i^0= 1、i^2=- 1、i^3=-i、i^4= 1

(-i)^0= 1，(-i)^2=- 1，(-i)^3=i，(-i)^4= 1

But 1 2 = (-1) 2 =1,so I and -i are generators, while1and-1are not generators.

2. (What do you mean by cycle? Does the period of a permutation being k mean that the k power of this permutation is a unit element and m (m

(a b is used to represent the b power of a)

A permutation t with an odd period k must be an even permutation; In fact, let the rotation of t be decomposed into

T = (a _1a _ 2 ... a _ p) (b _1b _ 2 ... q) ... (s_ 1 s_2...s_t)

Among them, the above rotation does not pay in pairs; Then any positive integer m has

t^m = (a_ 1 a_2...a_p)^m * (b_ 1 b_2...b_q)^m *...* (s_ 1 s_2...s_t)^m

And t m is the unit element if and only if (a _ 1 a _ 2...a _ p) m, (b _ 1 b _ 2...b _ q) m, ..., (s _ 1 s _ 2...s _ Therefore (A _ 1 A _ 2...A _ P), (B _ 1 B _ 2...B _ Q), ..., (S _ 1 S _

3. (use u for union)

Z=N U ( 1+N) U (2+N)

Where (1+n) = {...,-5,-2,1,4, ...}, (2+n) = {...,-4,-1,2, 5, 8, ...}

The three sets of n, 1+N and 2+N constitute all cosets of n.

4. obviously, any element a in h satisfies aH=Ha=H, so h is included in k; To verify the subgroup with K as G, it is only necessary to verify that any A and B belong to K and all A * B (- 1) belong to K; In fact, when both a and b belong to k.

Ah = ha

BH=Hb (multiply the left side of the set on both sides by b (- 1) and then by b (- 1) to get HB (- 1) = b (- 1) h).

therefore

ab^(- 1)h=ahb^(- 1)=hab^(- 1)

Explain that AB (- 1) belongs to K.

Finally, it is verified that H is a normal subgroup of K. In fact, any H belongs to H and K belongs to K, because

khk^(- 1)*h=khh*k^(- 1)=khk^(- 1)=hk*k^(- 1)=h

This shows that KHK (- 1) belongs to H, so H is a normal subgroup of K.

5、G={( 1)，( 12)，( 13)，(23)，( 123)，( 132)}

G first has trivial subgroups {( 1)} and g; For the nontrivial subgroup H, because the order of H is the divisor of the 6th order of G, it can only be 2,3; The second-order group and the third-order group are cyclic groups;

1) If H is a second-order group, its second-order generator must be one of (12), (13) and (23), so H has the following three possibilities:

H={( 1)，( 12)}

H={( 1)，( 13)}

H={( 1)，(23)}

2) If H is a third-order group, its third-order generator must be one of (123) and (132), so

h={( 1),( 123),( 123)^2=( 132)}

h={( 1),( 132),( 132)^2=( 123)}

(The two situations are the same. )

To sum up, there are four possibilities for H***, as shown in the above.