Thread: 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.

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'^-1g^-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).