問題：求模2+3i、3+6i的完全剩余系、既約剩余系。
(Z[i]/(4))^*=C_4×C_2，3個2階元，4個4階元
(Z[i]/(3))^*=C_8
(Z[i]/(2))^*=C_2
(Z[i]/(2+3i))^*=C_12
(Z[i]/(1+2i))^*=C_4
(Z[i]/(3+6i))^*=C_8×C_4
g++ -o UnMulTableE4 UnMulTableE4.cpp
./UnMulTableE4 4 0
模4的既約剩余系:1,i,3i,1+2i,2+1i,2+3i,3,3+2i
Φ(4)=8
1 2 3 4 5 6 7 8?
2 7 1 5 8 4 3 6?
3 1 7 6 4 8 2 5?
4 5 6 1 2 3 8 7?
5 8 4 2 7 1 6 3?
6 4 8 3 1 7 5 2?
7 3 2 8 6 5 1 4?
8 6 5 7 3 2 4 1?
./UnMulTableE4 3 0
模3的既約剩余系:1,i,2i,1+1i,1+2i,2,2+1i,2+2i
Φ(3)=8
1 2 3 4 5 6 7 8?
2 6 1 7 4 3 8 5?
3 1 6 5 8 2 4 7?
4 7 5 3 6 8 1 2?
5 4 8 6 2 7 3 1?
6 3 2 8 7 1 5 4?
7 8 4 1 3 5 2 6?
8 5 7 2 1 4 6 3
./UnMulTableE4 2 0
模2的既約剩余系:1,i
Φ(2)=2
1 2?
2 1?
./UnMulTableE4 1 2
模1+2i的既約剩余系:1,2,3,4
Φ(1+2i)=4
1 2 3 4?
2 4 1 3?
3 1 4 2?
4 3 2 1?
./UnMulTableE4 2 3
模2+3i的既約剩余系:1,2,3,4,5,6,7,8,9,10,11,12
Φ(2+3i)=12
1 2 3 4 5 6 7 8 9 10 11 12?
2 4 6 8 10 12 1 3 5 7 9 11?
3 6 9 12 2 5 8 11 1 4 7 10?
4 8 12 3 7 11 2 6 10 1 5 9?
5 10 2 7 12 4 9 1 6 11 3 8?
6 12 5 11 4 10 3 9 2 8 1 7?
7 1 8 2 9 3 10 4 11 5 12 6?
8 3 11 6 1 9 4 12 7 2 10 5?
9 5 1 10 6 2 11 7 3 12 8 4?
10 7 4 1 11 8 5 2 12 9 6 3?
11 9 7 5 3 1 12 10 8 6 4 2?
12 11 10 9 8 7 6 5 4 3 2 1?
./UnMulTableE4 3 6模3+6i的既約剩余系:1,i,2i,1+1i,2,2+1i,2+2i,3+2i,4,4+1i,4+2i,5+1i,5+2i,6+1i,7,7+1i,7+2i,8,8+2i,9+1i,9+2i,10+1i,10+2i,11,11+1i,12+1i,12+2i,13,13+2i,14,14+1i,14+2i
Φ(3+6i)=32
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32?
2 30 28 31 3 32 29 9 14 12 10 13 11 24 26 25 22 27 23 5 1 6 4 8 7 18 15 20 16 21 19 17?
3 28 24 29 14 10 6 18 27 23 19 4 31 15 21 17 13 2 25 9 5 11 7 26 22 1 30 8 32 20 16 12?
4 31 29 3 7 15 14 17 22 21 30 1 2 6 10 8 18 13 20 25 23 27 5 32 9 12 11 16 24 19 28 26?
5 3 14 7 9 11 22 26 18 19 31 23 4 27 30 32 12 1 16 8 20 13 25 15 17 21 2 24 6 28 29 10?
6 32 10 15 11 20 30 4 31 9 8 14 24 19 25 1 3 29 18 13 22 28 27 23 2 7 16 12 21 17 26 5?
7 29 6 14 22 30 27 12 13 20 28 5 3 11 19 26 1 4 8 17 25 2 9 10 18 23 31 32 15 16 24 21?
8 9 18 17 26 4 12 30 21 29 7 16 25 1 3 11 19 20 6 15 24 23 32 2 10 28 5 27 13 14 22 31?
9 14 27 22 18 31 13 21 1 16 29 25 7 2 28 10 23 5 32 26 8 4 17 30 12 20 3 15 11 24 6 19?
10 12 23 21 19 9 20 29 16 27 18 15 26 25 22 3 24 32 2 31 11 8 30 7 28 6 17 4 5 13 1 14?
11 10 19 30 31 8 28 7 29 18 26 27 15 16 17 5 14 6 1 4 13 24 2 25 3 22 32 23 20 12 21 9?
12 13 4 1 23 14 5 16 25 15 27 26 18 7 6 28 8 17 30 19 10 9 21 29 20 32 22 31 3 11 2 24?
13 11 31 2 4 24 3 25 7 26 15 18 27 29 32 20 9 22 21 23 12 14 1 16 5 17 6 19 28 10 30 8?
14 24 15 6 27 19 11 1 2 25 16 7 29 30 20 12 4 3 17 18 9 31 22 21 13 5 28 26 10 8 32 23?
15 26 21 10 30 25 19 3 28 22 17 6 32 20 9 4 29 24 13 2 27 16 11 5 31 14 8 1 23 18 12 7?
16 25 17 8 32 1 26 11 10 3 5 28 20 12 4 27 30 19 14 6 29 21 24 13 15 31 23 22 18 7 9 2?
17 22 13 18 12 3 1 19 23 24 14 8 9 4 29 30 20 25 15 10 32 5 26 31 21 16 7 11 2 6 27 28?
18 27 2 13 1 29 4 20 5 32 6 17 22 3 24 19 25 9 10 21 26 7 12 28 23 8 14 30 31 15 11 16?
19 23 25 20 16 18 8 6 32 2 1 30 21 17 13 14 15 10 3 29 31 26 28 22 24 11 12 7 9 4 5 27?
20 5 9 25 8 13 17 15 26 31 4 19 23 18 2 6 10 21 29 24 28 12 16 27 32 30 1 14 22 3 7 11?
21 1 5 23 20 22 25 24 8 11 13 10 12 9 27 29 32 26 31 28 30 17 19 14 16 15 18 3 7 2 4 6?
22 6 11 27 13 28 2 23 4 8 24 9 14 31 16 21 5 7 26 12 17 3 18 19 1 25 29 10 30 32 15 20?
23 4 7 5 25 27 9 32 17 30 2 21 1 22 11 24 26 12 28 16 19 18 20 6 8 10 13 29 14 31 3 15?
24 8 26 32 15 23 10 2 30 7 25 29 16 21 5 13 31 28 22 27 14 19 6 1 11 3 20 18 12 9 17 4?
25 7 22 9 17 2 18 10 12 28 3 20 5 13 31 15 21 23 24 32 16 1 8 11 26 19 4 6 27 29 14 30?
26 18 1 12 21 7 23 28 20 6 22 32 17 5 14 31 16 8 11 30 15 25 10 3 19 24 9 2 4 27 13 29?
27 15 30 11 2 16 31 5 3 17 32 22 6 28 8 23 7 14 12 1 18 29 13 20 4 9 24 21 19 26 10 25?
28 20 8 16 24 12 32 27 15 4 23 31 19 26 1 22 11 30 7 14 3 10 29 18 6 2 21 9 17 5 25 13?
29 16 32 24 6 21 15 13 11 5 20 3 28 10 23 18 2 31 9 22 7 30 14 12 27 4 19 17 26 25 8 1?
30 21 20 19 28 17 16 14 24 13 12 11 10 8 18 7 6 15 4 3 2 32 31 9 29 27 26 5 25 1 23 22?
31 19 16 28 29 26 24 22 6 1 21 2 30 32 12 9 27 11 5 7 4 15 3 17 14 13 10 25 8 23 20 18?
32 17 12 26 10 5 21 31 19 14 9 24 8 23 7 2 28 16 27 11 6 20 15 4 30 29 25 13 1 22 18 3

