text stringlengths 71 10k |
|---|
ded ideals. Theorem 16.10 Let I be an ideal of R. The factor group R/I is a ring with multiplication defined by (r + I)(s + I) = rs + I. Proof. We already know that R/I is an abelian group under addition. Let r+I and s+I be in R/I. We must show that the product (r+I)(s+I) = rs+I is independent of the choice of coset; th... |
system x ≡ a (mod m) x ≡ b (mod n) has a solution. If x1 and x2 are two solutions of the system, then x1 ≡ x2 (mod mn). Proof. The equation x ≡ a (mod m) has a solution since a + km satisfies the equation for all k ∈ Z. We must show that there exists an integer k1 such that a + k1m ≡ b (mod n). 258 CHAPTER 16 RINGS This... |
ove that 1R = 1S. 31. If we do not require the identity of a ring to be distinct from 0, we will not have a very interesting mathematical structure. Let R be a ring such that 1 = 0. Prove that R = {0}. 32. Let S be a subset of a ring R. Prove that there is a subring R of R that contains S. 33. Let R be a ring. Define th... |
ique integers q and r such that a = bq + r, where 0 ≤ r < b. The algorithm by which q and r are found is just long division. A similar theorem exists for polynomials. The division algorithm for polynomials has several important consequences. Since its proof is very similar to the corresponding proof for integers, it is... |
1(x) such that p |bj. Let α 1(x) and β obtained by reducing the coefficients of α1(x) and β1(x) modulo p. Since p | d, α 1(x) 1(x) is the zero polynomial and Zp[x] is an integral domain. Therefore, nor β d = 1 and the theorem is proven. 1(x) = 0 in Zp[x]. However, this is impossible since neither α 1(x)β 17.3 IRREDUCIBLE... |
and q(x) of polynomials. If d(x) = gcd(p(x), q(x)), find two polynomials a(x) and b(x) such that a(x)p(x) + b(x)q(x) = d(x). (a) p(x) = 7x3 + 6x2 − 8x + 4 and q(x) = x3 + x − 2, where p(x), q(x) ∈ Q[x] (b) p(x) = x3 + x2 − x + 1 and q(x) = x3 + x − 1, where p(x), q(x) ∈ Z2[x] (c) p(x) = x3 + x2 − 4x + 4 and q(x) = x3 +... |
ce [a1, b1] = [a2, b2] and [c1, d1] = [c2, d2], we know that a1b2 = b1a2 and c1d2 = d1c2. Therefore, (a1d1 + b1c1)(b2d2) = a1d1b2d2 + b1c1b2d2 = a1b2d1d2 + b1b2c1d2 = b1a2d1d2 + b1b2d1c2 = (b1d1)(a2d2 + b2c2). Lemma 18.3 The set of equivalence classes of S, FD, under the equivalence relation ∼, together with the operat... |
t in D that is not a unit. If a is irreducible, then we are done. If not, then there exists a factorization a = a1b1, where neither a1 nor b1 is a unit. Hence, a ⊂ a1. By Lemma 18.7, we know that a = a1; otherwise, a and a1 would be associates and b1 would be a unit, which would contradict our assumption. Now suppose t... |
n = s. Now rearrange the gi(x)’s so that fi(x) and gi(x) are associates for i = 1, . . . , n. Then there exist c1, . . . , cn and d1, . . . , dn in D such that (ci/di)fi(x) = gi(x) or cifi(x) = digi(x). The polynomials fi(x) and gi(x) are primitive; hence, ci and di are associates in D. Thus, a1 · · · am = ub1 · · · b... |
clusion, union, and intersection. Lattices are generalizations of order relations on algebraic spaces, such as set inclusion in set theory and inequality in the familiar number systems N, Z, Q, and R. Boolean algebras generalize the operations of intersection and union. Lattices and Boolean algebras have found applicat... |
∨ b) ∧ a] ∨ [(a ∨ b) ∧ c] = (a ∨ b) ∧ (a ∨ c). The converse follows directly from the Duality Principle. A Boolean algebra is a lattice B with a greatest element I and a smallest element O such that B is both distributive and complemented. The power set of X, P(X), is our prototype for a Boolean algebra. As it turns ou... |
w, a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c), is illustrated in Figure 19.5. If a is a switch, then a is the switch that is always open when a is closed and always closed when a is open. A circuit that is always closed is I in our algebra; a circuit that is always open is O. The laws for a ∧ a = O and a ∨ a = I are shown in Figu... |
