Bourbaki

Exercise 26, § 6

Algebra, I, 153

Bourbaki marks this one as harder than the rest.

Let GG be a finite group. GG is called supersolvable if there exists a composition series (Gi)0≤i≤n(G_i)_{0 \leq i \leq n} of GG consisting of normal subgroups of GG such that the quotients Gi/Gi+1G_i/G_{i+1} are cyclic. (a) Show that every subgroup, every quotient group and every finite product of supersolvable groups is supersolvable. (b) Show that nilpotent ⇒\Rightarrow supersolvable ⇒\Rightarrow solvable and that the converse implications are false.

(c) Suppose G is supersolvable. Show that, if G≠{1}G \neq \{1\}, there exists a normal subgroup C of G which is cyclic of prime order. Show that the derived group D(G) of G is nilpotent. Show that every subgroup of G distinct from G which is maximal is of prime index.

(d) A finite group G is bicyclic if there exist two elements a,ba, b of G such that, if CaC_a and CbC_b denote the cyclic subgroups generated by aa and bb respectively, then G=CaCb=CbCaG = C_a C_b = C_b C_a. Show that every bicyclic group is supersolvable. (Reduce it to proving that there exists in G a normal cyclic subgroup N≠{1}N \neq \{1\}. Attention may be confined to the case where Ca∩Cb={1}C_a \cap C_b = \{1\}; let m≥n>1m \geq n > 1 be the orders of aa and bb respectively; by considering the elements b−1akb^{-1} a^k, where 0≤k≤m−10 \leq k \leq m - 1, show that there exists an integer s such that as≠1a^s \neq 1 and b−1asb∈Cab^{-1} a^s b \in C_a; the set of asa^s with this property is the desired cyclic subgroup N.)

Solution

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Transcribed from the printed volume. Not translated.