Groups, Subgroups and Normal Subgroups

Hi there,

I'm going over some past papers for an exam but with no answers and have come a cropper on this one, any help appreciated!

Let be a subgroup of a group and let . Define

i) Prove that is a subgroup of .

ii) Let . Which of these statements is true

Give a proof of the statement that is always true

iii) Set . Prove that and that is a subgroup of that is contained in .

Pretty stuck except for ii) is obviously the right one (I hope!).

Re: Groups, subgroups and normal subgroups

(i) and (ii) should be straightforward.

Hint for (iii). Let denote the set of all left cosets of in Each element of induces a permutation of via for each left coset Let The mapping defined by is a homomorphism. Prove that is the kernel of this homomorphism.

Re: Groups, Subgroups and Normal Subgroups

i) let x,y be in H^{g}.

this means that x = ghg^{-1} for some h in H, and y = gh'g^{-1}, for some h' in H.

therefore, xy^{-1} = (ghg^{-1})(gh'g^{-1}) = ghh'^{-1}g^{-1},

and hh'^{-1} is in H (since H is a subgroup), therefore xy^{-1} is in H^{g}, so H^{g}

is a subgroup of G (by the one-step subgroup test).

ii) note that (H^{g})^{k} = k(H^{g})k^{-1} = k(gHg^{-1})k^{-1} = (kg)H(kg)^{-1} = H^{kg}

iii) first, suppose that a,b are in N. this means that a,b are in every gHg^{-1}, for any g in G. by part (i) above, each gHg^{-1}

is a subgroup, so ab^{-1} is in every gHg^{-1}, so ab^{-1} lies in N, thus N is a subgroup.

let x be any element of G, and let n be an element of N. note that N = ∩_{g} H^{g} = ∩_{xg} (H^{g})^{x}

since as g runs over all elements of G, so does xg.

but ∩_{xg} (H^{g})^{x} = ( ∩_{g} H^{g})^{x} = N^{x}, so N is normal.

(to see that for two subgroups H,K, that (H∩K)^{g} = H^{g}∩K^{g}, suppose that

x is in (H∩K)^{g}. then x = ghg^{-1}, where h is in H∩K. hence h is in H, so ghg^{-1} is in H^{g}.

but h is in K as well, so ghg^{-1} is in K^{g}. hence x = ghg^{-1} is in both H^{g} and K^{g},

so x is thus in their intersection. on the other hand suppose that x is in H^{g} and K^{g}. so

x = ghg^{-1} = gkg^{-1}, for some h in H, and k in K.

if ghg^{-1} = gkg^{-1}, then h = k, so h = k lies in H∩K, so x is in (H∩K)^{g}

the same proof holds for any arbitrary intersection of subgroups of G).