Homwork 5 M 373K Solutions Mark Lindbrg and Travis Schdlr 1. Prov that th ring Z/mZ (for m 0) is a fild if and only if m is prim. ( ) Proof by Contrapositiv: Hr, thr ar thr cass for m not prim. m 0: Whn m 0, Z/0 is Z, bcaus vry numbr is quivalnt to itslf (mod 0), but not to any othr. But Z dos not hav multiplicativ invrss, and so it cannot b a fild. m 1: Whn m 1, Z/Z is {0}, bcaus vry numbr is quivalnt to vry othr numbr (mod 1). But {0} dos not hav a non-zro multiplicativ invrs, bcaus 1 0, and so {0} is not a fild. m 4: Hr, w hav that m is composit, and thrfor thr ar at last 2 factors a, b of m such that ab m, but 1 < a, b < m. Sinc a, b < m, w hav that a 0 (mod m) and m 0 (mod m), but ab m 0 (mod m). Now considr just th lmnts a, b. W know that in a fild, thy must both hav an invrs. In particular, thr is som a 1 Z/mZ such that a 1 a 1 aa 1. But thn a 1 ab a 1 0 b 0, which contradicts th fact that 1 < b < m (hr not that, in any ring, a 0 0 sinc a 0 a (0 + 0) a 0 + a 0, thn cancl th a 0). Thrfor, a 1 cannot xist, and Z/mZ is not a fild. Thus, in any cas whr m is composit, Z/mZ is not a fild, so if Z/mZ is a fild, m must b prim. ( ) Lt m p b prim. Thn for any a Z/pZ, with 1 a p 1, w must hav that a has a multiplicativ invrs. Sinc p is prim, a p, and a n p n 1. But thn ach a n {1, 2, dots, p 1}, and so thr ar a finit numbr of possibilitis, and an infinit numbr of xponnts. Thus, w must hav som rpat. That is, thr ar som n > k 1 such that a n a k (mod p). Thn, by dfinition, p a n a k a k (a n k 1). But, sinc p is prim and p a, p a k, and so p (a n k 1) (*). Thus, by dfinition, a n k aa n k 1 1 (mod p), and thus, by dfinition, a n k 1 is a multiplicativ invrs of a, and in gnral, a n k 1 is a multiplicativ invrs of a in Z/pZ.Thus, if p is prim, Z/pZ is a fild. (*): Hr w us that p ab implis p a or p b. To s this, w prov th contrapositiv: if p a and p b, thn pz + az Z pz + bz; thn Z (pz + az)(pz + bz) p 2 Z + paz + pbz + abz pz + abz, hnc p ab. Thrfor, w provd that th ring Z/mZ (for m 0) is a fild if and only if m is prim. Bonus: Using th sam tchniqu, show that if M is any finit monoid, thn any lmnt a M obying th cancllation proprty, (ab ac) (b c), is invrtibl. So sinc this cancllation proprty holds for ā Z/mZ whnvr gcd(a, m) 1, w conclud that ā is invrtibl if and only if gcd(a, m) 1. (In fact, for a finit monoid, not ncssarily commutativ, th following ar quivalnt: a is lft invrtibl, a is right invrtibl, am M, Ma M, a obys th lft cancllation proprty, and a obys th right cancllation proprty). 2. Prov that, if K H G ar subgroups, thn if [G : H or [H : K is infinit, so is [G : K. Thus, [G : K [G : H[H : K still holds in this cas (w provd th finit cas in class). Hint: Find a natural surjction G/K G/H and injction H/K G/K. 1
