Sylow Theorems

Theorem

For every prime factor $p$ with multiplicity $n$ of the order of a finite group $G$, there exists a Sylow $p$-subgroup of $G$, of order $p^n$.

Theorem

Given a finite group $G$ and a prime number $p$, all Sylow $p$-subgroups of $G$ are conjugate to each other.

Theorem

Let $p$ be a prime factor with multiplicity $n$ of the order of a finite group $G$, so that $|G|=p^{n}m$ where $n>0$ and $(p,m)=1$. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then the following hold:

• $n_p$ divides $m$.
• $n_p\equiv 1\mathrm{mod}p$.
• $n_p=|G:N_G(P)|$, where $P$ is any Sylow $p$-subgroup of $G$.

Some Corollary

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

Link to original

Corollary

Let $M$ be a maximal subgroup of a $p$-group $G$, then $|G:M|=p$ and $M⊲G$.

\begin{proof} By ^ne0n10. \end{proof}

The following proposition is about the group action $G$ on $\{N_G(P)g:g\in G\}$.

Proposition

Let $G$ be a finite group and let $P\in {\mathrm{Syl}}_p(G)$. Consider $G$ acting on $\Omega =\{N_G(P)g:g\in G\}$, which induces a map $\phi :G\to \mathrm{Sym}(\Omega )$, then $\phi (P)$ on $\Omega$ has only one fixed point $N_G(P)$.

Furthermore, assume that $|P|=p$ and $\phi (P)$ acting on $\Omega$ has $k$ orbits, then for any $x\in N_G(P)-C_G(P)$, the number of fixed points of $\phi (x)$ is $⩽k$.

\begin{proof} i) If $N_G(P)xP=N_G(P)x$, then $P^x⩽N_G(P)$. Then $P^x=P$ and so $N_G(P)x=N_G(P)$.

ii) Otherwise, there at least exist two fixed points of $\phi (x)$, denoted by $N_G(P)a$ and $N_G(P)b$, in the same orbit of $\phi (P)$. So there exists $p\in P$ such that $N_G(P)a\phi (p)=N_G(P)b$. On the other hand, we have $N_G(P)a\phi (x)=N_G(P)a$ and $N_G(P)b\phi (x)=N_G(P)b$. It deduces that

$$N_G(P)a\phi (px)=N_G(P)b=N_G(P)a\phi (p)=N_G(P)a\phi (xp)$$

and so $[p,x]\in P$ fixes $N_G(P)a$. It deduces that $[p,x]=1$ and so $x\in C_G(p)=C_G(P)$, which is impossible. \end{proof}

Application

The key idea of the following propositions are similar:

• To show $G$ is (not) simple:
  • If $G$ has a subgroup $H$ of index $n$, then $G$ acting on $\{Hg:g\in G\}$ induces a normal subgroup $H_G⊲G$ and $G/H_G\lesssim S_n$. If $G$ is simple, then $H_G=\{1\}$ and $G\lesssim S_n$. Thus, there is a lower bound of index of subgroups of $G$.
  • Since $n_p=[G:N_G(P)]$ has lower bounder, we can determine $n_p$.
  • If $n_p$ is large enough, the intersection of Sylow $p$-subgroups may not trivial. Let $A=P_i\cap P_j$ for example. Then consider the index/order of $C_G(A)$ or $N_G(A)$.
  • ^56a6de describes the order of intersection of two Sylow $p$-subgroups.

Theorem

A simple group of order $60$ is isomorphic to $A_5$.

\begin{proof} Assume that $H⩽G$ with $[G:H]=m$, then consider $G$ acting on $\Omega =\{Hg:g\in G\}$. There exists $H_G⊲G$ such that $G/H_G\lesssim S_m$. If $m⩽4$, then $H_G\ne 1$ and $G$ is not simple, which is impossible. Therefore, $G$ does not has subgroups of index $⩽4$. We claim that $G$ has a subgroup of index $5$. If it holds, then $G/H_G\lesssim S_5$ with $H_G=\{1\}$, i.e., $G$ is a subgroup of $S_5$. Since $S_5$ has only one subgroup of order $60$, we have $G\simeq A_5$.

