### The Mizar article:

### Cyclic Groups and Some of Their Properties --- Part I

**by****Dariusz Surowik**

- Received November 22, 1991
Copyright (c) 1991 Association of Mizar Users

- MML identifier: GR_CY_1
- [ MML identifier index ]

environ vocabulary INT_1, REALSET1, FINSEQ_1, NAT_1, ORDINAL2, ARYTM, BINOP_1, FUNCT_1, VECTSP_1, QC_LANG1, SETWISEO, RLVECT_1, FINSEQ_2, GROUP_1, GROUP_4, RELAT_1, ARYTM_1, FUNCOP_1, FINSET_1, CARD_1, GROUP_2, TARSKI, ARYTM_3, FILTER_0, RLSUB_1, GRAPH_1, GR_CY_1, CARD_3; notation TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, ORDINAL2, ORDINAL1, NUMBERS, XCMPLX_0, XREAL_0, RELAT_1, FUNCT_1, REAL_1, FUNCT_2, SETWISEO, CARD_1, FINSET_1, BINOP_1, DOMAIN_1, INT_1, INT_2, NAT_1, RLVECT_1, GROUP_2, STRUCT_0, VECTSP_1, GROUP_1, GROUP_4, FUNCOP_1, FINSOP_1, SETWOP_2, FINSEQ_1; constructors WELLORD2, REAL_1, SETWISEO, DOMAIN_1, NAT_1, FUNCSDOM, GROUP_4, FINSOP_1, NAT_LAT, MEMBERED, XBOOLE_0; clusters SUBSET_1, FUNCT_1, CARD_1, INT_1, GROUP_1, GROUP_2, RELSET_1, STRUCT_0, FINSEQ_1, ARYTM_3, XREAL_0, MEMBERED, ZFMISC_1, XBOOLE_0, ORDINAL2; requirements NUMERALS, REAL, SUBSET, BOOLE, ARITHM; definitions GROUP_1, SETWISEO, FUNCT_1, BINOP_1, VECTSP_1, TARSKI; theorems AXIOMS, BINOP_1, INT_1, INT_2, NAT_1, REAL_1, FINSEQ_1, FINSEQ_2, FINSEQ_4, VECTSP_1, ZFMISC_1, FUNCT_2, FUNCOP_1, GROUP_1, GROUP_2, GROUP_4, TARSKI, FUNCT_1, CARD_1, WELLORD2, RLVECT_1, FINSOP_1, RELSET_1, XREAL_0, ORDINAL2, XBOOLE_0, XBOOLE_1, SQUARE_1, RELAT_1, XCMPLX_0, XCMPLX_1; schemes BINOP_1, NAT_1, FINSEQ_1, FINSEQ_2; begin reserve i1,i2,i3 for Element of INT; reserve j1,j2,j3 for Integer; reserve p,s,r,g,k,n,m,q,t for Nat; reserve x,y,xp,yp for set; reserve G for Group; reserve a,b for Element of G; reserve F for FinSequence of the carrier of G; reserve I for FinSequence of INT; ::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: Some properties of modulo function :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::: theorem Th1: m mod n = (n*k + m) mod n proof per cases by NAT_1:18; suppose A1: n > 0; m mod n = t implies (n*k + m) mod n = t proof assume m mod n = t; then consider q such that A2: m = n * q + t & t < n by A1,NAT_1:def 2; ex g st n*k + m = n*g + t & t < n proof take k + q; n*k + m = (n*k + n*q) + t & t < n by A2,XCMPLX_1:1; hence thesis by XCMPLX_1:8; end; hence thesis by NAT_1:def 2; end; hence thesis; suppose n = 0; hence thesis; end; theorem Th2: (p+s) mod n = ((p mod n)+s) mod n proof per cases by NAT_1:18; suppose n > 0; then (p + s) mod n = (n*(p div n) + (p mod n) + s) mod n by NAT_1:47 .=(n*(p div n) + ((p mod n) + s)) mod n by XCMPLX_1:1 .=((p mod n) + s) mod n by Th1; hence thesis; suppose A1: n = 0; hence (p+s) mod n = 0 by NAT_1:def 2 .= ((p mod n)+s) mod n by A1,NAT_1:def 2; end; theorem (p+s) mod n = (p + ( s mod n)) mod n by Th2; theorem Th4: k < n implies k mod n = k proof assume A1:k<n; ex p st k=n*p + k & k<n proof k=n*0 + k; hence thesis by A1; end; hence thesis by NAT_1:def 2; end; theorem Th5: n mod n = 0 proof per cases by NAT_1:18; suppose A1:n>0; ex p st n=n*p +0 & 0<n proof n=n*1+0; hence thesis by A1; end; hence thesis by NAT_1:def 2; suppose n = 0; hence thesis by NAT_1:def 2; end; theorem Th6: 0 = 0 mod n proof n = 0 or n > 0 by NAT_1:18; hence thesis by Th4,NAT_1:def 2; end; definition let n be natural number such that A1:n > 0; func Segm (n) -> non empty Subset of NAT equals :Def1: {p: p<n}; coherence proof set X={p:p<n}; 0 in X by A1; then reconsider X as non empty set; X c= NAT proof for x holds x in X implies x in NAT proof let x be set; assume x in X; then ex p being Element of NAT st p=x & p<n; hence thesis; end; hence thesis by TARSKI:def 3; end; hence thesis; end; end; Lm1: for n st n > 0 holds x in Segm (n) implies x is Nat; canceled 3; theorem Th10: for n,s being natural number st n > 0 holds s in Segm (n) iff s < n proof let n,s be natural number; assume A1: n>0; A2: s is Nat by ORDINAL2:def 21; s in { p : p<n } iff ex p st s=p & p<n; hence thesis by A1,A2,Def1; end; canceled; theorem for n being natural number st n > 0 holds 0 in Segm (n) by Th10; theorem Th13: Segm (1) = {0} proof now let x; thus x in Segm(1) implies x in { 0 } proof assume A1:x in Segm(1); Segm(1) = { p : p < 1 } by Def1; then consider p such that A2:x = p & (p < 1) by A1; p < 0 + 1 by A2; then p <= 0 by NAT_1:38; then p = 0 by NAT_1:18; hence thesis by A2,TARSKI:def 1; end; assume x in { 0 }; then x = 0 by TARSKI:def 1; hence x in Segm(1) by Th10; end; hence Segm(1) = { 0 } by TARSKI:2; end; ::::::::::::::::::::::::::::::::::::::::::::::::::::: :: Definition addint :: ::::::::::::::::::::::::::::::::::::::::::::::::::::: definition func addint -> BinOp of INT means: Def2: for i1,i2 being Element of INT holds it.(i1,i2) = addreal.(i1,i2); existence proof defpred P[Element of INT, Element of INT,set] means $3 = addreal.($1,$2); A1:for i1,i2 being Element of INT ex c being Element of INT st P[i1,i2,c] proof let i1,i2 be Element of INT; reconsider i1' = i1, i2' = i2 as Integer; reconsider c = i1'+i2' as Element of INT by INT_1:12; take c; i1 is Element of REAL & i2 is Element of REAL by XREAL_0:def 1; hence thesis by VECTSP_1:def 4; end; ex B being BinOp of INT st for i1,i2 being Element of INT holds P[i1,i2,B.(i1,i2)] from BinOpEx(A1); hence thesis; end; uniqueness proof let f1,f2 be BinOp of INT such that A2:for i1,i2 being Element of INT holds f1.(i1,i2)=addreal.(i1,i2) and A3:for i1,i2 being Element of INT holds f2.(i1,i2)=addreal.(i1,i2); for i1,i2 being Element of INT holds f1.(i1,i2)=f2.(i1,i2) proof let i1,i2 be Element of INT; thus f1.(i1,i2)=addreal.(i1,i2) by A2 .=f2.(i1,i2) by A3; end; hence thesis by BINOP_1:2; end; end; theorem Th14: for j1,j2 holds addint.