PROBLEMS ON ABSTRACT ALGEBRA

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1 PROBLEMS ON ABSTRACT ALGEBRA 1 (Putam 197 A). Let S be a set ad let be a biary operatio o S satisfyig the laws x (x y) = y for all x, y i S, (y x) x = y for all x, y i S. Show that is commutative but ot ecessarily associative. (Putam 197 B3). Let A ad B be two elemets i a group such that ABA = BA B, A 3 = 1 ad B 1 = 1 for some positive iteger. Prove B = 1. 3 (Putam 007 A5). Suppose that a fiite group has exactly elemets of order p, where p is a prime. Prove that either = 0 or p divides (Putam 011 A6). Let G be a abelia group with elemets, ad let {g 1 = e, g,..., g k } G be a (ot ecessarily miimal) set of distict geerators of G. A special die, which radomly selects oe of the elemets g 1, g,..., g k with equal probability, is rolled m times ad the selected elemets are multiplied to produce a elemet g G. Prove that there exists a real umber b (0, 1) such that 1 1 lim Prob(g = x) m b m is positive ad fiite. x G 5 (Putam 1990 B4). Let G be a fiite group of order geerated by a ad b. Prove or disprove: there is a sequece g 1, g, g 3,..., g such that (a) every elemet of G occurs exactly twice, ad (b) g i+1 equals g i a or g i b for i = 1,,...,. (Iterpret g +1 as g 1.) 6 (Putam 016 A5). Suppose that G is a fiite group geerated by the two elemets g ad h, where the order of g is odd. Show that every elemet of G ca be writte i the form m 1 h 1 m h m g g g r h r with 1 r G ad m, 1, m,,..., m r, r {1, 1}. (Here G is the umber of elemets of G.) 7 (Putam 1977 B6). Let H be a subgroup with h elemets i a group G. Suppose that G has a elemet a such that for all x i H, (xa) 3 = 1, the idetity. I G, let P be the subset of all products x 1 ax a x a, with a positive iteger ad the x i s i H. (a) Show that P is a fiite set. (b) Show that, i fact, P has o more tha 3h elemets. 8 (Putam 1984 B3). Prove or disprove the followig statemet: If F is a fiite set with two or more elemets, the there exists a biary operatio o F such that for all x, y, z i F, 1

2 (i) x z = y z implies x = y (right cacellatio holds), ad (ii) x (y z) = (x y) z (o case of associativity holds). 9 (Putam 1987 B6). Let F be the field of p elemets where p is a odd prime. Suppose S is a set of (p 1)/ distict ozero elemets of F with the property that for each a = 0 i F, exactly oe of a ad a is i S. Let N be the umber of elemets i the itersectio S {a : a S}. Prove that N is eve. 10 (Putam 1989 B). Let S be a oempty set with a associative operatio that is left ad right cacellative (xy = xz implies y = z, ad yx = zx implies y = z). Assume that for every a i S the set {a : = 1,, 3,... } is fiite. Must S be a group? 11 (Putam 199 B6). Let M be a set of real matrices such that (i) I M, where I is the idetity matrix; (ii) if A M ad B M, the either AB M or AB M, but ot both; (iii) if A M ad B M, the either AB = BA or AB = BA; (iv) if A M ad A / I, there is at least oe B M such that AB = BA. Prove that M cotais at most matrices. 1 (Putam 1996 A4). Let S be a set of ordered triples (a, b, c) of distict elemets of a fiite set A. Suppose that (1) (a, b, c) S if ad oly if (b, c, a) S; () (a, b, c) S if ad oly if (c, b, a) / S [for a, b, c distict]; (3) (a, b, c) ad (c, d, a) are both i S if ad oly if (b, c, d) ad (d, a, b) are both i S. Prove that there exists a oe-to-oe fuctio g from A to R such that g(a) < g(b) < g(c) implies (a, b, c) S. 13 (Putam 008 A6). Prove that there exists a costat c > 0 such that i every otrivial fiite group G there exists a sequece of legth at most c l G with the property that each elemet of G equals the product of some subsequece. (The elemets of G i the sequece are ot required to be distict. A subsequece of a sequece is obtaied by selectig some of the terms, ot ecessarily cosecutive, without reorderig them; for example, 4, 4, is a subsequece of, 4, 6, 4,, but,, 4 is ot.) 14 (Putam 009 A5). Is there a fiite abelia group G such that the product of the orders of all its elemets is 009? 15 (Putame 010 A5). Let G be a group, with operatio. Suppose that 1. G is a subset of R 3 (but eed ot be related to additio of vectors);. For each a, b G, either a b = a b or a b = 0 (or both), where is the usual cross product i R 3. Prove that a b = 0 for all a, b G.

