4 5 = So 2. No, as = ± and invariant factor 6. Solution 3 Each of (1, 0),(1, 2),(0, 2) has order 2 and generates a C

Transcription

1 Soluos (page 7) ρ Soluo = ρ as = So ρ, ρ s a Z bass of Z ad 5 4 ( m, m ) = (,7) = (,9) No, as = ± Soluo The 6 elemes of Z K are: g = K + e + e, g = K + e, g = K + e, 4g = K + e, 5g = K + e + e, 6g = K So Z K = g s cyclc wh geeraor g ad vara facor 6 Soluo Each of (, ),(, ),(, ) has order ad geeraes a C ype subgroup The elemes (, ),(, ),(, ),(, ) each have order 4 H = (, ) (, ) has somorphsm ype C C (, ) = (, ) ad (, ) = (, ) are C 4 ype subgroups (, ) ad G are subgroups of ype C ad C C4 respecvely H s eher (, ) or (, ), H s eher (, ) or (, ) Soluos (page 6) Soluo,,, (a) r r, c c are pared (b) r + r, c c are cougae The eros r, r + lr where l Z ( > ), leave rows,, uchaged Soluo ( ) D ad, for example, P = =, Q = 4 7 ( ) D = ad, for example, P =, Q = 5 ( ) D = ad, for example, P =, Q = Soluo ( ) D =, P =, Q = 5, x = (,,) ( ) D =, P =, Q =, x = ( l, l, l), l Z ( ) D =, P =, Q =, x = (,,) 4 6

2 Noe ha he eros r + r ad r + r are dffere! Soluo 4 7 ( ) D = (6, ), Q =, 7 6 =, 4 6 = 7, de Q =, a =, b = ( ) D = (,), Q =, deq =, a =, b = 6 46 ( ) D = (9, ), Q =, = 6, = 46, de Q =, a = 9, b = Soluo 5 (a) pply c + c, c c (b) pplyg he sequece o = ( a, b) gves ( b, a) pplyg o ( b, a) gves ( a, b) ad applyg o ( a, b) gves ( b, a) Oe of hese has o-egave eres (c) Use (b) o ge ( a, b ) wh a, b ssumg b > as we ca, he Eucldea algorhm apples c qc o ( a, b ) ad c qc o ( b, a ), where a = bq + r, a b >, ulmaely edg wh (, d ) or ( d,) = D Use par (a) f eeded o fsh The Q s he produc of he correspodg elemeary marces all of whch have deerma, ad so deq = a b (d) P = where ad bc = So gcd{ a, b } = ad ( a, b ) reduces o (,) usg oly ecos of c d ype ( ) So P o applyg hese ecos ad fally c ec, as deermas are e uchaged by such ecos (e) pply c c, c c, c + c, c c, c c Soluo 6 T T (a) P = P for eros of ypes ( ) ad ( ) Oherwse P, P correspod o r + lr, r + lr (b) s D s symmerc, rasposg P = DQ gves P T = Q D Hece T T T T T T T T T T ( Q P) = P Q = Q DQ = Q P showg Q P symmerc s Q P s a produc of elemeary marces, every square marx over Z ca be made symmerc by applyg eros (c) The sequece c c, c c, c c, r + 9r, reduces o D = dag(,46) Posmulplcao by Q T carres ou he above ecos, ad so premulplcao by Q P carres ou T 5 5 r + 9r, r r, r + r, r + r ad chages o Q P = 5 46 Soluos (page ) Soluo (a) T = T gves a = b b a

3 (b) e P = e + lek, e P = e ( ) Posmulply by : e P = e + lek, e row of P s row of + l(row k of ) lso e P = e, ha s row of P s row of for So P s he resul of applyg r + lr o k T T T T T T (c) Qe = e, Qe = e, Q e = e (, k) Premulply by : k k T T T T T T Qe = ek, Qe k = e, Qe = e, e colums ad k of Q are colums k ad respecvely of, colum of Q s colum of (, k) So Q s he resul of applyg c c o k (d) Le P ad Q deoe respecvely he s s ad dey marces over Z The PQ = showg ( ) for all s marces over Z Suppose B There are verble P ad Q over Z wh PQ = B The P B( Q ) = showg ( ) B B sce P ad are verble over Z Suppose B ad B C There are verble P, P, Q, Q over Z wh P Q = B ad P BQ = C The ( P P ) ( Q Q ) = C showg ( ) B ad B C C as P P ad QQ are verble over Z So s a equvalece relao Soluo D = dag( a, b, c) where ( a, b, c ) s (,,),(,,),(,,),(,,4),(,,6),(,,),(,,),(,,6) as de D = abc (a) Jus oe, D = dag(,,,,5) as 5 = 5 7 s a produc of dffere prmes (b) dag(,,,,), dag(,,,,5), dag(,,,5,), dag(,,,,) e four f s > Jus oe for s = Soluo c ( a a ) c for Soluo 4 (a) c c, r r, r + r4, c4 c, c c4, c4 c, r4 + 5r gvg D = dag(,5,5,5) (b) dag(,6,5,) dag(,,,) dag(,,,6) = D, each equvalece as () usg 5 elemeary operaos (c) Le he operao σ creae a ew dagoal marx from a gve oe wh o egave eres by replacg he h ad ( + ) s dagoal eres by her gcd ad lcm as () Each σ s he produc of a mos 5 elemeary operaos ad applyg σ, σ,, σ s o D produces S( D ) (d) The proof of () goes hrough uchaged he case l ad m boh egave Suppose l ad m have oppose sgs Le gcd{ l, m} = d = al + bm where a, b Z The followg sequece of elemeary operaos chages dag( l, m ) o Smh ormal form dag( d, lm d ) : c + ac, r + br, c c, c ( l d) c, r ( m d) r s gcd{8, 4} = 6 = ( )8 + ( )( 4) we see a = b = So c c 4 r r 4 c c c c 4 7 r + 4r 7 Q

4 4 Le he operao σ for < replace he (, ) ery ad he (, ) ery a dagoal marx by her gcd ad lcm respecvely Each σ s expressble as he produc of a mos 5 elemeary operaos The sequece σ, σ,, σs, σ,, σ s,, σ s s of s( s ) operaos σ creaes D Smh ormal form ad s he composo of a mos 5 s( s ) = (5 ) s( s ) elemeary operaos Soluo 5 (a) Usg he mehod of (9), s applcaos of (7) produce a marx B = ( b ) wh eb = e The s eros r b r for < s produce B wh e = eb, T T Be = e Now use duco o s If P = he he ero rs fally gves I (b) The sequece c c, c c, r 6 r, r reduces P o I The sequece c c, c c, r r, r r, c c, c c, r 6 r, r reduces Q o I (c) P = 6 (d) The eros r r, r r, r 6 r, r, r r, r r, r r, r r reduce Q o I (e) The ecos c c, c c, c c, c c, c, c 6 c, c c, c c reduce Q o I Soluo (a) B =, B =, B =, B4 =, B5 =, B 6, D 65 7 = 7 = 7 (b) a =, a = 9, a4 = 69 O dvdg a + by a he remader s a, e a+ = a ( a ) + a So gcd{ a+, a} = gcd{ a, a} = a s a = a (4a + ), a = a (4a + ),, a = (4a + ) o subsug for he facors + a, a,, a we oba he facorsao a+ = (4a + )(4a + ) (4a + ) Hece a a r+ = (4a + ) (4a r + + )(4a r + ) ; as each facor s cogrue o modulo 4a r so also s her produc a a r + Therefore a a r+ = q ra r + where q r s a posve mulple of 4 for r > = B = B = B = B

5 5 a pplyg he ecos c a c, c c o produces B = pplyg he eros a a a a a r a r, r r o B produces B = Suppose > r > ad ha Br a a s as saed Take r eve ad so a r+ Br = a r ( a a r+ ) ( a a r+ ) The mehod of (9) ells us o apply he Eucldea algorhm o he egers colum of Br : apply he ero r a r q r + r (gvg (,) ery a r ad leavg he oher eres uchaged) for r > followed by he eros r a r r, r r obag B r for r 4 The case r odd, r s T smply he marx raspose of hs ad so he duco s compleed Therefore B or B equals a a( a a) a 5 = a a a accordg as s odd or eve Eher c ( a ) c or r ( a ) r gves B = dag(, a ) Usg he mehod of (9), for wo ecos chage o B, wo eros chage B o B, hree ecos chage B o B, hree eros chage B o B 4,, hree elemeary operaos chage B o B ad fally us oe elemeary operao chages B o B furher 5 elemeary operaos chage B o S( ) = dag(, a), amely r r, c ( q ) c where a = a a = q +, c c, c c, r + ( a ) r So a oal of + ( ) = + elemeary operaos are eeded o carry ou he algorhm of () O he oher had he four elemeary operaos r + a r, r, r r, c c chage o S( ) = dag(, a ) So s a awkward marx as far as he algorhm () s cocered! Soluo 7 (a) Suppose ( P, Q),( P, Q ) Z( D) The PP D = PDQ = DQQ shows ( P, Q)( P, Q ) = ( PP, QQ ) Z( D) lso PD = DQ DQ = P D ( P, Q) = ( P, Q ) Z( D) s ID = DI we coclude ha Z( D ) s a subgroup of G (b) Suppose ( P, Q) Z( D) Comparg eres PD = DQ gves p d = d q for all, So gves ( d d ) p = q s p d = d q for we oba p = ( d d ) q (c) P = DQ ad P = DQ gve = ( P ) DQ = ( P ) DQ ad so DQ ( Q ) = P ( P ) D, ha s, ( P, Q )( P, Q ) Z( D) (Lookg ahead hs meas ha ( P, Q ) ad ( P, Q ) belog o he same lef cose of Z ( D ) G ) (d),, =, 5