計算模n的Dirichlet特征：
∵ φ(2) = 1、φ(3) = 2、φ(4) = 2、φ(5) = 4、φ(6) = 2、φ(10) = 4
∴只有1種模2特征。
^U(2):
χ1(n):1,
共有2種兩兩不同的模3特征、模4特征、模6特征。
^U(3)、^U(6):
χ1(n):1,1,
χ2(n):1,-1,
共有4種兩兩不同的模5特征、模10特征。
^U(5):{忠實表示：1=1,PrimitiveRootMod(5)=2=E(4),3=-E(4),4=-1}
χ1(n):1,1,1,1,
χ2(n):1,E(4),-E(4),-1,
χ3(n):1,-E(4),E(4),-1,
χ4(n):1,-1,-1,1,
^U(10):{忠實表示：1=1,PrimitiveRootMod(10)=3=E(4),9=-1,7=-E(4)}
χ1(n):1,1,1,1,
χ2(n):1,-1,1,-1,
χ3(n):1,E(4),-1,-E(4),
χ4(n):1,-E(4),-1,E(4),
^U(5)=C_4
i:=E(4);;X1:=[1,1,1,1];X2:=[1,i,-i,-1];X3:=[1,-i,i,-1];X4:=[1,-1,-1,1];XL:=[X1,X2,X3,X4];;for j in [1..4] do for k in [1..4] do J:=XL[j];K:=XL[k];JK:=[];for m in [1..4] do Add(JK,J[m]*K[m]);od;jk:=Position(XL,JK);Print(jk," ");od;Print("\n");od;
G4ElementToOrder(0)=1=ord(X1(n))
G4ElementToOrder(1)=4=ord(X2(n))
G4ElementToOrder(2)=4=ord(X3(n))
G4ElementToOrder(3)=2=ord(X4(n))
G4有1個1階元，1個2階元，2個4階元
^U(10)=C_4
i:=E(4);;X1:=[1,1,1,1];X2:=[1,-1,1,-1];X3:=[1,i,-1,-i];X4:=[1,-i,-1,i];XL:=[X1,X2,X3,X4];;for j in [1..4] do for k in [1..4] do J:=XL[j];K:=XL[k];JK:=[];for m in [1..4] do Add(JK,J[m]*K[m]);od;jk:=Position(XL,JK);Print(jk," ");od;Print("\n");od;
G4ElementToOrder(0)=1=ord(X1(n))
G4ElementToOrder(1)=2=ord(X2(n))
G4ElementToOrder(2)=4=ord(X3(n))
G4ElementToOrder(3)=4=ord(X4(n))
G4有1個1階元，1個2階元，2個4階元
U(5)={ZmodnZObj( 1, 5 ), ZmodnZObj( 2, 5 ) , ?ZmodnZObj( 3, 5 ), ZmodnZObj( 4, 5 )}
U(10)={ZmodnZObj( 1, 10 ), ZmodnZObj( 3, 10 ) , ?ZmodnZObj( 7, 10 ), ZmodnZObj( 9, 10 )}
gap> L1:=[1,E(4),-1,-E(4)];;List(L1,Order);
[ 1, 4, 2, 4 ]
gap> m:=5;;for n in [1..m-1] do if(Gcd(n,m)=1) then Print(n,"->",OrderMod(n,m),"階元\n");fi;od;
1->1階元
2->4階元
3->4階元
4->2階元
gap> m:=10;;for n in [1..m-1] do if(Gcd(n,m)=1) then Print(n,"->",OrderMod(n,m),"階元\n");fi;od;
1->1階元
3->4階元
7->4階元
9->2階元
問題1：設χ_1是模q_1的特征，χ_2是模q_2的特征，則χ_1χ_2 是模[q_1,q_2]的特征。當q_1≠q_2時，這里特征的乘法怎么定義？
定理：n階Abel群G有且僅有n個不同的特征，G的特征標表^G中的行、列經過重排后是唯一確定的，且^G=G。
χ1(n)是主特征(在G上恒等于1的函數)，χ2(n)、...、χ10(n)是非主特征(對G中某個a，它們有性質f(a)≠1)，特征之間是沒有固定順序的，某一特征χk(n)的分量χk(n=j)也沒有固定的順序，下面是一種遍歷所有特征χk(n)的順序：
特征乘法：把特征看成一個復向量，則兩個復向量對應位置的復數相乘得到的復向量就是兩個特征的乘積。
χ2(n)*χ2(n)=χ4(n)<=>PowerMod(2,2,11)=4
χ4(n)*χ2(n)=χ8(n)<=>PowerMod(2,3,11)=8
χ8(n)*χ2(n)=χ5(n)<=>PowerMod(2,4,11)=5
χ5(n)*χ2(n)=χ10(n)<=>PowerMod(2,5,11)=10
χ10(n)*χ2(n)=χ9(n)<=>PowerMod(2,6,11)=9
χ9(n)*χ2(n)=χ7(n)<=>PowerMod(2,7,11)=7
χ7(n)*χ2(n)=χ3(n)<=>PowerMod(2,8,11)=9
χ3(n)*χ2(n)=χ6(n)<=>PowerMod(2,9,11)=6
χ6(n)*χ2(n)=χ1(n)<=>PowerMod(2,10,11)=1
χ1(n)*χ2(n)=χ2(n)<=>PowerMod(2,11,11)=2=PowerMod(2,0,11)
χ2(n)^-1=χ6(n)<=>PowerMod(2,-1,11)=6
χ4(n)^-1=χ3(n)<=>PowerMod(4,-1,11)=3
χ8(n)^-1=χ7(n)<=>PowerMod(8,-1,11)=7
χ5(n)^-1=χ9(n)<=>PowerMod(5,-1,11)=9
χ10(n)^-1=χ10(n)<=>PowerMod(10,-1,11)=10
χ1(n)^-1=χ1(n)<=>PowerMod(1,-1,11)=1
G的特征的集合在特征的乘法下形成一個n階Abel群^G。^G的恒等元是主特征f_1，f_i的逆元是1/f_i。
這里χ10(n)*χ10(n)=χ1(n)，χ10(n)就是勒讓德符號，稱為模p=11的二次特征。
對于Abel群，特征標表所提供的關于群結構的信息是完全的（^G=G），但對于非Abel群則不是完全的。
利用原根與指數去構造模m=11的所有的狄利克雷特征。
gap> m:=11;;gi:=PrimitiveRootMod(m);;L:=[];;for n in [1..m-1] do Add(L,PowerMod(gi,n,m));od;List(L);Li:=[];;for n in [1..m-1] do Add(Li,Position(L,n)mod(m-1));od;List(Li);g:=E(m-1);;for j in [1..m-1] do Print("χ",j,"(n):");for nn in [0..m-2] do ni:=Position(L,nn+1);nj:=Li[j];Print((g^nj)^ni,",");od;Print("\n");od;
[ 2, 4, 8, 5, 10, 9, 7, 3, 6, 1 ]
[ 0, 1, 8, 2, 4, 9, 7, 3, 6, 5 ]
χ1(n):1,1,1,1,1,1,1,1,1,1,
χ2(n):1,-E(5)^3,E(5)^4,E(5),E(5)^2,-E(5)^2,-E(5),-E(5)^4,E(5)^3,-1,
χ3(n):1,E(5)^4,E(5)^2,E(5)^3,E(5),E(5),E(5)^3,E(5)^2,E(5)^4,1,
χ4(n):1,E(5),E(5)^3,E(5)^2,E(5)^4,E(5)^4,E(5)^2,E(5)^3,E(5),1,
χ5(n):1,E(5)^2,E(5),E(5)^4,E(5)^3,E(5)^3,E(5)^4,E(5),E(5)^2,1,
χ6(n):1,-E(5)^2,E(5),E(5)^4,E(5)^3,-E(5)^3,-E(5)^4,-E(5),E(5)^2,-1,
χ7(n):1,-E(5),E(5)^3,E(5)^2,E(5)^4,-E(5)^4,-E(5)^2,-E(5)^3,E(5),-1,
χ8(n):1,-E(5)^4,E(5)^2,E(5)^3,E(5),-E(5),-E(5)^3,-E(5)^2,E(5)^4,-1,
χ9(n):1,E(5)^3,E(5)^4,E(5),E(5)^2,E(5)^2,E(5),E(5)^4,E(5)^3,1,
χ10(n):1,-1,1,1,1,-1,-1,-1,1,-1,
gap> m:=11;;for n in [1..m-1] do i:=Legendre(n,m);Print(n,"->",i,"\n");od;
1->1
2->-1
3->1
4->1
5->1
6->-1
7->-1
8->-1
9->1
10->-1
2是模11的最小素數原根, g = 2。
gap> m:=11;;for n in [1..m-1] do Print(OrderMod(n,m),",");od;
1,10,5,5,5,10,10,10,5,2,
gap> PrimitiveRootMod(11);
2
gap> m:=11;;gi:=PrimitiveRootMod(m);;L:=[];;for n in [1..m-1] do Add(L,PowerMod(gi,n,m));od;List(L);for n in [1..m-1] do Print(Position(L,n)mod(m-1),",");od;
[ 2, 4, 8, 5, 10, 9, 7, 3, 6, 1 ]
0,1,8,2,4,9,7,3,6,5,
gap> m:=11;;gi:=PrimitiveRootMod(m);;for n in [1..m-1] do Print(LogMod(n,gi,m),",");od;
0,1,8,2,4,9,7,3,6,5,
gap> m:=11;;gi:=PrimitiveRootMod(m);;L:=[];;for n in [1..m-1] do Add(L,PowerMod(gi,n,m));od;List(L);Li:=[];;for n in [1..m-1] do Add(Li,Position(L,n)mod(m-1));od;List(Li);g:=E(m-1);;for j in [2,4] do Print("χ",j,"(n):");for nn in [0..m-2] do ni:=Position(L,nn+1);nj:=Li[j];Print((g^nj)^ni,",");od;Print("\n");od;
