Let \( P \) act by conjugation on the set \( {\operatorname{Syl}}_{p}\left( G\right) \) . Let \( {\mathcal{O}}_{1},\ldots ,{\mathcal{O}}_{s} \) be the orbits under this action ... | Yes |
Theorem 17. Let \( G \) be a group, \( S \) a set and \( \varphi : S \rightarrow G \) a set map. Then there is a unique group homomorphism \( \Phi : F\left( S\right) \rightarrow G \) such that the following diagram commutes: | \( \textit{Proof: Such a map }\Phi \textit{ must satisfy }\Phi \left( {{s}_{1}^{{\epsilon }_{1}}{s}_{2}^{{\epsilon }_{2}}\ldots {s}_{n}^{{\epsilon }_{n}}}\right) = \varphi {\left( {s}_{1}\right) }^{{\epsilon }_{1}}\varphi {\left( {s}_{2}\right) }^{{\epsilon }_{2}}\ldots \varphi {\left( {s}_{n}\right) }^{{\epsilon }_{n}... | Yes |
Corollary 18. \( F\left( S\right) \) is unique up to a unique isomorphism which is the identity map on the set \( S \) . | Proof: This follows from the universal property. Suppose \( F\left( S\right) \) and \( {F}^{\prime }\left( S\right) \) are two free groups generated by \( S \) . Since \( S \) is contained in both \( F\left( S\right) \) and \( {F}^{\prime }\left( S\right) \), we have natural injections \( S \hookrightarrow {F}^{\prime ... | Yes |
Proposition 1. Let \( R \) be a ring. Then\n\n(1) \( {0a} = {a0} = 0 \) for all \( a \in R \) .\n\n(2) \( \left( {-a}\right) b = a\left( {-b}\right) = - \left( {ab}\right) \) for all \( a, b \in R \) (recall \( - a \) is the additive inverse of\na).\n\n(3) \( \left( {-a}\right) \left( {-b}\right) = {ab} \) for all \( a... | Proof: These all follow from the distributive laws and cancellation in the additive group \( R \) . For example,(1) follows from \( {0a} = \left( {0 + 0}\right) a = {0a} + {0a} \) . The equality \( \left( {-a}\right) b = - \left( {ab}\right) \) in (2) follows from \( {ab} + \left( {-a}\right) b = \left( {a + \left( {-a... | No |
Proposition 2. Assume \( a, b \) and \( c \) are elements of any ring with \( a \) not a zero divisor. If \( {ab} = {ac} \), then either \( a = 0 \) or \( b = c \) (i.e., if \( a \neq 0 \) we can cancel the \( a \) ’s). In particular, if \( a, b, c \) are any elements in an integral domain and \( {ab} = {ac} \), then e... | Proof: If \( {ab} = {ac} \) then \( a\left( {b - c}\right) = 0 \) so either \( a = 0 \) or \( b - c = 0 \) . The second statement follows from the first and the definition of an integral domain. | Yes |
Corollary 3. Any finite integral domain is a field. | Proof: Let \( R \) be a finite integral domain and let \( a \) be a nonzero element of \( R \) . By the cancellation law the map \( x \mapsto {ax} \) is an injective function. Since \( R \) is finite this map is also surjective. In particular, there is some \( b \in R \) such that \( {ab} = 1 \), i.e., \( a \) is a uni... | Yes |
Proposition 4. Let \( R \) be an integral domain and let \( p\left( x\right), q\left( x\right) \) be nonzero elements of \( R\left\lbrack x\right\rbrack \) . Then\n\n(1) degree \( p\left( x\right) q\left( x\right) = \) degree \( p\left( x\right) + \) degree \( q\left( x\right) \),\n\n(2) the units of \( R\left\lbrack x... | Proof: If \( R \) has no zero divisors then neither does \( R\left\lbrack x\right\rbrack \) ; if \( p\left( x\right) \) and \( q\left( x\right) \) are polynomials with leading terms \( {a}_{n}{x}^{n} \) and \( {b}_{m}{x}^{m} \), respectively, then the leading term of \( p\left( x\right) q\left( x\right) \) is \( {a}_{n... | Yes |
(1) The image of \( \varphi \) is a subring of \( S \) . | Proof: (1) If \( {s}_{1},{s}_{2} \in \operatorname{im}\varphi \) then \( {s}_{1} = \varphi \left( {r}_{1}\right) \) and \( {s}_{2} = \varphi \left( {r}_{2}\right) \) for some \( {r}_{1},{r}_{2} \in R \) . Then \( \varphi \left( {{r}_{1} - {r}_{2}}\right) = {s}_{1} - {s}_{2} \) and \( \varphi \left( {{r}_{1}{r}_{2}}\rig... | Yes |
