1.5 Sylow Theorem

Motivation. Let $G$ be a finite abelian group. If $m\mid n$, then there exists some subgroup $H<G$ such that $|H|=m$.

Lemma

Let $H$ be a group with $|H|=p^n$. Suppose $H$ acts on finite set $S$. Let $S_0=\{x\in S:hx=x,\forall h\in H\}=S^H=$ fixed points. Then $|S_0|\equiv |S|\mathrm{mod}p$.

\begin{proof} Note that $|S|=|S_0|+{\sum }_{s\notin S_0}|\overline{s}|$. Since $|\overline{s}||H_s|=|H|$, then $|\overline{s}|\equiv 0\mathrm{mod}p$. It follows that $|S_0|\equiv |S|\mathrm{mod}p$. \end{proof}

Cauchy

Let $G$ be a finite group and $p\mid |G|$. Then there exists $a\in G$ such that $|a|=p$.

\begin{proof} Let $S=\{(a_1,\cdots ,a_p):a_i\in G,a_1\cdots a_p=1\}$. Then $|S|=n^{p-1}$ and $p\mid |S|$. Consider $\mathbb{Z}/p\mathbb{Z}\curvearrowright S$ such that $k\cdot (a_1,\cdots ,a_p)=(a_{k1},\cdots ,a_{kp})$. Also $S_0=\{(a,\cdots ,a):a\in G\}$ is divisible by $p$ by ^acf430. Since $(1,\cdots ,1)\in S_0$, then $S_0$ is non-empty and so it has a non-trivial element $(a,\cdots ,a)$. Therefore, we have that $a^p=1$ and so $|a|=p$. \end{proof}

Definition

Let $G$ be a group.

• If $\forall x\in G$, $|x|=p^a$, then we say $G$ is a $p$-group.
• Let $H<G$. If $H$ is a $p$-group, then $H$ is called a $p$-subgroup.

Corollary

Let $|G|<\infty$. Then $G$ is a $p$-group iff $|G|=p^n$ for some $n$.

\begin{proof} Assume that $|G|=p^n$. Then by Lagrange theorem $G$ is a $p$-group. Conversely, assume that $G$ is a $p$-group. If there is a $q\ne p$ such that $q\mid |G|$, then by ^9af89b there is a $a\in G$ with $|a|=q$, contradiction. \end{proof}

Corollary

Let $G$ be a finite $p$-group. Then its center $|C(G)|⩾p$.

Lemma

Let $G$ be a finite group, and let $H$ be a $p$-subgroup. Then $[N_G(H):H]\equiv [G:H]\mathrm{mod}p$.

\begin{proof} Let $S=\{xH:x\in G\}$. Consider $H$ acts on $S$ by $h(xH)=hxH$. Consider $S_0$. Note that $hxH=xH$ iff $x^{-1}hx\in H$ for all $h\in H$ iff $x\in N_G(H)$. Therefore, $|S_0|=[N_G(H):H]$. By ^acf430, $|S|\equiv |S_0|\mathrm{mod}p$. \end{proof}

Corollary

Let $H<G$ be a $p$-group. If $p\mid [G:H]$, then $N_G(H)⊋H$.

\begin{proof} By ^e0a58f. \end{proof}

First Sylow theorem

Let $G$ be a finite group with $|G|=p^n\cdot m$, $(p,m)=1$. Then $G$ contains a subgroup of order $p^i$ for each $1⩽i⩽n$. And each subgroup of order $p^i$ is normal in some subgroup of order $p^{i+1}$.

\begin{proof} By induction. When $i=1$, by ^9af89b. Assume that there exists $H<G$ such that $|H|=p^i$ with $1⩽i<n$. It suffices to find a subgroup with order $p^{i+1}$. Since $p\mid [G:H]$, then $p\mid |N_G(H)/H|$ by ^e0a58f. By ^9af89b, $N_G(H)/H$ has a subgroup of order $p$, which is of form $H_1/H$ with $H⊲H_1<N_G(H)$. Then $|H_1|=p^{i+1}$ and $H⊲H_1$. Now we finish the proof. \end{proof}

Second Sylow theorem

Let $|G|=p^n\cdot m$ and $H<G$ a $p$-group. Let $P<G$ be a Sylow $p$-subgroup. Then there exists $x\in G$ such that $H<xP{x}^{-1}$. In particular, any two Sylow $p$-subgroups are conjugate to each other.

\begin{proof} Let $S=\{gP:g\in G\}$. Note that $|S|=[G:P]$ and $p\nmid |S|$. Consider $H$ acts on $S$ by left translation. Then $gP\in S_0$ iff $hgP=gP$ for all $h\in H$ iff $g^{-1}hg\in P$ for all $h\in H$ iff $g^{-1}Hg<P$. Since $|S_0|\equiv |S|\mathrm{mod}S$, then $|S_0|$ is non-empty. Now we finish the proof. \end{proof}

Third Sylow theorem

Let $G$ be a finite group, and let $n_p=\text{mbox}numberofSylowp\text{mbox}-subgroups$. Then $n_p\mid |G|$ and $n_p=kp+1$ for some $k⩾0$.

\begin{proof} Let $P$ be a fixed Sylow $p$-subgroup. Let $S=\{\text{mbox}allSylowp\text{mbox}-subgroups\}$. By ^0acdfe, $S=\{g^{-1}Pg:g\in G\}$. Note that $G\curvearrowright S$ by conjugation. It follows that $S=|\overline{P}|=[G:N_G(P)]$. By Lagrange theorem, $n_p\mid |G|$. Consider $P$ acts on $S$. Then $Q\in S_0$ iff $xQ{x}^{-1}=Q$ for all $x\in P$ iff $P⩽N_G(Q)$. So $P,Q$ are two distinct Sylow $p$-subgroups of $N_G(Q)$ and so $P=Q$. Therefore, $S_0=\{P\}$ and $|S|\equiv |S_0|=1\mathrm{mod}p$ by ^acf430. \end{proof}

Theorem

Let $G$ be a finite group. If $P<G$ is a Sylow $p$-subgroup. Then $N_G(N_G(P))=N_G(P)$.

\begin{proof} Firstly, note that $N_G(N_G(P))⩾N_G(P)$. It suffices to show $N_G(P)⩾N_G(N_G(P))$. Since $P$ is a Sylow $p$-subgroup of $N_G(P)$, then $P$ is the unique Sylow $p$-subgroup of $N_G(P)$. Let $x\in N_G(N_G(P))$. Then $N_G(P{)}^x=N_G(P)$ and $P^x\subseteq N_G(P)$. Therefore, $P^x=P$ and $x\in N_G(P)$. \end{proof}

Remark. $P<G$ is a Sylow $p$-subgroup. Suppose $P⊲Q⊲R$, then $P⊲R$.