[ 2, 4, 8, 5, 10, 9, 7, 3, 6, 1 ]
[ 0, 1, 8, 2, 4, 9, 7, 3, 6, 5 ]
χ2(n):1,-E(5)^3,E(5)^4,E(5),E(5)^2,-E(5)^2,-E(5),-E(5)^4,E(5)^3,-1,
χ4(n):1,E(5),E(5)^3,E(5)^2,E(5)^4,E(5)^4,E(5)^2,E(5)^3,E(5),1,
gap> m:=11;;gi:=PrimitiveRootMod(m);;L:=[];;for n in [1..m-1] do Add(L,PowerMod(gi,n,m));od;List(L);;Li:=[];;for n in [1..m-1] do Add(Li,Position(L,n)mod(m-1));od;List(Li);;g:=E(m-1);;for j in [1..m-1] do Print("~χ",j,"(n):");for nn in [0..m-2] do ni:=Position(L,nn+1);nj:=Li[j];z:=(g^nj)^ni;z1:=ComplexConjugate(z);Print(z1,",");od;Print("\n");od;
~χ1(n):1,1,1,1,1,1,1,1,1,1,
~χ2(n):1,-E(5)^2,E(5),E(5)^4,E(5)^3,-E(5)^3,-E(5)^4,-E(5),E(5)^2,-1,
~χ3(n):1,E(5),E(5)^3,E(5)^2,E(5)^4,E(5)^4,E(5)^2,E(5)^3,E(5),1,
~χ4(n):1,E(5)^4,E(5)^2,E(5)^3,E(5),E(5),E(5)^3,E(5)^2,E(5)^4,1,
~χ5(n):1,E(5)^3,E(5)^4,E(5),E(5)^2,E(5)^2,E(5),E(5)^4,E(5)^3,1,
~χ6(n):1,-E(5)^3,E(5)^4,E(5),E(5)^2,-E(5)^2,-E(5),-E(5)^4,E(5)^3,-1,
~χ7(n):1,-E(5)^4,E(5)^2,E(5)^3,E(5),-E(5),-E(5)^3,-E(5)^2,E(5)^4,-1,
~χ8(n):1,-E(5),E(5)^3,E(5)^2,E(5)^4,-E(5)^4,-E(5)^2,-E(5)^3,E(5),-1,
~χ9(n):1,E(5)^2,E(5),E(5)^4,E(5)^3,E(5)^3,E(5)^4,E(5),E(5)^2,1,
~χ10(n):1,-1,1,1,1,-1,-1,-1,1,-1,
gap> m:=11;;gi:=PrimitiveRootMod(m);;L:=[];;for n in [1..m-1] do Add(L,PowerMod(gi,n,m));od;List(L);;Li:=[];;for n in [1..m-1] do Add(Li,Position(L,n)mod(m-1));od;List(Li);;g:=E(m-1);;for j in [1..m-1] do Print("|χ|",j,"(n):");for nn in [0..m-2] do ni:=Position(L,nn+1);nj:=Li[j];z:=(g^nj)^ni;z1:=ComplexConjugate(z);Print(Sqrt(z*z1),",");od;Print("\n");od;
|χ|1(n):1,1,1,1,1,1,1,1,1,1,
|χ|2(n):1,1,1,1,1,1,1,1,1,1,
|χ|3(n):1,1,1,1,1,1,1,1,1,1,
|χ|4(n):1,1,1,1,1,1,1,1,1,1,
|χ|5(n):1,1,1,1,1,1,1,1,1,1,
|χ|6(n):1,1,1,1,1,1,1,1,1,1,
|χ|7(n):1,1,1,1,1,1,1,1,1,1,
|χ|8(n):1,1,1,1,1,1,1,1,1,1,
|χ|9(n):1,1,1,1,1,1,1,1,1,1,
|χ|10(n):1,1,1,1,1,1,1,1,1,1,
容易看出，若χ是模q的特征，則~χ亦是模q的特征。
這里定義~χ(n)= ~(χ(n))，~χ稱為是χ的共軛特征，~χχ≡χ1，即~χ=χ^-1。
~χ1(n)=χ1(n)、~χ2(n)=χ6(n)、~χ3(n)=χ4(n)、~χ4(n)=χ3(n)、~χ5(n)=χ9(n)、~χ6(n)=χ2(n)、~χ7(n)=χ8(n)、~χ8(n)=χ7(n)、~χ9(n)=χ5(n)、~χ10(n)=χ10(n)
gap> for n in [1,3,7,9,11,13,17,19] do t:=1;if(n mod 4=3)then t:=-1;fi;i:=Legendre(n,5);Print(n,"->",i*t,"\n");od;
1->1
3->1
7->1
9->1
11->-1
13->-1
17->-1
19->-1
計算狄利克雷特征χ(a mod20)時進行了二次剩余符號(a/5)的計算
χ(1mod20)=(1/5)θ(1)=1·1=1
χ(3mod20)=(3/5)θ(3)=-1·-1=1
χ(7mod20)=(7/5)θ(7)=-1·-1=1
χ(9mod20)=(9/5)θ(9)=1·1=1
χ(11mod20)=(11/5)θ(11)=1·-1=-1
χ(13mod20)=(13/5)θ(13)=-1·1=-1
χ(17mod20)=(17/5)θ(17)=-1·1=-1
χ(19mod20)=(19/5)θ(19)=1·-1=-1
代數數域是代數數論的研究對象，代數數論的起源可追溯到高斯研究二平方和問題時涉及到的高斯整數環Z[i]=Z[ζ_4]和庫默爾研究費馬大定理時涉及到的分圓環Z[ζ_p]，它的大部分經典理論產生于19世紀。
Z[ω]=Z[ζ_3]
n次分圓域是多項式x^n-1的分裂域，因此是有理數域的伽羅瓦擴域。
這個擴張的次數等于歐拉φ函數：[Q(ζ_n):Q]=φ(n)
Gal(Q(ζ_n)/Q)=(Z_n)^×——這個群是偶數階交換群
【
定理：對任意n>2，2|φ(n)。
證明：
設n=p_1*p_2*…p_i, 則φ(n)=n*(p_1-1)(p_2-1)…(p_i-1)/(p_1*p_2*…p_i)=n*(1-1/p_1)*(1-1/p_2)…(1-1/p_i)，因為p_i-1是偶數或1，且必存在n的素因子p_i使得p_i-1是偶數，所以n>2時，φ(n)是偶數。
】?
ζ_n的所有伽羅瓦共軛是ζ_n^a，其中a遍歷模n的簡化剩余系（所有與n互質的剩余類）
ζ_3=(sqrt(3)i-1)/2
ζ_6=(sqrt(3)i+1)/2
——ζ_3是6次單位根，但不是本原6次單位根
分圓域Q(ζ_3)=分圓域Q(ζ_6)=二次域Q(sqrt(3))
> euler(3)
2
> euler(6)
2
> euler(8)
4
> euler(15)
8
> euler(16)
8
root@ubuntu:/home/cpptest# g++ -o CyclotomicPolynomial CyclotomicPolynomial.cpp
root@ubuntu:/home/cpptest# ./CyclotomicPolynomial 2
共有1個2次本原單位根:[-1.000000i]
Φ_2(x)=(x-ξ_1)=1x^0+1x^1=[1,1]
root@ubuntu:/home/cpptest# ./CyclotomicPolynomial 3
共有2個3次本原單位根:[-0.500000+0.866025i,-0.500000-0.866025i]
Φ_3(x)=(x-ξ_1)(x-ξ_2)=1x^0+1x^1+1x^2=[1,1,1]
root@ubuntu:/home/cpptest# ./CyclotomicPolynomial 4
共有2個4次本原單位根:[+1.000000i,-1.000000i]
Φ_4(x)=(x-ξ_1)(x-ξ_2)=1x^0+0x^1+1x^2=[1,0,1]
root@ubuntu:/home/cpptest# ./CyclotomicPolynomial 5
共有4個5次本原單位根:[0.309017+0.951057i,-0.809017+0.587785i,-0.809017-0.587785i,0.309017-0.951057i]
Φ_5(x)=(x-ξ_1)(x-ξ_2)(x-ξ_3)(x-ξ_4)=1x^0+1x^1+1x^2+1x^3+1x^4=[1,1,1,1,1]
root@ubuntu:/home/cpptest# ./CyclotomicPolynomial 6
共有2個6次本原單位根:[0.500000+0.866025i,0.500000-0.866025i]
Φ_6(x)=(x-ξ_1)(x-ξ_2)=1x^0+-1x^1+1x^2=[1,-1,1]
n次分圓多項式僅包含所有的n次單位原根，degΦ_n(x)=φ(n)
Φ_1(x)=x-1
Φ_2(x)=(x^2-1)/Φ_1(x)=x+1
Φ_3(x)=(x^3-1)/Φ_1(x)=x^2+x+1
Φ_4(x)=(x^4-1)/(Φ_1(x)Φ_2(x))=x^2+1
Φ_6(x)=(x^6-1)/(Φ_1(x)Φ_2(x)Φ_3(x))=x^2-x+1
Φ_12(x)=(x^12-1)/(Φ_1(x)Φ_2(x)Φ_3(x)Φ_4(x)Φ_6(x))=x^4-x^2+1
定理：多項式Φ_n(x)在Q[x]中不可約。
循環群C_n共有φ(n)個生成元素或本原n次單位根共有φ(n)個，這是一個純群論的定理。
Φ_n(x)=(x-ξ_1)(x-ξ_2)…(x-ξ_φ(n))稱為分圓多項式。意思是說求出它的一個根就可以把單位圓分成n等份了。
n=1時，生成元ξ=1，φ(1)=1，Φ_1(x)=x-1
n=2時，生成元ξ=-1，φ(2)=1，Φ_2(x)=x+1
n=3時，生成元ξ=(sqrt(3)i-1)/2，ξ^2=(-sqrt(3)i-1)/2，φ(3)=2，Φ_3(x)=(x-ξ)(x-ξ^2)=x^2+x+1
n=4時，生成元ξ=i，ξ^3=-i，φ(4)=2，Φ_4(x)=(x-i)(x+i)=x^2+1
分圓多項式是復數域到有限域的橋梁。
定理：x^n-1=∏[d|n]Φ_d(x)。
定理：Φ_n(x)是整系數多項式。
問題：不存在n，使得U(n)=C_8=GAP4[8,1]？
gap> for i in [1..500] do L:=IdGroup(Units(ZmodnZ(i)));;if L[1]=8 and L[2]=1 then Print("U(",i,")=[8,1]\n"); fi;od;
gap>