6 6 P P (e) Suppose D = dag( d, d r,, ) where dr >, > r Paro P = ad P P 4 Q Q Q = where P Q Q ad Q are r r marces The 4 ( P, Q) Z( D) ( P, Q ) Z( dag( d,, dr )), P =, Q =, P ad Q arbrary, P 4 ad Q 4 are verble ( r) ( r) marces over Z Soluos (page 4) Soluo (a) ( ) d = gcd{,gcd{75,75}} = gcd{, 5} = = ( ) d = gcd{gcd{4,66},gcd{54, }} = gcd{6,77} = = (b) Le d = gcd X, d = gcd X ad d = gcd{ d, d} For l X X eher d l or d l ad so ceraly d l Le d l for all l X X The d l for all l X ad so d d lso d l for all l X ad so d d Hece d d So d s he o-egave gcd of X X by (6), e gcd{gcd X, gcd X } = gcd{ d, d } = d = gcd X X (c) Wre d = gcd{ l, l,, l k } ad π = k The m = π d = l d = ( l d), e m as d l for k, showg ( ) Le m Z sasfy m for k So here are m Z wh m = m for k By (6) here are a, a,, ak Z wh d = al + al + + ak lk So m = m = a lm d + a lm d + + ak lkm d = al m + alm + + aklkkmk = aπ m d + aπ m d + + akπ mk d = ( am + am + + akmk ) π d as l = π for k So m m as am + am + + akmk Z showg ( ) Therefore m = lcm{,,, k } Soluo (a) K = as gcd{6,,49}= gcd{gcd{6, },49} = gcd{,49} = { k K : < k < } = { 9, 6,,,, 6,9} (b) d d d d d d d = qd ( q Z ) bd = bqd (for all b Z ) d d The deals of Z whch coa are geeraed by o-egave dvsors of ad so are,,, 5, 6,, 5, Z = ad p are he oly deals of Z coag p, as ad p are he oly o-egave dvsors of p The deals of Z coag 64 are ascedg order (c) Le k K The k = k K Suppose k K for some posve eger The ( + ) k = k + k K as K s closed uder addo So k K for all posve egers k = K lso ( ) k = k K as K s closed uder egao So mk K for all m Z ad hece K s a deal of Z

7 7 (d) We show ha K K sasfes he deal codos ( ) ( v) usg he fac ha K ad K dvdually sasfy hese codos Cosder k, k K K The k, k K ad k, k K s K ad K are closed uder addo we see k + k K ad k + k K Hece k + k K K, verfyg ( ) for K K s K ad K we see K K, verfyg ( ) for K K s k K ad k K we see k K K, verfyg ( ) for K K s bk K ad bk K we see bk K K for all b Z, verfyg ( v ) for K K Therefore K K s a deal of Z We show ha K + K sasfes he deal codos ( ) ( v) usg he fac ha K ad K dvdually sasfy hese codos Cosder k, k K + K There are k, k K ad k, k K wh k = k + k, k = k + k The k + k K ad k + k K ad so k + k = ( k + k) + ( k + k ) = ( k + k ) + ( k + k ) K + K, showg ha K + K s closed uder addo, e ( ) holds s K ad K we see = + K + K, verfyg ( ) s k K ad k K we oba k = ( k) + ( k) K + K, verfyg ( ) For b Z we kow bk K ad bk K Hece bk = b( k + k) = bk + bk K + K, verfyg ( v ) So K + K s a deal of Z By (5) here are o-egave egers d ad d wh K = d ad K = d lso by (5) here are o-egave egers d ad l wh K + K = d ad K K = l s K we see k = k + K + K for all k K, e K K + K, e d d By par (b) above d d I he same way d d ad so d s a commo dvsor of d ad d Le d be a commo dvsor of d ad d The d d ad d d by (b) above ad hece d = K + K = d + d d as he deal d s closed uder addo By (b) above d d Therefore d = gcd{ d, d} by (6) I a smlar way l = K K K = d showg d l by (b) above lso d l ad so l s a commo mulple of d ad d Le l be a commo mulple of d ad d The l d ad l d by (b) above Hece l d d = K K = l By (b) above l l Therefore l = lcm{ d, d } by Queso (c) above wh k = Soluo 5 6 (a) The umber of l mors s, e,5,,75,6 for l =,,,4,5 respecvely l l (b) ( ) g( ) =, g( ) =, S( ) = ( ) g( ) =, g( ) = 54, S( ) = 8 ( ) g( ) =, g( ) =, g( ) = 9, S( ) = (c)

8 = 9 = The mehod of () gves P = ad Q = sasfyg 4 PQ = dag(,,) = S( ) Q(ad ) P = dag(9,,) = ad S( ) Ierchagg rows ad, ad also colums ad, gves S(ad ) = dag(,, 9) (d) ( PQ )( Q(ad ) P = P( ad ) P = P(de ) IP = (de ) I Pre-mulplyg by ( PQ ) = ( dag( d,, d )) gves Q(ad ) P = dag(de d,,de d ) Takg deermas of PQ = S( ) gves de = ± d d ad as de > we see de = d d So he (, ) ery Q(ad ) P s d d d whch s he (, ) ery he dagoal marx ad S( ) Hece Q(ad ) P = ad S( ) whch dffers from S(ad ) oly ha he dagoal eres appear he oppose order, e ad has h vara facor d d d + Q(ad ) P = ad S( ) for de < ; oherwse o chage S( ) = dag( d,, d,), d > he case rak = s S( ) ad Q(ad ) P have zero produc, e ( ) (ad ) (ad ) ( ) S Q P = I = Q P S Comparg eres gves (ad ) (,,, ) Q P = dag x By () x = ± g(ad ) = ± g ( ) = ± g ( S( )) = ± dd d Hece S(ad ) = dag( dd d,,,,) I he case rak we have S( ) = dag( d,, d,,) ad so g ( ) = g ( S( )) = So ad = = S(ad ) (e) Take = ad le S( ) = dag( d, d) The by (d) above S(ad ) = dag( d, d) also So ad for all marces over Z Take = ad wre S( ) = dag( d, d, d) The S(ad ) = dag( dd, dd, dd) For d = we see ad as = = ad sce d = d = For d >, d = we have rak =, rak (ad ) = as d = ad so ad For d >, d = we see rak =, rak (ad ) = as d > ad so ad For d > he ad ad boh have rak Suppose ad The S( ) = S(ad ), e d = dd, d = dd gvg d = d = d = So s verble over Z Coversely each marx whch s verble over Z sasfes ad as ad s also verble over Z, e S( ) = I = S(ad ) Suppose The argume used he above paragraph geeralses o show ad eher = or s verble overz Soluo 4 (a)

