Basic properties of the p-core of a finite group

Let R be a normal p-subgroup (not necessarily Sylow) of G a finite group.

  1. Prove that R is contained in every Sylow p-subgroup of G.
  2. If S is another normal p-subgroup of G, prove that RS is also a normal p-subgroup of G.
  3. Let \mathcal{A} = \{ R \leq G \ |\ R \vartriangleleft G, R\ \mathrm{a}\ p\ \mathrm{subgroup} \}, and define O_p(G) = \langle \bigcup_{R \in \mathcal{A}} R \rangle. Prove that O_p(G) is the \leq-greatest normal p-subgroup of G. Prove further that O_p(G) = \bigcap_{P \in \mathsf{Syl}_p(G)} P.
  4. Prove that O_p(G/O_p(G)) = 1. (That is, G/O_p(G) has no nontrivial normal p-subgroups.)

  1. Let R \leq G be a normal p-subgroup. Now R \leq P for some Sylow p-subgroup P by Sylow’s Theorem. Also by Sylow, if Q \leq G is a second Sylow p-subgroup, then Q = gPg^{-1} for some g \in G. Then R = gRg^{-1} \leq gPg^{-1} = Q, so that R \leq Q.
  2. Let R and S be normal p-subgroups. If P is a Sylow p-subgroup, then R,S \leq P by part (1). Then RS \leq P is a p-subgroup, and moreover RS \leq G is normal.
  3. We need to show three things: normalcy, p-groupitude, and \leq-greatestness.
    1. Let g \in G. Then g O_p(G) g^{-1} = g \langle \bigcup_{R \in \mathcal{A}} R \rangle g^{-1} \langle g \left( \bigcup_{R \in \mathcal{A}} R \right) g^{-1} \rangle = \langle \bigcup_{R \in \mathcal{A}} gRg^{-1} \rangle = \langle \bigcup_{R \in \mathcal{A}} R \rangle = O_p(G). Thus O_p(G) is normal.
    2. If P is some Sylow p-subgroup of G, then R \leq P for each R \in \mathcal{A}. Thus \bigcup_{R \in \mathcal{A}} R \subseteq P, and we have O_p(G) \leq P; thus O_p(G) is a p-subgroup.
    3. Suppose S \leq G is some other normal p-subgroup; then S \leq O_p(G) by definition. Thus O_p(G) is the \leq-greatest normal p-subgroup of G.

    We showed that O_p(G) \leq P for all Sylow p-subgroups P, hence O_p(G) \leq \bigcap_{P \in \mathsf{Syl}_p(G)} P. To see the other inclusion, note that for each g \in G, conjugation by G permutes the Sylow p-subgroups of G. That is, g \left( \bigcap_{P \in \mathsf{Syl}_p(G)} P \right) g^{-1} = \bigcap_{P \in \mathsf{Syl}_p(G)} gPg^{-1} = = \bigcap_{Q \in \mathsf{Syl}_p(G)} Q; hence \bigcap_{P \in \mathsf{Syl}_p(G)} is a normal subgroup of G, hence is contained in O_p(G).

  4. Suppose N \leq G/O_p(G) is a nontrivial normal p-subgroup. By the Lattice Isomorphism Theorem, N = M/O_p(G) for some normal subgroup M \leq G with O_p(G) \leq M. By Lagrange’s Theorem, we have |N| = |M/O_p(G)| = |M|/|O_p(G)|, so that |M| = |N| \cdot |O_p(G)|. Then M is a normal p-subgroup of G, and we have M \leq O_p(G). But then N = 1 is trivial, a contradiction. So no such subgroup N exists.
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