(1) (The First Isomorphism Theorem for Rings) If \( \varphi : R \rightarrow S \) is a homomorphism of rings, then the kernel of \( \varphi \) is an ideal of \( R \), the image of \( \varphi \) is a subring of \( S \) and \( R/\ker \varphi \) is isomorphic as a ring to \( \varphi \left( R\right) \) . | Proof: This is just a matter of collecting previous calculations. If \( I \) is the kernel of \( \varphi \), then the cosets (under addition) of \( I \) are precisely the fibers of \( \varphi \) . In particular, the cosets \( r + I, s + I \) and \( {rs} + I \) are the fibers of \( \varphi \) over \( \varphi \left( r\ri... | Yes |
Proposition 9. Let \( I \) be an ideal of \( R \). (1) \( I = R \) if and only if \( I \) contains a unit. | Proof: (1) If \( I = R \) then \( I \) contains the unit 1 . Conversely, if \( u \) is a unit in \( I \) with inverse \( v \), then for any \( r \in R \)\n\n\[ r = r \cdot 1 = r\left( {vu}\right) = \left( {rv}\right) u \in I \]\n\nhence \( R = I \) . | Yes |
Corollary 10. If \( R \) is a field then any nonzero ring homomorphism from \( R \) into another ring is an injection. | Proof: The kernel of a ring homomorphism is an ideal. The kernel of a nonzero homomorphism is a proper ideal hence is 0 by the proposition. | Yes |
Proposition 11. In a ring with identity every proper ideal is contained in a maximal ideal. | Proof: Let \( R \) be a ring with identity and let \( I \) be a proper ideal (so \( R \) cannot be the zero ring, i.e., \( 1 \neq 0 \) ). Let \( \mathcal{S} \) be the set of all proper ideals of \( R \) which contain \( I \) . Then \( \mathcal{S} \) is nonempty \( \left( {I \in \mathcal{S}}\right) \) and is partially o... | Yes |
Proposition 12. Assume \( R \) is commutative. The ideal \( M \) is a maximal ideal if and only if the quotient ring \( R/M \) is a field. | Proof: This follows from the Lattice Isomorphism Theorem together with Proposition 9(2). The ideal \( M \) is maximal if and only if there are no ideals \( I \) with \( M \subset I \subset R \) . By the Lattice Isomorphism Theorem the ideals of \( R \) containing \( M \) correspond bijectively with the ideals of \( R/M... | Yes |
Proposition 13. Assume \( R \) is commutative. Then the ideal \( P \) is a prime ideal in \( R \) if and only if the quotient ring \( R/P \) is an integral domain. | Proof: This proof is simply a matter of translating the definition of a prime ideal into the language of quotients. The ideal \( P \) is prime if and only if \( P \neq R \) and whenever \( {ab} \in P \), then either \( a \in P \) or \( b \in P \) . Use the bar notation for elements of \( R/P \) : \( \bar{r} = r + P \) ... | Yes |
Corollary 14. Assume \( R \) is commutative. Every maximal ideal of \( R \) is a prime ideal. | Proof: If \( M \) is a maximal ideal then \( R/M \) is a field by Proposition 12. A field is an integral domain so the corollary follows from Proposition 13. | Yes |
Let \( R \) be an integral domain and let \( Q \) be the field of fractions of \( R \). If a field \( F \) contains a subring \( {R}^{\prime } \) isomorphic to \( R \) then the subfield of \( F \) generated by \( {R}^{\prime } \) is isomorphic to \( Q \). | Proof: Let \( \varphi : R \cong {R}^{\prime } \subseteq F \) be a (ring) isomorphism of \( R \) to \( {R}^{\prime } \). In particular, \( \varphi : R \rightarrow F \) is an injective homomorphism from \( R \) into the field \( F \). Let \( \Phi : Q \rightarrow F \) be the extension of \( \varphi \) to \( Q \) as in the... | Yes |
