Hall's Theorems

Definition

Let $G$ be a finite group of order $|G|=p_1^{f_1}\cdots p_t^{f_t}$, where $p_i$ are the prime divisors of $|G|$. Let $\pi (G)=\{p_1,\cdots ,p_t\}$. For a subset $\pi \subseteq \pi (G)$, a subgroup $H⩽G$ is called a Hall $\pi$-subgroup of $G$ if $\pi (H)=\pi$ and $\gcd (|H|,|G/H|)=1$.

Remark. Note that Hall subgroups do NOT always exist, and two Hall subgroups are NOT always conjugate.

• Let $G=A_5$. Suppose $H⩽G$ with $|H|=20$. Then $[G:H]$ is of size $3$. Since $G$ is simple, $G^{[G:H]}\cong G$ acting on $[G:H]$ is faithful. So we have $A_5=G⩽\mathrm{Sym}([G:H])\cong S_3$, which is impossible. Therefore, $\{2,5\}$-Hall subgroup does not exist.
• Let $G={\mathrm{PSL}}_2(11)$. This group has two subgroups $H\cong A_4$ and $K\cong D_{12}$, which are Hall $\{2,3\}$-subgroups of $G$ but not conjugate.

Theorem

Let $G$ be a finite group of order $|G|=p_1^{f_1}\cdots p_t^{f_t}$ where $p_i$ are prime. Then $G$ is solvable if and only if $G$ has Hall $\pi$-subgroups for each subset $\pi \subseteq \pi (G)$.

\begin{proof} If $|\pi |=1$, then $G$ is solvable as $p$-group has nontrivial center. If $|\pi |=2$, by Burnside’s Theorem $G$ is solvable. Now we assume that $|\pi |⩾3$ and assume inductively that this theorem holds for groups of which the order has at most $t-1$ prime divisors including multiplicity.

i) Firstly, assume that $G$ has Hall subgroups for all $\pi \subseteq \pi (G)$. We aim to show $G$ is a solvable group. Let $H_i$ be a Hall $\{p_i{\}}^{\prime }$-subgroup of $G$ for $1⩽i⩽t$, where $\{p_i{\}}^{\prime }=\pi (G)\setminus \{p_i\}$. By induction hypothesis, $H_i$ is solvable. There exists a prime divisor $p$ of $|H_1|$ such that $O_p(H_1)\ne 1$. Since $p$ divides $|H_2|$ or $|H_3|$, WLOG we may assume that $p$ divides $|H_2|$. Let $P$ be a Sylow $p$-subgroup of $H_2$. Then $P$ is a Sylow $p$-subgroup of $G$. Furthermore, by Sylow theorem $O_p(H_1)⩽P^g⩽H_2^g$ for some $g\in G$. We have $G=H_1{H}_2^g$, because

$$|H_1{H}_2^g|=\frac{|H_1||H_2^g|}{|H_1\cap H_2^g|}⩾\frac{(p_2^{f_2}\cdots p_t^{f_t})(p_1^{f_1}{p}_3^{f_3}\cdots p_t^{f_t})}{p_3^{f_3}\cdots p_t^{f_t}}⩾|G|.$$

Let $N=⟨O_p(H_1{)}^x:x\in G⟩$ be the normal closure of $O_p(H_1)$ in $N$. Then $N⊲G$. For any $x\in G$, there exist $x_1\in H_1$ and $x_2\in H_2^g$ such that $x=x_1{x}_2$. Then we have $O_p(H_1{)}^x=O_p(H_1{)}^{x_1{x}_2}=O_p(H_1{)}^{x_2}⩽H_2^g$ and so $N⩽H_2^g$. Thus $N$ is the proper normal subgroup of $G$.

For a Hall $\pi$-subgroup $H$ of $G$, we have that $N\cap H$ is a Hall $\pi$-subgroup of $N$ by computing $|HN|=|H||N|/|H\cap N|$, and $HN/N\cong H/(H\cap N)$ is a Hall subgroup of $G/N$. By induction hypothesis $N$ and $G/N$ are solvable, which yields that $G$ is solvable.

ii) Conversely, assume that $G$ is solvable. Take a subset $\pi \subseteq \pi (G)$ with $|\pi |=|\pi (G)|-1$, and it suffices to find a Hall $\pi$-subgroup of $G$. Let $M{⊲}_{\min }G$. Then $M={\mathbb{Z}}_p^d$ for some prime $p$ and $\overline{G}:=G/M$ is solvable. Let $\overline{H}$ be a Hall $\pi$-subgroup of $\overline{G}$. If $p\in \pi$, then the full preimage of $\overline{H}$ under $G\to \overline{G}$ is a Hall $\pi$-subgroup of $G$. Otherwise, if $p\notin \pi$, then $p\nmid |\overline{H}|$.