9 9 (b) T 9 T de BB = = de BY de Y B = (de B{,} ) + (de B{,} ) + (de B{,} ) = = + + = Y 7 ( ) T + + = de BB = Each deerma s zero So x = 4, y = 6 x y x y (c) de BB T = M M where M rus hrough he s mors of B Hece de BB T = each M = gs( B) = (d) BC = B C ad so de BC = de B C = (de B )(de C ) = = by (8) as colum l of B s zero ad row l of C s zero Soluo 5 (a) ( P) Y = P Y Y So g (( P ) ) = g ( ) by() l Y l Y Suppose ca be chaged o S( ) = D = dag( d, d,, d ) usg eros oly By (4) here s a s s verble marx P over Z wh P = D The gl ( Y ) = gl ( DY) = d, he produc of he vara facors d for Y where l = Y I parcular g ( { } ) = d for ad g( ) = d d d {,,, l} l for l For he coverse we use duco o ad () Suppose here s P verble over Z wh P = dag( d, d,, d ) where Y = {,,, } If d = he g ( { } ) = d = ad so { } = Y gvg P = dag( d, d,, d,) = S( ) So we may assume d Wre b for he (, ) ery P ad d = gcd{ b, b +,, bs } By (7) rasposed here are eros leavg row of P uchaged for < ad creag a marx P P wh oly oe o-zero ery, amely he (, ) ery d, row for s where P s a verble s s marx over Z s P P has oly oe o-zero mor, amely dd d d we see dd d d = g ( P P) = g ( ) = dd d d So d = d s g ( { } ) = d, he eros r ( b d ) r for < reduce all oher eres colum o zero whou chagg P PY So here s a verble s s marx P over Z wh P P P = dag( d, d,, d ) = S( ) whch complees he duco So s reducble o S( ) usg eros oly o applyg (4) o P P P (b) () No, as he colum gcds are,, 4 bu g ( ) ± 4 as g ( ) = de = 48 () Yes, as r r, r r, r r, r r produce S( ) = dag(,,4) (c) By () dd ds = gs ( ) = So each d = ad S( ) = ( I s ) where I s s he s s dey marx There s a verble s s marx P over Z ad a verble marx Q over Z P Q S( ) I s = = So Q P ( I ) ( P ) ( I ) wh ( ) s Y P = = = s where I I s s

10 s he ( s) ( s) dey marx So = ( ) So ca be reduced o ( ) I Q s submarx of Q cossg of s frs s rows P Q = Q s verble over Z ad sasfes I s S( ) = I s usg ecos oly by (4) lso s he 6 5 ( ) Reducg (6,5) o s Smh ormal form (,,) produces Q = ( ) Reducg o produces Q = Soluo 6 (a) s de ad de B are o-zero we see de B s also o-zero s de = ± d ( ) all he vara facors of (ad for he same reaso hose of B ad B ) are posve sce oe are zero By () wh r = s = we see d ( ) ad d ( B ) are coprme dvsors of d ( B ), sce d ( ) ad k k dk ( B ) are dvsors of de ad de B respecvely Hece dk ( ) dk ( B ) s a dvsor of dk ( B ) for k Wre d ( B) = µ d ( ) d ( B) From de( B) = de de B we deduce, o akg k k k k modul, d ( B) = d ( ) d ( B) So µ k = ad hece each µ k = as µ k s a posve k k k k= k= eger So dk ( B) = dk ( ) dk ( B) for k whch gves S( B) = S( ) S( B) (b) de = ± 8, de B = ± By (a) above, S( B) = S( ) S( B) = dag(,) k = Secodly wre S( B) = dag( d, d) The d, 6 d, 4 d by () ad dd = 6 4 So eher S( B) = dag(, 4) or S( B) = dag(4,) k l = = k k l (c) Le d ( ) = p p where he expoes are o-egave Le P dag( p,, p ), dag( p,, p ) for < < l, k k = k l l l = ( l,, l ) dag p p Q k The = l, dk ( ) = p ad S( ) = dag( p,, p ) for l By par (a) above mus have hs las propery as S( ) = S( ) S( ) S( l ) ad so all choces for are equvale (e s uque up o equvalece) 4 4 (d) = P dag(, 8) Q where P = ad Q = So ake = P dag(,4) = ad = dag(,7) Q = The = s 4 7 S( ) = S( ) S( ) = S( ) S( ) = S( ) we see ha ad are ecessarly equvale k

11 Soluos (page 58) Soluo (a) The addo able of he addve group Z 5 s: s (4) =, (4) = 4, (4) =, (4) =, 4(4) = every eleme of Z 5 s a eger mulple of 4, ad so 4 geeraes Z 5 I he same way each of,,, 4 geeraes Z 5 The wo subgroups of Z 5 are {} ad Z 5 self (b) The addo able of he Z module Z 6 s: () =, 7(4) = 7( 4) = 7() = 4, 5() + (4) = By () he submodules H of Z 6 are {}, {, }, {,, 4},Z 6 The correspodg submodules K of Z are 6,,, Geeraors of he submodules H are,,, respecvely geeraes Z 6 ad 5 geeraes Z 6 (c) ( ) 4, 7, are he eger mulples of 4 Z ( ) 5, 9,,8,, 6, are he eger mulples of 5 Z The orders of 4,5 are, 7 respecvely Each of he remag elemes geeraes Z as he oly submodules of Z are =, 4 = 7, 5 = ad Z = by () Soluo (a) gcd{9,89} = s has order 89 he Z module Z 89, by (7) he order of 9= 9() s 89 gcd{9, 89} = 89 So 9 geeraes Z 89 s gcd{5,89} = 7, he order of 5 s 89 7 = 7 ad so 5 does o geerae he Z module Z 89 (b) s has order he Z module Z, by (7) he order of m = m() Z s gcd{ m, } Hece m geeraes Z m has order Z gcd{ m, } = (c) 5,, 5,, 5 are he o-geeraors of Z 5 Yes, hey are precsely he elemes of he submodule 5 Z 5 coas 5 elemes r wh gcd{ r,5}, e 5 r By (b) above each of he remag 5 5 = elemes are geeraors of Z 5

12 (d) ( ) p, ( ) p p, ( ) p p, ( v ) p l l p Soluo (a) 4 4 () = = 6 = as 6 (mod) (mod) So 6 4 () = () () = 4 = = as 6 6 () = () () = ( ) = showg ha he order of sasfes Bu s o a dvsor of eher 4 = or 6 = sce 4 6 (), () Hece = as ad are he oly prme dvsors of The eger powers of are () =, () = 4, () = 8, () =, () = 6, () =, () =, () = 9, () = 5, () =, () = 7, () = These accou for all he elemes of Z ad so geeraes Z Usg (7) mulplcave oao, () l has order gcd{ l,} So () l geeraes Z () l has order gcd{ l,} = gcd{ l,} = There are four egers l wh hs propery amely,5, 7, Hece he elemes r whch geerae Z are 4 (b) does o geerae Z 7 as () = However sasfes ad so () = () () = ( 4) = 6 = Hece has order 6 as 5 7, () = 6, () = ad () = 7 4 () = 4 as 8 4(mod7) Z The 5 subgroups of Z () = () () = ( ) = bu () So = g geeraes 7 are {},{, },{ 4,,, 4},{ 8, 4,,,,, 4, 8} ad Z 7 self, beg geeraed by () =,() =,() = 4,() = 8 ad respecvely Z 7 has 8 geeraors amely 7, 6, 5,,, 5, 6, 7, he 8 elemes of Z 7 o he subgroup of order 8 (c) so 8 (mod 7) as 6 () = () Z 7 lso 8 6 () = () Bu order 6 ad so geeraes Z = 59 = 7 7 So 8 4 = ( ) 6 (mod 7) ad = ( ) 4 (mod 7) ad so () = () () = ( ) = Hece 7 H 6 Z has mulplcave H H 8 H 4 H H 9 6 H H H 6 Le H d deoe he subgroup of Z 7 geeraed by () d for each of he 9 posve dvsors d of 6 The Hd = d ad he lace of subgroups of Z 7 has dagram as show The geeraors of Z 7 are he 6 (8 + 6) = elemes Z 7 bu o eher H 8 or H

13 (d) 5 9(mod 4) ad so has order The cogrueces order 8 By (7) he eleme ad so oba (mod 4), ( 9) = 8 (mod 4) Hece 4 (mod 4) ad (6) = () () = = lso (mod 4) ad so = (mod 4) show ha has 4 = () has order gcd{, } = lso 4 5 (mod 4) 5 (mod 4) Hece 5 has order s 6 = we 4 4 (6) () () () = = = ad (6) = () () = = So 6 has order 4 ad so 6 geeraes Z 4 as Z 4 = 4 Soluo 4 (a) Le g, g G The ( g + g) θ = c( g + g) = cg + cg = ( g) θ + ( g) θ For g G, m Z, ( mg) θ = c( mg) = ( cm) g = ( mc) g = m( cg) = m(( g) θ ) So θ s Z lear s ( g) θ G = g here s a eger c wh ( g) θ = cg For g G here s m Z wh g = mg Hece ( g) θ = ( mg) θ = m(( g) θ ) = m( cg) = c( mg) = cg So here s a eger c as saed Le c Z sasfy ( g) θ = c g for all g G The ( c c ) g = cg c g = ( g) θ ( g) θ = showg ha c c Hece ( c c ), e c c (mod ), e c s uque modulo I parcular c s uque for = ad c s arbrary for = Suppose ha θ s a auomorphsm of G s θ s surecve here s a Z wh ( ag) θ = g, e cag = g, e ( ca ) g =, e ca, e ca = b for some b Z Hece ca b = showg gcd{ c, } = Coversely suppose gcd{ c, } = There are egers a, b wh ca b = Reversg he above seps gves ( ag) θ = g ad hece ( mag) θ = mg for all m Z, showg θ o be surecve Suppose ( mg) θ = ( m g) θ for some m, m Z The cmg = cm g ad so cm cm, e c( m m ) Hece m m as gcd{ c, } = So m m s s he order deal of g we coclude m m g auomorphsm beg becve ( ) =, e mg = m g showg ha θ s ecve So θ s a The addve group Z s geeraed by he eger wh order deal ; so = ad gcd{ c,} = c = ± So Z has exacly wo auomorphsms amely m m ad m m for all m Z For > he Z module Z s cyclc beg geeraed by wh order deal By he frs par every Z lear mappg θ : Z Z s of he form ( m) θ = cm for some eger c ad all m Z s c s uque modulo we may wre cm = c m uambguously I follows drecly from he frs par wh G =Z, g =, ha θ s a auomorphsm of Z gcd{ c, } = So he addve group Z 9 has 6 auomorphsms correspodg o he 6 verble elemes c of Z 9 amely,, 4, 5, 7, 8, e he elemes c wh gcd{ c,9} = Yes, all hese auomorphsms are powers of θ sce () =, () = 4, () = 8, () = 6 = 7, () = = 5, () = 64 =, e 4 geeraes he mulplcave group of verble elemes of Z 9 So ( m) θ = 8 m, ( m) θ = 7 m ec (b) s g =, applyg ϕ gves (( g) ϕ) = ( g) ϕ = () ϕ = showg, e Suppose frs ha θ : G G s Z lear ad ( g) θ = g The ( mg) θ = m(( g) θ ) = mg for all m Z ad so here s a mos oe such θ Cosder θ : G G gve by ( mg) θ = mg for all