Theorem 17. (Chinese Remainder Theorem) Let \( {A}_{1},{A}_{2},\ldots ,{A}_{k} \) be ideals in \( R \) . The map \( R \rightarrow R/{A}_{1} \times R/{A}_{2} \times \cdots \times R/{A}_{k}\; \) defined by \( \;r \mapsto \left( {r + {A}_{1}, r + {A}_{2},\ldots, r + {A}_{k}}\right) \) is a ring homomorphism with kernel \(... | Proof: We first prove this for \( k = 2 \) ; the general case will follow by induction. Let \( A = {A}_{1} \) and \( B = {A}_{2} \) . Consider the map \( \varphi : R \rightarrow R/A \times R/B \) defined by \( \varphi \left( r\right) = \left( {r{\;\operatorname{mod}\;A}, r{\;\operatorname{mod}\;B}}\right) \), where mod... | Yes |
Corollary 18. Let \( n \) be a positive integer and let \( {p}_{1}{}^{{\alpha }_{1}}{p}_{2}{}^{{\alpha }_{2}}\ldots {p}_{k}{}^{{\alpha }_{k}} \) be its factorization into powers of distinct primes. Then\n\n\[ \n\mathbb{Z}/n\mathbb{Z} \cong \left( {\mathbb{Z}/{p}_{1}{}^{{\alpha }_{1}}\mathbb{Z}}\right) \times \left( {\m... | If we compare orders on the two sides of this last isomorphism, we obtain the formula\n\n\[ \n\varphi \left( n\right) = \varphi \left( {{p}_{1}{}^{{\alpha }_{1}}}\right) \varphi \left( {{p}_{2}{}^{{\alpha }_{2}}}\right) \ldots \varphi \left( {{p}_{k}{}^{{\alpha }_{k}}}\right)\n\]\n\nfor the Euler \( \varphi \) -functio... | Yes |
Proposition 1. Every ideal in a Euclidean Domain is principal. More precisely, if \( I \) is any nonzero ideal in the Euclidean Domain \( R \) then \( I = \left( d\right) \), where \( d \) is any nonzero element of \( I \) of minimum norm. | Proof: If \( I \) is the zero ideal, there is nothing to prove. Otherwise let \( d \) be any nonzero element of \( I \) of minimum norm (such a \( d \) exists since the set \( \{ N\left( a\right) \mid a \in I\} \) has a minimum element by the Well Ordering of \( \mathbb{Z} \) ). Clearly \( \left( d\right) \subseteq I \... | Yes |
Proposition 3. Let \( R \) be an integral domain. If two elements \( d \) and \( {d}^{\prime } \) of \( R \) generate the same principal ideal, i.e., \( \left( d\right) = \left( {d}^{\prime }\right) \), then \( {d}^{\prime } = {ud} \) for some unit \( u \) in \( R \) . In particular, if \( d \) and \( {d}^{\prime } \) ... | Proof: This is clear if either \( d \) or \( {d}^{\prime } \) is zero so we may assume \( d \) and \( {d}^{\prime } \) are nonzero. Since \( d \in \left( {d}^{\prime }\right) \) there is some \( x \in R \) such that \( d = x{d}^{\prime } \) . Since \( {d}^{\prime } \in \left( d\right) \) there is some \( y \in R \) suc... | Yes |
Theorem 4. Let \( R \) be a Euclidean Domain and let \( a \) and \( b \) be nonzero elements of \( R \) . Let \( d = {r}_{n} \) be the last nonzero remainder in the Euclidean Algorithm for \( a \) and \( b \) described at the beginning of this chapter. Then\n\n(1) \( d \) is a greatest common divisor of \( a \) and \( ... | Proof: By Proposition 1, the ideal generated by \( a \) and \( b \) is principal so \( a, b \) do have a greatest common divisor, namely any element which generates the (principal) ideal \( \left( {a, b}\right) \) . Both parts of the theorem will follow therefore once we show \( d = {r}_{n} \) generates this ideal, i.e... | Yes |
Proposition 5. Let \( R \) be an integral domain that is not a field. If \( R \) is a Euclidean Domain then there are universal side divisors in \( R \) . | Proof: Suppose \( R \) is Euclidean with respect to some norm \( N \) and let \( u \) be an element of \( R - \widetilde{R} \) (which is nonempty since \( R \) is not a field) of minimal norm. For any \( x \in R \) , write \( x = {qu} + r \) where \( r \) is either 0 or \( N\left( r\right) < N\left( u\right) \) . In ei... | Yes |
