For any finite group let be the minimum number of generators of . (For example, if and only if is a nontrivial cyclic group, and .) Let be the maximum of the integers as ranges over all abelian subgroups of . (For example, and .) Define . ( is called the Thompson subgroup of .)
 Prove that is characteristic in .
 For each of the groups in the list , , , and , give the abelian subgroups which satisfy . [Hint: Use the lattices of subgroups which were given in §2.7.]
 Show that , , , and .
 Prove that if and , then . Deduce that if is a subgroup of a finite group and if for some , then .
 First we prove some lemmas.
Lemma 1: Let be a group, a subgroup, and an automorphism. Then . Proof: Suppose is a minimal generating set for ; that is, and . Note that . Since is bijective, . Thus . Now is also an automorphism of , so that . Hence .
Now to the main result. Suppose . Then . Now . Moreover, as ranges over all abelian subgroups of , so does . Thus . Hence is characteristic in .

 Every proper subgroup of is abelian. Moreover, every proper subgroup of is cyclic; thus , and every proper subgroup satisfies . Specifically, these subgroups are , , , and .
 Every proper subgroup of has order 2 or 4 and thus is abelian. Six of these subgroups are cyclic, and the remaining two are noncyclic groups of order 4, hence isomorphic to . Thus , and the subgroups such that are and .
 Every proper subgroup of of order 2 or 4 is abelian. Ten of these are cyclic, and the remaining four are noncyclic groups of order 4, so that . Of the three subgroups of order 8, two are nonabelian and one is cyclic. Hence . The subgroups which satisfy are , , , and .
 Every proper subgroup of of order 2 or 4 is abelian. Eight of these are cyclic, and the remaining two are noncyclic groups of order 4, so that . Of the three subgroups of order 8, two are nonabelian and one is cyclic. Hence . The subgroups which satisfy are and .
 From the subgroup lattices of these groups, we clearly have , , , and . Note that in , and . Thus .
 We begin with a lemma.
Lemma 2: If then . Proof: Let denote the set of abelian subgroups of . Clearly we have . Thus .
Lemma 3: If then . Proof: By Lemma 2 we have . Moreover, there exists such that ; hence .
Now to the main result. If , then . If is abelian and , then (Since ) and . Thus . Likewise, if is abelian and , then and . Hence .
Suppose now that is a finite group with . Suppose further that for some normal subgroup . Now is characteristic in by part 1, so that is normal in .