計算模16的Dirichlet特征：
gap> U16:=Units(Integers mod 16);;IdGroup(U16);
[ 8, 2 ]
U16=C_2×C_4
gap>g:=DirectProduct(CyclicGroup(2),CyclicGroup(4));;IdGroup(g);cl:=ConjugacyClasses(g);;L1:=List(cl,Representative);;L2:=List(cl,Centralizer);;L3:=List(L2,IdGroup);;L4:=List(cl,Size);;tbl:=CharacterTable( g );;Display( tbl );
[ 8, 2 ]
CT8

???? 2? 3? 3? 3? 3? 3 3? 3? 3

?????? 1a 2a 4a 2b 4b 2c 4c4d

X.1???? 1? 1? 1? 1? 1 1? 1? 1
X.2???? 1 -1? 1? 1 -1 -1? 1 -1
X.3???? 1? 1 -1? 1 -1? 1 -1 -1
X.4???? 1 -1 -1? 1? 1 -1 -1? 1
X.5???? 1? 1? A -1? A -1 -A -A
X.6???? 1 -1? A -1 -A? 1 -A? A
X.7???? 1? 1 -A -1 -A -1? A? A
X.8???? 1 -1 -A -1? A? 1? A -A

A = E(4)
? = Sqrt(-1) = i
C_2×C_2×C_2
gap>g:=DirectProduct(CyclicGroup(2),CyclicGroup(2),CyclicGroup(2));;IdGroup(g);cl:=ConjugacyClasses(g);;L1:=List(cl,Representative);;L2:=List(cl,Centralizer);;L3:=List(L2,IdGroup);;L4:=List(cl,Size);;tbl:=CharacterTable( g );;Display( tbl );
[ 8, 5 ]
CT9

???? 2? 3? 3? 3? 3? 3 3? 3? 3

?????? 1a 2a 2b 2c 2d 2e 2f2g

X.1???? 1? 1? 1? 1? 1 1? 1? 1
X.2???? 1 -1? 1? 1 -1 -1? 1 -1
X.3???? 1? 1 -1? 1 -1? 1 -1 -1
X.4???? 1 -1 -1? 1? 1 -1 -1? 1
X.5???? 1? 1? 1 -1? 1 -1 -1 -1
X.6???? 1 -1? 1 -1 -1? 1 -1? 1
X.7???? 1? 1 -1 -1 -1 -1? 1? 1
X.8???? 1 -1 -1 -1? 1? 1? 1 -1
gap>g:=CyclicGroup(8);;IdGroup(g);cl:=ConjugacyClasses(g);;L1:=List(cl,Representative);;L2:=List(cl,Centralizer);;L3:=List(L2,IdGroup);;L4:=List(cl,Size);;tbl:=CharacterTable( g );;Display( tbl);???????????????????????????????????????? [ 8, 1 ]
CT10

???? 2? 3?? 3? 3 3?? 3?? 3? 3?? 3

?????? 1a? 8a 4a 2a? 8b? 8c4b? 8dX.1???? 1?? 1? 1? 1? 1?? 1? 1?? 1
X.2???? 1? -1? 1? 1? -1? -1 1? -1
X.3???? 1?? A -1? 1? -A?? A-1? -A
X.4???? 1? -A -1? 1?? A? -A-1?? A
X.5???? 1?? B? A -1 -/B? -B -A? /B
X.6???? 1? -B? A -1? /B?? B -A -/B
X.7???? 1 -/B -A -1?? B? /B? A? -B
X.8???? 1? /B -A -1? -B -/B? A?? BA = E(4)
? = Sqrt(-1) = i
B = E(8)

計算模8的Dirichlet特征
∵ φ(8) = 4
∴共有4種兩兩不同的特征。
gap> U8:=GroupWithGenerators([ZmodnZObj(3,8),ZmodnZObj(5,8)]);;L:=Elements(U8);IdGroup(U8);for i1 in L do for i2 in L do Print(Position(L,i1*i2)," ");od;Print("\n");od;n:=Size(L);for i in [1..n] do Print(i,"->",L[i],",",Order(L[i]),"階元\n");od;
[ ZmodnZObj( 1, 8 ), ZmodnZObj( 3, 8 ), ZmodnZObj( 5, 8 ), ZmodnZObj( 7, 8 ) ]
[ 4, 2 ]
1 2 3 4
2 1 4 3
3 4 1 2
4 3 2 1
4
1->ZmodnZObj( 1, 8 ),1階元
2->ZmodnZObj( 3, 8 ),2階元
3->ZmodnZObj( 5, 8 ),2階元
4->ZmodnZObj( 7, 8 ),2階元
gap> L1:=[1,-1];X1:=[1,1,1,1];X2:=[1,-1,1,-1];X3:=[1,1,-1,-1];X4:=[1,-1,-1,1];XL:=[X1,X2,X3,X4];;for Xi in XL do for i1 in Xi do Print(Position(L1,i1)," ");od;Print("\n");od;
[ 1, -1 ]
[ 1, 1, 1, 1 ]
[ 1, -1, 1, -1 ]
[ 1, 1, -1, -1 ]
[ 1, -1, -1, 1 ]
1 1 1 1
1 2 1 2
1 1 2 2
1 2 2 1
非循環Abel群的同態像是？
Imφ1={1}=C_1
Imφ2=Imφ3=Imφ4={1,2}=C_2

C_2×C_2=D_2=V_4
1,1,1,1
1,1,-1,-1
1,-1,1,-1
1,-1,-1,1
gap>g:=DirectProduct(CyclicGroup(2),CyclicGroup(2));;IdGroup(g);cl:=ConjugacyClasses(g);;L1:=List(cl,Representative);;L2:=List(cl,Centralizer);;L3:=List(L2,IdGroup);;L4:=List(cl,Size);;tbl:=CharacterTable( g );;Display( tbl );
[ 4, 2 ]
CT1

???? 2? 2? 2? 2 2

?????? 1a 2a 2b 2c

X.1???? 1? 1? 1? 1
X.2???? 1 -1? 1 -1
X.3???? 1? 1 -1 -1
X.4???? 1 -1 -1? 1

計算模10的Dirichlet特征
∵ φ(10) = 4
∴共有4種兩兩不同的特征。
gap> UR4:=GroupWithGenerators([E(4)]);;IdGroup(UR4);L:=Elements(UR4);L1:=[1];;for a in L do if a<>1 then Add(L1,a);fi;od;for i1 in L1 do for i2 in L1 do Print(Position(L1,i1*i2)," ");od;Print("\n");od;n:=Size(L1);for i in [1..n] do Print(i,"->",L1[i],",",Order(L1[i]),"階元\n");od;
[ 4, 1 ]
[ -1, 1, -E(4), E(4) ]
1 2 3 4
2 1 4 3
3 4 2 1
4 3 1 2
4
1->1,1階元
2->-1,2階元
3->-E(4),4階元
4->E(4),4階元
gap> L1;
[ 1, -1, -E(4), E(4) ]
gap> A:=E(4);;X1:=[1,1,1,1];X2:=[1,-1,1,-1];X3:=[1,A,-1,-A];X4:=[1,-A,-1,A];XL:=[X1,X2,X3,X4];;for Xi in XL do for i1 in Xi do Print(Position(L1,i1)," ");od;Print("\n");od;
[ 1, 1, 1, 1 ]
[ 1, -1, 1, -1 ]
[ 1, E(4), -1, -E(4) ]
[ 1, -E(4), -1, E(4) ]
1 1 1 1
1 2 1 2
1 4 2 3
1 3 2 4
循環群的同態像是循環群：
Imφ1={1}=C_1
Imφ2={1,2}=C_2
Imφ3=Imφ4={1,2,3,4}=C_4