3 16. Let R be a ocommutative rig with idetity. Suppose that x, y are elemets of R such that 1 xy ad 1 yx are ivertible. (By the previous problem it suffice to assume that oly 1 xy is ivertible, but this is irrelevat.) Show that (1 + x)(1 yx) 1 (1 + y) = (1 + y)(1 xy) 1 (1 + x). (1) This problem illustrates that ocommutative high school algebra is a lot harder tha ordiary (commutative) high school algebra. Formally we have (1 yx) 1 = 1 + yx + yxyx + yxyxyx + ad similarly for (1 xy) 1. Thus both sides of (1) are formally equal to the sum of all alteratig words (products of x s ad y s with o two x s or y s appearig cosecutively). This makes the idetity (1) plausible, but our formal argumet is ot a proof. 17. Let G be a group of order 4 +, 1. Prove that G is ot a simple group, i.e., G has a proper ormal subgroup. 18. Let R satisfy all the axioms of a rig except commutativity of additio. Show that ax + by = by + ax for all a, b, x, y R. 19. Let G deote the set of all ifiite sequeces (a 1, a,... ) of itegers a i. We ca add elemets of G coordiate-wise, i.e., (a 1, a,... ) + (b 1, b,... ) = (a 1 + b 1, a + b,... ). Let Z deote the set of itegers. Suppose f : Z is a fuctio satisfyig f(x + y) = f(x) + f(y) for all x, y G. Let e i be the elemet of G with a 1 i positio i ad 0 s elsewhere. (a) Suppose that f(e i ) = 0 for all i. Show that f(x) = 0 for all x G. (b) Show that f(e i ) = 0 for all but fiitely may i. 0. Let G be a fiite group, ad set f(g) = #{(u, v) G G : uv = vu}. Fid a formula for f(g) i terms of the order of G ad the umber k(g) of cojugacy classes of G. (Two elemets x, y G are cojugate if y = axa 1 for some a G. Cojugacy is a equivalece relatio whose equivalece classes are called cojugacy classes.) 1 (difficult). Let be a odd positive iteger. Show that the umber of ways to write the idetity permutatio ι of 1,,..., as a product uvw = ι of three -cycles is ( 1)! /( + 1).. Let G be ay fiite group, ad let w G. Fid the umber of pairs (u, v) G G satisfyig w = uvu vuv. 3. Show that the umber of ways to write the cycle (1,,..., ) as a product of 1 traspositios is. For istace, whe = 3 we have (multiplyig permutatios left-to-right) three ways: (1,, 3) = (1, 3)(, 3) = (1, )(1, 3) = (, 3)(1, ). 4 (difficult). Let s i = (i, i + 1) S, i.e., s i is the permutatio of 1,,..., that trasposes i ad i+1 ad fixes all other j. Let f() be the umber of ways to write the permutatio, 1,..., 1 i the form s i1 s i s ip, where p =. For istace, 31 = s 1 s s 1 = s s 1 s, so f(3) =. Moreover, f(4) = 16. Show that f() is the umber of sequeces a 1,..., a p of 1 1 s, s,..., oe 1, such that i ay prefix a 1, a,..., a k, the umber of i + 1 s does ot exceed the umber of i s. For istace, whe = 3 there are the two sequeces 11 ad 11. 3

4 A explicit formula is kow for f(), but this is irrelevat here. 5 (difficult). I the otatio of the previous problem, show that i 1,i,...,i p i 1 i i p = p!, where the sum is over all sequeces i 1,..., i p for which, 1,..., 1 = s i1 s i s ip. For istace, whe = 3 we get = 3!. The oly kow proofs are algebraic. It would be iterestig to give a combiatorial proof. 4

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