form a vector space over R. Given vectors u = (u1, . . . , un) and v = (v1, . . . , vn) in Rn and α in R, we can define vector addition by u + v = (u1, . . . , un) + (v1, . . . , vn) = (u1 + v1, . . . , un + vn) and scalar multiplication by αu = α(u1, . . . , un) = (αu1, . . . , αun). Example 2. If F is a field, then F ... |
tors for V , then S is a basis for V . (b) If S = {v1, . . . , vn} spans V , then S is a basis for V . (c) If S = {v1, . . . , vk} is a set of linearly independent vectors for V with k < n, then there exist vectors vk+1, . . . , vn such that {v1, . . . , vk, vk+1, . . . , vn} is a basis for V . 12. Prove that any set o... |
, αn by F (α1, . . . , αn). If E = F (α) for some α ∈ E, then E is a simple extension of F . √ 2 and i are algebraic over Q since they are zeros of the Example 4. Both polynomials x2 − 2 and x2 +1, respectively. Clearly π and e are algebraic over the real numbers; however, it is a nontrivial fact that they are transce... |
5 i, 3√ 5 is the 5 is the real cube root of 5. We know that 5 i) : Q( 3√ √ 5i } is a basis for Q( 3 √ 5 )2} is a basis for Q( 3 √ 5 i) over Q( 3 5, 5 ). 5 ) over Q. Hence, a {1, √ 5 i is a zero of x6 + 5. We can show that this polynomial is Notice that 6 irreducible over Q using Eisenstein’s Criterion, where we let p =... |
-and-compass constructions from what is now high school geometry; that is, we are allowed to use only a straightedge and compass to solve them. The problems can be stated as follows. 1. Given an arbitrary angle, can one trisect the angle into three equal subangles using only a straightedge and compass? 21.3 GEOMETRIC C... |
nning of the twentieth century. Hilbert and Minkowski were both mathematicians at G¨ottingen University in Germany. G¨ottingen was truly one the most important centers of mathematical research during the last two centuries. The large number of exceptional mathematicians who studied there included Gauss, Dirichlet, Riem... |
ting field of xpn − x over Zp. 22.1 STRUCTURE OF A FINITE FIELD 361 Proof. Let f (x) = xpn − x and let F be the splitting field of f (x). Then by Lemma 22.4, f (x) has pn distinct zeros in F , since f (x) = pnxpn−1 − 1 = −1 is relatively prime to f (x). We claim that the roots of f (x) form a subfield of F . Certainly 0 a... |
n ideal in Z2[x] that contains xn − 1. By Theorem 17.12, we know that every ideal I in Z2[x] is a principal ideal, since Z2 is a field. Therefore, I = g(x) for some unique monic polynomial in Z2[x]. Since xn − 1 is contained in I, it must be the case that g(x) divides xn − 1. Consequently, every ideal C in Rn is of the ... |
rable over F , show that K is also separable over E. 18. Let E be an extension of a finite field F , where F has q elements. Let α ∈ E be algebraic over F of degree n. Prove that F (α) has qn elements. 19. Show that every finite extension of a finite field F is simple; that is, if E is a finite extension of a finite field F , ... |
: F (α)]. Consequently, there are [E : F ] = [E : F (α)][F (α) : F ] possible automorphisms of E that fix F , or |G(E/F )| = [E : F ]. Corollary 23.6 Let F be a finite field with a finite extension E such that [E : F ] = k. Then G(E/F ) is cyclic or order k. Proof. Let p be the characteristic of E and F and assume that th... |
F . 2. E is a splitting field over F of a separable polynomial. 3. F = EG for some finite group of automorphisms of E. Proof. (1) ⇒ (2). Let E be a finite, normal, separable extension of F . By the Primitive Element Theorem, we can find an α in E such that E = F (α). 23.2 THE FUNDAMENTAL THEOREM 385 Let f (x) be the minima... |
t will tell us whether or not a polynomial f (x) is solvable by radicals by examining the Galois group f (x). The easiest polynomial to solve by radicals is one of the form xn − a. As we discussed in Chapter 4, the roots of xn − 1 are called the nth roots of unity. These roots are a finite subgroup of the splitting field... |
sion of Q in which the extension field is contained. √ 2, 3 (d) G(Q( 2, i)/Q) √ √ (a) G(Q( √ (b) G(Q( 4 √ (c) G(Q( 30 )/Q) 5 )/Q) √ 3, 2, √ 5 )/Q) √ (e) G(Q( 6, i)/Q) 2. Determine the separability of each of the following polynomials. (a) x3 + 2x2 − x − 2 over Q (b) x4 + 2x2 + 1 over Q (c) x4 + x2 + 1 over Z3 (d) x3 + x... |