gap> C4:=Group([[1]],[[-1]],[[E(4)]],[[-E(4)]]);IdGroup(C4);s:=Elements(C4);for si in sdo Print("ord:");Print(Order(si));Print(",Tr:");Print(TraceMat(si));Print(",中心化子:");Print(IdGroup(Centralizer(C4,si)));Print("\n");od;
Group([ [ [ 1 ] ], [ [ -1 ] ], [ [ E(4) ] ], [ [ -E(4) ] ] ])
[ 4, 1 ]
[ 4, 1 ]
[ [ [ -1 ] ], [ [ 1 ] ], [ [ -E(4) ] ], [ [ E(4) ] ] ]
ord:2,Tr:-1,中心化子:[ 4, 1 ]
ord:1,Tr:1,中心化子:[ 4, 1 ]
ord:4,Tr:-E(4),中心化子:[ 4, 1 ]
ord:4,Tr:E(4),中心化子:[ 4, 1 ]
gap> C4m2:=Group([[1,0],[0,1]],[[-1,0],[0,-1]],[[0,1],[-1,0]],[[0,-1],[1,0]]);IdGroup(C4m2);s:=Elements(C4m2);forsi in s do Print("ord:");Print(Order(si));Print(",Tr:");Print(TraceMat(si));Print(",中心化子:");Print(IdGroup(Centralizer(C4m2,si)));Print("\n");od;
Group([ [ [ 1, 0 ], [ 0, 1 ] ], [ [ -1, 0 ], [ 0, -1 ] ], [ [ 0, 1 ], [ -1, 0 ]], [ [ 0, -1 ], [ 1, 0 ] ] ])
[ 4, 1 ]
[ [ [ -1, 0 ], [ 0, -1 ] ], [ [ 0, -1 ], [ 1, 0 ] ], [ [ 0, 1 ], [ -1, 0 ] ], [[ 1, 0 ], [ 0, 1 ] ] ]
ord:2,Tr:-2,中心化子:[ 4, 1 ]
ord:4,Tr:0,中心化子:[ 4, 1 ]
ord:4,Tr:0,中心化子:[ 4, 1 ]
ord:1,Tr:2,中心化子:[ 4, 1 ]
gap>g:=CyclicGroup(4);;IdGroup(g);cl:=ConjugacyClasses(g);;L1:=List(cl,Representative);;L2:=List(cl,Centralizer);;L3:=List(L2,IdGroup);;L4:=List(cl,Size);;tbl:=CharacterTable( g );;Display( tbl );
[ 4, 1 ]
CT2

???? 2? 2? 2? 2? 2

?????? 1a 4a 2a 4b

X.1???? 1? 1? 1? 1
X.2???? 1 -1? 1 -1
X.3???? 1? A -1 -A
X.4???? 1 -A -1? A

A = E(4)
? = Sqrt(-1) = i
gap> G:=GroupWithGenerators(Units(GaussianIntegers));;IdGroup(G);
[ 4, 1 ]
gap> G:=GroupWithGenerators(Units(Integers));;IdGroup(G);
[ 2, 1 ]

計算模9的Dirichlet特征
∵ φ(9) = 6
∴共有6種兩兩不同的特征。
gap> U9:=Units(Integers mod 9);;IdGroup(U9);C6:=GroupWithGenerators([ZmodnZObj(2,9)]);;L:=Elements(C6);;IdGroup(C6);for i1 in L do for i2 in L do Print(Position(L,i1*i2)," ");od;Print("\n");od;n:=Size(L);for i in [1..n] do Print(i,"->",L[i],",",Order(L[i]),"階元\n");od;
[ 6, 2 ]
[ 6, 2 ]
1 2 3 4 5 6
2 3 6 1 4 5
3 6 5 2 1 4
4 1 2 5 6 3
5 4 1 6 3 2
6 5 4 3 2 1
6
1->ZmodnZObj( 1, 9 ),1階元
2->ZmodnZObj( 2, 9 ),6階元
3->ZmodnZObj( 4, 9 ),3階元
4->ZmodnZObj( 5, 9 ),6階元
5->ZmodnZObj( 7, 9 ),3階元
6->ZmodnZObj( 8, 9 ),2階元
gap> UR6:=GroupWithGenerators([E(6)]);;IdGroup(UR6);L:=Elements(UR6);L1:=[1];;for a in L do if a<>1 then Add(L1,a);fi;od;for i1 in L1 do for i2 in L1 do Print(Position(L1,i1*i2)," ");od;Print("\n");od;n:=Size(L1);for i in [1..n] do Print(i,"->",L1[i],",",Order(L1[i]),"階元\n");od;
[ 6, 2 ]
[ -1, 1, -E(3), -E(3)^2, E(3)^2, E(3) ]
1 2 3 4 5 6
2 1 6 5 4 3
3 6 5 1 2 4
4 5 1 6 3 2
5 4 2 3 6 1
6 3 4 2 1 5
6
1->1,1階元
2->-1,2階元
3->-E(3),6階元
4->-E(3)^2,6階元
5->E(3)^2,3階元
6->E(3),3階元
gap> A:=E(3)^2;;L1:=[1,-1,A,-A,1/A,-1/A];X1:=[1,1,1,1,1,1];X2:=[1,-1,1,-1,1,-1];X3:=[1,A,1/A,1,A,1/A];X4:=[1,-A,1/A,-1,A,-1/A];X5:=[1,1/A,A,1,1/A,A];X6:=[1,-1/A,A,-1,1/A,-A];XL:=[X1,X2,X3,X4,X5,X6];;for Xi in XL do for i1 in Xi do Print(Position(L1,i1)," ");od;Print("\n");od;
[ 1, -1, E(3)^2, -E(3)^2, E(3), -E(3) ]
[ 1, 1, 1, 1, 1, 1 ]
[ 1, -1, 1, -1, 1, -1 ]
[ 1, E(3)^2, E(3), 1, E(3)^2, E(3) ]
[ 1, -E(3)^2, E(3), -1, E(3)^2, -E(3) ]
[ 1, E(3), E(3)^2, 1, E(3), E(3)^2 ]
[ 1, -E(3), E(3)^2, -1, E(3), -E(3)^2 ]
1 1 1 1 1 1
1 2 1 2 1 2
1 3 5 1 3 5
1 4 5 2 3 6
1 5 3 1 5 3
1 6 3 2 5 4
非循環Abel群的同態像是？
Imφ1={1}=C_1
Imφ2={1,2}=C_2
Imφ3=Imφ5={1,3,5}=C_3
Imφ4=Imφ6={1,2,3,4,5,6}=C_6

g:=CyclicGroup(6);;IdGroup(g);cl:=ConjugacyClasses(g);;L1:=List(cl,Representative);;L2:=List(cl,Centralizer);;L3:=List(L2,IdGroup);;L4:=List(cl,Size);;tbl:=CharacterTable( g );;Display( tbl );
[ 6, 2 ]
CT4

???? 2? 1?? 1? 1? 1 1?? 1
???? 3? 1?? 1? 1? 1 1?? 1

?????? 1a? 6a 3a 2a 3b? 6b

X.1???? 1?? 1? 1? 1 1?? 1
X.2???? 1? -1? 1 -1? 1? -1
X.3???? 1?? A /A? 1? A? /A
X.4???? 1? -A /A -1? A -/A
X.5???? 1? /A? A? 1 /A?? A
X.6???? 1 -/A? A -1 /A? -A