14 4 m Z Le m g = mg The ( m m) as m m sce ( m m ) g = s d we deduce d ( m m ) So ( m m ) g = as d s he order deal of g So m g = mg showg ha θ s uambguously defed lso θ s addve as ( mg + m g ) θ = (( m + m ) g ) θ = ( m + m ) g = mg + m g = ( mg ) θ + ( m g ) θ for m, m Z s ( m( m g)) θ = (( mm ) g) θ = ( mm ) g = m( m g ) = m(( m g) θ ) we see θ s Z lear (c) Wh g = g = () we oba ( + ) θ = () θ + () θ, e () θ = () θ + () θ as + = dd ()θ, he egave G of ()θ, o boh sdes obag = () θ + () θ = () θ + () θ + () θ = + () θ = () θ pply θ o g + g = ad use () o oba ( g) θ + ( g) θ = ( g + g) θ = () θ = whch meas ( g) θ = ( g) θ for all g G The eger m s he order deal of r mr = Z mr = q for some q Z m ( r gcd{ r, }) = q( gcd{ r, }) ( gcd{ r, }) m Therefore gcd{ r, } s he order deal of r Z For r Z here s a uque Z lear mappg θ : Zm Z wh () θ = r ( gcd{ r, }) m Hece ( gcd{ m, }) gcd{ r, } ad so ( gcd{ m, }) r as gcd{ r, } r Coversely ( gcd{ m, }) r ( gcd{ r, }) m he same way So here are gcd{ m, } choces for r Z amely r = l( gcd{ m, }) for l gcd{ m, } (d) ( g + g) θθ = (( g) θ + ( g) θ ) θ = ( g) θθ + ( g) θθ for all g, g G ad ( mg) θθ = (( mg) θ ) θ = ( m(( g) θ )) θ = m((( g) θ ) θ ) = m(( g) θθ ) for all m Z, g G So θθ s Z lear Suppose θ becve The ( g + g ) θ θ = g + g = ( g ) θ θ + ( g ) θ θ = (( g ) θ + ( g ) θ θ as θ s Z lear s θ s ) ( g + g ) θ = ( g ) θ + ( g ) θ for all g, g G lso ) ecve (( mg ) θ ) θ = ( mg ) θ θ = mg = m(( g ) θ θ ) = ( m(( g ) θ ) θ ad as θ s ecve ( mg ) θ = m(( g ) θ ) for all m Z, g G So θ s Z lear Le θ, ϕ, ψ be auomorphsms of G The θϕ ug by he above heory wh G = G = G ad θ = ϕ lso ( θϕ) ψ = θ ( ϕψ ) as composo of mappgs s assocave The dey ι : G G s ug ad ιθ = θ = θι for all θ ug For each θ ug we see θ ug ad θ θ = ι = θθ Hece ug s a group From (a) above u Z 9 s cyclc of order 6 wh geeraor θ However u Z 8 = { θ, θ, θ5, θ7} s o cyclc as θ = θ5 = θ7 = θ, he dey eleme of u Z 8 Soluo 5 (a) H s closed uder addo sce g + g = ( g + g ) as G s closed uder addo ( g, g G) H coas he zero of G as = + = H s closed uder egao sce g = ( g) as G s closed uder egao Le k, k K The ( k + k ) = k + k = + = So k + k K = ad so K ( k) = k = = So k K Therefore H ad K are subgroups of G ad hece are submodules of G, e mh H ad mk K for all m Z, h H, k K ( ) Take G =Z The H = {}, K =Z ad so H K ( ) Take G =Z 4 The H = {, } = K ( ) Take G =Z 8 The H = {,, 4, 6}, K = {, 4} ad so K H ( v ) Take G =Z 6 The H = {,, 4}, K = {, } ad so H K ad K H

15 5 (b) K s a Z module ad scalar mulplcao by elemes m of Z s uambguously defed by mk = mk as k = for all k K Hece K s a vecor space over Z For odd, K = {} ad dm K = For eve, K = {, } ad so dm K = Soluo 6 (a) ( mq + mq ) + ( m q + m q) = ( m + m ) q + ( m + m ) q So q, q s closed uder addo = q + q q, q ( mq + mq ) = ( m ) q + ( m ) q So q, q s closed uder egao m( mq + mq ) = ( mm ) q + ( mm ) q So q, q s closed uder eger mulplcao Therefore q, q s a submodule of he Z module Q (b) 6 = ( ), Hece 6, s = 9( 6), = 4( 6) we see, 6 sce m ( ) + m ( ) = (9m + 4 m )( 6) So 6 =, showg ha, s cyclc wh geeraor 6 (c) s gcd{ a, a } = ad gcd{ a, b } = we deduce ha gcd{ a, a b } = Smlarly gcd{ b, a b } gcd a b, a b = Therefore sa b + a b = for s, Z Hece = ad so ( ) q = gcd{ a, a }( sa b + a b )gcd{ b, b } b b = ( sa b + a b ) b b = sq + q q, q ad so q q, q s q q ( a gcd{ a, a})(lcm{ b, b} b ) q, q so q, q q Hece q, q q q q q = Z we deduce ha q ad = s cyclc wh geeraor q ad,, = q, q s also cyclc (d) s = gcd{6, 75} ad 7 = gcd{5, 56} we oba lcm{5,56} = = 8 So 8 geeraes 6 5, by (c) above s = gcd{,8} ad 5 = gcd{8,5} we see ha lcm{8,5} = 84 ad so 84 geeraes 8, 8 5 = 6 5,75 56, 8 5 Z 8 = ad so Z 6 5,75 56 Z 84 = = Z ad so Z 6 5,75 56,8 5 Soluo 7 (a) ( ) Le h, h H H The h, h H ( =,) ad so h + h H as H s closed uder addo So h + h H H showg ha H H s closed uder addo H ( =,) ad so H H h H ( =,) as H s closed uder egao ad so h H H Therefore H H s a subgroup of G ( ) ( h + h ) + ( h + h ) = ( h + h ) + ( h + h ) H + H for all h, h H ( =,) = + H + H ( h + h ) = ( h ) + ( h ) H + H So H + H s a subgroup of G ( ) Suppose o Pck h H, h H, h H, h H The h + h H H as H H s closed uder addo Bu h + h H mples h = h + ( h + h ) H (a coradco) ad h + h H mples h = ( h + h ) h H (a coradco) H H s eher H or H, boh of whch are subgroups of G (b) Subgroups of he addve group Z are cyclc, beg prcpal deals of he rg Z by (5) So H H =, H + H = More geerally m m ad m + m are cyclc beg geeraed by lcm{ m, m} = mm gcd{ m, m} ad gcd{ m, m } respecvely Soluo 8

