Thread: How to obtain isomorphic factor groups of a group from its ismomorphic subgroups?

Dear Friends,
If two normal subgroups H and K of a group G are ısomorphic then the factor groups G\H and G\K need not be isomorphic. My question is what are further conditions we need on H, K and probably on G that gaurantee the isomorphisms of G\H and G\K when H and K are isomorphic

i claim the following is a sufficient (do not know about necessary) condition on H,K and G:

there exists σ in Aut(G) with σ(H) = K.

we would hope that this defines an isomorphism:

σ_H/K from G/H to G/K, given by:

σ_H/K(gH) = σ(g)K.

first, we need to verify that σ_H/K is well-defined.

suppose gH = g'H. then g'^-1g is in H, so σ(g'^-1g) = (σ(g'))^-1σ(g) is in K, so:

σ(g)K = σ(g')K, thus σ_H/K(gH) = σ_H/K(g'H).

is σ_H/K a homomorphism?

well, σ_H/K((gH)(g'H)) = σ_H/K(gg'H) = σ(gg')K = σ(g)σ(g')K = (σ(g)K)(σ(g')K) = σ_H/K(g)σ_H/K(g').

is σ_H/K injective?

suppose σ_H/K(gH) = σ(g)K = K. then σ(g) is in K, so g = σ^-1(σ(g)) is in σ^-1(K) = H.

thus the the only coset of H that is in the kernel of σ_H/K is H.

is σ_H/K surjective?

suppose that we have ANY coset of K in G/K, say xK. let g = σ^-1(x). then σ_H/K(gH) = σ(g)K = σ(σ^-1(x))K = xK.

ta da!