(j1,j2)=j1+j2 proof let j1,j2; reconsider j1'=j1,j2'=j2 as Element of INT by INT_1:12; A1:j1 is Element of REAL & j2 is Element of REAL by XREAL_0:def 1; thus addint.(j1,j2)=addreal.(j1',j2') by Def2 .=j1+j2 by A1,VECTSP_1:def 4; end; theorem Th15: for i1 st i1 = 0 holds i1 is_a_unity_wrt addint proof let i1; assume A1:i1=0; A2: for i2 holds addint.(i1,i2) = i2 proof let i2; reconsider e=i1 as Integer; addint.(e,i2) =addint.(e,@i2) by GROUP_4:def 2 .= e +@i2 by Th14 .= i2 by A1,GROUP_4:def 2; hence thesis; end; for i2 holds addint.(i2,i1) = i2 proof let i2; reconsider e=i1 as Integer; addint.(i2,e) =addint.(@i2,e)by GROUP_4:def 2 .= @i2 + e by Th14 .= i2 by A1,GROUP_4:def 2; hence thesis; end; hence i1 is_a_unity_wrt addint by A2,BINOP_1:11; end; theorem Th16: the_unity_wrt addint = 0 proof reconsider zero=0 as Element of INT by INT_1:12; zero is_a_unity_wrt addint by Th15; hence thesis by BINOP_1:def 8; end; theorem Th17: addint has_a_unity proof reconsider O = 0 as Element of INT by INT_1:12; take O; thus thesis by Th15; end; theorem addint is commutative proof let i1,i2; thus addint.(i1,i2) = addint.(@i1,i2) by GROUP_4:def 2 .= addint.(@i1,@i2) by GROUP_4:def 2 .= @i2 + (@i1) by Th14 .= addint.(@i2,@i1) by Th14 .= addint.(i2,@i1) by GROUP_4:def 2 .= addint.(i2,i1) by GROUP_4:def 2; end; theorem addint is associative proof let i1,i2,i3; thus addint.(i1,addint.(i2,i3)) = addint.(@i1,addint.(i2,i3)) by GROUP_4:def 2 .= addint.(@i1,addint.(@i2,i3)) by GROUP_4:def 2 .= addint.(@i1,addint.(@i2,@i3)) by GROUP_4:def 2 .= addint.(@i1,@i2+@i3) by Th14 .= @i1+(@i2+@i3) by Th14 .= @i1+@i2+@i3 by XCMPLX_1:1 .= addint.((@i1+@i2),@i3) by Th14 .= addint.(addint.(@i1,@i2),@i3) by Th14 .= addint.(addint.(@i1,@i2),i3) by GROUP_4:def 2 .= addint.(addint.(@i1,i2),i3) by GROUP_4:def 2 .= addint.(addint.(i1,i2),i3) by GROUP_4:def 2; end; definition let F be FinSequence of INT; func Sum(F) -> Integer equals :Def3: addint $$ F; coherence; end; theorem Th20: Sum(I^<*i1*>) =Sum I +@i1 proof thus Sum(I^<*i1*>) = addint $$ (I^<*i1*>) by Def3 .= addint.(addint $$ I,i1) by Th17,FINSOP_1:5 .= addint.(Sum I,i1) by Def3 .= addint.(Sum I,@i1) by GROUP_4:def 2 .= Sum I + @i1 by Th14; end; theorem Sum <*i1*> = i1 proof set F = <*i1*>; thus Sum F = addint $$ F by Def3 .= i1 by FINSOP_1:12; end; theorem Th22: Sum (<*> INT) = 0 proof set F = <*> INT; thus Sum F = addint $$ F by Def3 .= 0 by Th16,Th17,FINSOP_1:11; end; Lm2: Product(((len(<*> INT)) |->a)|^(<*> INT))=a|^Sum(<*> INT) proof A1:(<*> the carrier of G)|^(<*> INT)=<*> the carrier of G by GROUP_4:27; len(<*> INT)|->a =0 |-> a by FINSEQ_1:32 .=<*> the carrier of G by FINSEQ_2: 72 ; then Product((len(<*> INT)|->a)|^(<*> INT))=1.G by A1,GROUP_4:11 .= a|^(Sum(<*> INT)) by Th22,GROUP_1:59; hence thesis; end; Lm3: for I being FinSequence of INT for w being Element of INT st Product(((len I)|->a)|^I)=a|^Sum I holds Product(((len (I^<*w*>))|->a)|^(I^<*w*>))=a|^Sum(I^<*w*>) proof let I; let w be Element of INT; A1:@(@w) =@(w) by GROUP_4:def 2 .=w by GROUP_4:def 2; assume A2:Product(((len I)|->a)|^I)=a|^Sum I; A3:len((len I) |->a) =len I by FINSEQ_2:69; A4:len <*a*> = 1 by FINSEQ_1:57 .= len <*w*> by FINSEQ_1:57; Product(((len (I^<*w*>))|->a)|^(I^<*w*>)) =Product(((len I+1)|->a)|^(I^<*w*>)) by FINSEQ_2:19 .= Product((((len I)|->a)^<*a*>)|^(I^<*w*>)) by FINSEQ_2:74 .= Product((((len I)|->a)|^I)^(<*a*>|^<*w*>)) by A3,A4,GROUP_4:25 .= Product((((len I)|->a)|^I))*Product(<*a*>|^<*@(@w)*>) by A1,GROUP_4:8 .= Product((((len I)|->a)|^I))*Product(<*a|^(@w)*>) by GROUP_4:28 .= (a|^Sum I)*(a|^(@w)) by A2,GROUP_4:12 .= a|^(Sum I+@w) by GROUP_1:63 .= a|^Sum(I^<*w*>) by Th20; hence thesis; end; canceled; theorem Th24: for I being FinSequence of INT holds Product(((len I)|->a)|^I) = a|^Sum I proof defpred P[FinSequence of INT] means Product(((len $1) |->a)|^($1))=a|^Sum($1); A1: P[<*> INT] by Lm2; A2: for p being FinSequence of INT for x being Element of INT st P[p] holds P[p^<*x*>] by Lm3; for p being FinSequence of INT holds P[p] from IndSeqD(A1,A2); hence thesis; end; ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: Finite groups and their some properties :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: theorem Th25: b in gr {a} iff ex j1 st b=a|^j1 proof A1:b in gr {a} implies ex j1 st b=a|^j1 proof assume b in gr {a}; then consider F,I such that A2: len F = len I and A3: rng F c= {a} and A4: Product(F |^ I ) = b by GROUP_4:37; A5:dom ((len F) |-> a) = Seg len F by FINSEQ_2:68 .= dom F by FINSEQ_1:def 3; for p st p in dom F holds F.p = ((len F)|-> a).p proof let p; A6: dom F = Seg len F by FINSEQ_1:def 3; assume A7: p in dom F; then F.p in rng F by FUNCT_1:def 5; then F.p = a by A3,TARSKI:def 1 .= ((len F)|-> a).p by A6,A7,FINSEQ_2:70; hence thesis; end; then A8: b= Product(((len I) |-> a)|^I) by A2,A4,A5,FINSEQ_1:17 .= a|^Sum I by Th24; take Sum I; thus thesis by A8; end; (ex j1 st b=a|^j1) implies b in gr {a} proof given j1 such that A9: b =a|^j1; consider n such that A10: j1 = n or j1 = -n by INT_1:8; now per cases by A10; suppose A11:j1=n; ex F,I st len F = len I & rng F c= {a} & Product(F |^ I) = b proof take F =(n |-> a); A12: len F = n by FINSEQ_2:69 .= len(n |->@ 1) by FINSEQ_2:69; F = Seg n --> a by FINSEQ_2:def 2; then A13: rng F c= {a} by FUNCOP_1:19; Product(F|^(n |->@ 1)) = Product(F|^(len F |->@ 1)) by FINSEQ_2:69 .=Product(n |-> a) by GROUP_4:31 .=b by A9,A11,GROUP_4:15; hence thesis by A12,A13; end; hence thesis by GROUP_4:37; suppose j1 = -n; then A14: b"=a|^(-(-n)) by A9,GROUP_1:70 .= a|^n; ex F,I st len F = len I & rng F c= {a} & Product(F |^ I) = b" proof take F =(n |-> a); A15: len F = n by FINSEQ_2:69 .