(c) 43 − 18i. (e) i. √ √ 16. (a) 17. (a) 3 + i. (c) −3. 2 cis(7π/4). (c) 2 √ 18. (a) (1 − i)/2. (c) 16(i − 22. (a) 292. (c) 1523. 27. |g ∩ h| = 1. 2 cis(π/4). (e) 3 cis(3π/2). √ 3 ). (e) −1/4. 31. The identity element in any group has finite order. Let g, h ∈ G have orders m and n, respectively. Since (g−1)m = e and (g... |
) 3, 4. 7. (2x + 1)2 = 1. 8. (a) Reducible. (c) Irreducible. 10. x2 + x + 8 = (x + 2)(x + 9) = (x + 7)(x + 4). 13. Z is not a field. 14. False. x2 + 1 = (x + 1)(x + 1). 16. Let φ : R → S be an isomorphism. Define φ : R[x] → S[x] by φ(a0 + a1x + · · · + anxn) = φ(a0) + φ(a1)x + · · · + φ(an)xn. 19. Define g(x) by g(x) = Φp... |
L using a publicly available DTD, and standard-conforming simple HTML, PostScript or PDF designed for human modification. Examples of transparent image formats include PNG, XCF and JPG. Opaque formats include proprietary formats that can be read and edited only by proprietary word processors, SGML or XML for which the D... |
or or publisher of that section if known, or else a unique number. Make the same adjustment to the section titles in the list of Invariant Sections in the license notice of the combined work. In the combination, you must combine any sections Entitled “History” in the various original documents, forming one section Enti... |
48, 166, 227 Cancellation law for groups, 47 for integral domains, 248 Cardano, Gerolamo, 282 Carmichael numbers, 113 Cauchy’s Theorem, 231 Cauchy, Augustin-Louis, 85 Cayley table, 44 Cayley’s Theorem, 148 Cayley, Arthur, 149 Center of a group, 55 of a ring, 265 Centralizer, 55 of a subgroup, 217 of an element, 167 Cha... |
fits in with the notation of Section 1.2b where we informally introduced an equivalence relation. An equivalence relation in X is intimately connected with a partition of X, i.e. a decom position of X into disjoint subsets of X such that every element of X belongs to some subset. Examples of partitions of {I, 2, 3, 4,... |
P is an image. (c) (a) 2a = 1; 5a = 1; 6a = 1. (b) 2, 5, 6 and 27 have no preimages. (c) 1 has every element of P as a preimage and no other element of P has a preimage. (a) 2a = 1; 5a = 1; 6a = 1. (b) 1 is a preimage of 2; 5, 6 and 7 have no preimages. (c) 2 has one preimage, namely 1. 1 has an infinite number of pre... |
1 --> 3. Also, 2 may have any of 3 images, either 1, 2 or 3. So we have in all 3 X 3 possibilities for the actions of mappings on 1 and 2. Then 3 can be sent into 1, 2 or 3, giving 3 X 3 X 3 = 27 possible mappings of {I, 2, 3} into itself. There are 3 X 2 X 1 = 6 possible one-to-one and onto mappings; for when we once ... |
3' = 1, since Sa ¥= t l . Hence a 0 (3 = a 0 (3' and sumption that a 0 (3 = a 0 (3' implies t(3 = 1 t l(3 = 2. Now if S E S, Sa 0 (3 = (Sa)(3 = 1 and (3 ¥= (3'. This contradicts the as (3 = (3'. Thus a must be onto. t(3' = 1 for all t E T, t ¥= tl Conversely, assume a is onto and we can find a set U and two mappings, (... |
looks like the usual multi plication tables. One often calls p. a multiplication in S. Thus when we talk about a multiplication /L in a set S, we mean that p. is a binary operation in S. There is a reverse procedure to the one described above. For example, suppose we start out with a table Then there is a natural way ... |
ation' ; then 1 . 1 = 1, 1· 2 = 3, 2· 2 = 1, 3· 2 = 2, etc. 26 Sec. 2.1) GROUPOIDS 27 These products look bizarre unless we recall that the notation employed is a shorthand version of (1,1)/.L = 1, (1,2),u = 3, (2,2)" = 1, (3,2),u = 2, etc. If we use the expression "the groupoid G," where G is a set, it is understood t... |
all mappings of {1, 2, 3} into itself contains 27 elements, and so is finite. In Example 3, (Q,o) is not finite as there are an infinite number of rational numbers. In Example 5 the set F is infinite. To show that F is not finite we construct an infinite number for all i = j. Therefore we have found an infinite number ... |