A = E(3)^2
? = (-1-Sqrt(-3))/2 = -1-b3

計算模11的Dirichlet特征
∵ φ(11) = 10
∴共有10種兩兩不同的特征。
gap>?U11:=Units(Integers mod 11);;IdGroup(U11);
[ 10, 2 ]
C_10=U11有1個1階元，1個2階元，4個5階元，4個10階元
Dirichlet characters modulo 11
k = 11, φ(k) = 10, (Z/kZ)*=C_10,?
w^5 = -1, w = exp(πi/5) =-E(5)^3=E(10)
gap> UR10:=GroupWithGenerators([E(10)]);;L:=Elements(UR10);;IdGroup(UR10);L1:=[1];;for a in L do if a<>1 then Add(L1,a);fi;od;for i1 in L1 do for i2 in L1 do Print(Position(L1,i1*i2)," ");od;Print("\n");od;n:=Size(L1);for i in [1..n] do Print(i,"->",L1[i],",",Order(L1[i]),"階元\n");od;
[ 10, 2 ]
1 2 3 4 5 6 7 8 9 10
2 1 10 9 8 7 6 5 4 3
3 10 9 8 7 1 2 6 5 4
4 9 8 7 1 10 3 2 6 5
5 8 7 1 10 9 4 3 2 6
6 7 1 10 9 8 5 4 3 2
7 6 2 3 4 5 8 9 10 1
8 5 6 2 3 4 9 10 1 7
9 4 5 6 2 3 10 1 7 8
10 3 4 5 6 2 1 7 8 9
10
1->1,1階元
2->-1,2階元
3->-E(5),10階元
4->-E(5)^2,10階元
5->-E(5)^3,10階元
6->-E(5)^4,10階元
7->E(5)^4,5階元
8->E(5)^3,5階元
9->E(5)^2,5階元
10->E(5),5階元
gap> X1:=[1,1,1,1,1,1,1,1,1,1];X2:=[1,w,-w^3,w^2,w^4,-w^4,-w^2,w^3,-w,-1];X3:=[1,w^2,-w,w^4,-w^3,-w^3,w^4,-w,w^2,1];X4:=[1,w^3,w^4,-w,w^2,-w^2,w,-w^4,-w^3,1];X5:=[1,w^4,w^2,-w^3,-w,-w,-w^3,w^2,w^4,1];X6:=[1,-1,1,1,1,-1,-1,-1,1,-1];X7:=[1,-w,-w^3,w^2,w^4,w^4,w^2,-w^3,-w,1];X8:=[1,-w^2,-w,w^4,-w^3,w^3,-w^4,w,w^2,-1];X9:=[1,-w^3,w^4,-w,w^2,w^2,-w,w^4,-w^3,1];X10:=[1,-w^4,w^2,-w^3,-w,w,w^3,-w^2,w^4,-1];XL:=[X1,X2,X3,X4,X5,X6,X7,X8,X9,X10];;for Xi in XL do for i1 in Xi do Print(Position(L1,i1)," ");od;Print("\n");od;
[ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ]
[ 1, -E(5)^3, E(5)^4, E(5), E(5)^2, -E(5)^2, -E(5), -E(5)^4, E(5)^3, -1 ]
[ 1, E(5), E(5)^3, E(5)^2, E(5)^4, E(5)^4, E(5)^2, E(5)^3, E(5), 1 ]
[ 1, -E(5)^4, E(5)^2, E(5)^3, E(5), -E(5), -E(5)^3, -E(5)^2, E(5)^4, 1 ]
[ 1, E(5)^2, E(5), E(5)^4, E(5)^3, E(5)^3, E(5)^4, E(5), E(5)^2, 1 ]
[ 1, -1, 1, 1, 1, -1, -1, -1, 1, -1 ]
[ 1, E(5)^3, E(5)^4, E(5), E(5)^2, E(5)^2, E(5), E(5)^4, E(5)^3, 1 ]
[ 1, -E(5), E(5)^3, E(5)^2, E(5)^4, -E(5)^4, -E(5)^2, -E(5)^3, E(5), -1 ]
[ 1, E(5)^4, E(5)^2, E(5)^3, E(5), E(5), E(5)^3, E(5)^2, E(5)^4, 1 ]
[ 1, -E(5)^2, E(5), E(5)^4, E(5)^3, -E(5)^3, -E(5)^4, -E(5), E(5)^2, -1 ]
1 1 1 1 1 1 1 1 1 1
1 5 7 10 9 4 3 6 8 2
1 10 8 9 7 7 9 8 10 1
1 6 9 8 10 3 5 4 7 1
1 9 10 7 8 8 7 10 9 1
1 2 1 1 1 2 2 2 1 2
1 8 7 10 9 9 10 7 8 1
1 3 8 9 7 6 4 5 10 2
1 7 9 8 10 10 8 9 7 1
1 4 10 7 8 5 6 3 9 2
循環群的同態像是循環群：
Imφ1={1}=C_1
Imφ2=Imφ4=Imφ8=Imφ10={1,2,3,4,5,6,7,8,9,10}=C_10
Imφ3=Imφ5=Imφ7=Imφ9={1,7,8,9,10}=C_5
Imφ6={1,2}=C_2
計算模20的Dirichlet特征：
gap> U20:=GroupWithGenerators([ZmodnZObj(3,20),ZmodnZObj(19,20)]);;L:=Elements(U20);IdGroup(U20);for i1 in L do for i2 in L do Print(Position(L,i1*i2)," ");od;Print("\n");od;n:=Size(L);for i in [1..n] do Print(i,"->",L[i],",",Order(L[i]),"階元\n");od;
[ ZmodnZObj( 1, 20 ), ZmodnZObj( 3, 20 ), ZmodnZObj( 7, 20 ), ZmodnZObj( 9, 20 ), ZmodnZObj( 11, 20 ),
? ZmodnZObj( 13, 20 ), ZmodnZObj( 17, 20 ), ZmodnZObj( 19, 20 ) ]
[ 8, 2 ]
1 2 3 4 5 6 7 8
2 4 1 3 6 8 5 7
3 1 4 2 7 5 8 6
4 3 2 1 8 7 6 5
5 6 7 8 1 2 3 4
6 8 5 7 2 4 1 3
7 5 8 6 3 1 4 2
8 7 6 5 4 3 2 1
8
1->ZmodnZObj( 1, 20 ),1階元
2->ZmodnZObj( 3, 20 ),4階元
3->ZmodnZObj( 7, 20 ),4階元
4->ZmodnZObj( 9, 20 ),2階元
5->ZmodnZObj( 11, 20 ),2階元
6->ZmodnZObj( 13, 20 ),4階元
7->ZmodnZObj( 17, 20 ),4階元
8->ZmodnZObj( 19, 20 ),2階元
gap> i:=E(4);;L1:=[1,i,-1,-i];X1:=[1,1,1,1,1,1,1,1];X2:=[1,i,-i,-1,-1,-i,i,1];X3:=[1,-1,-1,1,1,-1,-1,1];X4:=[1,-i,i,-1,-1,i,-i,1];X5:=[1,1,1,1,-1,-1,-1,-1];X6:=[1,i,-i,-1,1,i,-i,-1];X7:=[1,-1,-1,1,-1,1,1,-1];X8:=[1,-i,i,-1,1,-i,i,-1];XL:=[X1,X2,X3,X4,X5,X6,X7,X8];;for Xi in XL do for i1 in Xi do Print(Position(L1,i1)," ");od;Print("\n");od;
[ 1, E(4), -1, -E(4) ]
[ 1, 1, 1, 1, 1, 1, 1, 1 ]
[ 1, E(4), -E(4), -1, -1, -E(4), E(4), 1 ]
[ 1, -1, -1, 1, 1, -1, -1, 1 ]
[ 1, -E(4), E(4), -1, -1, E(4), -E(4), 1 ]
[ 1, 1, 1, 1, -1, -1, -1, -1 ]
[ 1, E(4), -E(4), -1, 1, E(4), -E(4), -1 ]
[ 1, -1, -1, 1, -1, 1, 1, -1 ]
[ 1, -E(4), E(4), -1, 1, -E(4), E(4), -1 ]
1 1 1 1 1 1 1 1
1 2 4 3 3 4 2 1
1 3 3 1 1 3 3 1
1 4 2 3 3 2 4 1
1 1 1 1 3 3 3 3
1 2 4 3 1 2 4 3
1 3 3 1 3 1 1 3
1 4 2 3 1 4 2 3
非循環Abel群的同態像是？
Imφ1={1}=C_1
Imφ2=Imφ4=Imφ6=Imφ8={1,2,3,4}=C_4
Imφ3=Imφ5=Imφ7={1,3}=C_2

gap> U20:=Units(Integers mod20);;IdGroup(U20);g:=U20;;cl:=ConjugacyClasses(g);;L1:=List(cl,Representative);;L2:=List(cl,Centralizer);;L3:=List(L2,IdGroup);;L4:=List(cl,Size);;tbl:=CharacterTable( g );;Display( tbl );
[ 8, 2 ]
CT6

???? 2? 3? 3? 3? 3? 3 3? 3? 3

?????? 1a 4a 4b 2a 2b 4c 4d 2c

X.1???? 1? 1? 1? 1? 1 1? 1? 1
X.2???? 1 -1 -1? 1? 1 -1 -1? 1
X.3???? 1 -1 -1? 1 -1? 1? 1 -1
X.4???? 1? 1? 1? 1 -1 -1 -1 -1
X.5???? 1? A -A -1? 1? A -A -1
X.6???? 1 -A? A -1? 1 -A? A -1
X.7???? 1 -A? A -1 -1? A -A? 1
X.8???? 1? A -A -1 -1 -A? A? 1

