In a finite solvable group with no nontrivial normal subgroups relatively prime to p, the p-core contains its centralizer

Let p be a prime dividing the order of the finite solvable group G. Assume G has no nontrivial normal subgroups of order prime to p. Let P = O_p(G) be the largest normal p-subgroup of G (cf. this previous exercise). Prove that C_G(P) \leq P; i.e., that C_G(P) = Z(P).

If G is simple, then G \cong Z_p for some prime p. In this case, P = G (by definition), and we have C_G(P) = Z(P) = G.

Suppose G is not simple. Then nontrivial normal subgroups of G exist. Let K \leq G be a minimal nontrivial normal subgroup; by this previous exercise, K is an elementary abelian q-group for some prime q. By our hypothesis, in fact q = p. In particular, P = O_p(G) is nontrivial.

Let N = C_G(P), and let \pi denote the set of primes dividing |N| except p. (p indeed does divide N since, as a p-group, Z(P) is nontrivial.) Note that Z(P) \leq C_G(P) is normal and that C_G(P) \leq N_G(P) = G is normal. Letting H be a Hall \pi-subgroup of N and Q a Sylow p-subgroup, we have N = QH by Lagrange. Note that Z(P) \leq Q. Suppose this inclusion is strict; that is, there are elements in Q which are not also in Z(P). Then these elements are not in O_p(G). However, note that if xO_p(G) \in G/O_p(G). (@@@)

We now consider Z(P) H. Since Z(P) \leq P, H \leq C_G(Z(P)). Now if zh \in Z(P) H, note that (zh)H(zh)^{-1} = zhHh^{-1}z^{-1} = zHz^{-1} = Hzz^{-1} = H. Thus H \leq Z(P) H is normal; since H \leq Z(P) H \leq N is a Hall subgroup, it is in fact the unique subgroup of Z(P) H of order |H|. Thus H is characteristic in Z(P) H.

Recall that C_G(P) \leq N_G(P) = G is normal, so that H is normal in G. Since |H| is prime to p by construction, H = 1. Thus C_G(P) = Z(P).

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  • K.-S. Liu  On September 28, 2010 at 4:29 am

    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.

    • nbloomf  On September 28, 2010 at 9:03 pm

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

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

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