e are equal to 2 and the other is equal to 2 or 3, then (ab)e = 4 = a(be) because 2· 2 = 4 and 2· 3 = 2 = 3' 2. The follow ing calculations take care of the remaining cases: 3(3 • 33' 3)3 3(2 • 33' 2)3 2(3 • 32 • 3)3 3(3 • 2) = 3 • 2 = 2 = (1 • 2) = (3 • 3)2 1 is the identity element of G. The only elements which have ... |
eaves every element of jO'I = 1, M{1,2} unchanged; hence the ficst row and first column are easily written down. Since j = 1,2, and if u E M, 1 ,", then k(UUl) = (ku)ul = 1; thus 0'0'1 = Ul' Hence the second column consists of Ul' '-, ' Similarly UU3 = U:l, so the last eoiumn consists of U:l' We must still calculate 0'... |
e homomorphisms? (a) 1 ... a, 2 ... b, 3'" e (b) 1 ... a, 2'" a, 3'" a (e) 1 ..... a, 2 ... b, 3 ..... b (d) 1 ... b, 2 ... e, 3'" e (e) 1 ... b, 2 ... b, 3'" b (I) 1'" e, 2'" a, 3 ... b Solution: We use u to indicate the mapping in each case. (a) 1u2u = ab = a and (1' 2)u = 2u = b. u is not a homomorphism. (b) 1u2u = ... |
pimorphism from the groupoid G to the groupoid H, then H shares some of the properties of G. Theorem 2.6: Let 0 be an epimorphism from the groupoid G to the groupoid H. Then (a) if G is a groupoid with an identity 1, so is Hand 1() is the identity of H. Furthermore if f is an inverse of g in G, then (() is an inverse o... |
mutative and which are associative? 2.61. Define the following binary operation + in R+, the non-negative real numbers: a + b = the maximum of a and b, a, bE R+. Does (R+, +) have an identity? What elements have inverses? 2.62. Let (G, *) be the groupoid of Problem 2.56. What is the identity of (G, *)? Find an infinite... |
2 + 5b 2 a -b a2 + 5b2 + a2 + 5b2 R and so 1 _ ,----;; E S. The associativity of multiplication in S follows from associativity of multi a + bv-5 plication of complex numbers. 3.5. Let m be any fixed positive integer and let S = {O, 1, 2, ... , m -I}. Define a binary operation in S by aob a 0 b = r a+b if a+b<m if a +... |
he multiplication table for S3 is 0'1 0'1 0'2 ! 7'3 7'1 7'2 0'2 0'2 ! 0'1 7'2 7'3 7'1 1'1 1'2 1'3 1'1 1'2 1'3 1'2 1'3 1'1 1'3 ! 0'1 0'2 1'1 0'2 ! 0'1 1'2 0'1 ' 0'2 ! ! 0'1 0'2 1'1 7'2 7'3 The reader should check some of the entries, Note that 0'11'1 = 1'2 and 1'10'1 = 7'3' so that 0'11'1 # 7 10'1' Hence S3 is not commu... |
' is even, then a - l is even too, since (J'(J'-l = ~ is even. Thus if a, T E An, l E An. An is therefore a subgroup of 8 n. It is called the alternating group of degree n. fIT- As an illustration let us find the multiplication table for A 4 • From the list of elements of 84 given in Problem 3.21 we determine that the ... |
g and thus fla is a permutation of R2. Now (x, Y)flafl/; I = (ax, aY)fl/; I = (~x, ~ Y) = (x, Y)fla/b' Hence the set of all fla is a sub fla/b since group of SR2. c. Isometries of the plane Let E be the set R2 = R x R. If (XA, YA) = A, (XB, YB) = B are two elements of E, we define the distance between A and B as V(XA -... |
and C = (0,1), d(A(1, C(1) = d(B, C) = V2 ~ d(A, C). Hence (1 is not an isometry. d. Isometries are products of reflections, translations and rotations We will prove in this section that an isometry is determined uniquely by its action on any three points not all on a straight line. This enables us to prove that every... |
wing elements: (a) (b) (J1, the identity mapping of I, K(J1 = K, L(J1 = L, 1I1(J1 = M. (J2' the reflection in KN, K(J2 = K, L(J2 = M, M(J2 = L. Hence IIsl:=O 2. Let (J E Is. Since d(M, L) = V2 and the only two points of KLM which are a distance V2 apart are Land M, then either Sec. 3.4J GROUPS OF ISOMETRIES 75 (a) Mu =... |
e can show that u(-d, b, c, -a) u(a, b, c, d) = ,. Hence u(-d,b,c,-a) = u(a,b,c,d)-l. 3045. Prove u(a, b, c, d) E S"E for any choice of a, b, c, d such that ad - bc ¥ O. Solution: In Problem 3.44 we have seen that each u(a, b, c, d) has an inverse. By Theorem 2.4, page 36, any mapping of a set into itself which has an ... |
eserve distance whereas automorphisms of groupoids preserve groupoid multiplication. As the analog to Theorem 3.5, page 67, we have Theorem 3.13: The set A of all automorphisms of a groupoid G is a subgroup of SG, the symmetric group on G. Proof: I. t, the identity mapping, belongs to A; hence A =F ~. II. If a, f3 E A ... |