Proof: Assum [H : K is infinit. Now H/K by dfinition is all costs {hk h H}, and g G/K is, by dfinition, all costs {gk g G}. Sinc H G, thn h H h G. Thus, for any hk H/K, hk G/K. Thrfor, w can dfin f : H/K G/K by f(hk) hk. Thn for any ak, bk H/K, ak f(ak) f(bk) ak, and so by dfinition, f is injctiv. Thus, [G : K [H : K, bcaus vry lmnt in [H : K maps to a uniqu lmnt in [G : K. Now assum [G : H is infinit. With th sam dfinitions abov, w s that for any ah G/H, thr is som ak G/K so w can dfin f : G/H G/K, such that f(ak) ah, and thn f is a surjction by dfinition. Thus, [G : K [G : H, and so [G : K must b infinit as wll. Thus, bcaus if ithr [G : H or [H : K is infinit, thn [G : K is infinit, thn [G : K [G : H[H : K for all infinit cass. 3. Show that th invrsion map is a bijction G/H H\G, so that th numbr of lft costs always quals th numbr of right costs (vn if G is infinit). Hr G/H is th st of lft costs and H \G is th st of right costs. (Hint: (ah) 1 Ha 1 so this maps lft costs to right costs and vic-vrsa, and not that invrsion twic is th idntity.) Lt f b th invrsion map, so f(ah) (ah) 1 Ha 1 for all a G. Thus, f : G/H H \G. Now, lt f(ah) f(bh) Ha 1 Hb 1. Applying th invrsion principl again, w s that (Ha 1 ) 1 ah (Hb 1 ) 1 bh, and so w hav shown injctivity. Now lt Hc H \ G. Thn (Hc) 1 c 1 H G/H, and f(c 1 H) (c 1 H) 1 Hc, so all lmnts hav an invrs imag, and f is surjctiv. Thus, bcaus f is injctiv and surjctiv, f is bijctiv, and so th numbr of lft costs always quals th numbr of right costs. 4. Show that, if gcd(r, s) 1, thn Z/r Z/s Z/rs. Show that this dos not hold if gcd(r, s) 1. Hint for th first part: thr ar two possibl maps: ithr (ā, b) sa + rb in th forward dirction, or ā (ā, ā) in th rvrs dirction; howvr ths ar not invrs to ach othr: th composition of th two dirctions is multiplication by r + s. Hint for th scond part: Show that th ordr of vry lmnt in Z/r Z/s is a factor of lcm(r, s). (Why do th two maps givn bfor no longr dfin isomorphisms in this cas?) Considr th mapping φ : Z/r Z/s Z/rs, (ā, b) sa + rb. Thn w s that φ((a, b) + (c, d)) φ((a + c, b + d)) φ((a + c, b + d)) s(a + c) + r(b + d) sa + sc + rb + rd sa + rb + sc + rd φ((a, b)) + φ((c, d)) and so φ is a group homomorphism. (Not that it is not a ring homomorphism, although th othr possibl map, Z/rsZ Z/rZ Z/sZ, ā (ā, ā), is a ring homomorphism, which on can show is an isomorphism if gcd(r, s) 1 similarly to what w do hr.) 2
Considr φ((0, 0)) s0 + r0 0, so (0, 0) K, th krnl. Now considr any (a, b) such that φ((a, b)) sa + rb 0. Thn, by dfinition, sa + rb 0 (mod rs), so rs sa + rb. Thn s sa + rb, and sinc s sa, s rb. Sinc gcd(r, s) 1, w hav that s b. Similarly, w can s that r a. Thn a 0 (mod r) and b 0 (mod s), Thn a 0 in r, and b 0 in s, so (a, b) (0, 0), and so w s that (0, 0) is th only lmnt of th krnl, which, by a thorm from th book, shows that φ is injctiv. Now, not that th max ordr lmnt of Z/r Z/s is ordr lcm(r, s), and in Z/rs th max ordr lmnt is ordr rs. In th cas whr gcd(r, s) 1, sinc rs gcd(r, s) lcm(r, s), w hav that lcm(r, s) rs, and so th max ordr lmnts hav th sam ordr. Thn w not that Z/r Z/s rs, by th dfinition of how th cartsian product works, and Z/rs rs, by th dfinition of th Z/rs, which will hav rs