Solvable Groups' Composition Representations and Factors (3.4.8)

Dummit and Foote Abstract Alegbra, section 3.4 exercise 8:

G a finite group. Prove that the following are logically equivalent:
(i) G is solvable
(ii) G has a chain of subgroups: 1 = H₀ ≤ H₁ ≤ H₂ ... ≤ H_s = G such that H_i / H_i-1 is a cyclic group, 0≤i≤s
(iii) All composition factors of G have prime order
(iv) G has a chain of subgroups: 1 = N₀ ≤ N₁ ≤ N₂ ... ≤ N_s = G such that N_i is normal in G, and N_i / N_i-1 is abelian, 0≤i≤s

Proof:

(ii) ⇒ (iii) : Since H_i / H_i-1 is cyclic, assume | H_i / H_i-1 | is of prime order. Since prime-ordered cyclic groups are without nontrivial, proper subgroups, H_i / H_i-1 would be simple. Therefore, assume | H_i / H_i-1 | = pn for some prime p and n∈Z⁺. Then by Cauchy's Theorem there exists xH_i-1∈H_i / H_i-1 such that | xH_i-1 | = p. Since H_i / H_i-1 is cyclic and thus abelian, < xH_i-1 > = < x > / H_i-1 is a normal subgroup of H_i / H_i-1, and 1 / H_i-1 = H_i-1 / H_i-1 is clearly a normal subgroup of < x > / H_i-1. By the Lattice Isomorphism Theorem, < x > is a normal subgroup of H_i, and H_i-1 is a normal subgroup of < x >. Furthermore, < x > / H_i-1 is cyclic, and H_i / < x > is cyclic (since H_i is cyclic), so the initial conditions can be applied to these two groups to seal an infinite descent-style induction proof that eventually only prime-ordered quotient groups will remain.

 

(iii) ⇒ (ii) : If G has such a composition series, then since H_i / H_i-1 is of prime order, it must be cyclic. We have proved (ii) ⇔ (iii)

(ii) ⇒ (i) : If these quotients are cyclic, then by nature they are abelian, and thus fulfill the requirement of a solvable group.

(i) ⇒ (ii) : If G has such a chain of subgroups, then obtain the composition series by splitting the non-simple factors thus: If G_i is normal in G_i-1, then nontrivial, proper N / G_i-1 normal in G_i / G_i-1 exists, so N normal in G_i exists by the Lattice Isomorphism Theorem. Similarly, since G_i-1 / G_i-1 = 1 / G_i-1 is clearly normal in N / G_i-1, G_i-1 is normal in N. Now to prove that these two new potential quotients are abelian: Since N / G_i-1 is a subgroup of a priori abelian G_i / G_i-1, N / G_i-1 is abelian. G_i / G_i-1 abelian implies:

∀x,y∈G_i, xG_i-1yG_i-1 = yG_i-1xG_i-1 ⇒
xy∈yxG_i-1 ⇒
xy∈yxN (since G_i-1 ≤ N) ⇒
xNyN = yNxN ⇒ G_i / N is abelian

Using the familiar infinite descent, we must eventually arrive at a composition series of simple abelian factors/quotients. By exercise number one, these are isomorphic to prime cyclic subgroups, and are thereby cyclic. We have proved (i) ⇔ (ii)

(iv) ⇒ (i) : Implied by definition of a solvable group.

(i) ⇒ (iv) : Let M be the minimal normal subgroup after the trivial group within the solvable series (in special cases it may be that M=G). Take the next smallest part of the link of the composition series of M (existent by part 1 of the Jordan-Hölder Theorem) as N, which is normal in M. Then since M is solvable due to being a subgroup of a solvable group (exercise 5), and since (i) ⇒ (iii), we have that | M / N | is prime. Thus M / N is cyclic, and therefore abelian. xNyN = yNxN ⇒ x^-1y^-1xy∈N, ∀x,y∈M.

Now assume that M isn't abelian, so there exist m,n∈M such that m^-1n^-1mn = [m,n] ≠ 1. Set V = < [x,y] | ∀x,y∈M >. Now, g[x,y]g^-1 = gx^-1g^-1gy^-1g^-1gxg^-1gyg^-1 = [gxg^-1,gyg^-1] = [x₁,y₁]∈V, so we have that V is a normal subgroup in G. Since we have proved that V ≤ N and N < M, we have V is a normal subgroup of strictly smaller order than M, which violates the minimality of M. Therefore, M is abelian, so M / 1 is abelian.

Now, the induction step. If Q is the next minimal normal subgroup of G after another normal subgroup T, then not only is Q / T a normal subgroup of G / T, but Q / T is the minimal normal subgroup after the trivial group of G / T by extrapolation of the Lattice Isomorphism Theorem. Therefore, Q / T is abelian. We have proved (i) ⇔ (iv), and thus the proof is complete.

∎