one more try with more info

Suppose a simple group G of order 660 is a subgroup of the symmetric group S11 ( here S is with 11 as a subscript ) and x = {1,2,3,4,5,6,7,8,9,10,11}( a permutation like 1 goes to 2 and so on, I think ) in G. If P equals the span of x ( or ) determine the permutations which generate NGP ( here G is written as the subscript of N (the normalizer of P in G, I think)). Hint if you know the order of the normalizer N of can you find the generators for N ? Give a solid math arguement why this is correct and please keep the answer as simple as possible at the same time. thank you

Ideals

Let R be a ring, and suppose that I and J are ideals of R. Let I + J= {x+y: xEI, yEJ} . Prove that I + J is an ideal of R.

Factor Group/ Torsion Group

Heres my problem. Consider the group (reals under addition) and its normal subgroups Z (integers) and Q (rationals0. (These are normal because R is abelian, of course.) (i) Find an element of Q/Z of order 350. (ii) Show that Q/Z is the torsion subgroup of R/Z. This problem is quite straightforward if you use the definitions and stay focussed; in particular, pay attention to the definition of a rational number. (iii) Show that R/Q is torsion free. Think carefully about the elements of R that are not in Q here.

permutation groups

Here's my problem: Let (i1, i2, . . . , ik) be a k-cycle (k less or equal to n) element of Sn and let sigma be an element of Sn. (i) Find a precise expression for sigma * (i1, i2, . . . , ik)* sigma-inverse. Hint: experiment a little, perhaps, then take a guess and prove it. (ii) Describe precisely the set {sigma * (1, 2, . . . , k) * sigma-inverse | sigma is an element of Sn}.

Kernel and Homomorphism

Here's my problem: If A and B are subsets of a group G, define AB = {ab | a 2 A, b 2 B}. Now suppose phi: G -> G0 is a homomorphism of groups and N = Ker(phi) is its kernel. (i) If H is a subgroup of G, show that HN = NH. (Warning: this is an equation of sets; proceed accordingly; do not assume that G is abelian.) (ii) Show that phi-inverse[phi[H]] = HN. Thanks.