quivalncy classs {0, 1,..., rs 1}. Thn, w apply problm 1(c) from homwork 3, and sinc Z/r Z/s Z/rs finit, injctivity implis surjctivity, and so w hav a bijctiv homomorphism. That is, φ is an isomorphism. rs gcd(r,s) Now, considr th cas whr gcd(r, s) > 1. Thn lcm(r, r) < rs, and so th maximum ordr of any lmnt in Z/r Z/s must b < rs, and so it cannot b isomorphic to th cyclic group Z/rs. 5. Chaptr 2, Miscllanous problm M.3. Classify groups of ordr 6 by analyzing th following thr cass: (i) G contains an lmnt of of ordr 6. Lt th lmnt b x. Thn th group must hav {1, x, x 2, x 3, x 4, x 5 }, and so x is a gnrator, and G is th cyclic group of ordr 6. (ii) G contains an lmnt of ordr 3 but non of ordr 6. Lt th lmnt b x. Thn th group H : x {1, x, x 2 } is a subgroup of indx two, so it must b normal (cf. Exrcis 7 blow): for any othr lmnt y / H, w must hav Hy yh G H, th st of lmnts of G not in H. Tak such a y / H. Thn y 2 H is a cost; w claim it is H. If not, y 2 H yh, and canclling, yh H, a contradiction. Thus y 2 H, and y has ordr two. Thus y 2 1 x 3. Sinc y was an arbitrary lmnt not in H, w actually s all lmnts not in H hav ordr two. So (xy) 2 1 as wll, or altrnativly, yx x 1 y 1 x 2 y. Thus x and y oby all th rlations of S 3 givn in (2.2.6). Sinc ths rlations dfin S 3, w gt a homomorphism S 3 G snding x S 3 to x G and y S 3 to y G. This is surjctiv sinc G is clarly gnratd by x and y. Sinc G 6 S 3, it must also b injctiv, hnc S 3 G is an isomorphism. (iii) All lmnts of G hav ordr 1 or 2. In this cas, all lmnts othr than th idntity must b of ordr 2, and so G must b of th form {1, a, b, c, d, }. Thn vry lmnt is its own invrs. But thn ab (ab) 1 b 1 a 1 ba, and so th group must b ablian. Thn w s that w hav a subgroup formd by S {1, a, b, ab}, whr ab is som othr lmnt of G (c, d, or ). But thn S 4, which contradicts Lagrang s thorm, which tlls us that S must divid G 6. Thus, thr can b no groups of ordr 6 of this typ. 3
a b 6. Chaptr 2, Exrcis 12.5. Lt G b th group of uppr triangular ral matrics, with a and d diffrnt from zro. For ach of th following substs, dtrmin whthr or not S is a subgroup, and whthr or not S is a normal subgroup. If S is a normal subgroup, idntify th quotint group G/S. (i) S is th subst dfind by b 0. I 1 S. a 0 If A S, a, d 0, A 1 1 d 0 ad S. 0 a a 0 b 0 ab 0 S, so closd undr multiplication. 0 c c Thus, S is a subgroup of G. [ b c a 0 1 [ c b b ba cd 1 c 1 b b a ac+cd 1, and so S is not a normal subgroup of 0 0 0 G. (ii) S is th subst dfind by d 1. I 1 S. [ a b If A, a 0, thn A 0 1 1 1 1 b 1 b a a a S, so invrss ar in S. 0 a 0 1 a b c d ac ad + b, so closd undr multiplication. 0 1 0 1 0 1 [ c d a b 1 [ c c ca cb + d 1 1 c c a ad+cb+d 1, and so S is a 0 0 1 0 0 0 0 1 normal subgroup of G. a b x y ax ay + b W s that, and so as x, y vary across R, w will 0 1 r s gt a cost of th form, r, s R, r 0. Thus, w s that as d spans R, d 0, this will span G, maning that w hav xamind all costs, and and so w can idntify th costs by d R, d 0. (iii) S is th subst dfind by a d. I 2 S. a b If A S, a 0, thn A 0 a 1 1 a b S, and A a 2 0 a so th invrss ar prsnt. a b c d ac ad + bc S, so closd undr multiplication. 