= len(n |->@ 1) by FINSEQ_2:69; F = Seg n --> a by FINSEQ_2:def 2; then A16: rng F c= {a} by FUNCOP_1:19; Product(F|^(n |->@ 1)) = Product(F|^(len F |->@ 1)) by FINSEQ_2:69 .=Product(n |-> a) by GROUP_4:31 .= b" by A14,GROUP_4:15; hence thesis by A15,A16; end; then b" in gr {a} by GROUP_4:37; then b"" in gr {a} by GROUP_2:60; hence thesis by GROUP_1:19; end; hence thesis; end; hence thesis by A1; end; theorem Th26: G is finite implies a is_not_of_order_0 proof assume A1:G is finite; ex n st n<> 0 & a|^n = 1.G proof deffunc F(Nat) = a|^$1; consider F being FinSequence such that A2: len F = ord G +1 and A3: for p st p in Seg (ord G +1) holds F.p = F(p) from SeqLambda; A4: dom F = Seg (ord G + 1) by A2,FINSEQ_1:def 3; A5: for y st y in rng F holds ex n st y=a|^n proof let y; assume y in rng F; then consider x such that A6: x in dom F and A7: F.x=y by FUNCT_1:def 5; reconsider n=x as Nat by A6; take n; thus thesis by A3,A4,A6,A7; end; rng F c= the carrier of G proof for x holds x in rng F implies x in the carrier of G proof let y; assume y in rng F; then ex n st y= a|^n by A5; hence thesis; end; hence thesis by TARSKI:def 3; end; then reconsider F'=F as Function of Seg (ord G +1),the carrier of G by A4, FUNCT_2:def 1,RELSET_1:11; consider c being finite set such that A8: c = the carrier of G & ord G = card c by A1,GROUP_1:def 14; A9:Card ord G = Card the carrier of G by A8,CARD_1:def 11; A10:Card Seg(ord G+1) = Card (ord G +1) by FINSEQ_1:76; ord G < ord G +1 by REAL_1:69; then Card the carrier of G <` Card Seg (ord G +1) by A9,A10,CARD_1:73; then consider x,y such that A11:x in Seg (ord G +1) and A12:y in Seg (ord G +1) and A13:x <> y and A14: F'.x = F'.y by FINSEQ_4:80; reconsider p=x,n=y as Nat by A11,A12; now per cases by A13,REAL_1:def 5; suppose A15: n>p; then reconsider t = n-p as Nat by INT_1:18; reconsider e=0 as Integer; take t; A16: t <>0 by A15,SQUARE_1:11; F'.p =a|^p by A3,A11; then a|^n=a|^p by A3,A12,A14; then a|^(n+-p)=a|^p *a|^(-p) by GROUP_1:64; then a|^(n+-p)=a|^(p+-p) by GROUP_1:64; then a|^t =a|^(p+-p) by XCMPLX_0:def 8; then a|^t=a|^e by XCMPLX_0:def 6; then a|^t=1.G by GROUP_1:59; hence thesis by A16; suppose A17: p>n; then reconsider t = p-n as Nat by INT_1:18; take t; reconsider e=0 as Integer; A18: t <>0 by A17,SQUARE_1:11; F'.p =a|^p by A3,A11; then a|^n=a|^p by A3,A12,A14; then a|^(p+-n)=a|^n *a|^(-n) by GROUP_1:64; then a|^(p+-n)=a|^(n+-n) by GROUP_1:64; then a|^t =a|^(n+-n) by XCMPLX_0:def 8; then a|^t=a|^e by XCMPLX_0:def 6; then a|^t=1.G by GROUP_1:59; hence thesis by A18; end; hence thesis; end; hence thesis by GROUP_1:def 10; end; theorem Th27: G is finite implies ord a = ord gr {a} proof assume A1:G is finite; then A2:gr {a} is finite by GROUP_2:48; deffunc F(Nat) = a|^$1; consider F being FinSequence such that A3: len F = ord a and A4: for p st p in Seg ord a holds F.p = F(p) from SeqLambda; A5: dom F = Seg ord a by A3,FINSEQ_1:def 3; a is_not_of_order_0 by A1,Th26; then A6:ord a <> 0 by GROUP_1:def 11; then A7:ord a >0 by NAT_1:19; A8: a in rng F proof ex x st x in dom F & F.x=a proof set x'=1; A9:x' in dom F proof (ord a)+x' > 0+x' by A7,REAL_1:53; then (ord a) >= x' by NAT_1:38; hence thesis by A5,FINSEQ_1:3; end; then F.x'=a|^x' by A4,A5 .=a by GROUP_1:44; hence thesis by A9; end; hence thesis by FUNCT_1:def 5; end; A10: rng F = the carrier of gr {a} proof A11:for y st y in rng F holds ex n st y=a|^n proof let y; assume y in rng F; then consider x such that A12: x in dom F and A13: F.x=y by FUNCT_1:def 5 ; reconsider n=x as Nat by A12; take n; thus thesis by A4,A5,A12,A13; end; A14: rng F c= the carrier of gr {a} proof for x holds x in rng F implies x in the carrier of gr {a} proof let y; assume y in rng F; then consider n such that A15:y= a|^n by A11; y in gr {a} by A15,Th25; hence thesis by RLVECT_1:def 1; end; hence thesis by TARSKI:def 3; end; the carrier of gr {a} c= rng F proof ex H being strict Subgroup of G st the carrier of H = rng F proof reconsider car=rng F as non empty set by A8; A16:dom(the mult of G)= [:the carrier of G,the carrier of G:] by FUNCT_2:def 1; the carrier of gr {a} c= the carrier of G by GROUP_2:def 5; then A17: car c= the carrier of G by A14,XBOOLE_1:1; then [:car,car:] c= [:the carrier of G,the carrier of G:] by ZFMISC_1:119 ; then A18:dom((the mult of G)|[:car,car:])=[:car,car:] by A16,RELAT_1:91; rng((the mult of G)|[:car,car:]) c= car proof for y st y in rng((the mult of G)|[:car,car:]) holds y in car proof set f=(the mult of G)|[:car,car:]; let y; assume y in rng(f); then consider x such that A19:x in dom(f) and A20: f.x=y by FUNCT_1:def 5; consider xp,yp such that A21:[xp,yp]=x by A18,A19,ZFMISC_1:102; xp in car by A18,A19,A21,ZFMISC_1:106; then consider p such that A22: xp = a|^p by A11; yp in car by A18,A19,A21,ZFMISC_1:106; then consider s such that A23: yp = a|^s by A11; A24: y = (the mult of G).[a|^p,a|^s] by A19,A20,A21,A22,A23,FUNCT_1:70 .= (the mult of G).(a|^p,a|^s) by BINOP_1:def 1 .= a|^p*(a|^s) by VECTSP_1:def 10 .= a|^(p+s) by GROUP_1:48; a|^(p+s) in car proof ex x st x in dom F & F.x=a|^(p+s) proof per cases by NAT_1:18; suppose A25:p+s=0; A26:ord a in dom F by A5,A6,FINSEQ_1:5; A27:a|^(p+s)=1.G by A25,GROUP_1:43 .= a|^(ord a) by GROUP_1:82 .= F.ord a by A4,A5,A26; take ord a; thus thesis by A5,A6,A27,FINSEQ_1:5; suppose p+s>0; thus thesis proof set t=(p+s) mod (ord a); A28:t < ord a by A7,NAT_1:46; now per cases by NAT_1:18; suppose A29:t=0; A30:ord a in dom F by A5,A6,FINSEQ_1:5; A31:a|^(p+s)=a|^((ord a)*((p+s) div (ord a)) + ((p+s) mod (ord a))) by A7, NAT_1:47 .=a|^((ord a)*((p+s) div (ord a)))*(a|^ ((p+s) mod (ord a))) by GROUP_1:48 .=a|^(ord a)|^((p+s) div (ord a))*(a|^ ((p+s) mod (ord a))) by GROUP_1:50 .=(1.G)|^((p+s) div (ord a))*(a|^ ((p+s) mod (ord a))) by GROUP_1:82 .