facts we see that if m (n "" 0) is any element in Q, then m a r= (m.!') a = mala = m' 1:.. = m. Therefore a is the i~entity mapping and is the only possible automorphism of Q. The automorphism group of Q is of order one. ±r ±r n n n n n 88 3.62. GROUPS AND SUBGROUPS [CHAP. 3 Find the automorphism group of F = {a + bV2 ... |
number of integers z}. Let B' = {x I x = (0, b), bE B}, C' {x I x = (0, c), c E C}. Prove that B' is a subgroup of Wand C' is a subgroup of B'. (Hard.) 3.84. Using the notation of the preceding problem, let W = {x I x = (m, c), where mE Z and c E C}. Prove W is a subgroup of W. (Hard.) SYMMETRIC GROUPS AND ALTERNATING ... |
on F and <f>(J == identity mapping on G, then (J and <f> are isomorphisms of F onto G and of G onto F respectively. Solution: (J is one-to-one, for if x(J == y(J, then x(J<f> == y(J<f>. But (J<f> is the identity on F. Hence x == y. Similarly <f> is one-to-one. Next let U E G; then U<f> E F. U¢(J == U; hence U is the i... |
are arbitrary integers. Now if it were true that aDam = am, then multiplication by aO leaves am unchanged. Hence we have only one choice in extending the exponent notation and retaining the law (4.1), namely putting aO = 1, the identity. Now if m = -n where n > 0, m + n = O. Because we want (4.1) to be satisfied, we mu... |
G and a nonzero integer n such that there are two elements a, bEG with an = bn but a ¥ b. Solution: Since G is abelian, so (ab-1)S = as(b-1)s = 1. Since G = Up(x) and the order of G is m, then ab- 1 = XT for some r, and (XT)S = 1. Hence x TS = 1 and m divides rs. But sand m are co-prime; then m divides r, say r = qm. ... |
s for finite groups. (ii) Cosets of a subgroup sometimes enable us to construct a new group from an old. We can also see how a group G is built up from one of its subgroups H and the group constructed from the cosets of H. (iii) The fundamental idea of a homomorphism can be re-interpreted in terms of the idea of a grou... |
d, a, b), e.g. if a = b = c = 1, d = 0, then a(I, 1, 0, 1) oF a(I, 0,1,1). Thus the left and right cosets of H in G do not coincide. We ask: when do the right and left cosets of a subgroup H in a group G coincide? Suppose every left coset of H is also a right coset of H in G. Let a E G. aH contains a, as does Ha. Sinc... |
roups In Section 4.3a we mentioned that the concept of a coset sometimes gives rise to a new group. This occurs when, and only when, the group is normal. Let G be a group and N <J G. Let us denote by GIN (read as "G over N", or "G factor N", or "the factor group of G by N") the set of right co sets of N in G. We turn G... |
bers defined by XfJ = 1 if x is even, and XfJ = -1 if x is odd. Find the kernel of fJ and examine the claim G/(Ker fJ) == GfJ. Solution: Ker fJ = {x I xfJ = I} = {x I x is even} and G/(Ker fJ) = {Ker fJ, Ker fJ + I}, so Ker fJ + 1 (Ker fJ + 1) + (Ker fJ + 1) = Ker fJ; hence Ker fJ + 1 is of order 2 generates G/(Ker fJ)... |
E H. Hence giO(gjO)-1 E S, i.e. kik j- l E S, from which Ski = Skj and i = j. Thus we have shown that all the co sets in (4.5) are distinct. Let g E G. Then gO E K and so gO E Ski for some integer i. Hence gO = Ski with l = s. Consequently ggi l is in the pre s E S. Consider x = ggil. xO = gO(giO)-1 = skiki image of S,... |
are B! = (MIN)A! = {Av A 3} and B2 = (MIN)A2 = {A!AZ' A 3A z} = {A 2, A 4 }. plication in the group (GIN)/(MIN) is calculated in the usual way for cosets, e.g., B2B2 = (MIN)A 2(MIN)A 2 = (MIN)(A 2A 2) = (MIN)A! = B! (A3A2' for example, is calculated as follows: A3A2 = (rN)(u2N) = (ru2)N = A 4.) Multi as A2A2 = (U2N)(U2... |
stigated the simplest class of groups, the cyclic groups. We know that there are cyclic groups of all orders, we know their subgroups, we know that they have as homomorphic images only cyclic groups, and we know whether any cyclic group G has as homomorphic image a given cyclic group. Furthermore the sub groups of cycl... |
oups of order 3 and s31 IGI. But (1 + 3k) I 15 implies k = O. Therefore G has one and only one subgroup of order 3. Similarly G has one and only one subgroup of order 5. These subgroups must be cyclic (Problem 4.48, page 110). Let HI = {I, a, a 2 the subgroup of order 5. H l nH2 = {I}, because an element #1 cannot have... |