Now we prove the claim. Recall that $n_p=|G:N_G(P)|$, we have $n_p⩾5$ for any $p\in \{2,3,5\}$. It deduces that $n_3=10$, $n_2\in \{5,15\}$ and $n_5=6$.

• If $n_2=5$, $N_G(P)$ with $P\in \mathrm{Sy}{\mathrm{l}}_2(\mathrm{G})$ is what we desire.
• Now assume $n_2=15$.
  • If any $2$ Sylow $2$-subgroups intersect trivially, then there are $1+3\times 15=46$ elements of order $2^k$, contradicting with $n_5=6$.
  • Thus there exist Sylow $2$-groups $P_1$ and $P_2$ such that $P_1\cap P_2=A$ with $|A|=2$. Then $C_G(A)⩾⟨P_1,P_2⟩$ and so $C_G(A)⩾P_1$ and $|C_G(A)|>4$. It deduces that $|C_G(A)|⩾12$. On the other hand, $[G:C_G(A)]⩾4$ yields $|C_G(A)|⩽15$. Thus $|C_G(A)|=12$ and it is a subgroup of index $5$.

Now we finish the proof. \end{proof}

Proposition

Group of order $144$ is not simple.

\begin{proof} Assume that $G$ with $|G|=144$ is simple. For any $H⩽G$ and $[G:H]=m$, there is $G\lesssim S_m$ and so $m⩾6$.

Note that $n_2\in \{1,3,9\}$ and $n_3\in \{1,4,16\}$. As $n_p=[G:N_G(P)]$, we have $n_2=9$ and $n_3=16$.

• If any two Sylow $3$-subgroups intersect trivially, then there are $8\times 16+1=129$ elements of order $3^k$ and then $n_2=1$, which is impossible.
• Hence there exist Sylow $3$-subgroups $P_1$ and $P_2$ such that $P_1\cap P_2=D$ and $|D|=3$. Then consider the group $H:=N_G(D)⩾⟨P_1,P_2⟩$. Assume the number of Sylow $3$-subgroups of $H$ is $m_3$. Then $m_3\equiv 1(\mathrm{mod}3)$ and $m_3⩾2$. It deduces that $m_3⩾4$. Notice that $|H|=|N_H(P)|\cdot m_3⩾|P|\cdot m_3=9\times 4=36$ and so $[G:H]⩽4$. It is a contradiction.

Now we finish the proof. \end{proof}

In ^24a4c2 and ^9a9397, the key is to consider the intersection of two Sylow $p$-subgroups. Here is a more refined result, which is used to prove ^f95b09.

Theorem

Let $P_1,\cdots ,P_n$ be all Sylow $p$-subgroups of $G$. If for any $i\ne j$ there is $|P_i:P_i\cap P_j|⩾p^d$, then $n\equiv 1(\mathrm{mod}{p}^d)$.

\begin{proof} Let $\Omega =\{P_1,\cdots ,P_n\}$ be the set of Sylow $p$-subgroups. Consider $P_n$ acts on $\{P_1,\cdots ,P_{n-1}\}$ by conjugation. Note that $P_i^x\ne P_n$ for any $x\in P_n$, so the action is well-defined. Let $H_i=P_n{}_{P_i}$ be the point stabilizer of $P_i$, then $H⩽N_G(P_i)\cap P_n$. Since $N_G(P_i)$ has only one Sylow $p$-subgroup $P_i$, there is $N_G(P_i)\cap P_n⩽P_i$ and so $N_G(P_i)\cap P_n=P_i\cap P_n$. Hence, $|P_n:P_n\cap P_i|=o(P_i)⩾p^d$ equals to the length of the orbit containing $P_i$ and so each orbit length is $k{p}^d$ for some $k\in {\mathbb{N}}_{+}$. Therefore, $p^d\mid n-1$ and $n\equiv 1(\mathrm{mod}{p}^d)$. \end{proof}

Proposition

Group of order $432=2^4\cdot 3^3$ is not simple.

\begin{proof} Assume that $G$ is simple. Note that $n_3\in \{4,16\}$. If $n_3=4$, $G$ has a subgroup of index $4$ and $G\lesssim S_4$, which is impossible. Thus $n_3=16$. Since $n_3\not\equiv 1(\mathrm{mod}{3}^2)$, there exist Sylow $3$-subgroups $P_i$ and $P_j$ such that $|P_i:P_i\cap P_j|=3$. Then $|P_i\cap P_j|=9$.

