Normalizer of p-subgroup

non-Sylow Subgroup

Lemma

The number of right cosets of $P$ contained in $PxP$ is equal to $|P|/|P^x\cap P|$.

\begin{proof} Let $S$ be the set of right cosets of $P$ contained in $PxP$. Then for every $s\in S$, there exists $y\in xP$ such that $Py=s$. Suppose $y_1=x{p}_1$ and $y_2=x{p}_2$ satisfy $P{y}_1=P{y}_2$, then $p_1{p}_2^{-1}\in x^{-1}Px=P^x$. Therefore, for a fixed $y\in xP$, there are $|P^x\cap P|$ elements $y_i$ in $xP$ such that $Py=P{y}_i$. Therefore, the number of right cosets of $P$ contained in $PxP$ is equal to $|P|/|P^x\cap P|$. \end{proof}

Theorem

Let $G$ be finite, and let $P$ be a non-Sylow $p$ subgroup of $G$, then $P<N_G(P)$.

\begin{proof} Decompose $G$ as $G={\cup }_{x\in G}PxP$. Note that

$$|G|=|P1P|+\cdots +|P{x}_{n}P|,$$

and so

$$|G|/|P|=1+|P|/|P^{x_1}\cap P|+\cdots +|P|/|P^{x_n}\cap P|.\text{\hspace{1em}(*)}$$

Since $P$ is not a Sylow $p$-group, then LHS of $(\ast )$ is divisible by $p$. Thus there exists at least one $i$ such that $|P^{x_i}\cap P|=|P|$. It yields $P^{x_i}=P$ and so $x_i\in N_G(P)$. Therefore, $P<N_G(P)$ by $x_i\notin P$. \end{proof}

Corollary

Let $G$ be a $p$-group.

• If $H<G$, then $H<N_G(H)$.
• If $H{<}_{\max }G$, then $H⊲G$ and $|G|/|H|=p$.
• If $N$ is a normal subgroup of $G$ and $|N|=p$, then $N⩽Z(G)$.

\begin{proof} i) Since $H<G$, then $H$ is not a Sylow subgroup of $G$ and so $H<N_G(H)$ by ^1c3218.

ii) By i), we obtain that $H<N_G(H)$ and so $N_G(H)=G$. Thus $H⊲G$ and $G/H$ is simple. Therefore, $G/H\cong {\mathbb{Z}}_p$ by here and $|G|/|H|=p$.

iii) By NC lemma, $G/C_G(N)⩽\mathrm{Aut}(N)={\mathbb{Z}}_{p-1}$. Thus $C_G(N)=G$, that is, all elements in $G$ commutes with $N$. So $N⩽Z(G)$. \end{proof}

Sylow Subgroup

Theorem

Let $G$ be a finite group and $P<G$ be a Sylow p-subgroup. Let $N_G(P)$ be the normalizer of $P$ in $G$ . Let $H<G$ be a subgroup containing $N_G(P)$. Prove that $N_G(H)=H$.

\begin{proof} For any $g\in N_G(H)$, there is

$$g^{-1}Pg⩽g^{-1}Hg=H<G.$$

Since $g^{-1}Pg$ and $P$ are Sylow subgroups of $H$, by Sylow theorem there exists $h\in H$ such that $g^{-1}Pg=h^{-1}Ph$. Then $g{h}^{-1}\in H$ and so $g\in H$. Therefore, $N_G(H)=H$. \end{proof}

Corollary

Let $P$ be a Sylow subgroup of $G$. Then $N_G(P)=N_G\left(N_G(P)\right)$.

\begin{proof} By ^d0505d. \end{proof}

Theorem

Let $G$ be a finite group and $p$ be a prime. Let $H$ be a p-subgroup of $G$. Then

$$[G:H]\equiv [N_G(H):H]\mathrm{mod}p.$$

\begin{proof} Let a group $G$ act on a finite set $X$; then

$$|X|=|\mathrm{Fix}(X)|+\sum [G:C(x)]$$

where $C(x)=\{g\in G:gx=x\}$ for every $x\in X$, and the sum is over a system of representatives of all distinct nontrivial $G$-orbits on $X$.

In this case let $H$ act on $[G:H]$. If $Hg$ is fixed for all $h\in H$, we have

$$Hgh=Hg$$

which means that

$$gh{g}^{-1}\in H$$

and clearly this holds for all $h\in H$ if and only if $g\in N_G(H)$. Since $Hg=H{g}_1$ iff $g_1^{-1}g\in H$, the number of fixed cosets is $\left[N_G(H):H\right]$.

Since $H$ is a $p$-group, $[H:C(x)]$ is divided by $p$. Now we finish the proof. \end{proof}