h2 implies NH(A)hi = N H(A)h2. then hi E NfJ(A)h2, If NH(A)hi Therefore for some n E NH(A). i.e. hi = nh2 NH(A)h2, because n-iAn = A by definition of NH(A). Hence NH(A)hi = NH(A)h2 (t is clearly onto, so the proof is complete. h;lAhi = h;:lAh2. Most of our arguments are concerned with sets whose elements are subsets of... |
is complete. 138 FINITE GROUPS [CHAP. 5 The following gives a simple formula for the normalizer of a Sylow p-subgroup P in a subgroup H of G, where IHI is a power of p. It will be used in the proof of the second Sylow theorem. l .. emma 5.10: If G is a finite group, P a Sylow p-subgroup of G, and H is a subgroup of G ... |
1 are not isomorphic. Solution: Let G be the group defined in Problem 5.20 and let H be the group defined in Problem 5.21, for p an odd prime. Then by Problem 5.22, if g E G, gP = 1. But if G "'" H, it follows from Problem 5.24 that hp = 1 for all hE H. But by Problem 5.23, (0, l)p '7'= 1. Therefore G is not isomorphic... |
= IHI, because hlki = h2ki if and only if hi = hz. Therefore IHKI = IHkl1 + IHk21 + ... + IHknl since n = IK\/IHnKI. IHKI = n IHI = IHIIKI IHnKI To illustrate the use of Proposition 5.18, let G be a group of order 28 and HI and H2 subgroups of G of orders 7 and 4 respectively. HI n H 2 = {1}, because an element in HI ... |
ut Hand K are cyclic groups of order 2 and p respectively. Thus by Problem 5.34, G is cyclic of order 2p. (ii) Let K = gp(a) wher.e aP = 1. Since K is the only subgroup of order p, b (/:. K implies b2 = 1. Clearly, G = K u bK. Hence G consists of the distinct elements 1, a, a2 Now if i = 0,1, .. . ,p-l, then , ••• , aP... |
GROUPS OF LOW ORDER 153 (i) Let F be the Sylow 2-subgroup and T the Sylow 3-subgroup of G. Then F <J G and T <J G, since a Sylow p-subgroup is a normal subgroup if it is unique (Problem 5.7, page 133). Furthermore, Fn T = {I} since any element in the intersection must have order dividing 3 and 4 and so must be the ide... |
ups in Chapter 3. Therefore D4 must be isomorphic to the group given in Table 5.2. (ii) A4 is a non-abelian group of order 12. Hence it is either isomorphic to the group of Table 5.3 or 5.4 or to D3 X C2 (see page 155). As can be seen from the multiplication table for A4 given in Chapter 3, page 63, A4 has exactly thre... |
s a solvable series for 8 3, since [H: {,}] = 3 and Now H = {" (11) (12} 116, H <l 8 3, Thus {,} C;;; H C;;; 8 3 and so 8 3 is solvable. 5.45. Show that 8 4 is solvable. Solution: The alternating group A4 is a subgroup of order 12 in 8 4, Then [84 : A4] = 2 and A4 <l 8 4 by Problem 4.69, page 116. We have seen in Probl... |
roup of G, and G(i+ 1) = (G(i))'. Prove that G is solvable if and only if G(n) = {l} for some integer n. Solution: Let G(n) = {l}. Then {l} = G(n) k ... k GO) k G is a subnormal series for G and G(i) IG(i+ 1) is abelian. Hence G is solvable. Now let G be solvable. Then there exists a subnormal series {l} = Hr k ... k H... |
another is determined. In the next few sections we shall prove that if n > 5, An is simple. The groups An are not all the simple groups and indeed there is no classification of simple groups as yet. This is one of the basic problems of finite group theory. The question of how a group G is built from Hand K if G is an e... |
s if 7r is even, and the product of an odd number of trans positions if 7r is odd. Our first task is to show that any transposition is odd. We have already noted that (1,2) is odd (Section 3.3d, page 64). We will use the following lemma. Lemma 5.29: Let () E Sn and let (ai, ... , am) be a cycle. Then (}-l(al, .. . ,am ... |
n)' If m = 2, let (3 = (aI' a2, aa). Then 01- 1(301 = (a101, a2"', aa"') = (a2, av b) where b = aaOl and b is an integer different from a1 and a2' No matter what b is, 01- 1(301 "'" f3 since a2(3 = as but a201-1(301 = a1" Henc!! 01 ~ Z(Sn)' Thus no non-trivial element of Sn belongs to Z(Sn)' 5.68. Prove that An' Sn and... |
