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Since G is a finite p-group and B is a proper subgroup of G, we have [B.sup.G] [not equal to] G.
We claim that [A.sub.ab]/[[bar.C].sub.A] [not equal to] {1}: By Theorem 8, gp([C.sub.A], [[delta].sub.2] A) is a proper subgroup of A since [C.sub.A] is.
We have < x > Q is a proper subgroup of G and so < x > Q =< x > x Q, i.e., <x> [less than or equal to] [N.sub.G](Q).
So, this group is not a direct product of proper subgroups.
Lemma 2.4.[3]Suppose that G is a group which is not p-nilpotent but whose proper subgroups are all p-nilpotent for some prime p.
Among results that have no analog in any existing books are groups admitting an irredundant covering by a few proper subgroups, characterizing abelian and minimal nonabelian groups, and groups containing a soft subgroup.
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