Homework 3 Solution Chapter 3.

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1 Homework 3 Solution Chpter 3 2 Let Q e the group of rtionl numers under ddition nd let Q e the group of nonzero rtionl numers under multiplition In Q, list the elements in 1 2 In Q, list the elements in 1 2 In Q, 1 2 {n 1 2 n Z} {, 2, 3 2, 1, 1 2, 0, 1 2, 1, 3 2, 2, } {n 2 n Z} In Q, 1 2 { ( 1 2 ) n n Z} { 1 2 n n Z} {, 4, 2, 1, 1 2, 1 4, } {2n n Z} 4 Prove tht in ny group, n element nd its inverse hve the sme order If n, n e So ( 1 ) n ( n ) 1 e 1 e Therefore 1 n y the definition of order By the sme reson, ( 1 ) 1 1 So we otin 1 Now suppose tht We need to show tht 1 s well If not, 1 n for some n > 0 Then e ( 1 ) n ( n ) 1, or equivlently, n e 1 e Therefore n nd rise ontrdition Hene 1 6 In the group Z 12 find,, nd + for eh se () 6, 2 6, , + 4, 3 6, 4 8, 5 10, , 2 ( + ) 16 4, 3 ( + ) () 3, 8 3, 2 6, 3 9, (see ())

2 () 5, 4 5, 2 10, , 4 8, , 6 6, 7 11, , 9 9, , 11 7, , 2 8, If is n element of group G nd 7, show tht is the ue of some element of G If 7, 7 e Then So is the ue of 5 ( 5 ) ( 7 ) 2 e 2 18 Suppose tht is group element nd 6 e Wht re the possiilities for? Provide resons for your nswer If 6 e, then must e t most 6 So only 1, 2, 3, 4, 5, 6 re ll possiilities Furthermore, if 4 e, then 2 2 e ( 4 ) 2 e 2 e So 4 is impossile Similrly, if 5 e, then 3 3 e 2 3 ( 6 ) 2 15 ( 5 ) 3 e 3 e So 5 is impossile s well In summry, only 1, 2, 3, 6 re possile (They re ll divisors of 6 Of ourse, it is not n ident) 19 If is group element nd hs infinite order, prove tht m n when m n Suppose tht m n Without lose of generlity, we my ssume tht m n Then m n m ( n ) 1 m ( m ) 1 e If m > n, then from m n e, we know tht < Therefore m n 24 Suppose n is n even positive integer nd H is sugroup of Z n Prove tht either every memer of H is even or extly hlf of the memers of H re even Suppose tht H is sugroup of Z n If H onsists of even integers, then there is nothing to prove Now ssume tht there is H, whih is n odd integer Let E {x H x is even} nd O {x H x is odd} We need to show tht the numer of elements of E is equl to tht of O Tke mp f : H H, given y f(x) x + mod n in Z n We lim tht if x is even, then f(x) is odd nd if x is odd, then f(x) is even This is trivil if x+ < n If x + n, then x + mod n is odd if nd only if x + is odd euse n is n even integer So the mp f hs property tht f(e) O nd f(o) E 2

3 Moreover, f is injetive Indeed, if f(x) f(y), then x + y + so y nelltion, x y So from f(e) O, we know tht the numer of elements of E is less thn or equl to tht of O Similrly, from f(o) E, the numer of elements of O is less thn or equl to tht of E Thus they hve the sme numer of elements 31 For eh divisor k > 1 of n, let U k (n) {x U(n) x mod k 1} List the elements of U 4 (20), U 5 (20), U 5 (30), nd U 10 (30) Prove tht U k (n) is sugroup of U(n) Let H {x U(10) x mod 3 1} Is H sugroup of U(10)? Suppose tht k is divisor of n Step 1 U k (n) Beuse 1 U k (n), U k (n) Step 2, U k (n) U k (n) U 4 (20) {1, 9, 13, 17}, U 5 (20) {1, 11} U 5 (30) {1, 11}, U 10 (30) {1, 11} If, U k (n), then mod k 1 nd mod k 1, or equivlently, xk + 1 nd yk + 1 for two integers x, y Z By using division lgorithm, if we write qn + r with 0 r < n, then r in U(n) Also n uk for some u Z Now r qn (xk + 1)(yk + 1) qn xyk 2 + xk + yk + 1 quk k(xyk + x + y qu) + 1 so r 1 mod k Thus U k (n) s well Step 3 U k (n) 1 U k (n) In U(n), 1 is the solution Z n of x 1 mod n For suh, pn + 1 for some p Z Sine mod k 1, qk + 1 Then pn + 1 (qk + 1) qk + Also n uk for some u Z Now Thus 1 mod k nd U k (n) qk + pn + 1 puk + 1 If n 10 nd k 3, then H {1, 7} But / H So H is not losed under the multiplition nd H is not sugroup of U(10) 32 If H nd K re sugroups of G, show tht H K is sugroup of G (Cn you see tht the sme proof shows tht the intersetion of ny numer of sugroups of G, finite or infinite, is gin sugroup of G?) Step 1 H K Beuse e H nd e K, e H K Thus H K Step 2, H K 1 H K 3