Proposition 6. Let \( R \) be a Principal Ideal Domain and let \( a \) and \( b \) be nonzero elements of \( R \) . Let \( d \) be a generator for the principal ideal generated by \( a \) and \( b \) . Then\n\n(1) \( d \) is a greatest common divisor of \( a \) and \( b \)\n\n(2) \( d \) can be written as an \( R \) -l... | Proof: This is just Propositions 2 and 3. | No |
Proposition 7. Every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal. | Proof: Let \( \left( p\right) \) be a nonzero prime ideal in the Principal Ideal Domain \( R \) and let \( I = \left( m\right) \) be any ideal containing \( \left( p\right) \) . We must show that \( I = \left( p\right) \) or \( I = R \) . Now \( p \in \left( m\right) \) so \( p = {rm} \) for some \( r \in R \) . Since ... | Yes |
Corollary 8. If \( R \) is any commutative ring such that the polynomial ring \( R\left\lbrack x\right\rbrack \) is a Principal Ideal Domain (or a Euclidean Domain), then \( R \) is necessarily a field. | Proof: Assume \( R\left\lbrack x\right\rbrack \) is a Principal Ideal Domain. Since \( R \) is a subring of \( R\left\lbrack x\right\rbrack \) then \( R \) must be an integral domain (recall that \( R\left\lbrack x\right\rbrack \) has an identity if and only if \( R \) does). The ideal \( \left( x\right) \) is a nonzer... | Yes |
Proposition 9. The integral domain \( R \) is a P.I.D. if and only if \( R \) has a Dedekind-Hasse norm. | Proof: Let \( I \) be any nonzero ideal in \( R \) and let \( b \) be a nonzero element of \( I \) with \( N\left( b\right) \) minimal. Suppose \( a \) is any nonzero element in \( I \), so that the ideal \( \left( {a, b}\right) \) is contained in \( I \) . Then the Dedekind-Hasse condition on \( N \) and the minimalit... | No |
Proposition 11. In a Principal Ideal Domain a nonzero element is a prime if and only if it is irreducible. | Proof: We have shown above that prime implies irreducible. We must show conversely that if \( p \) is irreducible, then \( p \) is a prime, i.e., the ideal \( \left( p\right) \) is a prime ideal. If \( M \) is any ideal containing \( \left( p\right) \) then by hypothesis \( M = \left( m\right) \) is a principal ideal. ... | Yes |
Proposition 12. In a Unique Factorization Domain a nonzero element is a prime if and only if it is irreducible. | Proof: Let \( R \) be a Unique Factorization Domain. Since by Proposition 10, primes of \( R \) are irreducible it remains to prove that each irreducible element is a prime. Let \( p \) be an irreducible in \( R \) and assume \( p \mid {ab} \) for some \( a, b \in R \) ; we must show that \( p \) divides either \( a \)... | Yes |
Proposition 13. Let \( a \) and \( b \) be two nonzero elements of the Unique Factorization Domain \( R \) and suppose\n\n\[ a = u{p}_{1}{}^{{e}_{1}}{p}_{2}{}^{{e}_{2}}\cdots {p}_{n}{}^{{e}_{n}}\;\text{ and }\;b = v{p}_{1}{}^{{f}_{1}}{p}_{2}{}^{{f}_{2}}\cdots {p}_{n}{}^{{f}_{n}} \]\n\nare prime factorizations for \( a ... | Proof: Since the exponents of each of the primes occurring in \( d \) are no larger than the exponents occurring in the factorizations of both \( a \) and \( b, d \) divides both \( a \) and \( b \) . To show that \( d \) is a greatest common divisor, let \( c \) be any common divisor of \( a \) and \( b \) and let \( ... | Yes |
Corollary 15. (Fundamental Theorem of Arithmetic) The integers \( \mathbb{Z} \) are a Unique Factorization Domain. | Proof: The integers \( \mathbb{Z} \) are a Euclidean Domain, hence are a Unique Factorization Domain by the theorem. | Yes |
