An element of a group is called a nongenerator if for all proper subgroups , is also proper. Prove that is the set of all nongenerators in .
[Not finished.] Let . Now let be a proper subgroup. If is contained in some maximal subgroup , then we have , so that is proper as desired. Suppose now that is not contained in any maximal subgroup of . Suppose further that . (@@@)
[Old proof, assumes is finite.]
Let . By a lemma we proved for a previous theorem, if is proper then is proper. Now and , so that is proper. Thus is a nongenerator.
Let be a nongenerator of . If has no maximal subgroups, then , and we have . Suppose now that is maximal; in particular, is proper, so that is proper. Thus , so that . Since is contained in every maximal subgroup, .