=(1.G) *(a|^ ((p+s) mod (ord a))) by GROUP_1:42 .=a|^0 by A29,GROUP_1:def 4 .= 1.G by GROUP_1:43 .= a|^(ord a) by GROUP_1:82 .= F.ord a by A4,A5,A30; take x = ord a; thus x in dom F & F.x=a|^(p+s) by A5,A6,A31,FINSEQ_1:5; suppose t>0; then t+1 > 0+1 by REAL_1:53; then A32:t >= 1 by NAT_1:38; take x = t; A33:t in dom F by A5,A28,A32,FINSEQ_1:3; a|^(p+s)=a|^((ord a)*((p+s) div (ord a)) + ((p+s) mod (ord a))) by A7,NAT_1 :47 .=a|^((ord a)*((p+s) div (ord a))) *(a|^ ((p+s) mod (ord a))) by GROUP_1:48 .=a|^(ord a)|^((p+s) div (ord a)) *(a|^ ((p+s) mod (ord a))) by GROUP_1:50 .=(1.G)|^((p+s) div (ord a)) *(a|^ ((p+s) mod (ord a))) by GROUP_1:82 .=(1.G) *(a|^ ((p+s) mod (ord a))) by GROUP_1:42 .=a|^ ((p+s) mod (ord a)) by GROUP_1:def 4; hence x in dom F & F.x=a|^(p+s) by A4,A5,A33; end; hence thesis; end; thus thesis; end; hence thesis by FUNCT_1:def 5; end; hence thesis by A24; end; hence thesis by TARSKI:def 3; end; then reconsider op=(the mult of G)|[:car,car:] as BinOp of car by A18,FUNCT_2: def 1,RELSET_1:11; set H=HGrStr(#car,op#); H is Group proof A34: for f,g being Element of H , f',g' being Element of G st f = f' & g = g' holds f'*g' = f*g proof let f,g be Element of H; let f' ,g' be Element of G; assume A35: f = f' & g =g'; set A = [:car,car:]; A36:f*g = ((the mult of G) | A).(f,g) by VECTSP_1:def 10 .= ((the mult of G) | A).[f,g] by BINOP_1:def 1; f'*g' = (the mult of G).(f',g') by VECTSP_1:def 10 .= (the mult of G).[f',g'] by BINOP_1:def 1; hence thesis by A35,A36,FUNCT_1:72; end; A37:for f,g,h being Element of H holds (f * g) * h = f * (g * h ) proof let f,g,h be Element of H; reconsider f'=f as Element of G by A17,TARSKI:def 3; reconsider g'=g as Element of G by A17,TARSKI:def 3; reconsider h'=h as Element of G by A17,TARSKI:def 3; A38: f'*g' = f*g by A34; A39: g * h = g' * h' by A34; (f*g)*h =(f'*g')*h' by A34,A38 .= f'*(g'*h') by VECTSP_1:def 16 .=f*(g*h) by A34,A39; hence thesis; end; ex e being Element of H st for h being Element of H holds h * e = h & e * h = h & ex g being Element of H st h * g = e & g * h = e proof a|^(ord a) in car proof ex x st x in dom F & F.x=a|^(ord a) proof set x'=(ord a); A40: x' in Seg ord a by A6,FINSEQ_1:5; then F.x'=a|^x' by A4; hence thesis by A5,A40; end; hence thesis by FUNCT_1:def 5; end; then reconsider e=1.G as Element of H by GROUP_1:82; take e; for h being Element of H holds h * e = h & e * h = h & ex g being Element of H st h * g = e & g * h = e proof let h be Element of H; reconsider h'=h as Element of G by A17,TARSKI:def 3; thus h * e = h proof h*e = h' *(1.G) by A34 .= h' by GROUP_1:def 4; hence thesis; end; thus e * h = h proof e*h = (1.G)*h' by A34 .= h' by GROUP_1:def 4; hence thesis; end; thus ex g being Element of H st h * g = e & g * h = e proof ex n st h=a|^n & (1 <= n & ord a >= n) proof consider x such that A41: x in dom F and A42: F.x=h by FUNCT_1:def 5; reconsider n=x as Nat by A41; take n; thus thesis by A4,A5,A41,A42,FINSEQ_1:3; end; then consider n such that A43:h=a|^n and A44: ord a >= n; a|^(ord a - n) in car proof ex x st x in dom F & F.x =a|^(ord a - n) proof per cases by A44,REAL_1:def 5; suppose ord a = n; then A45: ord a - n =0 by XCMPLX_1:14; set x = ord a; A46: x in dom F by A5,A6,FINSEQ_1:5; then F.x=a|^x by A4,A5 .= 1.G by GROUP_1:82 .= a|^(ord a - n) by A45,GROUP_1:43; hence thesis by A46; suppose A47:ord a > n; then reconsider x = ord a - n as Nat by INT_1:18; take x; x in Seg ord a proof reconsider r1=ord a ,r2=n as Real; A48:x >= 1 proof r1 - r2 > r2 -r2 by A47,REAL_1:54; then r1-r2 > 0 by XCMPLX_1:14; then x >= 0 +1 by NAT_1:38; hence x >= 1; end; x <= ord a proof r1 <= r1 + r2 by NAT_1:29; hence thesis by REAL_1:86; end; hence thesis by A48,FINSEQ_1:3; end; hence thesis by A3,A4,FINSEQ_1:def 3; end; hence thesis by FUNCT_1:def 5; end; then reconsider g =a|^(ord a - n) as Element of H; A49:n+(ord a-n) =ord a proof reconsider n'=n ,z = ord a as Integer; reconsider n''=n' ,z' =z as Real; n+(ord a-n) = (z' + n'') - n'' by XCMPLX_1:29 .= z' + (n'' - n'') by XCMPLX_1:29 .= z'+0 by XCMPLX_1:14 .= z'; hence thesis; end; A50: h * g = e proof h * g =a|^n *(a|^(ord a - n)) by A34,A43 .=a|^(ord a) by A49,GROUP_1:64 .= e by GROUP_1:82; hence thesis; end; g * h = e proof g * h =a|^(ord a - n) * (a|^n) by A34,A43 .=a|^(ord a) by A49,GROUP_1:65 .= e by GROUP_1:82; hence thesis; end; hence thesis by A50; end; thus thesis; end; hence thesis; end; hence thesis by A37,GROUP_1:5; end; then reconsider H1=H as Group; the carrier of gr {a} c= the carrier of G by GROUP_2:def 5; then the carrier of H1 c= the carrier of G by A14,XBOOLE_1:1; then H1 is Subgroup of G by GROUP_2:def 5; hence thesis; end; then consider H being strict Subgroup of G such that A51: the carrier of H = rng F; {a} c= the carrier of H by A8,A51,ZFMISC_1:37; then gr {a} is Subgroup of H by GROUP_4:def 5; hence the carrier of gr {a} c= rng F by A51,GROUP_2:def 5; end; hence thesis by A14,XBOOLE_0:def 10; end; F is one-to-one proof let x,y; assume that A52: x in dom F and A53: y in dom F and A54: F.x = F.y and A55:x<>y; reconsider p=x as Nat by A52; reconsider s=y as Nat by A53; reconsider s'=s,p'=p,z= ord a as Real; A56:F.p=a|^p by A4,A5,A52; A57:F.s=a|^s by A4,A5,A53; per cases; suppose A58:p<=s; then A59:p<s by A55,REAL_1:def 5; reconsider r1=s-p as Nat by A58,INT_1:18; A60: r1=s+-p by XCMPLX_0:def 8; 1.G=a|^s*(a|^p)" by A54,A56,A57,GROUP_1:def 5; then a|^0=a|^s*(a|^p)" by GROUP_1:43; then a|^0=a|^s*(a|^(-p)) by GROUP_1:71; then a|^0=a|^(s+(-p)) by GROUP_1:64; then A61:1.G=a|^r1 by A60,GROUP_1:43; A62: r1 <> 0 by A59,SQUARE_1:11; A63:p >0 proof p>=1 by A5,A52,FINSEQ_1:3; hence thesis by AXIOMS:22; end; A64:r1 < ord a proof A65: s'<=z by A5,A53,FINSEQ_1:3; z<z+p' by A63,REAL_1:69; then