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

• Nov 20th 2012, 04:07 AM
raed
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
• Nov 21st 2012, 04:16 PM
Deveno
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!