Equivalent characterizations of solvability for groups

Let G be a finite group. Prove that the following are equivalent.

  1. G is solvable.
  2. G has a subnormal series 1 = H_1 \vartriangleleft \cdots \vartriangleleft H_k = G such that H_{i+1}/H_i is cyclic for 1 \leq i < k.
  3. All composition factors in any composition series of G have prime order.
  4. G has a subnormal series 1 = H_1 \vartriangleleft \cdots \vartriangleleft H_k = G such that H_i is normal in G and H_{i+1}/H_i is abelian for 1 \leq i < k.

  5. We begin with some lemmas.

    Lemma 1: Let G be a finite group and N \leq G a normal subgroup such that G/N is abelian. Then there exists a chain of subgroups N = K_1 \vartriangleleft K_2 \vartriangleleft \cdots \vartriangleleft K_\ell = G such that K_{i+1}/K_i has prime order for all 1 \leq i < \ell. Proof: G/N is a finite group, so by a previous exercise there exists a composition series 1/N = K_1/N \vartriangleleft K_2/N \vartriangleleft \cdots \vartriangleleft K_\ell/N = G/N such that (K_{i+1}/N)/(K_i/N) is simple for all 1 \leq i < \ell. Now because G/N is abelian, each K_i/N is abelian. We saw in a previous exercise that the only abelian simple groups are (up to isomorphism) \mathbb{Z}/(p) for primes p, each (K_{i+1}/N)/(K_i/N) has prime order. Using the Lattice Isomorphism Theorem we have a chain of subgroups N = K_1 \vartriangleleft \cdots \vartriangleleft K_\ell = G and for all 1 \leq i < \ell, K_{i+1}/K_i \cong (K_{i+1}/N)/(K_i/N) \cong \mathbb{Z}/(p) by the Third Isomorphism Theorem. \square

    Lemma 2: If G is simple and solvable, then G \cong \mathbb{Z}/(p) for some prime p. Proof: If G is simple, then the only subnormal series of G is 1 \vartriangleleft G. Then G is abelian; we saw previously that every abelian simple group is isomorphic to \mathbb{Z}/(p) for some prime p. \square

    Lemma 3: If 1 = H_0 \vartriangleleft H_1 \vartriangleleft \cdots \vartriangleleft H_k = G is a composition series where G is solvable and finite, then H_{i+1}/H_i is abelian for all 0 \leq i < k. Proof: We proceed by induction on k. If k = 0, then G = 1 and the conclusion that H_{i+1}/H_i is abelian for all 0 \leq i < 0 holds vacuously. Suppose now that for all composition series of length n such that G = H_n is solvable we have all composition factors abelian, and let 1 = H_0 \vartriangleleft \cdots \vartriangleleft H_n \vartriangleleft H_{n+1} = G be a composition series such that G is solvable. Now H_n \leq G is solvable by a previous exercise, and 1 = H_0 \vartriangleleft \cdots \vartriangleleft H_n is a composition series for H_n of length n. By the induction hypothesis, we have H_{i+1}/H_i abelian for all 0 \leq i < n. Moreover, H_{n+1}/H_n is simple and solvable by a previous exercise, so that H_{n+1}/H_n \cong \mathbb{Z}/(p) is abelian. By induction the conclusion holds for all composition series which terminate with a solvable group. \square

    Lemma 4: Let G be a group and H \leq G a subgroup. Then \bigcap_{g \in G} gHg^{-1} is a normal subgroup of G. Proof: Suppose x,y \in \bigcap_{g \in G} gHg^{-1}. Then for all g \in G, we have x = gag^{-1} and y = gbg^{-1} for some a,b \in H. Now (gag^{-1})(gb^{-1}g^{-1}) = g(ab^{-1})g^{-1} \in gHg^{-1}; by the Subgroup criterion, \bigcap_{g \in G} gHg^{-1} is a subgroup of G. Now let h \in G; then h(\bigcap_{g \in G} gHg^{-1})h^{-1} = \bigcap_{g \in G} hgHg^{-1}h^{-1} = \bigcap_{k^\prime \in G} kHk^{-1}; hence this subgroup is normal in G. \square