s'<z+p' by A65,AXIOMS:22; then s'-p'<z+p'-p' by REAL_1:54; then s'-p'<z+p'+-p' by XCMPLX_0:def 8; then s'-p'<z+(p'+-p') by XCMPLX_1:1; then s'-p'<z+0 by XCMPLX_0:def 6; hence thesis; end; a is_not_of_order_0 by A1,Th26; hence thesis by A61,A62,A64,GROUP_1:def 11; suppose A66:s<=p; then A67:s<p by A55,REAL_1:def 5; reconsider r2=p-s as Nat by A66,INT_1:18; A68: r2=p+-s by XCMPLX_0:def 8; 1.G=a|^p*(a|^s)" by A54,A56,A57,GROUP_1:def 5; then a|^0=a|^p*(a|^s)" by GROUP_1:43; then a|^0=a|^p*(a|^(-s)) by GROUP_1:71; then a|^0=a|^(p+(-s)) by GROUP_1:64; then A69:1.G=a|^r2 by A68,GROUP_1:43; A70: r2 <> 0 by A67,SQUARE_1:11; A71:s >0 proof s>=1 by A5,A53,FINSEQ_1:3; hence thesis by AXIOMS:22; end; A72:r2 < ord a proof A73: p'<=z by A5,A52,FINSEQ_1:3; z<z+s' by A71,REAL_1:69; then p'<z+s' by A73,AXIOMS:22; then p'-s'<z+s'-s' by REAL_1:54; then p'-s'<z+s'+-s' by XCMPLX_0:def 8; then p'-s'<z+(s'+-s') by XCMPLX_1:1; then p'-s'<z+0 by XCMPLX_0:def 6; hence thesis; end; a is_not_of_order_0 by A1,Th26; hence thesis by A69,A70,A72,GROUP_1:def 11; end; then A74: Seg ord a,the carrier of gr {a} are_equipotent by A5,A10,WELLORD2:def 4; consider ca being finite set such that A75: ca = the carrier of gr {a} & ord gr {a} = card ca by A2,GROUP_1:def 14; reconsider So = Seg ord a as finite set; card So = card ca by A74,A75,CARD_1:21; hence thesis by A75,FINSEQ_1:78; end; theorem Th28: G is finite implies ord a divides ord G proof assume A1: G is finite; then ord a = ord gr {a} by Th27; hence ord a divides ord G by A1,GROUP_2:178; end; theorem Th29: G is finite implies a|^ord G = 1.G proof assume G is finite; then ord a divides ord G by Th28; then consider t being Nat such that A1: ord G = ord a * t by NAT_1:def 3; a|^ord G = a |^ ord a |^ t by A1,GROUP_1:50 .= (1.G) |^ t by GROUP_1:82 .= 1.G by GROUP_1:42; hence thesis; end; theorem Th30: G is finite implies (a|^n)" = a|^(ord G - (n mod ord G)) proof assume A1: G is finite; set q = ord G; set p'=n mod q; q >=1 by A1,GROUP_1:90; then A2: q > 0 by AXIOMS:22; then n mod q <=q by NAT_1:46; then reconsider q'=q -( n mod q) as Nat by INT_1:18; a|^n*(a|^(q-(n mod q)))=a|^(n+q') by GROUP_1:48 .= a|^((q*(n div q)+(n mod q))+q') by A2,NAT_1:47 .= a|^(q*(n div q)+(q'+(n mod q))) by XCMPLX_1:1 .= a|^(q*(n div q)+((q +(-(n mod q)))+(n mod q))) by XCMPLX_0:def 8 .= a|^(q*(n div q)+(q +(-p'+p'))) by XCMPLX_1:1 .= a|^(q*(n div q)+q +(-p'+p')) by XCMPLX_1:1 .= a|^(q*(n div q)+q)*(a|^(-p'+p')) by GROUP_1:64 .= a|^(q*(n div q)+q)*(a|^(-p')*(a|^p')) by GROUP_1:65 .= a|^(q*(n div q)+q)*((a|^p')"*(a|^p')) by GROUP_1:71 .= a|^(q*(n div q)+q)*(1.G) by GROUP_1:def 5 .= a|^(q*(n div q)+q*1) by GROUP_1:def 4 .= a|^(q*(n div q+1)) by XCMPLX_1:8 .= a|^q|^(n div q+1) by GROUP_1:50 .= (1.G)|^(n div q+1) by A1,Th29 .= 1.G by GROUP_1:42; hence thesis by GROUP_1:20; end; theorem Th31: for G being strict Group holds ord G > 1 implies ex a being Element of G st a <> 1.G proof let G be strict Group; assume that A1: ord G > 1 and A2: for a being Element of G holds a = 1.G; for a being Element of G holds a in (1).G proof let a be Element of G; a = 1.G by A2; then a in {1.G} by TARSKI:def 1; then a in the carrier of (1).G by GROUP_2:def 7; hence thesis by RLVECT_1:def 1; end; then G = (1).G by GROUP_2:71; hence contradiction by A1,GROUP_2:81; end; theorem for G being strict Group holds G is finite & ord G = p & p is prime implies for H being strict Subgroup of G holds H = (1).G or H = G proof let G be strict Group; assume that A1: G is finite and A2: ord G = p and A3: p is prime; let H be strict Subgroup of G; A4:H is finite by A1,GROUP_2:48; ord H divides p by A1,A2,GROUP_2:178; then ord H = 1 or ord H = p by A3,INT_2:def 5; hence thesis by A1,A2,A4,GROUP_2:82,85; end; theorem Th33: HGrStr(#INT,addint#) is associative Group-like proof set G = HGrStr (# INT, addint #); A1: now let h,g be Element of G, A,B be Integer; assume h = A & g = B; hence h * g = addint.(A,B) by VECTSP_1:def 10 .= A + B by Th14; end; thus for h,g,f being Element of G holds (h * g) * f = h * (g * f) proof let h,g,f be Element of G; reconsider A=h,B = g, C = f as Integer by INT_1:12; A2: h * g = A + B by A1; A3: g * f = B + C by A1; thus (h * g) * f = A + B + C by A1,A2 .= A + (B + C) by XCMPLX_1:1 .= h * (g * f) by A1,A3; end; reconsider e=0 as Element of G by INT_1:12; take e; let h be Element of G; reconsider A = h as Integer by INT_1:12; thus h * e = A + 0 by A1 .= h; thus e * h = 0 + A by A1 .= h; reconsider g = - A as Element of G by INT_1:12; take g; thus h * g = A + (- A) by A1 .= e by XCMPLX_0:def 6; thus g * h = (- A) + A by A1 .= e by XCMPLX_0:def 6; end; definition func INT.Group -> strict Group equals : Def4: HGrStr(#INT,addint#); coherence by Th33; end; definition let n such that A1:n > 0; func addint(n) -> BinOp of Segm(n) means :Def5: for k,l being Element of Segm(n) holds it.(k,l) = (k+l) mod n; existence proof defpred P[Element of Segm n,Element of Segm n,set] means $3 = ($1+$2) mod n; A2: for k,l being Element of Segm(n) ex c being Element of Segm(n) st P[k,l,c] proof let k,l be Element of Segm(n); reconsider k'=k,l'=l as Nat; ((k'+l') mod n) < n by A1,NAT_1:46; then reconsider c = (k'+l') mod n as Element of Segm(n) by A1,Th10; take c; thus thesis; end; ex o being BinOp of Segm n st for a,b being Element of Segm n holds P[a,b,o.(a,b)] from BinOpEx(A2); hence thesis; end; uniqueness proof let i1,i2 be BinOp of Segm(n) such that A3:for k,l being Element of Segm(n) holds i1.(k,l)=((k+l) mod n) and A4:for k,l being Element of Segm(n) holds i2.(k,l)=((k+l) mod n); for k,l being Element of Segm(n) holds i1.(k,l)=i2.(k,l) proof let k,l be Element of Segm(n); thus i1.(k,l)=((k+l) mod n) by A3 .=i2.