Corollary 16. Let \( R \) be a P.I.D. Then there exists a multiplicative Dedekind-Hasse norm on \( R \) . | Proof: If \( R \) is a P.I.D. then \( R \) is a U.F.D. Define the norm \( N \) by setting \( N\left( 0\right) = 0 \) , \( N\left( u\right) = 1 \) if \( u \) is a unit, and \( N\left( a\right) = {2}^{n} \) if \( a = {p}_{1}{p}_{2}\cdots {p}_{n} \) where the \( {p}_{i} \) are irreducibles in \( R \) (well defined since t... | Yes |
Lemma 17. The prime number \( p \in \mathbb{Z} \) divides an integer of the form \( {n}^{2} + 1 \) if and only if \( p \) is either 2 or is an odd prime congruent to 1 modulo 4 . | Proof: The statement for \( p = 2 \) is trivial since \( 2 \mid {1}^{2} + 1 \) . If \( p \) is an odd prime, note that \( p \mid {n}^{2} + 1 \) is equivalent to \( {n}^{2} = - 1 \) in \( \mathbb{Z}/p\mathbb{Z} \) . This in turn is equivalent to saying the residue class of \( n \) is of order 4 in the multiplicative gro... | Yes |
Let \( n \) be a positive integer and write\n\n\[ n = {2}^{k}{p}_{1}^{{a}_{1}}\ldots {p}_{r}^{{a}_{r}}{q}_{1}^{{b}_{1}}\ldots {q}_{s}^{{b}_{s}} \]\n\nwhere \( {p}_{1},\ldots ,{p}_{r} \) are distinct primes congruent to 1 modulo 4 and \( {q}_{1},\ldots ,{q}_{s} \) are distinct primes congruent to 3 modulo 4 . Then \( n ... | The first statement in the corollary was proved above. Assume now that \( {b}_{1},\ldots ,{b}_{s} \) are all even. For each prime \( {p}_{i} \) congruent to 1 modulo 4 write \( {p}_{i} = {\pi }_{i}\overline{{\pi }_{i}} \) for \( i = 1,2,\ldots, r \), where \( {\pi }_{i} \) and \( \overline{{\pi }_{i}} \) are irreducibl... | Yes |
Proposition 2. Let \( I \) be an ideal of the ring \( R \) and let \( \left( I\right) = I\left\lbrack x\right\rbrack \) denote the ideal of \( R\left\lbrack x\right\rbrack \) generated by \( I \) (the set of polynomials with coefficients in \( I \) ). Then\n\n\[ R\left\lbrack x\right\rbrack /\left( I\right) \cong \left... | Proof: There is a natural map \( \varphi : R\left\lbrack x\right\rbrack \rightarrow \left( {R/I}\right) \left\lbrack x\right\rbrack \) given by reducing each of the coefficients of a polynomial modulo \( I \) . The definition of addition and multiplication in these two rings shows that \( \varphi \) is a ring homomorph... | Yes |
Theorem 3. Let \( F \) be a field. The polynomial ring \( F\left\lbrack x\right\rbrack \) is a Euclidean Domain. Specifically, if \( a\left( x\right) \) and \( b\left( x\right) \) are two polynomials in \( F\left\lbrack x\right\rbrack \) with \( b\left( x\right) \) nonzero, then there are unique \( q\left( x\right) \) ... | Proof: If \( a\left( x\right) \) is the zero polynomial then take \( q\left( x\right) = r\left( x\right) = 0 \) . We may therefore assume \( a\left( x\right) \neq 0 \) and prove the existence of \( q\left( x\right) \) and \( r\left( x\right) \) by induction on \( n = \) degree \( a\left( x\right) \) . Let \( b\left( x\... | Yes |
Corollary 4. If \( F \) is a field, then \( F\left\lbrack x\right\rbrack \) is a Principal Ideal Domain and a Unique Factorization Domain. | Proof: This is immediate from the results of the last chapter. | No |
Proposition 5. (Gauss’ Lemma) Let \( R \) be a Unique Factorization Domain with field of fractions \( F \) and let \( p\left( x\right) \in R\left\lbrack x\right\rbrack \) . If \( p\left( x\right) \) is reducible in \( F\left\lbrack x\right\rbrack \) then \( p\left( x\right) \) is reducible in \( R\left\lbrack x\right\r... | Proof: The coefficients of the polynomials on the right hand side of the equation \( p\left( x\right) = A\left( x\right) B\left( x\right) \) are elements in the field \( F \), hence are quotients of elements from the Unique Factorization Domain \( R \) . Multiplying through by a common denominator for all these coeffic... | Yes |