would like a briefer account of abelian groups may refer to Sec tions 6.la, 6.lc and 6.3. This will bring him quickly to the fundamental theorem of abelian groups, i.e. every finitely generated abelian group is the direct sum of cyclic groups. 177 178 ABELIAN GROUPS [CHAP. 6 6.1 PRELIMIN ARIES Here we will practice ex... |
inite, then S is the Sylow p-subgroup of G. For if P is any subgroup of G of order a power of p, by the definition of S, P ~ S. So every Sylow p-subgroup of G is contained in S. S itself is of order a power of p (Problem 5.6, page 132). Since the order of a Sylow p-subgroup is the maximal power of p dividing the order ... |
e the result. c. The homomorphic property of direct sums and free abelian groups Let G = A EEl B and let H be a group which contains isomorphic copies A and B of A (but not necessarily that H = A EEl B). and B respectively. Suppose that H = A + B What connection, if any, is there between G and H? It turns out that H is... |
B))* = {,}, and as the elements of (aut (A))* commute with the elements of (aut (B))*, we have (aut (A))* + (aut (B))* = (aut (A))* EB (aut (B))*, by Theorem 5.16, page 144. Now let ()IA induces an auto morphism ()A on A, for A() must go into a subgroup of order 9 and by Sylow's theorem there is only one subgroup of or... |
we shall prove, constitute a large class of abelian groups. Problems 6.34. Use Theorem 6.10 to prove that an abelian group of order pq, where p and q are different primes, is the direct sum of a cyclic group of order p and a cyclic group of order q. Solution: Let G be of order pq. Then by Theorem 6.10, G = Gp EB Gq ; f... |
ml> m2, nl, n2 integers, then Thus every set of two elements is dependent. Accordingly the rank of Q is 1. 6.39. Show that the p-Priifer group has no independent set consisting of two elements. Solution: Let x, y be elements of G, a p-Priifer group, x -:;6 0, Y -:;6 O. Then x, y E Cr, say, for some r (see Example 1, p... |
oup (Section 6.1c). By Lemma 6.15, F has a basis Cl, ••• , Cn such that R = gp(UICl, ••. , Uncn) for some nonnegative integers Ul, .•• , Un. We now apply Corollary 6.17 to conclude that G """ FIR is the direct sum of cyclic groups. Corollary 6.19: If G is finitely generated, it is the direct sum of a finite number of i... |
nd G 2" H. 6.49. Find up to isomorphism all abelian groups cf order 1800. Solution: Observ~ that 1800 = 233252. So an abelian group of order 1800 is a direct sum of a group of order 23 , a group of order 32 and a group of order 52. The possible types of !l group of order 23 are (23 ; 0), (2 2,2; 0), (2,2,2; 0). Thus th... |
finite abelian group G is isomorphic to a subgroup of G. Therefore we know the types of homomorphic images of finite abelian groups. Problems 6.57. Let G = A EEl B and let C, D be subgroups of A, B respectively. Show that C + D = C EEl D. (This can obviously be generalized to the direct sum of any number of groups.) So... |
m that KlnD = {OJ. For if rXl + kED, r E {O, 1, ... , w-1}, and k E K, then' D (f) K = rXl + (D (f) K) = r(xl + (D (f) K» Since Xl + (D (f) K) = x + (D (f) K), we must have r = O. So kED and thus k = O. Therefore KlnD = {OJ. But K is maximal. This contradiction shows that our original assumption, i.e. G ¥= D (f) K, is ... |
direct sum of infinite cyclic groups Ci we choose one copy of the rationals Qi for each (i E I): F = ~ Ci• Now i E I. Let K = ~ Qi and let di ;6 0 be iEI iEI chosen in each Qi' F is clearly isomorphic to gp({di liE I}). But ~ Q i is divisible since each Qi is divisible. The result follows. i E I (b) If G is any group, ... |
rated. DIVISIBLE GROUPS 6.98. Show that a divisible abelian group has no subgroup of finite index. 6.99. If G is a nondivisible abelian group, then G has a subgroup of prime index. (Hint: Use the fol lowing theorem (not proved in this book): An abelian group G for which nG = {O}, n of= 0, is the direct sum of cyclic gr... |
epresentation is of degree 6. (ii) This representation is of degree 3. Notice that in (i) and (ii) we have two representations of the same group, namely the symmetric group on {I, 2, 3}, of different degrees. (iii) This representation is of infinite degree. Notice that here G is cyclic of order 2. Hence there are repre... |