(k,l) by A4; end; hence thesis by BINOP_1:2; end; end; theorem Th34: for n st n > 0 holds HGrStr(#Segm(n),addint(n)#) is associative Group-like proof let n; assume A1:n>0; set G = HGrStr (# Segm(n), addint(n) #); A2: now let h,g be Element of G, A,B be Element of Segm(n); assume h = A & g = B; hence h * g = (addint(n)).(A,B) by VECTSP_1:def 10 .= (A + B) mod n by A1,Def5; end; thus for h,g,f being Element of G holds (h * g) * f = h * (g * f) proof let h,g,f be Element of G; reconsider A = h, B = g, C = f as Element of Segm(n); A3: h * g = (A + B) mod n by A2; A4: g * f = (B + C) mod n by A2; thus (h * g) * f = (((A + B) mod n) + C) mod n by A2,A3 .= ((A + B) + C) mod n by Th2 .=(A + (B + C)) mod n by XCMPLX_1:1 .=(A + ((B + C) mod n)) mod n by Th2 .= h * (g * f) by A2,A4; end; reconsider e=0 as Element of G by A1,Th10; take e; let h be Element of G; reconsider A = h as Element of Segm(n); reconsider A as Nat; A5:A < n by A1,Th10; thus h * e = (A + 0) mod n by A2 .= h by A5,Th4; thus e * h = (0 + A) mod n by A2 .= h by A5,Th4; consider p such that A6: n=A+p by A5,NAT_1:28; A7: p <= n by A6,NAT_1:37; (p mod n) < n by A1,NAT_1:46; then reconsider g = p mod n as Element of G by A1,Th10; take g; reconsider B=g as Element of Segm(n); thus h * g =e proof now per cases by A7,REAL_1:def 5; suppose A8:p = n; then n + 0 =A + n by A6; then A9:A = 0 by XCMPLX_1:2; h * g = (A +B) mod n by A2 .= (0 * n) mod n by A8,A9,Th5 .= e by GROUP_4:101; hence thesis; suppose A10:p < n; h * g = (A +B) mod n by A2 .= n mod n by A6,A10,Th4 .= e by Th5; hence thesis; end; hence thesis; end; thus g * h = e proof now per cases by A7,REAL_1:def 5; suppose A11:p = n; then n + 0 =A + n by A6; then A = 0 by XCMPLX_1:2; then g * h = (n mod n + 0) mod n by A2,A11 .= (0 * n) mod n by Th5 .= e by GROUP_4:101; hence thesis; suppose A12:p < n; g * h = (A + B) mod n by A2 .=n mod n by A6,A12,Th4 .= e by Th5; hence thesis; end; hence thesis; end; end; definition let n such that A1: n > 0; func INT.Group(n) -> strict Group equals :Def6: HGrStr(#Segm(n),addint(n)#); coherence by A1,Th34; end; theorem Th35: 1.INT.Group = 0 proof A1: now let h,g be Element of INT.Group, A,B be Integer; assume h = A & g = B; hence h * g = addint.(A,B) by Def4,VECTSP_1:def 10 .= A + B by Th14; end; reconsider e = 0 as Element of INT by INT_1:12; reconsider e as Element of INT.Group by Def4; for h being Element of INT.Group holds h * e = h & e * h = h proof let h be Element of INT.Group; reconsider A = h as Integer by Def4,INT_1:12; A2: h * e = A + 0 by A1 .= h; e * h = 0 + A by A1 .= h; hence thesis by A2; end; hence thesis by GROUP_1:10; end; theorem Th36: for n st n>0 holds 1.INT.Group(n) = 0 proof let n; assume A1: n>0; then A2: INT.Group(n) = HGrStr(#Segm(n),addint(n)#) by Def6; A3: now let h,g be Element of INT.Group(n), A,B be Element of Segm(n); assume h = A & g = B; hence h * g = (addint(n)).(A,B) by A2,VECTSP_1:def 10 .= (A + B) mod n by A1,Def5; end; reconsider e = 0 as Element of Segm(n) by A1,Th10; reconsider e as Element of INT.Group(n) by A2; for h being Element of INT.Group(n) holds h * e = h & e * h = h proof let h be Element of INT.Group(n); reconsider A = h as Element of Segm(n) by A2; reconsider A as Nat; A4:A < n by A1,Th10; A5: h * e = (A + 0) mod n by A3 .= h by A4,Th4; e * h = (0 + A) mod n by A3 .= h by A4,Th4; hence thesis by A5; end; hence thesis by GROUP_1:10; end; definition let h be Element of INT.Group; func @'h -> Integer equals : Def7: h; coherence by Def4,INT_1:12; end; definition let h be Integer; func @'h -> Element of INT.Group equals h; coherence by Def4,INT_1:12; end; theorem Th37: for h being Element of INT.Group holds h" = -@'h proof A1: now let h,g be Element of INT.Group, A,B be Integer; assume h = A & g = B; hence h * g = addint.(A,B) by Def4,VECTSP_1:def 10 .= A + B by Th14; end; let h be Element of INT.Group; A2:h=@'h by Def7; reconsider g=-@'h as Element of INT.Group by Def4,INT_1:12; h *g =@'h + -@'h by A1,A2 .=1.INT.Group by Th35,XCMPLX_0:def 6; hence thesis by GROUP_1:20; end; reserve G1 for Subgroup of INT.Group, h for Element of INT.Group; Lm4:now let h,g be Element of INT.Group, A,B be Integer; assume h = A & g = B; hence h * g = addint.(A,B) by Def4,VECTSP_1:def 10 .= A + B by Th14; end; theorem Th38: for h st h=1 holds for k holds h|^k = k proof let h; assume A1:h=1; defpred P[Nat] means h|^$1=$1; h|^0=0 by Th35,GROUP_1:43; then A2: P[0]; A3: for k st P[k] holds P[k+1] proof let k; assume h|^k = k; then h|^k * h = k + 1 by A1,Lm4; hence thesis by GROUP_1:49; end; for k holds P[k] from Ind(A2,A3); hence thesis; end; theorem Th39: for h,j1 st h=1 holds j1 = h |^ j1 proof let h,j1; assume A1:h=1; consider k such that A2:j1 = k or j1 = -k by INT_1:8; now per cases by A2; suppose j1=k; hence thesis by A1,Th38; suppose A3: j1=-k; reconsider k'=k as Integer; reconsider k' as Element of INT.Group by Def4,INT_1:12 ; h|^j1 = (h|^k)" by A3,GROUP_1:71 .= (k')" by A1,Th38 .= -@'k' by Th37 .