Group Core

Definitions

Definition

For a group $G$, the normal core of a subgroup $H$ is the largest normal subgroup of $G$ that is contained in $H$, i.e. ${\mathrm{Core}}_G(H)={\cap }_{g\in G}{g}^{-1}Hg$. Moreover, if ${\mathrm{Core}}_G(H)=1$, we say $H$ is core-free.

Definition

For a prime $p$, the $p$-core of a finite group is defined to be its largest normal $p$-subgroup, denoted by $O_p(G)$.

Definition

The solvable radical of $G$ is defined to be the largest solvable normal subgroup, and is denoted $O_{\infty }(G)$.

Remark. “Core” means the largest normal subgroup with some properties, like containing some subgroup, being a $p$-group, or solvable.

Core-free and Faithful Coset Action

Let $G$ be a finite group, and let $\Omega :=[G:H]$ where $H<G$. Then

• $G_{(\Omega )}={\mathrm{Core}}_G(H)$, and
• $G/{\mathrm{Core}}_G(H)\cong G^{\Omega }$.

Lemma

If $H⩽G$ is core-free, then $G$ acting on $[G:H]$ is faithful.

\begin{proof} If there is a $g\in G$ such that $H{g}_{i}g=H{g}_i$ for all $g_i\in G$, then $g^G\in H$ and $g\in {\cap }_{g_i\in G}{H}^{g_i}={\mathrm{Core}}_G(H)=\{1\}$. Therefore, $g=1$ and so the action is faithful. \end{proof}

Exercises.

• ^hhebb9
• ^xorw21

Properties of $p$-core

Proposition

Note that $O_p(G)$ is the intersection of all Sylow $p$-groups of $G$, and the following holds:

• $O_p(G)$ is the normal core of every Sylow $p$-subgroup of the group $G$.
• $O_p(G){⊲}_{\mathrm{char}}G$ and $\overline{G}:=G/O_p(G)$ is such that $O_p(\overline{G})=1$.
• If $\pi (G)=\{p_1,\cdots ,p_k\}$, then its Fitting subgroups is $O_{p_1}(G)\times \cdots \times O_{p_k}(G)$.

\begin{proof} i) Easy.

ii) Since $O_p(G)$ is the intersection of all Sylow $p$-subgroups, then $O_p(G){⊲}_{\mathrm{char}}G$. Suppose $\overline{P_1},\cdots ,\overline{P_k}$ are all Sylow $p$-subgroups of $\overline{G}$, and suppose $P_i$ are preimage of $\overline{P_i}$. Then ${\cap }_{i=1}^k{P}_i⩽O_p(G)$ and so ${\cap }_{i=1}^k\overline{P_i}=1$, i.e., $O_p(\overline{G})=1$.

iii) See here. \end{proof}

Proposition

If $G$ is solvable, then there exists a prime $p$ such that $O_p(G)\ne 1$.

\begin{proof} Let $M$ be a minimal normal subgroup of $G$. Then $M\cong {\mathbb{Z}}_p^d$ for some prime $p$ and some positive integer $d$. Thus $O_p(G)$ is not trivial. \end{proof}