0 a 0 c 0 ac Thus, S is a subgroup of G. [ c d a b 1 [ c c ca cb + da 1 1 c c 1 0 0 a 0 0 a 0 normal subgroup of G. [ a b x y ax ay + bx W s that 0 x x [ r s gt a cost of th form dr 0 a a b S, 0 a a ad+cb+da, and so S is a 0 a. As x, y vary across R, x 0, w will, r, s R, r 0. Hr w s that, as d/a spans R, 4
d/a 0, this will span G, maning that w hav xamind all costs, and so w can again idntify th costs by d/a R, d/a 0. 7. Chaptr 2, Exrcis 8.10. Prov that vry subgroup of indx 2 is a normal subgroup, and show by xampl that a subgroup of indx 3 nd not b normal. Lt H G such that [H : G 2. Sinc H is a group, and closd, h 1, h 2 H, h 1 h 2 H, and so h 1 H H Hh 2, and H is a lft and right cost of itslf. Sinc [H : G 2, thr is only on othr right cost and on othr lft cost. Thus, for g G, if g H, gh H, and so gh must b th othr lft cost. Similarly, Hg must b th othr right cost, and w s that sinc both ar G H (st diffrnc), thy ar quivalnt. Now considr th st S 3 {1, x, x 2, y, xy, x 2 y} with x 3 1, y 2 1, and yx x 2 y as on pag 42 of Artin. Now considr th subgroup A {1, y}. Sinc A 2 and S 3 6, by th counting formula, [S 3 : A 3. But xa {x, xy} and Ax {x, yx x 2 y}, and w s that thy ar not th sam. 8. Chaptr 2, Exrcis 10.3. Lt G and G b cyclic groups of ordrs 12 and 6, gnratd by lmnts x and y, rspctivly, and lt φ : G G b th map dfind by φ(x i ) y i. Exhibit th corrspondnc rfrrd to in th Corrspondnc Thorm xplicitly. W s that φ(x a x b ) φ(x a+b ) y a+b y a y b φ(x a )φ(x b ), and so φ is a homomorphism. Thn w s that for y i G, which mans that 0 i 5, x i G, and so vry lmnt of G has a primag, and so w hav a surjctiv homomorphism. W s that φ( G ) φ(x 6 ) G, and this is not tru for any othr lmnts, sinc y i G if i 0 (mod 6). Thus, K { G, x 6 }. Thn th subgroups ar of G which contain X ar K, x 3 { G, x 3, x 6, x 9 }, x 2 { G, x 2, x 4, x 6, x 8, x 10 }, and G. Th corrsponding subgroups of G ar { G }, y 3 { G, y 3 }, y 2 { G, y 2, y 4 }, and G. Th bijction corrsponds thm in this ordr, and it is as givn in th txtbook, th imag undr φ. 9. Chaptr 2, Exrcis 11.4. In ach of th following cass, dtrmin whthr or not G is isomorphic to th product group H K. Justify your answrs! (a) G R, H {±1}, K {positiv ral numbrs}. W hav shown bfor that H and K ar groups. It is obvious that H K {1}, and w s that 1K is th positiv rals and 1K is th ngativ rals, and sinc R dos not includ 0, HK G. Sinc multiplication is commutativ ovr th rals, gag 1 gg 1 a a, and so w s that all subgroups of G ar normal, and in particular H and K ar normal. Thrfor, w can apply part (d) of proposition 2.11.4 from pag 65 of Artin, and so f : H X G, th multiplication map, is an isomorphism. (b) G {invrtibl uppr triangular 2 2 matrics}, H {invrtibl diagonal matrics}, K {uppr triangular matrics with diagonal ntris 1}. Not: It is apparnt that H and K must both b n n matrics. Sinc it must hold tru for any n, w must hav that it also holds for n 2, so w will considr (and show a countrxampl) for this cas. This will man that H has th form 1 c and K has th form, for a, b, c R, a, b 0. 0 1 [ a 0 0 b 5
