Solvable groups

Definition 17.18.1 The commutator of two elements x and y in a group is G indicated by [x, y] is the element

xyx−1y−1

The commutator subgroup or derived subgroup of a group G is defined as the subgroup generated by all commutators of elements of G and denoted by G '.

G ' := ⟨xyx−1y−1 | x,y ∈ G⟩

The commutator [x, y], coincides with the identity element of G if and only if xy = yx. G is thus abelian if and only if G ' = {e}.

Notice how the inverse of a commutator is still a commutator, thus G ', is composed by products of commutators. Every element of G ' has the form a1h1a2h2 ⋅⋅⋅ atht, where aj has the form xyx−1y−1, h = ±1 and t is a positive integer.

Let's calculate the derived subgroup S3 ', of S3. Notice how any commutator is an even permutation, so that S3 ' ⊆ A3 '. The only even permutations on 3 elements are id, and the cyclic permutations (1, 2, 3) and (1, 3, 2), then we have

(1,2) id (1,2)−1 id−1 = id ∈ S3 '
(1,2)(1,2,3)(1,2)−1(1,2,3)−1 = (1,2,3) ∈ S3 '
(1,2)(1,3,2)(1,2)−1(1,3,2)−1 = (1,3,2) ∈ S3 '

Thus A3 ⊆ S3 '. Thus the derived subgroup of S3 is the alternating subgroup A3.

Proposition 17.18.2 Let G a group. Then

  1. The derived subgroup G ' is a normal subgroup and the quotient G/G ' is abelian;

  2. if N ⊲̲ G is abelian if and only if G' ≤ N, that is G ' is the smallest normal subgroup of G such that the quotient is abelian;

  3. if N is a normal subgroup of G, then also N ', is a normal subgroup of G.

Proof. (a) We have to prove that for every x ∈ G ' and every g ∈ G it results gxg−1 ∈ G'. It suffices to show that the conjugate of a commutator is still a commutator. Let x = aba−1b−1. Then

gxg−1 = g(aba−1b−1)g−1 = ga g−1 g b g−1g a−1 g−1g b−1g−1
= (gag−1)(gbg−1)(ga−1g−1)(gb−1g−1)
= (gag−1)(gbg−1)(ga−1g−1)−1(gbg−1)−1 ∈ G '

We provr now that G/G ' is abelian; let G'x and G'y two elements of G/G '. Then G'x ⋅ G' y = G' xy = G ' [y,x] yx = G'yxy−1x−1xy = G' yx = G'y ⋅ Gx.

(b) It suffices to notice that

Nxy = Nyx   ⇐⇒   Nxyx−1y−1 = N   ⇐⇒   xyx−1y−1 ∈ N.

(c) It suffices to show that gnmn−1m−1g ∈ N ' ∀ g ∈ G and every n, m ∈ N. Indeed

gnmn−1m−1g = gng−1 gmg−1 gn−1g−1gm−1g−1
= gng−1 gmg−1 gn−1g−1 gm−1g−1
= ∈ N   ∈ N   ∈ N   ∈ N  
= gng−1gmg−1 (gng−1)−1 (gmg−1)−1 ∈ N '.  □

Starting now from G ' we can define the commutator subgroup of G ', i.e. (G')': we indicate as G(2). It is a normal subgroup of G hence of G' as well. More generally we define the commutator subgroup G(n) of order n as the group (G(n − 1)) '. Every G(k) is a normal subgroup of G and it is such that G(k−1)/G(k) is abelian.

Definition 17.18.3 A group G is said solvable if it is possible to find a chain of subgroups,

G = M0 ⊃ M1 ⊃ M2 ⊃ ⋅⋅⋅ ⊃ Ms = {e}

with each Mi is normal in Mi − 1 and with abelian quotients Mi − 1/Mi.

Examples of Solvable groups 17.18.4.

  1. Every group G is solvable: it suffices to consider the chain made by the two subgroup G and {e}.

  2. The symmetric group S3 is solvable; Consider the chain

    S3 ⊃ A3 ⊃ {e}

  3. The symmetric group S4 is solvable; Consider the chain

    S4 ⊃ A4 ⊃ V ⊃ {e}

    with V = {id, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}.

  4. Every group with order pq (with p and q primes) is solvable. If p = q, G is an abelian group (see Theorem 7.17.7). We can thus suppose p < q. By Sylow's theorems, there exists a normal Sylow q-subgroup N. The it suffices to take the chain

    G ⊃ N ⊃ {e}

    Observe that N is normal in G and {e} is normal in N. Since |G/N| = p and |N/{e}| = q, both quotient groups have prime order and are therefore abelian. ■

A non-abelian simple group, obviuosly cannot be solvable. But all abelian groups are trivially solvable —a subnormal series being given by just the group itself and the trivial group

Proposition 17.18.4 A group G is solvable if and only if there exists a m ≥ 0 such that G(m) = {e}.

