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
Re: How to obtain isomorphic factor groups of a group from its ismomorphic subgroups?
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:
σH/K from G/H to G/K, given by:
σH/K(gH) = σ(g)K.
first, we need to verify that σH/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 σH/K(gH) = σH/K(g'H).
is σH/K a homomorphism?
well, σH/K((gH)(g'H)) = σH/K(gg'H) = σ(gg')K = σ(g)σ(g')K = (σ(g)K)(σ(g')K) = σH/K(g)σH/K(g').
is σH/K injective?
suppose σH/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 σH/K is H.
is σH/K surjective?
suppose that we have ANY coset of K in G/K, say xK. let g = σ-1(x). then σH/K(gH) = σ(g)K = σ(σ-1(x))K = xK.
ta da!