16 6 (a) For = we have s = ( g + g) + g = g + ( g + g) by he assocave law Take > ad suppose ducvely he resul o be rue for all ordered ses of less ha elemes of G Each summao of g, g,, g order decomposes h + h for some wh < where h s a summao of g, g,, g order ad h s a summao of g+, g+,, g order By duco h = s ad h = s where s = ( (( g+ + g+ ) + g+ ) ) + g = s + g say Hece h + h = s + ( s + g) = ( s + s ) + g s s + s s a summao of g, g,, g we deduce s + s = s by duco Therefore h + h = s + g = s whch complees he duco Each summao of g, g,, g order s equal o s So he geeralsed assocave law of addo holds (b) By he commuave law g + g = g + g Take > ad suppose he resul s rue for all ses of less ha elemes of G Each summao of g, g,, g decomposes h + h for some wh < where h s a summao of g, X, X = ad h s a summao of g, Y, Y =, X Y = Ierchagg h ad h f ecessary, we may assume Y By duco h = h + g where s a summao of g for Y /{ } ad so h h + h = s by duco The duco s compleed by h + h = h + ( h + g ) = ( h + h ) + g = s + g = s (c) For m by (b) above m( g + g) = mg + mg o addg up he m elemes g, g,, g ( =,) wo ways For m < wre m = The m( g + g ) = ( g + g ) = g + ( g ) = mg + mg If mm = he ( m + m ) g = m g + mg By symmery we may assume m m For m >, m > usg (a) above wh g = g, ( m + m ) g = sm + m = sm + sm = m g + m g For m = <, m = < we have ( m + m ) g = ( + ) g = g + ( g) = m g + mg For m >, m = <, m + m >, ( m + m ) g = sm + m = sm s = m g g = m g + m g For m >, m = <, m + m = <, ( m + m ) g = g = s = s = ( s s ) = s s = m g g = m g + m g Now m m m ( mm ) g = = m ( mg) for mm = For m >, m >, by (a) above, ( mm ) g = s = m ( mg) Hece for m = <, m = <, m m ( m m ) g = (( )( )) g = ( ) g = ( g) = ( )( g) = m ( m g) m >, m = <, For ( m m ) g = ( m ) g = (( m ) g) = ( m ( g)) = m ( g) = m ( m g) m = <, m >, For ( m m ) g = ( m ) g = (( m ) g) = ( ( m g)) = ( )( m g) = m ( m g)

17 7 Soluos (page 7) Soluo (a) K = K + = {, 4}, K + = {, 5}, K + = {, 6}, K + = {,7} K + geeraes G K as r( K + ) = K + r for r < 4 G K has somorphsm ype C 4 (b) K = K + = {,, 6,9}, K + = {, 4,7,}, K + = {,5,8,} K + geeraes G K as r( K + ) = K + r, r < G K has somorphsm ype C (c) K = {8,, 6, } So K = 4 G K = G K = 4 4 = 6 K + geeraes G K whch has somorphsm ype C 6 (d) K = d where d = gcd{ m, } G K = G K = ( d) = d G K s cyclc beg geeraed by K + ad of somorphsm ype C d Soluo (a) The cyclc subgroup d has order d by (7), ad so d has dex ( d) = d Z Coversely le K be a subgroup of dex d Z So K = d By () K = d ad so he addve abela group Z has a uque subgroup of dex d (b) d has dex d Z as he coses of d Z are d + r for r < d Every o-zero subgroup K of he addve group Z s a deal of he rg Z, ad so s of he ype d, where d s a posve eger, by (5) So d s he oly subgroup of Z havg dex d No, bu s he oly subgroup of fe dex Z (c) Le G = g Every cose of K G s of he form K + mg = m( K + g) for some m Z So G K = K + g s cyclc Soluo (a) Z + has order as Z + Z, ( Z + ) = Z + Z, bu ( Z + ) = Z + = Z he zero eleme of Q Z Smlarly Z has order 8 lso Z + m, where gcd{ m, } =,, has order So every eleme of QZ has fe order (b) a( Z + m ) K Bu a( Z + m ) = Z + am = Z b + = Z + So Z + K Smlarly Z + K There are egers a, b wh a + b = d Hece b ( Z + ) + a ( Z + ) = Z + b + a = Z + d So Z + d K So d by he maxmaly of Hece d So = d ad Therefore q = ad Z + m = Z + qm = qm ( Z + ) So K = Z + (c) K Z + s a subgroup of QZ havg order Coversely le K be a subgroup of QZ wh = By (b) above K = Z + s s s+ (d) s ( Z + l ) + ( Z + m ) = Z + ( l + m ) K we see ha K s closed uder addo K coas Z + (pu l = ) he zero eleme of QZ, ad K s closed uder egao (replace l by l ) So K s a subgroup of QZ Each o-zero eleme of K s uquely expressble Z + l s s, l odd, l <, s > So K has a fe umber of elemes, represeaves beg,, 4, 4, 8, 8, 5 8, 7 8, ec K s o cyclc as QZ coas o elemes of fe order The fe subgroups of K are H H H s where H s s = Z + Le H

18 8 s be a subgroup of K ad le S = { s : Z + H} If S s bouded above, he H = H where = max{ s : s S} If S s ubouded he S s he se of all o-egave egers ad H = K s he oly fe subgroup of K Soluo 4 (a) Omg subscrps, he elemes g Z Z 4 for are (, ), (, ), (, ), (, ), (, ),(, ),(, ), (, ), (, ), (, ), (, ), (, ), e all elemes of Z Z 4 So g geeraes he addve abela group Z Z 4 whch s herefore cyclc of somorphsm ype C (b) Wre g = (, ) for =,, ad le g = (, ) The 8 o-zero elemes of G are ± g, ± g, ± g, ± g s g, g = g, g = for we see ha g ad g have order The 4 subgroups of order are g for G s a Z module, ad as every o-zero eleme has order, we see ha G s a Z module, e G s a vecor space over Z G = g, g has dmeso ad s he eral drec sum of ay wo dffere dmesoal subspaces, e G = g g = g g = g g g g g g g g = = = The addve abela group G has somorphsm ype C C (c) Wre d = gcd{, } The l( g, g) = ( lg, lg) = (( d) g,( d) g ) = (, ) = So ( g, g ) has fe order say where l lso ( g, g ) = (, ) ad so g = ad g = Hece ad So = q ad ( d) ( d) q o dvdg q hrough by d s gcd{ d, d } = we deduce ( d) q So ( d) q, e l Hece l = So he order of ( g, g ) s l = lcm{, } (d) Wre s = gcd{ s, m}, = gcd{, } By (7) he orders of s = s( ) ad = ( ) are m s ad respecvely By (c) above ( s, ) has order m s as gcd{ m s, } = Bu m m s = m s = s = = I he case m = 7, = 8 here are 6 = φ(7) choces for s 7 ad 4 = φ(8) choces for 8 Hece Z7 Z 8 has 6 4 = 4 geeraors, e here are 4 elemes of order 56 hs group (e) s m( g + h) = mg + mh = + m = we see ha g + h has fe order l where l m Now l(g + h) = ad so lg = lh Hece l g = ( lh) = lh = l = showg ha he order m of g s a dvsor of l, e m l s gcd{ m, } = we deduce m l I he same way we oba l ad so m l usg gcd{ m, } = aga Therefore m = l Noe ha G = K G K = m Replacg ϕ Exercses, Queso 4(b) by he aural homomorphsm η :G G K, we see ha he order s of h s a dvsor of he order of ( h ) η = K + h So h = ( s ) h has order By he above g + h has order m, as g has order m where K = g Therefore g + h geeraes G, e G = g + h s cyclc (f) Le g, g, g G ad g, g, g G ddo G G s assocave as (( g, g ) + ( g, g )) + ( g, g ) = ( g + g, g + g ) + ( g, g ) = (( g + g ) + g,( g + g ) + g ) = ( g + ( g + g ), g + ( g + g )) = ( g, g ) + ( g + g, g + g ) = ( g, g ) + (( g, g ) + ( g, g )) m m