Proof. Suppose that G(m) = {e} for some m, and let Mi = G(i), consider the chain of subgroups

G = M0 ⊃ M1 ⊃ M2 ⊃⋅⋅⋅ ⊃ Ms = {e}

since every Mi is normal in G it is also normal in Mi − 1. In addition

Mi − 1/Mi = G(i − 1)/ G(i) = G(i − 1) / G(i − 1) '

since every quotient is abelian, G is solvable;

Conversely, suppose G is solvable. There there's a finite chain of subgroups each normal in the next, with abelian quotients

G = M0 ⊃ M1 ⊃ M2 ⊃ ⋅⋅⋅ ⊃ Ms = {e}

such that for every i, Mi ⊲̲ Mi − 1 and Mi − 1/Mi is abelian. But then from Proposition 17.18.2 (b) it results Mi − 1 ⊆ Mi. We have the following situation

M1 ⊃ M0 ' = G '
M2 ⊃ M1 ' = G(2)   (from the precedent inclusion taking the derivative)
⋅⋅⋅
Mi ⊃ Mi − 1 ' ⊃ G(s)

from the last relation it follows G(s) = {e}.  □

Corollary 17.18.5 If a group is solvable, so is also every subgroup and every homomorphic image.

Theorem 17.18.6 If n ≥ 5, then Sn is not solvable.

Proof. We first show that if H, N are two subgroups of Sn such that N ⊆ H and N is normal in H, if H contains every 3-cycles and if H/N is abelian, then N contains every 3-cycles. To see this, let a, b, c, d, e, f be five distinct integers between 1 and n ≥ 5, and let

σ = (abc)   and   τ = (def)

Then

στσ−1τ−1 = (a, b, c)(d, e, f)(c, b, a)(f, e, d) = (e, d, a)

But N contains all comutators of elements of H since H/N was assumed abelian; Notice also that since the choice of a, b, c, d, e, f was arbitrary, we see that the cycles (e, d, a) all lie in N for all choices of distinct e, d, a, or since it contains a 3-cycle it contains all the conjugate class, thereby proving what we wanted.

Now suppose that we have a chain of subgroups

Sn = G0 ⊃ G1 ⊃ Gi ⊃ ⋅⋅⋅ ⊃ Gm = {e}

where each Gi+1 is a normal subgroup of Gi and Gi − 1/Gi is abelian. Since Sn contains every 3-cycles, we conclude that G1 contains every 3-cycles. By induction on i we conclude that Gm = {e} contains every 3-cycle, which is impossible. Hence such a chain of subgroups cannot exists, and our theorem is proved.  □

Exercises

  1. Prove that a solvable group is simple if and only if is cyclic of prime order.

  2. Prove that for n ≥ 5 the alternating group A5 is simple.

  3. Using the result of the previous exercise, prove that for n ≥ 5 the symmetric group Sn is not solvable.

  4. Let G be a group, and let N be a normal subgroup of G. Then G is solvable if and only if N and G/N are solvable.

  5. Check whether the group with order 91 is solvable.

Solutions

  1. 1st solution. G being solvable and nontrivial means it has proper subgroup N with G/N abelian. But G simple means N = {e}. Now G/{e} is abelian: {e}x * {e}y = {e}xy = {e}y * {e}x = {e} * yx so that {e⋅xy} = {xy} and {e⋅yx} = {yx}, but {xy} = {yx} as set equality, implies (since each set has exactly one element in it) xy = yx in G. Alternatevely use the First Isomorphism Theorem: the identity map i: G ⟶ G is a group homomorphism, that is surjective with kernel {e}, thus i(G) = G =~ G/{e}. So that means the only subgroups of G at all are G and {e}. If G is simple and abelian then G is cyclic as well: because if you take any nonidentity g ∈ G, and if the cyclic subgroup ⟨g⟩ is not all of G, then it is a normal subgroup of G (because G is abelian) that is different from {e} and G, not possible when G is simple. By Theorem 7.9.8 a simple abelian group has prime order. Hence composition factors of G have prime orders. The converse is an immediate consequence of the definition of solvability.  ■

    2nd solution. Assume G is a solvable simple group. If G is a solvable nontrivial group, then G' ≠ G, for otherwise all terms of the derived seriees would be G, and so it will never terminate to {e}. Since G is assumed to be simple, G ' = {e}, and so G is abelian. Since an abelian simple group is prime cyclic (Theorem 7.17.18), the result follows.  ■.

