Let be a prime dividing the order of the finite solvable group . Assume has no nontrivial normal subgroups of order prime to . Let be the largest normal -subgroup of (cf. this previous exercise). Prove that ; i.e., that .

If is simple, then for some prime . In this case, (by definition), and we have .

Suppose is not simple. Then nontrivial normal subgroups of exist. Let be a minimal nontrivial normal subgroup; by this previous exercise, is an elementary abelian -group for some prime . By our hypothesis, in fact . In particular, is nontrivial.

Let , and let denote the set of primes dividing except . ( indeed does divide since, as a -group, is nontrivial.) Note that is normal and that is normal. Letting be a Hall -subgroup of and a Sylow -subgroup, we have by Lagrange. Note that . Suppose this inclusion is strict; that is, there are elements in which are not also in . Then these elements are not in . However, note that if . (@@@)

We now consider . Since , . Now if , note that . Thus is normal; since is a Hall subgroup, it is in fact the unique subgroup of of order . Thus is characteristic in .

Recall that is normal, so that is normal in . Since is prime to by construction, . Thus .

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## Comments

I wound like to know why Z(P) is a Sylow p-subgroup of N. Is it possible that there is a p-subgroup which is larger than Z(P) but commuting with P? Thank you very much.

Good question. I’ll have to think about this.

In the meantime, I’ll mark this solution “Incomplete”.