19 9 The zero eleme of G G s (, ) sce (, ) + ( g, g) = ( + g, + g) = ( g, g) The egave of ( g, g ) s ( g, g) as ( g, g) + ( g, g) = ( g + g, g + g) = (, ) ddo G G s commuave as ( g, g ) + ( g, g ) = ( g + g, g + g ) = ( g + g, g + g ) = ( g, g ) + ( g, g ) So G G s a addve abela group Cosder α : G G G G defed by ( g, g) α = ( g, g) for all g G, g G The α : G G G G Soluo 5 (a) s r 7(mod) he possbles for r wh r < 4 are 7,8, 9, 4,5, 6, 7,84,95,6,7,8,9, whereas for r 6(mod ) he ls s 6,9,, 45,58, 7,84,97,,,6 So r = 84 leravely = 6 5 ad so r = = 59 = 84 (b) I a feld, he soluos of x = x are ad s Z ad Z are felds, he soluos of x = x he rg Z Z are (, ),(, ), (, ), (, ) Usg () he soluos of Z are,78, 66,, as 78 (mod), 78 (mod ) ec 4 (c) The soluos of x = he rg x = x Z Z are (, ), (, ), (, ), (, ) Usg he rg somorphsm α : Z4 Z Z of () we oba he soluos ±, ± of Z 4 The soluos of x = x are,, Z ad Z The pars (, ), (, ) Z Z correspod o 65, 77 Z 4 Usg (b) above, he soluos of x x = = x Z 4 are, ±, ±, ± 65, ± 66 (d) Usg Exercses, Queso 4(b) he umber of Z lear θ : Z Z5 Z 5 s 5, as for each r Z 5 here s a uque Z lear θ wh (, 5 ) θ = r sce he (addve) order of r s a dvsor of he order 5 of (, 5 ) Of hese 8 = φ(5) are group somorphsms (hose wh gcd{ r,5} = ) ad us oe ( r = ) s a rg somorphsm Soluo 6 (a) Suppose Z = m + m + + m There are egers a, a,, a wh = am + am + + am Le d = gcd{ m, m,, m } The d m for all wh So d ad so d = Coversely suppose d = There are egers a as above ad hece m = ma m + ma m + + ma m showg Z = m + m + + m as mam m for s gcd{5,6, 4} = ad gcd{5, 6,8} = he aswers are No ad Yes (b) Suppose o he corary ha he addve group Z has o-rval subgroups H ad H such ha Z = H H s H ad H are deals of he rg Z, by (5) here are posve egers ad wh H = ad H = Bu = + = + ( ), e he eger zero s expressble wo dffere ways as a sum of egers from H ad H So Z s decomposable (c) Suppose h + h = where h H, h H The h = h showg h H as H s closed uder egao So h H H = {} ad hece h = Therefore + h = gvg h = So H, H are depede submodules of G Suppose gve submodules H of G as saed for

20 ad suppose h + h + + h + h = where h H Replacg H, H he frs par by H + H + + H, H we deduce h + h + + h = ad h = So h = h = = h = by he depedece of H, H,, H Hece he submodules H, H,, H, H are depede as h = for Each eleme of H H H ca be expressed uquely he form h + h + + h wh h H There are H choces for each h ad so H H H = H H H (d) ypcal eleme of G = Z Z 9 s ( r, s 9) where r, s 9 There are 8 = 6 = φ(9) elemes ( r, s 9) of order 9 as here are choces for r ad φ (9) choces for s by 4( d) above The remag 8 o-rval elemes have order as all elemes of G have orders whch are facors of 9 Each cyclc subgroup of order 9 coas φ (9) elemes of order 9 ad so G coas 8 φ (9) = 8 6 = such subgroups, amely (, 9 ), (, 9 ), (, 9 ) Smlarly G has 8 φ () = 8 = 4 (cyclc) subgroups of order amely (, 9), (, 6 9), (, 9 9), (, 9) Each of he cyclc subgroups H of order 9 coas us oe subgroup of order amely (, 9 ) So for each H here are subgroups H of order wh H H = {} There are = 9 such depede pars H, H ad for each G = H H as H H = H H = 9 = 7 = G (e) Suppose k + k + + k = where k K for s k H for ad H, H,, H are depede, we see k = k = = k = By (4) he submodules K, K,, K are also depede So K = K K K Suppose K + h = K + h where h, h H There are uque elemes h, h H for wh h = h + h + + h ad h = h + h + + h s h h K here are uque elemes k K for wh h h = k + k + + k Bu h h = ( h h ) + ( h h ) + + ( h h ) whch s he oly way of expressg h h as a sum of elemes, oe from each H, s K H we deduce h h = k, e K + h = K + h for So α s uambguously defed Cosder ow ay h, h H ad le h = h + h + + h, h = h + h + + h where h, h H for The (( K + h) + ( K + h )) α = ( K + ( h + h )) α = ( K + ( h + h ),, K + ( h + h )) = ( K + h,, K + h ) + ( K + h,, K + h ) = ( K + h) α + ( K + h ) α showg α o be addve Each uple ( K + h, K + h,, K + h ) ca be wre ( K + h) α where h = h + h + + h So α s surecve Suppose ( K + h) α = ( K + h ) α The h h = k K for ddg hese equaos gves h h = ( h h ) + ( h h ) + + ( h h ) = k + k + + k K showg K + h = K + h So α s ecve Therefore α : H K ( H K) ( H K) ( H K )

21 Soluos (page 9) Soluo (a) ( ) Wre K = kerθ ad le k, k K The ( k + k ) θ = ( k) θ + ( k ) θ = + = showg ha k + k K, e K s closed uder addo lso ( k) θ = ( k) θ = = ad () θ = showg ha k K ad K s ( mk) θ = m(( k) θ ) = m = for m Z, we coclude ha mk K ad so K s a submodule of he Z module G Suppose K = {} ad le g, g G sasfy ( g) θ = ( g) θ The ( g g) θ = ( g) θ ( g) θ = ( g) θ ( g) θ = showg g g K So g g =, e g = g ad θ s ecve Coversely suppose ha θ s ecve ad le k K The ( k) θ = = () θ So k = by he ecvy of θ gvg K = {} ( ) Le g, g mθ The g = ( g) θ ad g = ( g) θ for some g, g G The g + g = ( g) θ + ( g) θ = ( g + g) θ mθ as g + g G lso g = ( g) θ = ( g) θ mθ as g G s = () θ mθ ad mg = m(( g) θ ) = ( mg) θ mθ for all m Z, we coclude ha mθ s a submodule of he Z module G Yes, mθ = G s he same as θ beg surecve (b) s (, ) θ = 4 = we see ha (, ) kerθ Yes, (, ) = (,) kerθ ad (, 4) = (, ) kerθ s m(, ) kerθ for all m Z we see (, ) kerθ Suppose ( l, m) kerθ The 4l m = ad so m = l Hece ( l, m) = ( l, l) = l(, ) (,) ad so ker θ (, ) We coclude ker θ = (, ) ll eve egers belog o mθ as (, m) θ = m s ( l, m) θ = ( l m) s eve we see mθ = whch s fe cyclc Yes as (ker θ + (, )) ɶ θ = (, ) θ = 4 = 8 ad (ker θ + (7, )) ɶ θ = (7,) θ = 4 7 = 8 By (6) ɶ θ : Z Z ker θ m θ ad so Z Z kerθ s fe cyclc wh geeraor ker θ + (,) (c) ( ) The addve group Z8 = Z 8 has subgroups,, 4, 8 ad he homomorphc mages of Z 8 are Z8, Z8, Z8 4, Z 8 8 whch are cyclc of somorphsm ypes C, C, C4, C 8 respecvely ( ) The addve group Z = Z has subgroups,,, 4, 6, ad he homomorphc mages of Z are Z, Z, Z, Z 4, Z 6, Z whch are cyclc of somorphsm ypes C, C, C, C4, C6, C respecvely ( ) The addve group Z = Z where > has subgroups d where d > ad d by () So a ypcal homomorphc mage of Z s Z d whch s cyclc of somorphsm ype C d ( v ) The addve group Z has subgroups d where d by (5) So Z d s a ypcal homomorphc mage of Z Z d s cyclc of somorphsm ype C d ( v ) The Kle 4 group Z Z = u, v has subgroups {}, u, v, u + v, Z Z The homomorphc mages of Z Z are Z Z {}, Z Z u, Z Z v, Z Z u + v, Z ZZ Z