Corollary 6. Let \( R \) be a Unique Factorization Domain, let \( F \) be its field of fractions and let \( p\left( x\right) \in R\left\lbrack x\right\rbrack \) . Suppose the greatest common divisor of the coefficients of \( p\left( x\right) \) is 1 . Then \( p\left( x\right) \) is irreducible in \( R\left\lbrack x\rig... | Proof: By Gauss’ Lemma above, if \( p\left( x\right) \) is reducible in \( F\left\lbrack x\right\rbrack \), then it is reducible in \( R\left\lbrack x\right\rbrack \) . Conversely, the assumption on the greatest common divisor of the coefficients of \( p\left( x\right) \) implies that if it is reducible in \( R\left\lb... | Yes |
Corollary 8. If \( R \) is a Unique Factorization Domain, then a polynomial ring in an arbitrary number of variables with coefficients in \( R \) is also a Unique Factorization Domain. | Proof: For finitely many variables, this follows by induction from Theorem 7, since a polynomial ring in \( n \) variables can be considered as a polynomial ring in one variable with coefficients in a polynomial ring in \( n - 1 \) variables. The general case follows from the definition of a polynomial ring in an arbit... | Yes |
Proposition 9. Let \( F \) be a field and let \( p\left( x\right) \in F\left\lbrack x\right\rbrack \) . Then \( p\left( x\right) \) has a factor of degree one if and only if \( p\left( x\right) \) has a root in \( F \), i.e., there is an \( \alpha \in F \) with \( p\left( \alpha \right) = 0 \) . | Proof: If \( p\left( x\right) \) has a factor of degree one, then since \( F \) is a field, we may assume the factor is monic, i.e., is of the form \( \left( {x - \alpha }\right) \) for some \( \alpha \in F \) . But then \( p\left( \alpha \right) = 0 \) . Conversely, suppose \( p\left( \alpha \right) = 0 \) . By the Di... | Yes |
A polynomial of degree two or three over a field \( F \) is reducible if and only if it has a root in \( F \) . | This follows immediately from the previous proposition, since a polynomial of degree two or three is reducible if and only if it has at least one linear factor. | Yes |
Proposition 11. Let \( p\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} \) be a polynomial of degree \( n \) with integer coefficients. If \( r/s \in \mathbb{Q} \) is in lowest terms (i.e., \( r \) and \( s \) are relatively prime integers) and \( r/s \) is a root of \( p\left( x\right) \),... | Proof: By hypothesis, \( p\left( {r/s}\right) = 0 = {a}_{n}{\left( r/s\right) }^{n} + {a}_{n - 1}{\left( r/s\right) }^{n - 1} + \cdots + {a}_{0} \) . Multiplying through by \( {s}^{n} \) gives\n\n\[ 0 = {a}_{n}{r}^{n} + {a}_{n - 1}{r}^{n - 1}s + \cdots + {a}_{0}{s}^{n}. \]\n\nThus \( {a}_{n}{r}^{n} = s\left( {-{a}_{n -... | Yes |
Proposition 12. Let \( I \) be a proper ideal in the integral domain \( R \) and let \( p\left( x\right) \) be a nonconstant monic polynomial in \( R\left\lbrack x\right\rbrack \) . If the image of \( p\left( x\right) \) in \( \left( {R/I}\right) \left\lbrack x\right\rbrack \) cannot be factored in \( \left( {R/I}\righ... | Proof: Suppose \( p\left( x\right) \) cannot be factored in \( \left( {R/I}\right) \left\lbrack x\right\rbrack \) but that \( p\left( x\right) \) is reducible in \( R\left\lbrack x\right\rbrack \) . As noted at the end of the preceding section this means there are monic, nonconstant polynomials \( a\left( x\right) \) a... | Yes |