A = -E(4)
? = -Sqrt(-1) = -i
模20的狄利克雷特征
http://www.di-mgt.com.au/cgi-bin/dirichlet.cgi?k=20&submit=+Go+
Dirichlet characters modulo 20
k = 20, φ(k) = 8, (Z/kZ)*=C4 x C2, generators = 3,19
G={1,3,7,9,11,13,17,19}, λ = 4, w2 = -1, w = exp(πi/2)
X(n) mod 20 n 1 3 7 9 11 13 17 19
X1(n) 1 1 1 1 1 1 1 1
X2(n) 1 i -i -1 -1 -i i 1
X3(n) 1 -1 -1 1 1 -1 -1 1
X4(n) 1 -i i -1 -1 i -i 1
X5(n) 1 1 1 1 -1 -1 -1 -1
X6(n) 1 i -i -1 1 i -i -1
X7(n) 1 -1 -1 1 -1 1 1 -1
X8(n) 1 -i i -1 1 -i i -1
X(n) mod 20
U(20)的特征標群^U(20)的凱萊表：
gap> i:=E(4);;X1:=[1,1,1,1,1,1,1,1];;X2:=[1,i,-i,-1,-1,-i,i,1];;X3:=[1,-1,-1,1,1,-1,-1,1];;X4:=[1,-i,i,-1,-1,i,-i,1];;X5:=[1,1,1,1,-1,-1,-1,-1];;X6:=[1,i,-i,-1,1,i,-i,-1];;X7:=[1,-1,-1,1,-1,1,1,-1];;X8:=[1,-i,i,-1,1,-i,i,-1];;XL:=[X1,X2,X3,X4,X5,X6,X7,X8];;for j in [1..8] do for k in [1..8] do J:=XL[j];K:=XL[k];JK:=[];for m in [1..8] do Add(JK,J[m]*K[m]);od;jk:=Position(XL,JK);Print(jk," ");od;Print("\n");od;
1 2 3 4 5 6 7 8
2 3 4 1 6 7 8 5
3 4 1 2 7 8 5 6
4 1 2 3 8 5 6 7
5 6 7 8 1 2 3 4
6 7 8 5 2 3 4 1
7 8 5 6 3 4 1 2
8 5 6 7 4 1 2 3
gap> m:=20;;gi:=PrimitiveRootMod(m);
fail
G8ElementToOrder(0)=1=ord(X1(n))
G8ElementToOrder(1)=4=ord(X2(n))
G8ElementToOrder(2)=2=ord(X3(n))
G8ElementToOrder(3)=4=ord(X4(n))
G8ElementToOrder(4)=2=ord(X5(n))
G8ElementToOrder(5)=4=ord(X6(n))
G8ElementToOrder(6)=2=ord(X7(n))
G8ElementToOrder(7)=4=ord(X8(n))
^U(20)=[8,2]=U(20)有1個1階元，3個2階元，4個4階元，0個8階元