W v sn bfor that H and K ar groups, but not that thy ar also ablian: a 0 c 0 ac 0 c 0 a 0 0 b 0 bd 0 b bcaus ac ca and db bd bcaus th rals ar ablian undr multiplication. Thrfor, H is an ablian group. Similarly: 1 a 1 b 1 a + b 1 b 1 a. 0 1 0 1 0 1 0 1 0 1 Thrfor, K is also an ablian group. Thn, w s that for any two lmnts in H K, (a, b)(c, d) (ac, bd) (ca, db) (c, d)(a, b), and so w s that H K is an ablian group. But s that: and [ d 0 f a b d 0 c 0 f a b 0 c ad a + bf 0 cf da db + c 0 fc and sinc db + c a + bf in gnral, G is not ablian. Now lt x, y H K. Thn xy yx, and so if thr was an isomorphism f : H K G, f(x)f(y) f(xy) f(yx) f(y)f(x), and thn G would also hav to b ablian, which is a contradiction. Thrfor, by countrxampl, thr can not xist any such isomorphism. (c) G C, H {unit circl}, K {positiv ral numbrs}. W hav shown bfor that H and K ar groups. Th only ral numbrs with r 1 ar ±1, and 1 K, so H K {1}. Sinc any non-zro complx numbr, and thrfor any complx numbr in C G, can b xprssd as r iθ for r > 0 and 0 θ < 2π, and iθ H, thn r iθ KH, and so KH G. Sinc th complx numbrs undr multiplication ar an ablian group, by th sam rasoning as part (a), w s that all subgroups, and in particular H and K, ar normal. Thrfor, w can again apply th sam proposition as in part (a), and f : H X G, th multiplication map, is an isomorphism. 10. Chaptr 2, Exrcis 12.2. In th gnral linar group GL 3 (R), considr th substs 1 1 0 H 0 1, and K 0 1 0, whr rprsnts an arbitrary ral numbr. Show that H is a subgroup of GL 3, that K is a normal subgroup of H, and idntify th quotint group H/K. Dtrmin th cntr of H. 1 a b W s that I 3 H by obsrvation. Now lt h 0 1 c. Thn h 1 6
1 a ac b 0 1 c H, so invrss ar prsnt. Thn 1 a b 1 d 1 d + a + af + b 0 1 c 0 1 f 0 1 f + c and so H is closd undr multiplication. Thrfor, H is a group. 1 0 a 1 0 a W s that I 3 K by obsrvation. Now lt k 0 1 0. Thn k 1 0 1 0 H, and so invrss ar prsnt. Thn 1 0 a 1 0 b 1 0 b + a 0 1 0 0 1 0 0 1 0 so K is closd undr multiplication. Thrfor, K is a group. Now obsrv that 1 a b 1 0 f 1 a ac b 1 a f + b 1 a ac b 1 0 f 0 1 c 0 1 0 0 1 c 0 1 c 0 1 c 0 1 0 so K is a normal subgroup of H. 1 a b 1 0 f 1 a f + b Considr 0 1 c 0 1 0 0 1 c. Whn w lt a, b, c R b fixd, 1 a r and f R varis, w s that w gt a cost of th form 0 1 c, whr r R can b any lmnt. Thus, w s that as w vary a and c, w will span H, and ach on givs a uniqu subgroup, and so w conclud that (a, c) idntify th subgroup. Mor 1 a b prcisly, th map snding 0 1 c to (a, c) R R is a surjctiv homomorphism with krnl K, so by th first isomorphism thorm, H/K R R. By a problm on a prvious homwork, w hav sn that only matrics of th form a 0 0 0 a 0 commut with all lmnts of GL 3, and thrfor th only lmnt of th 0 0 a 1 0 0 cntr of H is I 3 0 1 0. 7