Classification of Cyclic-Group-Only Orders (4.5.52-56)

Dummit and Foote Abstract Algebra, section 4.5, exercises 52-56:

52. Suppose G is a finite simple group in which every proper subgroup is abelian. Prove M∩N=1 for distinct maximal subgroups M and N.
53. Prove that if G is a non-abelian group wherein every proper subgroup is abelian then G is not simple. [Let G be a counterexample to this assertion and use exercise 4.3.24 to show that G has more than one conjugacy class of maximal subgroups. Use the method of 4.3.23 to count elements within these conjugates and show that it is greater than | G |.]
54-55. Prove: n is a product of distinct primes p₁p₂...p_r where p_i doesn't divide p_j-1 for all i, j ⇔ All groups of order n are cyclic.
56. Prove that if G is a finite group in which every proper subgroup is abelian, then G is solvable.

Proof: (52) Due to M and N's maximalities, we have N_G(M) = M and N_G(N) = N. Since M and N are abelian, we have that M ≤ N_G(M∩N) and N ≤ N_G(M∩N). Since M ≠ N and M is not a subgroup of N, M has at least one element not in N, so N < < M, N > = G. Therefore we have < M, N > = G ≤ N_G(M∩N), so that M∩N is a normal subgroup. Since G is simple and M∩N ≠ G, we have M∩N = 1.

(53) First note that if M is a maximal subgroup, then gMg^-1 is a maximal subgroup for all g∈G, since if gMg^-1 < H, then M < (g^-1)H(g^-1)^-1. (*)

Consider the set of elements within the conjugates of a maximal subgroup M. According to 4.3.24, we have an element within G that is not within this set. Call it x and proceed by placing the group < x > within a maximal subgroup N (we know that < x > ≠ G, else G is cyclic). We have by the previous exercise that none of the nontrivial elements within the conjugates of M overlap with those of N, since gMg^-1∩fNf^-1 ≠ 1 ⇒ gMg^-1 = fNf^-1 (since these are both maximal by (*) ) ⇒ N = (f^-1g)M(f^-1g)^-1, even though x∈N was defined to not be within any of the conjugates of M. Since the distinct conjugates of M and N all have non-overlapping elements as constituents, and since the number of distinct conjugates of M and N are respectively | G : M | and | G : N | (recall that since G is simple and M and N are maximal, N_G(M) = M and N_G(N) = N), we obtain (| M | - 1)| G : M | + (| N | - 1)| G : N | = 2| G | - | G : M | - | G : N | nonidentity elements. We can't have M = 1 or N = 1, else M ≤ N or N ≤ M, and if | M | > 1 and | N | > 1, then the above equation produces at least | G | + 1 elements including the identity, a contradiction.

(Note that this argument only required the non-cyclicity of G, not necessary the non-abelianness as proposed in the exercise)

(54) (1 ⇒ 2) Let G be a group and | G | = n. We can tell that the order of every subgroup of G is subject to the characteristic prime decomposition above, so that every proper subgroup of G is cyclic by induction and by the preceding exercise (and footnote) G is either cyclic or not simple. If G is not simple, then find a nontrivial proper normal subgroup N, which in these circumstances is cyclic. Proposition 16 implies that | Aut(N) | = φ(| N |) = (p₁ - 1)(p₂ - 1)...(p_k - 1), where the the primes in the decomposition of | G | have been reordered without a loss of generality. Simultaneously, | N_G(N) / C_G(N) | = | G / C_G(N) | = | G |/| C_G(N)| divides (p₁ - 1)(p₂ - 1)...(p_k - 1). If C_G(N) ≠ G, part of this this would imply that at least one prime p_i divides (p₁ - 1)(p₂ - 1)...(p_k - 1). But since p_i doesn't divide p_j-1 for all i, j and p_i is a prime, we have arrived at a contradiction, so that C_G(N) = G, i.e. N ≤ Z(G), and in particular Z(G) > 1. We have that | G / Z(G) | < | G |, so that G / Z(G) is cyclic by induction, and by 3.1.26 G is abelian. By Cauchy's Theorem take x₁, x₂, ..., x_r such that | x_i | = p_i, and by an inductive extrapolation of the process of 4.4.2, < x₁x₂...x_r > = G so that G is cyclic.

(55) (2 ⇒ 1) Disprove the existence of a contradiction by first identifying p and q within the prime decomposition of n such that p | q - 1. Reset n = p^aq^bm where neither p nor q divides m.

By the ending statements regarding groups of semiprime order on page 143, there exists a non-abelian group K of order pq. Let M = Z_p^a-1q^b-1m. Thus K × M is a group of order p^aq^bm = n, and it is non-abelian (take noncommuting x,y∈K and m∈M, so that (x,m)(y,m)≠(y,m)(x,m)) and thus non-cyclic. This violates the premises, and completes the implication.

(56) First assume that G is abelian. Then G / {1} is abelian, thus G is solvable. So assume that G is not abelian, and by exercise 53 there is a nontrivial proper normal subgroup N of G. Any proper subgroup R of G / N is isomorphic to H / N for some proper subgroup H of G, and since H is abelian in G, we have R is abelian in G / N. In this fashion, every proper subgroup of G / N is abelian, and inductively G / N is solvable. N is abelian and thus N / {1} is abelian and N is solvable, thus by the example theorem on page 105 G is solvable.

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