Proposition 13. (Eisenstein’s Criterion) Let \( P \) be a prime ideal of the integral domain \( R \) and let \( f\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \) be a polynomial in \( R\left\lbrack x\right\rbrack \) (here \( n \geq 1 \) ). Suppose \( {a}_{n - 1},\ldots ,{a}_{1},{a}_{0... | Proof: Suppose \( f\left( x\right) \) were reducible, say \( f\left( x\right) = a\left( x\right) b\left( x\right) \) in \( R\left\lbrack x\right\rbrack \), where \( a\left( x\right) \) and \( b\left( x\right) \) are nonconstant polynomials. Reducing this equation modulo \( P \) and using the assumptions on the coeffici... | Yes |
Corollary 14. (Eisenstein’s Criterion for \( \mathbb{Z}\left\lbrack x\right\rbrack \) ) Let \( p \) be a prime in \( \mathbb{Z} \) and let \( f\left( x\right) = {x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \in \mathbb{Z}\left\lbrack x\right\rbrack, n \geq 1 \) . Suppose \( p \) divides \( {a}_{i} \) f... | Proof: This is simply a restatement of Proposition 13 in the case of the prime ideal \( \left( p\right) \) in \( \mathbb{Z} \) together with Corollary 6 . | Yes |
Proposition 15. The maximal ideals in \( F\left\lbrack x\right\rbrack \) are the ideals \( \left( {f\left( x\right) }\right) \) generated by irreducible polynomials \( f\left( x\right) \) . In particular, \( F\left\lbrack x\right\rbrack /\left( {f\left( x\right) }\right) \) is a field if and only if \( f\left( x\right)... | Proof: This follows from Proposition 7 of Section 8.2 applied to the Principal Ideal Domain \( F\left\lbrack x\right\rbrack \) . | Yes |
Proposition 16. Let \( g\left( x\right) \) be a nonconstant element of \( F\left\lbrack x\right\rbrack \) and let\n\n\[ g\left( x\right) = {f}_{1}{\left( x\right) }^{{n}_{1}}{f}_{2}{\left( x\right) }^{{n}_{2}}\cdots {f}_{k}{\left( x\right) }^{{n}_{k}} \]\n\nbe its factorization into irreducibles, where the \( {f}_{i}\l... | Proof: This follows from the Chinese Remainder Theorem (Theorem 7.17), since the ideals \( \left( {{f}_{i}{\left( x\right) }^{{n}_{i}}}\right) \) and \( \left( {{f}_{j}{\left( x\right) }^{{n}_{j}}}\right) \) are comaximal if \( {f}_{i}\left( x\right) \) and \( {f}_{j}\left( x\right) \) are distinct (they are relatively... | Yes |
Proposition 17. If the polynomial \( f\left( x\right) \) has roots \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{k} \) in \( F \) (not necessarily distinct), then \( f\left( x\right) \) has \( \left( {x - {\alpha }_{1}}\right) \cdots \left( {x - {\alpha }_{k}}\right) \) as a factor. In particular, a polynomial of d... | Proof: The first statement follows easily by induction from Proposition 9. Since linear factors are irreducible, the second statement follows since \( F\left\lbrack x\right\rbrack \) is a Unique Factorization Domain. | No |
Proposition 18. A finite subgroup of the multiplicative group of a field is cyclic. In particular, if \( F \) is a finite field, then the multiplicative group \( {F}^{ \times } \) of nonzero elements of \( F \) is a cyclic group. | Proof: We give a proof of this result using the Fundamental Theorem of Finitely Generated Abelian Groups (Theorem 3 in Section 5.2). A more number-theoretic proof is outlined in the exercises, or Proposition 5 in Section 6.1 may be used in place of the Fundamental Theorem. By the Fundamental Theorem, the finite subgrou... | No |
Corollary 19. Let \( p \) be a prime. The multiplicative group \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ \times } \) of nonzero residue classes \( {\;\operatorname{mod}\;p} \) is cyclic. | Proof: This is the multiplicative group of the finite field \( \mathbb{Z}/p\mathbb{Z} \) . | No |
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