- #1

Deveno

Science Advisor

- 906

- 6

G solvable iff G/N and N solvable.

now, I was able to do so, but I wasn't very happy with my proof.

==> the way I approached this was to form the images of the Gk/G(k+1) under the canonical homomorphism G-->G/N to prove G/N solvable. my reasoning was, that by the 3rd isomorphism theorem (Gk/N)/(G(k+1)/N) is isomorphic to Gk/G(k+1), so the normal series for G induced a normal series for G/N.

then for N, i used Nk = N∩Gk. proving normality was messy, though. one thing that i wondered about was, if you have a homomorphism φ, can you use arguments involving sets like xNx^-1 just as if N was an element of G? i think so, but i am not sure.

<== this way was easier, i just used the normal series for N, and then pre-images of the normal series for G/N. the asymmetry in the proof methods didn't sit well with me, though. and again, there was this ambiguity of "sets" versus "elements".

do any of you shining lights know a more elegant approach?

sorry if this isn't terribly clear. i can give more details if you wish.