Solvable Groups

  • Thread starter Deveno
  • Start date
  • #1
Deveno
Science Advisor
906
6
I was asked recently to prove that if N is normal in G:

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.
 

Answers and Replies

  • #2
22,089
3,292
Your proof looks OK. This would be the way I would prove it. I don't know if there's a better way 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.
Can you explain what exactly you mean here? In general, you should be very careful with things like this. But maybe what you're trying to do is correct.
 
  • #3
Deveno
Science Advisor
906
6
well, here is an example:

if i set H = G/N, and set Hk to be the k-th entry in a normal series for H, then as i defined Gk to be φ^-1(Hk), proving G(k+1) normal in Gk got really messy. see, i want to show that gxg^-1 is in G(k+1) whenever g is in Gk, and x is in G(k+1). essentially i want to lift information about hyh^-1 (where φ(x) = y, φ(g) = h) back to G. it would be a lot simpler if i could just say hyh^-1 in H(k+1) implies gxg^-1 in G(k+1) without talking at the "element" level. that is it would be easier to argue using φ(g(G(k+1))g^-1) = h(H(k+1))h^-1 (φ(g(G(k+1))g^-1) = φ(g)φ(G(k+1))φ(g)^-1 in other words, regarding φ as a lattice homomorphism of some sort).

this especially got to be a pain showing the cosets G(k+1)(xy) and G(k+1)(yx) were equal, for x,y in Gk. this would be simpler if i could just say that
φ^-1(H(k+1)ab) = (φ^-1(H(k+1))(φ^-1(a)φ^-1(b))), rather than taking elements x,y in Gk to a = φ(x), b = φ(y) in Hk, and then showing ab is in H(k+1)(ba), and then going back to G to show that xy is in G(k+1)(yx), and then doing it again to show that yx is also in G(k+1)(xy).

see, i know that φ induces a bijection between subgroups of G containing N, and subgroups of G/N, but φ also preserves multiplication, and we can define multiplicative sets like KN, so to what extent, and under what restrictions on K and N can we say φ(KN) = φ(K)φ(N) and more importantly,
when φ^-1(AB) = φ^-1(A)φ^-1(B)?
 
  • #4
22,089
3,292
well, here is an example:

if i set H = G/N, and set Hk to be the k-th entry in a normal series for H, then as i defined Gk to be φ^-1(Hk), proving G(k+1) normal in Gk got really messy. see, i want to show that gxg^-1 is in G(k+1) whenever g is in Gk, and x is in G(k+1). essentially i want to lift information about hyh^-1 (where φ(x) = y, φ(g) = h) back to G. it would be a lot simpler if i could just say hyh^-1 in H(k+1) implies gxg^-1 in G(k+1) without talking at the "element" level. that is it would be easier to argue using φ(g(G(k+1))g^-1) = h(H(k+1))h^-1 (φ(g(G(k+1))g^-1) = φ(g)φ(G(k+1))φ(g)^-1 in other words, regarding φ as a lattice homomorphism of some sort).

this especially got to be a pain showing the cosets G(k+1)(xy) and G(k+1)(yx) were equal, for x,y in Gk. this would be simpler if i could just say that
φ^-1(H(k+1)ab) = (φ^-1(H(k+1))(φ^-1(a)φ^-1(b))), rather than taking elements x,y in Gk to a = φ(x), b = φ(y) in Hk, and then showing ab is in H(k+1)(ba), and then going back to G to show that xy is in G(k+1)(yx), and then doing it again to show that yx is also in G(k+1)(xy).
I see what you want to do. But doing things on the element-level is always safe and is less prone to mistakes. I would not suggest working with sets, because it's easier to prove stuff that isn't true. Also, anything you want to do with sets, has to be justified anyway on element level. So you'll need to prove it anyway...

see, i know that φ induces a bijection between subgroups of G containing N, and subgroups of G/N, but φ also preserves multiplication, and we can define multiplicative sets like KN, so to what extent, and under what restrictions on K and N can we say φ(KN) = φ(K)φ(N) and more importantly,
when φ^-1(AB) = φ^-1(A)φ^-1(B)?
To my knowledge, this is always true for normal subgroups. The reason is that KN is the "join" of K and N, and phi is a lattice isomorphism. Thus the equality (both of them) always hold. You could also prove this directly...
 
  • #5
Deveno
Science Advisor
906
6
well, what if N is a normal subgroup, but K is just a set (which is the case in the matter at hand)? the main motivation is cosets of N.
 
  • #6
lavinia
Science Advisor
Gold Member
3,237
624
I was asked recently to prove that if N is normal in G:

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.
Not sure. Your proof looks good. Here is a shot at simplification.

Show that the commutator subgroup of G/H is isomorphic to the commutator subgroup of G mod the commutator subgroup of H.

One gets a sequence of exact sequences of commutator subgroups.

1 -> H_i -> G_i -> G_i/H_i ->1

if H_i is trivial and G_i/H_i is trivial then G_i must be trivial.

If G_i is trivial H_i must be trivial and so G_i/H_i must also be trivial.

Is there a mistake here?
 
  • #7
Deveno
Science Advisor
906
6
i like that. homomorphisms preserve commutators, so [Hg1,Hg2] = [φ(g1), φ(g2)] = φ[g1,g2], and
([Hk,Hk])([g1,g2] = ([φ(Gk),φ(Gk)])[g1,g2] <--> [φ(Gk)g1,φ(Gk)g2] = [(Hk)g1,(Hk)g2] = φ([g1,g2]) is the desired isomorphism, no?
 
Last edited:

Related Threads on Solvable Groups

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
6
Views
6K
Replies
2
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
2
Views
3K
Top