Exercise 4.5.37 (Intersection of all Sylow $p$-subgroups of $G$)

Proof. Let $R$ be a normal $p$-subgroup of $G$.

(a)

Let $H$ be any Sylow $p$-subgroup of $G$. By the conjugacy part of Sylow’s Theorem, there is some $g\in G$ such that $R\subseteq \mathit{𝑔𝐻}{g}^{-1}$, thus $g^{-1}\mathit{𝑅𝑔}\subseteq H$. Moreover $R⊴G$, thus $g^{-1}\mathit{𝑅𝑔}=R$, so $R\subseteq H$.

This shows that $R$ is contained in every Sylow $p$-subgroup of $G$.

(b)

Let $S$ is another normal $p$-subgroup of $G$. Since $S⊴G$, $\mathit{𝑅𝑆}$ is a subgroup of $G$.

Let $H$ be a fixed Sylow $p$-subgroup of $G$, of order $p^{\alpha }$. By part (a), $R\subseteq H$ and $S\subseteq H$, thus

$$\mathit{𝑅𝑆}\le H.$$

By Lagrange’s Theorem, $|\mathit{𝑅𝑆}|$ divides $|H|=p^{\alpha }$, therefore $\mathit{𝑅𝑆}$ is a $p$-subgroup of $G$.

If $h\in \mathit{𝑅𝑆}$, and $g\in G$, then $h=\mathit{𝑟𝑠}$, where $r\in R$ and $s\in S$. Since $R⊴G$ and $S⊴G$, then

$$\mathit{𝑔h}{g}^{-1}=\mathit{𝑔𝑟𝑠}{g}^{-1}=(\mathit{𝑔𝑟}{g}^{-1})(\mathit{𝑔𝑠}{g}^{-1})\in \mathit{𝑅𝑆},$$

so $\mathit{𝑅𝑆}⊴G$.

If $R,S$ are normal $p$-subgroup of $G$, then $\mathit{𝑅𝑆}$ is also a normal $p$-subgroup of $G$.

(c)

Let $O_p(G)$ be the subgroup generated by all normal $p$-subgroups of $G$. Let $R_1,R_2,\dots ,R_k$ be all normal $p$ subgroups of $G$ (there are finitely many subgroups of $G$). Then $$O_p(G)=R_1{R}_2\cdots R_k:$$

by part (b), $R_1{R}_2\cdots R_k$ is a normal $p$-subgroup of $G$, and it contains $R_1,R_2,\dots ,R_k$, so it contains all normal $p$-subgroups of $G$. Moreover, if $K$ is a subgroup of $G$ which contains all normal $p$-subgroups of $G$, then $K$ contains $R_1{R}_2\cdots R_k$, so $R_1{R}_2\cdots R_k$ is the subgroup of $G$ generated by all normal $p$-subgroups of $G$.

So $R_i\le O_p(G)$ for all $i$, and $O_p(G)$ is a normal $p$-subgroup of $G$ (thus $O_p(G)=R_j$ for some index $j$), this shows that $O_p(G)$ is the largest normal $p$-subgroup of $G$ and $O_p(G)$ (for inclusion), and the greatest element of a (partially) ordered set is unique.

Finally, since $O_p(G)$ is a $p$-subgroup, it is contained in every Sylow subgroup $H$ of $G$, so

$$O_p(G)\subseteq \underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap }H.$$

Conversely, $\underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}H$ is contained in some Sylow $p$-subgroup of $G$, so is a $p$-subgroup of $G$.

If $g\in G$, then

$$\psi \{\begin{array}{ccc}\hfill \mathit{𝑆𝑦}{l}_p(G)\hfill & \hfill \to \hfill & \mathit{𝑆𝑦}{l}_p(G)\hfill \\ \hfill H\hfill & \hfill \mapsto \hfill & \mathit{𝑔𝐻}{g}^{-1}\hfill \end{array}$$

is bijective, so $\psi$ permutes the Sylow $p$-subgroups of $G$, therefore

$$\begin{array}{llllll}\hfill g\left(\underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap }H\right)g^{-1}& =\underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap }\mathit{𝑔𝐻}{g}^{-1}& \hfill & & \hfill & \\ \hfill & =\underset{K\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap }K& \hfill (K=\mathit{𝑔𝐻}{g}^{-1}=\psi (H))& & \hfill & & \hfill \\ \hfill & =\underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap }H.& \hfill & & \hfill & \end{array}$$

This shows that $\underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}H⊴G$.

Since $O_p(G)$ is the largest normal $p$-subgroup of $G$, $\underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}H\subseteq O_p(G)$, so

$$O_p(G)=\underset{H\in \mathit{𝑆𝑦}{l}_p(G)}{\bigcap }H.$$

(d)

Let $\bar{G}=G∕O_p(G)$, and let $$\pi :G\to \bar{G}=G∕O_p(G)$$

be the natural projection.

Consider any normal $p$-subgroup $\bar{H}$ of $\bar{G}$, and put $H={\pi }^{-1}(\bar{H})$. Then $\pi (H)=\bar{H}$ and $H⊴G$. Moreover $|H|=|\bar{H}|\cdot |O_p(G)|$ is a power of $p$, so $H$ is a normal $p$-subgroup of $G$. By part (c),

$$H\subseteq O_p(G),$$

therefore

$$\bar{H}=\pi (H)=\{\bar{1}\}.$$

This shows that $\bar{G}$ has no nontrivial normal $p$-subgroup. Since $O_p(\bar{G})$ is a normal $p$-subgroup of $\bar{G}$,

$$O_p(\bar{G})=\{\bar{1}\}.$$