  2. We use much the same strategy as in Exercise 1. Suppose that {e} ≠ N ⊲̲ An. Our strategy will be as follows:

    1. First, prove that N contains a 3-cycle.

    2. Second prove that then it contains all 3-cycles, and since the 3-cycles generate An, we conclude that N = An.

    (a). We shall prove that N must contain a 3-cycle. Let σ be a non-identity element in N that permutes k integers in the set S = {1, 2, ..., n} leaving fixed n − k, which is the maximum number of symbols that can be fixed by a permutation in N.

    If σ moves exactly 3 digits then it is a 3-cycle and we are done. Suppose then that σ moves more than 3 symbols. Assuming k ≥ 4, (σ can't move exactly four symbols, because that would force σ to have the form (a b c d) which is not even). We decompose σ into a product of disjoint cycles:

    σ = σ1 σ2 σ3 ··· σm.

    If σ moves more than 3 digits, there can be two cases:

    1. The decomposition of σ contains a k-cycle with k ≥ 3 and σ moves more than 3 digits.

      σ = (1 2 3 ..)ρ  (1st case)

      with ρ equal to a product of disjoint cycles not containing the symbols appearing in γ

    2. σ is a product of at least two disjoint transpositions.

      σ = (1 2) (3 4) ..   (2nd case)

    In the first case σ being even and not being a 3-cycle, moves at least two other symbols besides the three listed ones, say 4 and 5: If σ permutes exactly 4 integers, then σ = (1, 2, 3, a) for some integer a ≠ 1, 2, 3, and σ is odd, thus not in N. Let β = (3, 4, 5) ∈ An (because n ≥ 5), then βσβ−1 ∈ N, because N ⊲̲ An. Proposition 7.5.3 shows that σ' = βσβ−1 = γ'ρ' = (1, 2, 4, ···)ρ' with ρ' not containing any of the symbols appearing in γ'. Thus σ'σ−1 moves 3 into 4 and cannot be the identity. But σ'σ−1 is in N and fixes all symbols other than 1, 2, 3, 4, 5 that are fixed by σ. In addition, σ'σ−1 fixes 2, and none of 1, 2, 3, 4, 5 is fixed by σ. Thus σ'σ−1 is a member of N other than the identity that fixes fewer symbols than σ, and we have arrived at a contradiction.

    The second case is that σ is a product σ = (1, 2)(3, 4) ··· of disjoint transpositions. There must be at least two factors since σ is even. Put β = (1, 2)(4, 5), the symbol 5 existing since the group An in question has n ≥ 5. Then σ' = (1, 2)(3, 5) ···. Since σ'σ−1 carries 4 into 5, σ'σ−1 is a member of N other than the identity. It fixes all symbols other than 1, 2, 3, 4, 5 that are fixed by σ, and in addition it fixes 1 and 2. Thus σ'σ−1 fixes more symbols than σ does, and again we have arrived at a contradiction.

    (b) From the argument above N must contain a 3-cycle τ. We know that any two 3-cycles in Sn are conjugate. Hence τ is conjugate to any other 3-cycle in 𝓢n. Since N is normal and τ ∈ N each of these conjugates must also be in N. Therefore N contains all 3-cycles in Sn. We know from Lemma 7.4.19 that for n ≥ 3, the alternating group An is generated by 3-cycles. Thus N = An.  ■

  3. The symmetric group Sn is not solvable for n ≥ 5, because otherwise its normal subgroup An would be solvable (Corollary 17.18.5). Suppose the following to be the compoisition series for the simple An group

    An ⊃ {e}

    An is not solvable for n ≥ 5 because the quotient An/{e} is not abelian since An ≃ An/{e} is neither. Hence the only solvable simple groups are abelian.  ■

  4. Let

    N = N0 ⊃ N1 ⊃ Ni ⊃ ⋅⋅⋅ ⊃ Nm = {e}
    G/N = G0 ⊃ G1 ⊃ ⋅⋅⋅ ⊃ Gk = N/N = {eN} = {N}

    the solvable series for N and G/N. Here each Gi is a subgroup of G containing N. Since Gi − 1/N is normal in Gi/N, each Gi is normal in Gi − 1. Now

    G = G0 ⊃ ⋅⋅⋅ ⊃ Gk = N

    Appending the series for N

    G = G0 ⊃ ⋅⋅⋅ ⊃ Gk = N = N0 ⊃ N1 ⊃ ⋅⋅⋅ ⊃ Nm = {e}. ■

  5. Let |G| = 91 = 13 ⋅ 7, the number of Sylow 7-subgroup must be a divisor of 13, that is either 13 or 1. Since it must also be conguent to 1 modulo 7, we conclude there is only a Sylow 7-subgroup, we name it P. We have a subnormal series

    G ⊃ P ⊃ {e}

    Observe that P is normal in G and {e} is normal in P. Since |G/P| = 13 and |P/{e}| = 7, both quotient groups have prime order and are therefore abelian. Thus the group G is solvable.  ■

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