    Lemma 5: Let G be a finite nonsimple group such that every composition factor of every composition series of G has prime order and let M \leq G be a minimal nontrivial normal subgroup. Then M is abelian. Proof: Let M be such a subgroup of G. By a previous exercise, there exists a composition series of G which contains M as a term, say 1 = K_0 \vartriangleleft K_1 \vartriangleleft \cdots \vartriangleleft K_\alpha = M \vartriangleleft \cdots \vartriangleleft K_\ell = G. Let N = K_{\alpha - 1}; by our hypothesis, H_\alpha/H_{\alpha - 1} has prime order p and thus is isomorphic to the abelian simple group \mathbb{Z}/(p). Thus for all x,y \in M, we have (xy)N = (yx)N, hence x^{-1}y^{-1}xy \in N. Now let x,y \in M and g^{-1} \in G be arbitrary. Because M is normal in G, we have g^{-1}xg = \overline{x} \in M and g^{-1}yg = \overline{y} \in M. By the previous point, \overline{x}^{-1}\overline{y}^{-1}\overline{x}\overline{y} \in N. Thus g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg = g^{-1}x^{-1}y^{-1}xyg \in N, hence x^{-1}y^{-1}xy \in gNg^{-1}. Now by Lemma 4, \bigcap_{g \in G} gNg^{-1} < M is normal in G. Because M is a minimal nontrivial normal subgroup of G and N is strictly contained in M, we have \bigcap_{g \in G} gNg^{-1} = 1, hence x^{-1}y^{-1}xy = 1, hence xy = yx. Thus M is abelian. \square

    Now to the main result. We prove the following chains of implications: (1) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1) and (1) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1).

    Note that (3) \Rightarrow (2) because every group of prime order is cyclic and (2) \Rightarrow (1) because every cyclic group is abelian. (4) \Rightarrow (1) is trivial.

    (1) \Rightarrow (3) Let G be solvable and let 1 = H_1 \vartriangleleft \cdots \vartriangleleft H_k = G be an arbitrary composition series of G. By Lemma 3, all composition factors are abelian simple groups. We saw in a previous exercise that every such group is isomorphic to \mathbb{Z}/(p) for some prime p, thus all composition factors have prime order.

    (3) \Rightarrow (4) Let G be a finite group with the property that every composition factor of every composition series has prime order.

    If |G| = 1, then G = 1 and it is vacuously true that G has a normal series with all intermediate quotients abelian.

    Suppose that every finite group H with |H| \leq n has a normal series with all intermediate quotients abelian, and supose G is a group of order n+1.

    If G is simple, then G has only one composition series, namely 1 \vartriangleleft G, and so G has prime order by the hypothesis. Thus G \cong \mathbb{Z}/(p) for some prime p and G is abelian; so a normal series with all intermediate quotients abelian exists.

    If G is not simple, then the set A of nontrivial normal subgroups of G is not empty. Now A contains a minimal element M which, by Lemma 5, is abelian. By the induction hypothesis, since G/M is a finite group of order less than or equal to n, we have a normal series 1 = H_0/M \vartriangleleft \cdots \vartriangleleft H_k/M = G/M where (H_{i+1}/M)/(H_i/M) is abelian. By the Third Isomorphism Theorem, H_{i+1}/H_i is abelian, and by the Lattice Isomorphism Theorem, H_i \leq G is normal. Thus 1 \vartriangleleft M = H_0 \vartriangleleft \cdots \vartriangleleft H_k = G is a normal series for G with all intermediate quotients abelian.

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Comments

  • Gobi Ree  On December 1, 2011 at 1:13 am

    Another approach to 1=>3: Let G be solvable. It has a subnormal series with abelian factor groups. By Schreier refinement theorem, it has a refinement to a composition series, The factor groups of this series are abelian since they are quotients of abelian groups. So these are simple abelian, thus have prime orders by a previous exercise(3.4#1).

  • gunjan  On December 26, 2011 at 2:56 pm

    very nice

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