## 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.