22 ad hese are of somorphsm ypes C C, C, C, C, C respecvely (d) The dey homomorphsm ι : G G, gve by ( g) ι = g for all g G, has ker ι = {} ad mι = G So ɶ ι : G {} G by (6) The rval homomorphsm ο :G G, gve by ( g) ο = for all g G, has kerο = G ad m ο = {} So G G {} by (6) The proeco π : G G G gve by ( g, g) π = g for all ( g, g ) G G s a homomorphsm s mπ = G we see drecly ha G s a homomorphc mage of G G Smlarly π :G G G gve by ( g, g) π = g for all ( g, g ) G G s a homomorphsm wh mπ = G ad so G s a homomorphc mage of G G The mappg θ : G G G, gve by ( g, g ) θ = g g for all g, g G s a homomorphsm s kerθ = K ad mθ = G we deduce ( G G ) K G by (6) ad so he aswer s: Yes! (e) s ( g ( g) θ ) θ = ( g) θ ( g) θ = ( g) θ ( g) θ = we see g ( g) θ kerθ s ( g) θ mθ he equao g = ( g ( g) θ )) + ( g) θ shows G = kerθ + mθ To show ha kerθ ad mθ are depede submodules of G suppose k + l = where k ker θ, l mθ The l = ( g) θ for some g G pplyg θ o k + ( g) θ = gves ( k + ( g) θ ) θ = () θ = ad so ( k) θ + ( g) θ = whch gves + ( g) θ =, e l = Hece k + = ad so k = Therefore kerθ ad mθ are depede submodules of G ad so G = kerθ mθ by (5) For G = Z Z 4 we have l m = m l m = l m m l m = m m l = m l m Z Z as (, ) θ (, ) θ (, ( )) (, ) (, ) l = Z ad l l 4 4 = Z So θ s dempoe By speco ker θ = (, ) s cyclc of order ad m θ = (, ) s cyclc of order 4 Soluo (a) ( ) Cosder h, h H There are h, h H wh h = ( h ) θ, h = ( h ) θ The h + h = ( h ) θ + ( h ) θ = ( h + h ) θ H sce h + h H lso h = ( h ) θ = ( h ) θ H ad = () θ H as h, H So H s a subgroup of G s ker θ H s he kerel of θ ad H H s he mage of θ H, applyg (6) o θ gves H (ker θ H) H H ( ) Le h, h H The h + h H as ( h + h ) θ = ( h ) θ + ( h ) θ H sce ( h ) θ,( h ) θ H lso h, H as ( h ) θ = ( h ) θ H ad () θ = H So H s a subgroup of G I hs case he kerel of θ s kerθ ad he mage of θ s H mθ Replacg θ by θ (6) ow H H H gves H kerθ H mθ (b) s (,) θ = = ad (,) θ = = we see v = (,), v = (,) kerθ Suppose mv + mv = = (, ) for m, m Z The m + m = ad m = : so m = m = showg v, v o be Z depede Le ( l, m) kerθ The l m = l m = k Hece ( l, m) = mv + kv showg ha v, v geerae kerθ So, Z ad so here s k Z wh v v s a Z bass of kerθ s = (,)θ ad = (, )θ we see ha Z = mθ By (5) we have ɶ θ : Z Z kerθ Z ad so kerθ Z Z has somorphsm ype C s (,) kerθ we see kerθ Z Z s s prme, by Lagrage s heorem here are o subgroups H wh

23 {} H Z By (7) here are o subgroups H wh kerθ H Z Z, e kerθ s a maxmal subgroup of Z Z (c) s (, ) θ = + ( ) = ad (, 4) θ = 4 = we oba v = (, ), v = (,4) kerθ s (b) above, v, v are Z depede Le ( l, m) kerθ The l + m = Z 4 ad so here s k Z wh l + m = 4k Hece ( l, m) = lv + kv showg ker θ = v, v So kerθ has Z bass v, v s ( l,) θ = l Z 4 for l =,,, we see mθ =Z 4 By (5) we oba ( Z Z) kerθ Z 4 ad so ( Z Z ) kerθ has somorphsm ype C 4 s Z 4 has exacly subgroups H, amely 4,,, by (7) here are correspodg subgroups of Z Z coag kerθ, amely ker θ, H, Z Z, where H = {( l, m) Z Z : l + m } has Z bass (, ),(,) (d) s ( l, m) kerθ l = l, m = 4 m ( l, m Z ) ( l, m) = l ( e ) + m (4 e ) ( l, m) e,4e we see kerθ = e,4e, e kerθ has Z bass e, 4e For H = (, ) we have ( l, m) H ( l, m) θ (, ) ( l, m) {(,),(,)} l arbrary (ay eger), ρ m = 4m So ρ = (,), ρ = (,4) s a Z bass of H ad = = has vara facors ρ 4 d =, d = 4 G H = H + (, ) s cyclc of order 4 ad so of somorphsm ype C C4 = C4 Smlarly for H = (, ) we have = ; so d = d = ad G H = H + (, ), H + (, ) whch has somorphsm ype C C For H = (, ) we see = ; so d =, d = ad G H = H + (, ) whch has somorphsm ype C C = C (e) s θ η s surecve, beg he composo of wo surecve mappgs, we see mθη = G H lso ker θ η = { h G : ( h) θ η = H } as H s he zero eleme of G H So ker θ η = { h G : ( h) θ H } = H pplyg (6) o θ η gves θ η :G H G H Soluo (a) There are k, k K such ha r = r + k, r = r + k Therefore r r = ( r + k )( r + k ) = r r + k where k = r k + kr + k k s K s a deal of R we see ha r k, k r, k k K, ad so k K as K s closed uder addo Hece r r r r (mod K) ad so K + r r = K + r r by (9), showg ha cose mulplcao s uambguously defed Wre r = K + r ad he R K has bary operaos r + r = r + r ad ( r )( r ) = r r where r, r R Now ( R K, + ) s a abela group by () Le r, r, r R The (( r )( r ))( r ) = ( r r )( r ) = ( r r ) r = r ( r r ) = ( r )( r r ) = ( r )(( r )( r )) showg ha cose mulplcao s assocave Cose mulplcao s dsrbuve because (( r ) + ( r ))( r ) = ( r + r )( r ) = ( r + r ) r = r r + r r = r r + r r = ( r )( r ) + ( r )( r ) ad smlarly ( r )(( r ) + ( r )) = ( r )( r ) + ( r )( r ) lso ( e)( r) = er = r = re = ( r)( e) for all r R, ad so R K s a rg wh eleme e = K + e

24 4 By () η s addve s ( r r ) η = r r = ( r )( r ) = ( r ) η ( r ) η for all r, r R ad ( e) η = e we see η s a rg homomorphsm lso mη = R K ad kerη = K (b) By Queso(a)( ) above, mθ s a subgroup of ( R, + ) s ( r ) θ ( r ) θ = ( r r ) θ for all r, r R we see mθ s closed uder mulplcao The eleme e of R belogs o mθ as e = ( e) θ So mθ s a subrg of R ad hece mθ s self a rg By Queso(a)( ) above, kerθ s a subgroup of ( R, + ) Cosder r R, k K ; he ( rk) θ = ( r) θ ( k) θ = ( r) θ = ad so rk K = kerθ Smlarly kr K ad so K s a deal of R Kerels of rg homomorphsms are deals By (6) ɶ θ : R K mθ s a somorphsm of addve abela groups lso (( r )( r )) ɶ θ = ( r r ) ɶ θ = ( r r ) θ = ( r ) θ ( r ) θ = ( r ) ɶ θ ( r ) ɶ θ for all r, r R So ɶ θ s a rg somorphsm as ( e) ɶ θ = e Therefore ɶ θ : R K mθ (c) By (b) above kerθ s a deal of he rg Z By (5) here s a o-egave eger d wh kerθ = d By (b) above ɶ θ : Z d mθ, showg ha he rgs Z d = Z d are, up o somorphsm, he (rg) homomorphc mages of Z (d) For r, r R we see ( r + r ) θθ = (( r ) θ + ( r ) θ ) θ = ( r ) θθ + ( r ) θθ ad ( r r ) θθ = (( r ) θ ( r ) θ ) θ = ( r ) θθ ( r ) θθ showg ha θθ s addve ad mulplcave s ( e) θ = e ad ( e ) θ = e we see ( e) θθ = e where e, e, e are he elemes of R, R, R So θθ : R R s a rg homomorphsm Suppose ha θ s a rg somorphsm Cosder r, r R ad wre r = ( r ) θ, r = ( r ) θ The ad so (( r ) θ + ( r ) θ ) θ = ( r + r ) θ = ( r ) θ + ( r ) θ = r + r = ( r + r ) θ θ ( r ) θ + ( r ) θ = ( r + r ) θ as θ s ecve, ad θ s addve Smlarly r θ r θ θ = r r θ = r θ r θ = r r = r r θ θ ad so ( r ) θ ( r ) θ = ( r r ) θ θ s mulplcave s θ = we coclude θ : R R s a rg (( ) ( ) ) ( ) ( ) ( ) ( ) as θ s addve, ad ( e ) e somorphsm Take R = R = R ad suppose θ, θ are becve, e suppose θ, θ u R By he above heory θθ, θ u R s he dey mappg ι R of R belogs o u R we see ha u R s a group ( s a subgroup of he group of all becos of R R ) (e) By Exercses, Queso 4(f ) he drec sum R R of he addve groups of R ad R s self a addve group I order o verfy he rg axoms volvg mulplcao, cosder r, r, r R ad r, r, r R The ( r, r )(( r, r ) + ( r, r )) = ( r, r )( r + r, r + r ) = ( r ( r + r ), r ( r + r )) = ( r r + r r, r r + r r ) = ( r r, r r ) + ( r r, r r ) = ( r, r )( r, r ) + ( r, r )( r, r ) whch shows ha oe dsrbuve law holds R R The oher dsrbuve law holds R R ad ca be verfed he same way The assocave law of mulplcao holds R R as (( r, r )( r, r ))( r, r ) = ( r r, r r )( r, r ) = (( r r ) r,( r r ) r ) = ( r ( r r ), r ( r r )) = ( r, r )( r r, r r ) = ( r, r )(( r, r )( r, r )) usg hs law he rgs R ad R Le e ad e deoe he elemes of R ad R respecvely The ( e, e )( r, r ) = ( er, er ) = ( r, r ) = ( r e, re ) = ( r, r )( e, e ) whch shows ha R R has eleme ( e, e ) Therefore R R s a rg