= j1 by A3,Def7; hence thesis; end; hence thesis; end; ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :: Definition of cyclic group :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Lm5: ex a being Element of INT.Group st the HGrStr of INT.Group = gr {a} proof reconsider h = 1 as Element of INT by INT_1:12; reconsider h as Element of INT.Group by Def4; take h; {h} c= the carrier of (Omega).INT.Group & (for G1 being strict Subgroup of INT.Group st {h} c= the carrier of G1 holds (Omega).INT.Group is Subgroup of G1 ) proof A1: {h} c= the carrier of INT.Group; for G1 st {h} c= the carrier of G1 holds (Omega). INT.Group is Subgroup of G1 proof let G1; assume A2:{h} c= the carrier of G1; the carrier of (Omega).INT.Group c= the carrier of G1 proof let x; assume that A3: x in the carrier of (Omega).INT.Group and A4: not x in the carrier of G1; x in (Omega).INT.Group by A3,RLVECT_1:def 1; then x in INT.Group by GROUP_2:def 8; then x in INT by Def4,RLVECT_1:def 1; then reconsider x'=x as Integer by INT_1:12; h in {h} by TARSKI:def 1; then reconsider h''=h as Element of G1 by A2; A5: (h'')|^x' in the carrier of G1; (h'')|^x'=h|^x' by GROUP_4:4; hence contradiction by A4,A5,Th39; end; hence thesis by GROUP_2:66; end; hence thesis by A1,GROUP_2:def 8; end; then (Omega).INT.Group = gr {h} by GROUP_4:def 5; hence thesis by GROUP_2:def 8; end; definition let IT be Group; attr IT is cyclic means :Def9: ex a being Element of IT st the HGrStr of IT=gr {a}; end; definition cluster strict cyclic Group; existence proof INT.Group is cyclic by Def9,Lm5; hence thesis; end; end; theorem Th40: (1).G is cyclic proof 1.G in {1.G} by TARSKI:def 1; then reconsider PG=1.G as Element of (1).G by GROUP_2:def 7; take PG; {PG} c= the carrier of (Omega).(1).G & ( for G1 being strict Subgroup of (1).G st {PG} c= the carrier of G1 holds (Omega).(1).G is Subgroup of G1 ) proof A1: {PG} = the carrier of (1).G by GROUP_2:def 7; ( for G1 being Subgroup of (1).G st {PG} c= the carrier of G1 holds (Omega).(1).G is Subgroup of G1 ) proof let G1 be Subgroup of (1).G; assume A2:{PG} c= the carrier of G1; the carrier of (Omega).(1).G =the carrier of (1).G by GROUP_2:def 8; then the carrier of (Omega). (1).G c=the carrier of G1 by A2,GROUP_2:def 7; hence thesis by GROUP_2:66; end; hence thesis by A1,GROUP_2:def 8; end; then (Omega).(1).G = gr {PG} by GROUP_4:def 5; hence thesis by GROUP_2:def 8; end; theorem Th41: G is cyclic Group iff ex a being Element of G st for b being Element of G ex j1 st b=a|^j1 proof thus G is cyclic Group implies ex a being Element of G st for b being Element of G ex j1 st b=a|^j1 proof assume G is cyclic Group; then consider a being Element of G such that A1: the HGrStr of G = gr {a} by Def9; take a; for b being Element of G ex j1 st b=a|^j1 proof let b be Element of G; b in gr {a} by A1,RLVECT_1:def 1; hence thesis by Th25; end; hence thesis; end; thus (ex a being Element of G st for b being Element of G ex j1 st b = a|^j1) implies G is cyclic Group proof given a being Element of G such that A2:for b being Element of G ex j1 st b = a|^j1; for b being Element of G holds b in gr {a} proof let b be Element of G; ex j1 st b=a|^j1 by A2; hence thesis by Th25; end; then the HGrStr of G = gr {a} by GROUP_2:71; hence thesis by Def9; end; end; theorem Th42: G is finite implies (G is cyclic Group iff ex a being Element of G st for b being Element of G ex n st b=a|^n) proof assume A1: G is finite; thus G is cyclic Group implies ex a being Element of G st for b being Element of G ex n st b=a|^n proof assume G is cyclic Group; then consider a being Element of G such that A2: for b being Element of G ex j2 st b=a|^j2 by Th41; take a; let b be Element of G; consider j2 such that A3: b = a|^j2 by A2; consider n such that A4:j2=n or j2=-n by INT_1:8; per cases by A4; suppose j2=n; hence thesis by A3; suppose A5:j2=-n; ord G >=1 by A1,GROUP_1:90; then ord G > 0 by AXIOMS:22; then n mod ord G <=ord G by NAT_1:46; then reconsider q'=ord G -( n mod ord G) as Nat by INT_1:18; take q'; b=(a|^n)" by A3,A5,GROUP_1:71 .= a|^q' by A1,Th30; hence thesis; end; given a being Element of G such that A6: for b being Element of G ex n st b=a|^n; for b being Element of G ex j2 st b=a|^j2 proof let b be Element of G; consider n such that A7:b=a|^n by A6; reconsider n as Integer; take n; thus thesis by A7; end; hence thesis by Th41; end; theorem Th43: for G being strict Group holds G is finite implies ( G is cyclic Group iff ex a being Element of G st ord a =ord G ) proof let G be strict Group; assume A1: G is finite; thus G is cyclic Group implies ex a being Element of G st ord a =ord G proof assume G is cyclic Group; then consider a being Element of G such that A2: G = gr {a} by Def9; take a; thus thesis by A1,A2,Th27; end; given a being Element of G such that A3:ord a =ord G; ex a being Element of G st G = gr {a} proof consider a being Element of G such that A4: ord a = ord G by A3; take a; ord gr {a} = ord G by A1,A4,Th27; hence thesis by A1,GROUP_2:85; end; hence thesis by Def9; end; theorem for H being strict Subgroup of G holds G is finite & G is cyclic Group implies H is cyclic Group proof let H be strict Subgroup of G; assume that A1: G is finite and A2:G is cyclic Group; consider a such that A3: for b being Element of G ex n st b = a|^n by A1,A2,Th42; A4: H is finite by A1,GROUP_2:48; then A5: ord H >= 1 by GROUP_1:90; per cases by A5,REAL_1:def 5; suppose ord H =1; then H = (1).G by A4,GROUP_2:82; hence thesis by Th40; suppose ord H > 1; then consider h being Element of H such that A6:h <>1.H by Th31; defpred P[Nat] means $1 > 0 & a|^$1 is Element of H; h is Element of G by GROUP_2:51; then consider n such that A7: h = a|^n by A3; n <> 0 proof assume n = 0; then h = 1.G by A7,GROUP_1:59 .= 1.H by GROUP_2:53; hence contradiction by A6; end; then n > 0 by NAT_1:19; then A8: ex n st P[n] by A7; ex h1 being Element of H st for b being Element of H ex s st b = h1|^s proof ex k st P[k] & for n st P[n] holds k <= n from Min(A8); then consider n such that A9: (n>0 & a|^n is Element of H) and A10: for k st (k>0 & a|^k is Element of H) holds n <= k; consider h1 being Element of H such that A11: h1 = a|^n by A9; take h1; for b being Element of H ex s st b = h1|^s proof let b be Element of H; b is Element of G by GROUP_2:51; then consider p such that A12: b = a|^p by A3; consider s,r such that A13: p = (n*s) + r and A14:r < n by A9,NAT_1:42; A15: r =0 proof assume r<>0; then A16: r >0 by NAT_1:19; A17: a|^p = a|^(n*s) * (a|^r) by A13,GROUP_1:48; h1|^s=a|^n|^s by A11,GROUP_4:3; then (h1|^s)"=(a|^n|^s)" by GROUP_2:57 .