By ^qes1qo, $P_i\cap P_j$ is a maximal subgroup of $P_i$, so $N_{P_i}(P_i\cap P_j)=P_i$. It deduces that $N_G(P_i\cap P_j)⩾⟨P_i,P_j⟩$. Since $N_G(P_i\cap P_j)$ has at least $4$ Sylow $3$-subgroups, we have $|N_G(P_i\cap P_j)|⩾4\cdot N_{N_G(P_i\cap P_j)}(P_i)⩾4\cdot 3^3=108$ and $[G:N_G(P_i\cap P_j)]⩽4$. Then $G$ has a subgroup of index $⩽4$ and $G\lesssim S_4$, which is impossible. \end{proof}

Proposition

Group of order $180=2^2\cdot 3^2\cdot 5$ is not simple.

\begin{proof} Assume that $G$ is simple. Then by Sylow theorems and Burnside’s transfer theorem, one can get $n_2=15$, $n_3=10$ and $n_5=6$. There is an embedding $G\hookrightarrow S_6$ with $[S_6:G]=4$, which induces a group homomorphism $\phi :S_6\to S_4$ and $\phi (S_6)$ is transitive. Since $\ker \phi \in \{\{e\},A_6,S_6\}$, $\phi$ does not exist. Hence $G$ is not simple.

Alternating proof. Since $n_5=6$, there is $N_G(P)=30$ for any Sylow $5$-subgroup $P$. Note that $\mathrm{Aut}(P)\simeq C_4$ and $N_G(P)/C_G(P)\lesssim C_4$, then we know $C_G(P)=15$. Since a group of order $15$ is cyclic, $G$ has an element of order $15$. However, $G\lesssim S_6$ does not have element of order $15$. \end{proof}

Proposition

Group of order $1008$ is not simple.

\begin{proof} Assume $G$ is simple. Note that $n_7\in \{8,36\}$.

If $n_7=8$, then $|N_G(P)|=2\cdot 3^2\cdot 7$ and so $|C_G(P)|=21$ by NC lemma. Since $C_G(P)\simeq C_{21}$, $G$ has an element of order $21$. However, $n_7=8$ yields $G\lesssim S_8$, which does not contain an element of order $21$, contradiction.

If $n_7=36$, then $G\lesssim S_{36}$, $|N_G(P)|=28$ and $|C_G(P)|\in \{14,28\}$. By Burnside’s transfer theorem, $|C_G(P)|\ne |N_G(P)|$, otherwise $G$ has a normal $p^{\prime }$-subgroup. Hence $C_G(P)\simeq C_{14}$ and we assume $C_G(P)=⟨x⟩$. Then $x^2\in P$ can be written as a product of five $7$-cycles. Since $G$ is simple, there is $G\lesssim A_{36}$ and so $x$ is a product of two $14$-cycles and one $7$-cycles. It deduces that $x^7\in N_G(P)$ has $8$ fixed points. By ^x4xxhg, $x^7\in C_G(P)$ is the unique element in $N_G(P)\cap Cl(x^7)$, where $Cl(x^7)$ is the conjugacy class of $x^7$ in $G$. Then by Intersection of Conjugacy Class and Subgroup, there is

$$Cl(x)=\frac{[G:N_G(P)]|Cl(x^7)\cap N_G(P)|}{\mathrm{fix}(x^7)}=\frac{36\cdot 1}{8}\notin \mathbb{Z}$$

which is impossible. \end{proof}