25 5 (f) By Exercses, Queso 7( a)( ) ad ( ) boh K L ad K + L are addve abela groups Cosder r R ad m K L The m K ad m L s K s a deal of R we see rm, mr K s L s a deal of R we see rm, mr L So rm, mr K L ad K L s a deal of R Cosder r R, m K + L The m = k + l where k K ad l L So rm = r( k + l) = rk + rl K + L sce rk K ad rl L as before lso mr = ( k + l) r = kr + lr K + L sce kr K ad lr L So K + L s a deal of R For r, r R usg addo ad mulplcao he rgs R K, R L ad R K R L ( r + r ) α = ( r + r + K, r + r + L) = (( r + K) + ( r + K),( r + L) + ( r + L)) = ( r + K, r + L) + ( r + K, r + L) = ( r ) α + ( r ) α ad ( r r ) α = ( r r + K, r r + L) = (( r + K)( r + K),( r + L)( r + L)) = ( r + K, r + L)( r + K, r + L) = ( r ) α ( r ) α Le e be he eleme of R s ( e) α = ( e + K, e + L) s he eleme of R K R L we see ha α s a rg homomorphsm The eleme of R K R L s ( K, L ) s ( r) α = ( K, L) ( r + K, r + L) r K, r L r K L we see kerα = K L Now we use K + L = R o fd mα : here are elemes k K ad l L wh k + l = e Cosder a arbrary eleme ( s + K, + L) of R K R L ad so s, R Wre r = sl + k The r s = r se = s( l e) + k = s( k) + k = ( s) k K ad so r + K = s + K lso r = r e = sl + ( k e) = sl + ( l) = ( s ) l L ad so r + L = + L Therefore ( r) α = ( r + K, r + L) = ( s + K, + L) ad mα = R K R L By (b) above ɶ α : R ( K L) R K R L s a rg somorphsm where ( r + K L) ɶ α = ( r + K, r + L) for all r R Soluo 4 (a) Suppose ha K s ormal G Le g G ad cosder kg Kg where k K The kg g( g kg) gk = as g kg K So Kg gk Replacg g by g he ormaly codo gves gkg K for all k K Cosder gk gk The gk = ( gkg ) g Kg So gk Kg ad hece Kg = gk Coversely suppose Kg = gk for all g G For k K, g G we have kg Kg ad so kg gk There s k K wh kg = gk Hece k K, g G g kg k K =, e g kg K for all For g S he permuao g σ g s eve usg he rules of pary, as g ad g have he same pary ad σ s eve So g σ g σ showg ha σ s ormal S s () τ =,() τ = ad () σ τσ = () τσ = () σ = we see σ τσ τ So τ s o a ormal subgroup of S Suppose Kg = Kg ad Kg = Kg Usg he above heory we oba Kg g = Kg g = g g K = g g K = Kg g showg ha cose mulplcao s uambguously defed So G K s closed uder cose mulplcao Le e deoe he dey eleme of G ad le g, g, g, g G The ( KgKg ) Kg = K( gg ) g = Kg( gg) = Kg( KgKg) showg ha cose mulplcao s assocave s KeKg = Keg = Kg = Kge = KgKe we see ha K = Ke s he dey eleme of

26 6 G K s Kg Kg = Kg g = Ke = Kgg = KgKg we see ha G K s a group (b) G = {,, 4, 7, 8,,, 4} For K = {, 4} we have Kg s he verse of Kg So K = {, 8}, K 7 = {7, }, K = {, 4} whch paro G The mulplcao able of G K s K K K 7 K K K K K 7 K K K K K K 7 K 7 K 7 K K K K K K 7 K K For example K K 7 = K 4 = K The paer he able s he same as ha he addo able of he Kle 4 group Z Z So G K has somorphsm ype C C For K = {, 4} we have K = {, }, K 4 = {4, }, K 8 = {7, 8} whch paro G The mulplcao able of G K s K K K 4 K 8 K K K K 4 K 8 K K K 4 K 8 K K 4 K 4 K 8 K K K 8 K 8 K K K 4 ad so G K s cyclc beg geeraed by K So G K has somorphsm ype C 4 ( ) Takg K = {, 4} ad K = {, 4} we see ha K K as boh K ad K are cyclc of order However G K ad G K are o somorphc So K K does o mply G K G K Le K = {, 4,, 4} whch s a Kle 4 group The wo coses K ad K = {, 7, 8, } paro G The mulplcao able of G K s K K K K K K K K ad G K s of somorphsm ype C Le K = {,, 4, 8} whch s cyclc of order 4 The wo coses K ad K 7 = {7,,, 4} paro G The mulplcao able of G K s ad G K has somorphsm ype C K K 7 K K K 7 K 7 K 7 K

27 7 ( ) Takg K = {, 4,, 4} ad K = {,, 4, 8} we see ha G K ad G K are somorphc alhough K ad K are o somorphc So G K G K does o mply K K (c) Wre x = ( e) θ The x = ( e) θ ( e) θ = ( e ) θ = ( e) θ = x s x G here s x G wh xx = e Hece e = xx = x x = x, e ( e) θ = e pplyg θ o g g = e = gg gves ( g ) θ ( g) θ = e = ( g) θ ( g ) θ showg ha ( g ) θ s he verse of ( g) θ, e ( g ) θ = (( g) θ ) for all g G s ( e) θ = e we see e K Le k, k K The ( kk ) θ = ( k) θ ( k) θ = e e = e showg kk K s ( k ) θ (( k) θ ) e = = = e we see k K Therefore K = kerθ s a subgroup of G s ( e) θ = e we see e mθ s ( g) θ ( g) θ = ( gg ) θ mθ for g, g G ad so mθ s closed uder mulplcao s (( g) θ ) = ( g ) θ mθ for all g G we see mθ s a subgroup of G Le k K, g G The ( g kg) θ = ( g ) θ ( k) θ ( g) θ = (( g) θ ) e ( g) θ = (( g) θ ) ( g) θ = e showg ha g kg K So K = kerθ s ormal G Kerels of group homomorphsms are ormal subgroups s ( kg) θ = ( k) θ ( g) θ = e ( g) θ = ( g) θ all elemes of he cose Kg are mapped by θ o he same eleme ( g) θ So ɶ θ :G K mθ defed by ( Kg ) ɶ θ = ( g ) θ s uambguous ad surecve Suppose ( Kg) ɶ θ = ( Kg) ɶ θ The ( g ) θ = ( g ) θ ad so ( g g ) ( g ) (( g ) ) ( g ) (( g ) ) e θ = θ θ = θ θ = showg gg = k K So g = kg ad hece Kg = Kg, ha s, ɶ θ s ecve s (( Kg)( Kg)) ɶ θ = ( Kgg ) ɶ θ = ( gg ) θ = ( g) θ ( g) θ = ( Kg) ɶ θ ( Kg) ɶ θ we see ha ɶ θ s a group somorphsm ad so ɶ θ : G K mθ, he frs somorphsm heorem for groups Le, GL ( R) where R s a o-rval commuave rg The de,de U ( R) ad ( ) θ = de = de de = ( ) θ ( ) θ by he mulplcave propery (8) of deermas So θ : GL ( R) U ( R) s a group homomorphsm For u U ( R) he dagoal marx U = dag( u, e, e,, e), where e s he eleme of R, s verble over R ad ( U ) θ = de U = u So θ s surecve, e m θ = U ( R) lso SL ( R ) s a ormal subgroup of GL ( R ) wh GL ( R) SL ( R) U ( R) by he frs somorphsm heorem above for groups Takg R =Z we have U ( Z ) = Z = p ad so p p p SL Z GL Z p p p p p p p ( p) = ( p) ( ) = ( )( ) ( ) ( ) (d) Le g, g, g G ad g, g, g G The (( g, g )( g, g ))( g, g ) = ( g g, g g )( g, g ) = (( g g ) g,( g g ) g ) = ( g ( g g ), g ( g g )) = ( g, g )( g g, g g ) = ( g, g )(( g, g )( g, g )) showg ha compoewse mulplcao o G G s assocave The par ( e, e ) cossg of he dey elemes e of G ad e of G s he dey eleme of G G because ( g, g)( e, e ) = ( ge, ge ) = ( g, g) = ( e g, eg ) = ( e, e )( g, g) for all ( g, g) G G The verse of ( g, g ) s ( g, g ) as ( g, g )( g, g ) = ( g g, g g ) = ( e, e ) = ( g g, g g ) = ( g, g )( g, g )