= (a|^(n * s ))" by GROUP_1:50; then reconsider g= (a|^(n*s))" as Element of H; reconsider b=a|^p as Element of H by A12; (a|^(n*s))" * (a|^p) = ((a|^(n*s))" *(a|^(n*s)))*(a|^r) by A17,VECTSP_1:def 16 .= 1.G *(a|^r) by GROUP_1:def 5 .= a|^ r by GROUP_1:def 4; then g*b = a|^r by GROUP_2:52; hence contradiction by A10,A14,A16; end; take s; a|^p = a|^n|^s by A13,A15,GROUP_1:50 .= h1|^s by A11,GROUP_4:3; hence thesis by A12; end; hence thesis; end; hence thesis by A4,Th42; end; theorem for G being strict Group holds G is finite & ord (G) = p & p is prime implies G is cyclic Group proof let G be strict Group; assume that A1: G is finite and A2: ord (G) = p and A3: p is prime; ex a being Element of G st ord a = p proof assume A4: for a being Element of G holds ord a <> p; A5: now let a be Element of G; ord a divides p by A1,A2,Th28; then ord a = 1 or ord a = p by A3,INT_2:def 5; hence ord a = 1 by A4; end; A6:the carrier of G c= the carrier of (1).G proof for x being set holds x in the carrier of G implies x in the carrier of (1).G proof let x be set; assume x in the carrier of G; then reconsider x'=x as Element of G; ord x' = 1 by A5; then x' = 1.G by GROUP_1:85; then x' in {1.G} by TARSKI:def 1; hence thesis by GROUP_2:def 7; end; hence thesis by TARSKI:def 3; end; the carrier of (1).G c= the carrier of G by GROUP_2:def 5; then the carrier of G = the carrier of (1).G by A6,XBOOLE_0:def 10; then G = (1).G by GROUP_2:70; then ord G = 1 by GROUP_2:81; hence contradiction by A2,A3,INT_2:def 5; end; hence thesis by A1,A2,Th43; end; theorem Th46: for n st n>0 ex g being Element of INT.Group(n) st for b being Element of INT.Group(n) ex j1 st b = g|^j1 proof let n; assume A1:n>0; then A2: INT.Group(n) = HGrStr(#Segm(n),addint(n)#) by Def6; A3: now let h,g be Element of INT.Group(n), A,B be Element of Segm(n); assume h = A & g = B; hence h * g = (addint(n)).(A,B) by A2,VECTSP_1:def 10 .= (A + B) mod n by A1,Def5; end; 0 +1 < n+1 by A1,REAL_1:53; then A4:n >=1 by NAT_1:38; now per cases by A4,REAL_1:def 5; suppose n=1; then the carrier of INT.Group(n) = {1.INT.Group(n)} by A2,Th13,Th36 .= the carrier of (1).INT.Group(n) by GROUP_2:def 7; then INT.Group(n) = (1).INT.Group(n) by GROUP_2:70; then INT.Group(n) is cyclic Group by Th40; hence thesis by Th41; suppose n>1; then reconsider g'= 1 as Element of Segm(n) by A1,Th10; reconsider g=g' as Element of INT.Group(n) by A2; for b being Element of INT.Group(n) ex j1 st b = g|^j1 proof let b be Element of INT.Group(n); reconsider b'=b as Nat by A1,A2,Lm1; defpred P[Nat] means g|^$1 = $1 mod n; g|^0 = 1.INT.Group(n) by GROUP_1:43 .= 0 by A1,Th36 .= 0 mod n by Th6; then A5:P[0]; A6: for k st P[k] holds P[k+1] proof let k; k mod n < n by A1,NAT_1:46; then reconsider e=(k mod n) as Element of Segm(n) by A1,Th10; assume A7: g|^k = k mod n; g|^(k+1) =g|^k*(g|^1) by GROUP_1:48 .=g|^k*g by GROUP_1:44 .= (e +g') mod n by A3,A7 .=(k+1) mod n by Th2; hence thesis; end; A8: for k holds P[k] from Ind(A5,A6); A9: b'<n by A1,A2,Th10; A10: g|^b'=b' mod n by A8 .= b by A9,Th4; reconsider b' as Integer; take b'; thus thesis by A10; end; hence thesis; end; hence thesis; end; definition cluster cyclic -> commutative Group; coherence proof let G; assume A1:G is cyclic; for a,b being Element of G holds a * b = b * a proof let a,b be Element of G; ex e being Element of G st (ex j2 st a=e|^j2 & ex j3 st b=e|^j3) proof consider e being Element of G such that A2:for d being Element of G ex j3 st d=e|^j3 by A1,Th41; take e; A3:ex j2 st a=e|^j2 by A2; ex j3 st b=e|^j3 by A2; hence thesis by A3; end; then consider e being Element of G,j2,j3 such that A4:( a = e|^j2 & b = e|^j3); a * b = e |^(j3 + j2) by A4,GROUP_1:63 .= b * a by A4,GROUP_1:63; hence thesis; end; hence thesis by VECTSP_1:def 17; end; end; canceled; theorem Th48: INT.Group is cyclic proof reconsider h = 1 as Element of INT by INT_1:12; reconsider h as Element of INT.Group by Def4; take h; {h} c= the carrier of (Omega).INT.Group & ( for G1 being strict Subgroup of INT.Group st {h} c= the carrier of G1 holds (Omega).INT.Group is Subgroup of G1 ) proof A1: {h} c= the carrier of (Omega).INT.Group proof {h} c= the carrier of INT.Group; hence thesis by GROUP_2:def 8; end; for G1 st {h} c= the carrier of G1 holds (Omega).INT.Group is Subgroup of G1 proof let G1; assume A2:{h} c= the carrier of G1; the carrier of (Omega).INT.Group c= the carrier of G1 proof for x holds x in the carrier of (Omega).INT.Group implies x in the carrier of G1 proof let x; assume that A3: x in the carrier of (Omega).INT.Group and A4: not x in the carrier of G1; x in (Omega).INT.Group by A3,RLVECT_1:def 1; then x in INT.Group by GROUP_2:def 8; then x in INT by Def4,RLVECT_1:def 1; then reconsider x'=x as Integer by INT_1:12; h in {h} by TARSKI:def 1; then reconsider h''=h as Element of G1 by A2; A5: (h'')|^x' in the carrier of G1; (h'')|^x'=h|^x' by GROUP_4:4; hence contradiction by A4,A5,Th39; end; hence the carrier of (Omega). INT.Group c= the carrier of G1 by TARSKI:def 3; thus thesis; end; hence thesis by GROUP_2:66; end; hence thesis by A1; end; then (Omega).INT.Group = gr {h} by GROUP_4:def 5; hence thesis by GROUP_2:def 8; end; theorem Th49: for n st n > 0 holds INT.Group(n) is cyclic Group proof let n; assume n > 0; then ex g being Element of INT.Group(n) st for b being Element of INT.Group(n) ex j1 st b = g|^j1 by Th46; hence thesis by Th41; end; theorem INT.Group is commutative Group proof INT.Group is cyclic Group by Th48; hence thesis; end; theorem for n st n>0 holds INT.Group(n) is commutative Group by Th49;

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