n有無原根的群論含義：(Z/nZ)^*是φ(n)階非循環Abel群<=>n沒有原根
gap> L:=[];;L1:=[];;for n in [1..30] do Un:=Units(Integers mod n);;Cm:=CyclicGroup(Phi(n));Print(IdGroup(Un),",",IdGroup(Cm),"\n");if IdGroup(Un)=IdGroup(Cm) then Add(L,n);else Add(L1,n);fi;od;Print("有原根的:");for n in L do Print(n,",");od;Print("\n沒有原根的:");for n in L1 do Print(n,",");od;
[ 1, 1 ],[ 1, 1 ]
[ 1, 1 ],[ 1, 1 ]
[ 2, 1 ],[ 2, 1 ]
[ 2, 1 ],[ 2, 1 ]
[ 4, 1 ],[ 4, 1 ]
[ 2, 1 ],[ 2, 1 ]
[ 6, 2 ],[ 6, 2 ]
[ 4, 2 ],[ 4, 1 ]
[ 6, 2 ],[ 6, 2 ]
[ 4, 1 ],[ 4, 1 ]
[ 10, 2 ],[ 10, 2 ]
[ 4, 2 ],[ 4, 1 ]
[ 12, 2 ],[ 12, 2 ]
[ 6, 2 ],[ 6, 2 ]
[ 8, 2 ],[ 8, 1 ]
[ 8, 2 ],[ 8, 1 ]
[ 16, 1 ],[ 16, 1 ]
[ 6, 2 ],[ 6, 2 ]
[ 18, 2 ],[ 18, 2 ]
[ 8, 2 ],[ 8, 1 ]
[ 12, 5 ],[ 12, 2 ]
[ 10, 2 ],[ 10, 2 ]
[ 22, 2 ],[ 22, 2 ]
[ 8, 5 ],[ 8, 1 ]
[ 20, 2 ],[ 20, 2 ]
[ 12, 2 ],[ 12, 2 ]
[ 18, 2 ],[ 18, 2 ]
[ 12, 5 ],[ 12, 2 ]
[ 28, 2 ],[ 28, 2 ]
[ 8, 2 ],[ 8, 1 ]
有原根的:1,2,3,4,5,6,7,9,10,11,13,14,17,18,19,22,23,25,26,27,29,
沒有原根的:8,12,15,16,20,21,24,28,30,【注意：在某些算法中，認為1是沒有原根的】
U(8) ={[1]_8,[3]_8,[5]_8,[7]_8}=K_4≠C_4，<=>8沒有原根
U(15)=U(3)×U(5)=C_2×C_4，<=>15沒有原根
GAP4[24,15]=C_6×K_4=U84，沒有24階元，所以84沒有原根
//1個1階元，7個2階元，2個3階元，14個6階元
GAP4[24,9]=C_6×C_4=U35=U39，沒有24階元，所以35、39沒有原根
//1個1階元，3個2階元，2個3階元，4個4階元，6個6階元，8個12階元
gap> U84:=Units(Integers mod 84);;IdGroup(U84);
[ 24, 15 ]
gap> U39:=Units(Integers mod 39);;IdGroup(U39);
[ 24, 9 ]
gap> U35:=Units(Integers mod 35);;IdGroup(U35);
[ 24, 9 ]
20、21、28、32、40、48沒有原根：
G8ElementToOrder(0)=1
G8ElementToOrder(1)=4
G8ElementToOrder(2)=4
G8ElementToOrder(3)=2
G8ElementToOrder(4)=2
G8ElementToOrder(5)=4
G8ElementToOrder(6)=4
G8ElementToOrder(7)=2
GAP[8,2]=C_2×C_4=U20有1個1階元，3個2階元，4個4階元，0個8階元
12階群有5個：GAP4[ 12, 5 ]=Z_2+Z_2+Z_3=K_4+Z_3=C_3×K_4=C_2×C_6=C_2×C_2×C_3，GAP4[ 12, 2 ]=Z_12=Z_4+Z_3，GAP4[ 12, 4 ]=D_6=D_3×C_2=D_12，GAP4[ 12, 3 ]=A_4，GAP4[ 12, 1 ]=T=Q_12
GAP4[12,5]=C2C2C3=U21=U28有1個1階元，3個2階元，2個3階元，0個4階元，6個6階元，0個12階元
gap> U21:=Units(Integers mod 21);;IdGroup(U21);
[ 12, 5 ]
gap> U28:=Units(Integers mod 28);;IdGroup(U28);
[ 12, 5 ]
GAP4[16,5]=G16_3=C_2×C_8=U32有1個1階元，3個2階元，4個4階元，8個8階元，0個16階元
GAP4[16,10]=G16_4=C_2×C_2×C_4=U40=U48有1個1階元，7個2階元，8個4階元，0個8階元，0個16階元
16階非循環阿貝爾群：K_4×C_4=U40=U48≠GAP4[16,14]=K_4×K_4≠C_8×C_2=U32≠GAP4[16,2]=C_4×C_4
gap> G:=DirectProduct(CyclicGroup(2),CyclicGroup(8));;IdGroup(G);
[ 16, 5 ]
gap>G:=DirectProduct(CyclicGroup(2),CyclicGroup(2),CyclicGroup(4));;IdGroup(G);
[ 16, 10 ]
gap>G:=DirectProduct(CyclicGroup(2),CyclicGroup(2),CyclicGroup(2),CyclicGroup(2));;IdGroup(G);
[ 16, 14 ]
gap> G:=DirectProduct(CyclicGroup(4),CyclicGroup(4));;IdGroup(G);
[ 16, 2 ]
gap> L:=[];;L1:=[];;for n in [31..99] do Un:=Units(Integers mod n);;Cm:=CyclicGroup(Phi(n));Print(IdGroup(Un),",",IdGroup(Cm),"\n");if IdGroup(Un)=IdGroup(Cm) then Add(L,n);else Add(L1,n);fi;od;Print("有原根的:");for n in L do Print(n,",");od;Print("\n沒有原根的:");for n in L1 do Print(n,",");od;
[ 30, 4 ],[ 30, 4 ]
[ 16, 5 ],[ 16, 1 ]
[ 20, 5 ],[ 20, 2 ]
[ 16, 1 ],[ 16, 1 ]
[ 24, 9 ],[ 24, 2 ]
[ 12, 5 ],[ 12, 2 ]
[ 36, 2 ],[ 36, 2 ]
[ 18, 2 ],[ 18, 2 ]
[ 24, 9 ],[ 24, 2 ]
[ 16, 10 ],[ 16, 1 ]
[ 40, 2 ],[ 40, 2 ]
[ 12, 5 ],[ 12, 2 ]
[ 42, 6 ],[ 42, 6 ]
[ 20, 5 ],[ 20, 2 ]
[ 24, 9 ],[ 24, 2 ]
[ 22, 2 ],[ 22, 2 ]
[ 46, 2 ],[ 46, 2 ]
[ 16, 10 ],[ 16, 1 ]
[ 42, 6 ],[ 42, 6 ]
[ 20, 2 ],[ 20, 2 ]
[ 32, 16 ],[ 32, 1 ]
[ 24, 9 ],[ 24, 2 ]
[ 52, 2 ],[ 52, 2 ]
[ 18, 2 ],[ 18, 2 ]
[ 40, 9 ],[ 40, 2 ]
[ 24, 15 ],[ 24, 2 ]
[ 36, 5 ],[ 36, 2 ]
[ 28, 2 ],[ 28, 2 ]
[ 58, 2 ],[ 58, 2 ]
[ 16, 10 ],[ 16, 1 ]
[ 60, 4 ],[ 60, 4 ]
[ 30, 4 ],[ 30, 4 ]
[ 36, 14 ],[ 36, 2 ]
[ 32, 16 ],[ 32, 1 ]
[ 48, 20 ],[ 48, 2 ]
[ 20, 5 ],[ 20, 2 ]
[ 66, 4 ],[ 66, 4 ]
[ 32, 16 ],[ 32, 1 ]
[ 44, 4 ],[ 44, 2 ]
[ 24, 9 ],[ 24, 2 ]
[ 70, 4 ],[ 70, 4 ]
[ 24, 15 ],[ 24, 2 ]
[ 72, 2 ],[ 72, 2 ]
[ 36, 2 ],[ 36, 2 ]
[ 40, 9 ],[ 40, 2 ]
[ 36, 5 ],[ 36, 2 ]
[ 60, 13 ],[ 60, 4 ]
[ 24, 9 ],[ 24, 2 ]
[ 78, 6 ],[ 78, 6 ]
[ 32, 21 ],[ 32, 1 ]
[ 54, 2 ],[ 54, 2 ]
[ 40, 2 ],[ 40, 2 ]
[ 82, 2 ],[ 82, 2 ]
[ 24, 15 ],[ 24, 2 ]
[ 64, 26 ],[ 64, 1 ]
[ 42, 6 ],[ 42, 6 ]
[ 56, 8 ],[ 56, 2 ]
[ 40, 14 ],[ 40, 2 ]
[ 88, 2 ],[ 88, 2 ]
[ 24, 9 ],[ 24, 2 ]
[ 72, 36 ],[ 72, 2 ]
[ 44, 4 ],[ 44, 2 ]
[ 60, 13 ],[ 60, 4 ]
[ 46, 2 ],[ 46, 2 ]
[ 72, 9 ],[ 72, 2 ]
[ 32, 36 ],[ 32, 1 ]
[ 96, 2 ],[ 96, 2 ]
[ 42, 6 ],[ 42, 6 ]
[ 60, 13 ],[ 60, 4 ]
有原根的:31,34,37,38,41,43,46,47,49,50,53,54,58,59,61,62,67,71,73,74,79,81,82,83,86,89,94,97,98,
沒有原根的:32,33,35,36,39,40,42,44,45,48,51,52,55,56,57,60,63,64,65,66,68,69,70,72,75,76,77,78,80,84,85,
87,88,90,91,92,93,95,96,99,
gap> for n in [31..99] do if PrimitiveRootMod(n)=fail then Print(n,",");fi;od;
32,33,35,36,39,40,42,44,45,48,51,52,55,56,57,60,63,64,65,66,68,69,70,72,75,76,77,78,80,84,85,87,88,90,91,92,
93,95,96,99,
GAP命令為：
gap> for i in [1..100] do Print("U(",i,")=",IdGroup(Units(ZmodnZ(i))),"\n"); od;
U(1)=[ 1, 1 ]
U(2)=[ 1, 1 ]
U(3)=[ 2, 1 ]
U(4)=[ 2, 1 ]
U(5)=[ 4, 1 ]
U(6)=[ 2, 1 ]
U(7)=[ 6, 2 ]
U(8)=[ 4, 2 ]
U(9)=[ 6, 2 ]
U(10)=[ 4, 1 ]
U(11)=[ 10, 2 ]
U(12)=[ 4, 2 ]
U(13)=[ 12, 2 ]
U(14)=[ 6, 2 ]
U(15)=[ 8, 2 ]
U(16)=[ 8, 2 ]
U(17)=[ 16, 1 ]
U(18)=[ 6, 2 ]
U(19)=[ 18, 2 ]
U(20)=[ 8, 2 ]
U(21)=[ 12, 5 ]
U(22)=[ 10, 2 ]
U(23)=[ 22, 2 ]
U(24)=[ 8, 5 ]
U(25)=[ 20, 2 ]
U(26)=[ 12, 2 ]
U(27)=[ 18, 2 ]
U(28)=[ 12, 5 ]
U(29)=[ 28, 2 ]
U(30)=[ 8, 2 ]
U(31)=[ 30, 4 ]
U(32)=[ 16, 5 ]
U(33)=[ 20, 5 ]
U(34)=[ 16, 1 ]
U(35)=[ 24, 9 ]
U(36)=[ 12, 5 ]
U(37)=[ 36, 2 ]
U(38)=[ 18, 2 ]
U(39)=[ 24, 9 ]
U(40)=[ 16, 10 ]
U(41)=[ 40, 2 ]
U(42)=[ 12, 5 ]
U(43)=[ 42, 6 ]
U(44)=[ 20, 5 ]
U(45)=[ 24, 9 ]
U(46)=[ 22, 2 ]
U(47)=[ 46, 2 ]
U(48)=[ 16, 10 ]
U(49)=[ 42, 6 ]
U(50)=[ 20, 2 ]
U(51)=[ 32, 16 ]
U(52)=[ 24, 9 ]
U(53)=[ 52, 2 ]
U(54)=[ 18, 2 ]
U(55)=[ 40, 9 ]
U(56)=[ 24, 15 ]
U(57)=[ 36, 5 ]
U(58)=[ 28, 2 ]
U(59)=[ 58, 2 ]
U(60)=[ 16, 10 ]
U(61)=[ 60, 4 ]
U(62)=[ 30, 4 ]
U(63)=[ 36, 14 ]
U(64)=[ 32, 16 ]
U(65)=[ 48, 20 ]
U(66)=[ 20, 5 ]
U(67)=[ 66, 4 ]
U(68)=[ 32, 16 ]
U(69)=[ 44, 4 ]
U(70)=[ 24, 9 ]
U(71)=[ 70, 4 ]
U(72)=[ 24, 15 ]
U(73)=[ 72, 2 ]
U(74)=[ 36, 2 ]
U(75)=[ 40, 9 ]
U(76)=[ 36, 5 ]
U(77)=[ 60, 13 ]
U(78)=[ 24, 9 ]
U(79)=[ 78, 6 ]
U(80)=[ 32, 21 ]
U(81)=[ 54, 2 ]
U(82)=[ 40, 2 ]
U(83)=[ 82, 2 ]
U(84)=[ 24, 15 ]
U(85)=[ 64, 26 ]
U(86)=[ 42, 6 ]
U(87)=[ 56, 8 ]
U(88)=[ 40, 14 ]
U(89)=[ 88, 2 ]
U(90)=[ 24, 9 ]
U(91)=[ 72, 36 ]
U(92)=[ 44, 4 ]
U(93)=[ 60, 13 ]
U(94)=[ 46, 2 ]
U(95)=[ 72, 9 ]
U(96)=[ 32, 36 ]
U(97)=[ 96, 2 ]
U(98)=[ 42, 6 ]
U(99)=[ 60, 13 ]
U(100)=[ 40, 9 ]
數論中的偶數階Abel群的階
totient(2)=1
totient(3)=2
totient(4)=2
totient(5)=4
totient(6)=2
totient(7)=6
totient(8)=4
totient(9)=6
totient(10)=4
totient(11)=10
totient(12)=4
totient(13)=12
totient(14)=6
totient(15)=8
totient(16)=8
totient(17)=16
totient(18)=6
totient(19)=18
totient(20)=8
totient(21)=12
totient(22)=10
totient(23)=22
totient(24)=8
totient(25)=20
totient(26)=12
totient(27)=18
totient(28)=12
totient(29)=28
totient(30)=8
totient(31)=30
totient(32)=16
totient(33)=20
totient(34)=16
totient(35)=24
totient(36)=12
totient(37)=36
totient(38)=18
totient(39)=24
totient(40)=16
totient(41)=40
totient(42)=12
totient(43)=42
totient(44)=20
totient(45)=24
totient(46)=22
totient(47)=46
totient(48)=16
totient(49)=42
totient(50)=20
totient(51)=32
totient(52)=24
totient(53)=52
totient(54)=18
totient(55)=40
totient(56)=24
totient(57)=36
totient(58)=28
totient(59)=58
totient(60)=16
totient(61)=60
totient(62)=30
totient(63)=36
totient(64)=32
totient(65)=48
totient(66)=20
totient(67)=66
totient(68)=32
totient(69)=44
totient(70)=24
totient(71)=70
totient(72)=24
totient(73)=72
totient(74)=36
totient(75)=40
totient(76)=36
totient(77)=60
totient(78)=24
totient(79)=78
totient(80)=32
totient(81)=54
totient(82)=40
totient(83)=82
totient(84)=24
totient(85)=64
totient(86)=42
totient(87)=56
totient(88)=40
totient(89)=88
totient(90)=24
totient(91)=72
totient(92)=44
totient(93)=60
totient(94)=46
totient(95)=72
totient(96)=32
totient(97)=96
totient(98)=42
totient(99)=60

總結

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