4 If, H K, then, H nd, K Sine H nd K re oth sugroups, 1 H nd 1 K Thus 1 H K Therefore y sugroup test 2, H K G Note tht the sme proof holds for ritrry numer of sugroups (even for infinitely mny!) 34 Let G e group, nd let G Prove tht C() C( 1 ) If x C(), x x Then x 1 x 1 x nd x 1 1 x 1 1 x Thus x C( 1 ) Therefore C() C( 1 ) By pplying the sme ide for 1, we hve C( 1 ) C(( 1 ) 1 ) C() Thus C() C( 1 ) 40 In the group Z, find () 8, 14 ; () 8, 13 ; () 6, 15 ; (d) m, n ; (e) 12, 18, 45 We would like to show generl sttement (for (d)): m, n gd(m, n) Let d gd(m, n) Then m d nd n d for some, Z So m d d nd n d d Therefore m, n d On the other hnd, if d gd(m, n), then there re two integers x, y Z suh tht d xm + yn So d x m + y n m, n Hene d m, n Thus we hve d m, n () 8, 14 () 8, 13 () 6, 15 (d) m, n (e) 12, 18, 45 So gd(8, 14) 2 8, 14 2 gd(8, 13) 1 8, 13 1 Z gd(6, 15) 3 6, 15 3 m, n gd(m, n) gd(12, 45) 3 12, , 18, , 18, , 45 12, 18, , 45 12, 18, 45 12, 18,

5 42 If H is sugroup of G, then y the entrlizer C(H) of H we men the set {x G xh hx for ll h H} Prove tht C(H) is sugroup of G Step 1 C(H) Beuse eh he for ll h H, e C(H) Step 2, C(H) C(H) For ll h H, h h h euse, C(H) So C(H) Step 3 C(H) 1 C(H) For ll h H, h h Then h 1 h 1 h nd 1 h 1 h 1 h 1 for ll h H So 1 C(H) By sugroup test 1, C(H) G 51 Let e group element of order n, nd suppose tht d is positive divisor of n Prove tht d n/d Suppose tht n kd Then ( d ) k dk n e So d k n/d If d m, then ( d ) m dm e so dm n nd d m n/d Therefore d n/d 75 Let H e sugroup of group G Prove tht the set HZ(G) {hz h H, z Z(G)} is sugroup of G Step 1 HZ(G) Sine oth H nd Z(G) re sugroups of G, e H nd e Z(G) So e ee HZ(G) nd HZ(G) Step 2, HZ(G) 1 HZ(G) From, HZ(G), h 1 z 1 nd h 2 z 2 where h 1, h 2 H nd z 1, z 2 Z(G) Then 1 h 1 z 1 (h 2 z 2 ) 1 h 1 z 1 z2 1 h 1 2 h 1 z 1 h 1 2 z 1 2 h 1 h 1 2 z 1z2 1 Beuse h 1 h 1 2 H nd z 1 z 1 2 Z(G), 1 h 1 h 1 2 z 1z 1 2 HZ(G) By sugroup test 2, HZ(G) G 79 Let G GL(2, R) ( ) () Find C ( ) If C, then + + d d d + + d 5

6 So + +, +d, +d, nd These onditions re equivlent to nd +d Conversely, for G with nd +d, ( ) then y the sme eqution, C Therefore ( ) { } C G + d, ( ) () Find C ( ) If C, then 0 1 d So nd d Conversely, for G with nd d, then ( ) y the sme eqution, C Therefore ( ) { } C G d, () Find Z(G) k 0 Note tht for ny slr mtrix with k 0 is in Z(G), euse k 0 k k k 0 k kd { } k 0 So k 0 Z(G) Conversely, if Z(G), then ( ) so C Similrly, 0 1 6

7 so ( C d ) Therefore + d, nd d From + d nd d, 0 0 Hene Beuse this mtrix is in G GL(2, R), 0 0 So we hve { } k 0 Z(G) k 0 nd hene, Z(G) { k 0 k 0 } 7