entially the same except for the elements they act on. Give a definition which will make this idea of "essentially the same" precise. Solution: Let F be a permutation group on a set X and let G be a permutation group on a set Y. We say that F and G are isomorphic as permutation groups if there exists a one-to-one onto ... |
, if (h, x) = (1,1), (h, X)gl() = (a1, gl' l(g 117)) = (al,g2' l(g 217)) and 1 (g117) = 1 (g217) (7.19) from which Using equation (7.5) with x = 1, we see that gl = l(glP) = al,gl (1(g117)), aLQ1 = al,g2 g2 = 1 (g2P) = al,g2(1(g217)) Using (7.19) we conclude that gl = g2' Thus () is one-to-one. This completes the proof... |
a2 ), (2) gp(a, bab- l , b2 ), (3) gp(ab,a2ba- l,a2 ) 7.19. Let G = gp(a, b, e) and let N be a normal subgroup of G of index 3 with GIN =: gp(Na). Suppose N contains band e. Find a set of generators for N in terms of a, band e. Solution: Choose S = {a, b, e} and X = {I, a, a2} • . Then the elements ax,s, with x E X and... |
ique. Let g E G, x E X; then gx belongs to some coset of H in G, say the coset yH where y E X. Therefore gx = yh for some hE H. Now h is uniquely determined by g and x; we denote h by mg,x. Thus gx = ymg,x The elements mg,x correspond to the elements aX,g introduced in Section 7.5b. (We use mg,x instead ofax,g because ... |
he automorphism (h E H) of H, then a is itself a homomorphism of K into the group of automorphisms of H. (This explains why we used a left transversal in Section 7.7(a), namely so that the mapping a be a homomorphism.) We have only to prove that (kk')a = kak'a (7.29) To verify (7.29), let us take an arbitrary element h... |
ve that (amlbn1. am2 bn2 )T = aml+m2bml+m2+nl+n2 = amlbml+nl. am2bm2+n2 = (amlbnl)T(am2bn2)T Clearly Ker T = {1}, and so T is one-to·one. It is easy to check that T is also onto. Thus T is an automorphism. Note that aTP = ab p = a, so that TP acts as the identity on a and b, which form a set of generators of H. Hence T... |
nd of homomorphism of a group into an abelian subgroup, called the transfer. We then used the transfer to prove that the derived group of a group whose center is of finite index is finite. 244 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 Supplementary Problems PERMUTATION REPRESENTATIONS, COSET REPRESENTATIONS, FROBENIUS THE... |
ed by X. If x E X, consider gp({x}). This is a cyclic group. Now x· x· ...• x = x r, with r a positive integer, is a reduced product in X. Hence xr = x" where rand 8 are positive integers implies r = 8. Thus gp({x}) is infinite and is infinite cyclic and the result follows. 8.3. Prove that if G is freely generated by X... |
t = X~l •.• x:n where Xi E X and €i = ±1, then A to = (Xlf)) I . • . (Xnf)) n whether or not X~l •.• x:n is a reduced product. If n = 1, this is true by the definition of O. Assume it is true for all positive integers n < k and consider t = X~l •.• x:n when n = k. If this is a reduced product then to = (Xlf))f j • • •... |
tant in topology and analysis where groups arise in just this way, as the "groups of certain presentations". First we need a definition. If S is any subset of a group, then the normal closure of S is defined to be the intersection of all normal subgroups of G containing S. Clearly the normal closure of S is a normal su... |
N,Nb}. As A is a normal subgroup of FIN, AB is a subgroup of FIN which contains both Na and Nb. It follows that AB = FIN. Thus IFINI "" IAIIBI "" 2·2 = 4. On the other hand there is a homomorphism (J of F onto the direct product of two cyclic groups of order 2. Clearly Ker (J contains a2, b2 and [a, b]. Hence Ker (J d ... |
ist free groups of rank n for each positive integer n. 260 8.37. FREE GROUPS AND PRESENTATIONS [CHAP. 8 Let F be freely generated by a and b. Prove that the subgroup of F generated by aba3 and a2 b is freely generated by aba:l and a 2b. (Hard.) Solution: Let Y = {aba3 • a2b} and let Yl = aba 3 , Y2 = a2b. Consider a re... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.