Define $X:=M.\overline{H}$ as the full preimage of $\overline{H}$ under $G\to \overline{G}$. If $X<G$, then by induction, $X$ has a Hall $\pi$-subgroup $Y$ which is isomorphic to $\overline{H}$, and $Y$ is a Hall $\pi$-subgroup of $G$. So now we suppose that $G=X=M.\overline{H}$.

Since $\overline{H}$ is solvable, there exists a prime $q\in \pi (\overline{H})$ such that $\overline{Q}=O_q(\overline{H})\ne 1$. Let $M.\overline{Q}$ be the full preimage of $\overline{Q}$ under $G\to \overline{G}$, and let $Q$ be a Sylow $q$-subgroup of $M.\overline{Q}$. Remark that $Q\cong \overline{Q}$ and $M.\overline{Q}=M:Q⊲G$, because $\overline{Q}⊲\overline{H}$. For any $x\in N_M(Q)$ and $y\in Q$, we have $xy{x}^{-1}{y}^{-1}\in M\cap Q=1$ and so $N_M(Q)=C_M(Q)$.

Assume that $N_G(Q)\cap M=N_M(Q)=C_M(Q)=1$. With this assumption, we claim that $|N_G(Q)|⩾|N_{\overline{H}}(\overline{Q})|$. Since $\overline{Q}⊲\overline{H}$, there is $\overline{H}=N_{\overline{H}}(\overline{Q})$. Take an element $\overline{g}\in \overline{H}=N_{\overline{H}}(\overline{Q})$ and a preimage $g$ under $G\to \overline{G}$. Then $M:Q=(M:Q{)}^g=M(g^{-1}Qg)$ and so $g^{-1}Qg\subseteq M:Q$. Since $g^{-1}Qg$ is a Sylow $q$-subgroup of $M:Q$, there exists $x\in M$ such that $x^{-1}(g^{-1}Qg)x=Q$ by Sylow theorem, that is, $gx\in N_G(Q)$. If there is another $x^{\prime }$ such that $g{x}^{\prime }\in N_G(Q)$ with $x^{\prime }\in M$, then $x^{-1}{x}^{\prime }=(gx{)}^{-1}(g{x}^{\prime })\in N_G(Q)\cap M=1$ and it yields $x=x^{\prime }$. Therefore, there is a unique $x\in M$ such that $gx\in N_G(Q)$. Define

$$\phi :N_{\overline{H}}(\overline{Q})\to N_G(Q),\overline{g}\mapsto gx.$$

Suppose $g_1{x}_1=g_2{x}_2$. Then $\overline{g_2^{-1}{g}_1}=\overline{x_2{x}_1^{-1}}=\overline{1}$ yields that $\overline{g_1}=\overline{g_2}$ and so $\phi$ is injective. Thus $|N_{\overline{H}}(\overline{Q})|⩽|N_G(Q)|$ and we prove the claim.

If $|\overline{H}|=|N_{\overline{H}}(\overline{Q})|=|N_G(Q)|$, then $N_G(Q)$ is a Hall $\pi$-subgroup of $G$. Otherwise, if $|N_{\overline{H}}(\overline{Q})|<|N_G(Q)|$, then

$$N_G(Q)=N_G(Q)/(N_G(Q)\cap M)\cong N_G(Q)M/M⩽G/M=\overline{H}=N_{\overline{H}}(\overline{Q}),$$

which is impossible.

Now we assume that $C_M(Q)=N_M(Q)=N_G(Q)\cap M\ne 1$. Let $P:=C_M(Q)\ne 1$. Note that $P{⊲}_{\mathrm{char}}Z(M:Q){⊲}_{\mathrm{char}}M:Q⊲G$ yields that $P⊲G$. Since $P⩽M{⊲}_{\min }G$ and $P\ne 1$, we have $P=M$ and so $M:Q=M\times Q⊲G$. Thus $Q$ is a normal subgroup of $G$. Let $\overline{K}$ be a Hall $\pi$-subgroup of $G/Q$. Then the full preimage of $\overline{K}$ under $G\to G/Q$ is a Hall $\pi$-subgroup of $G$. \end{proof}

Remark. The proof of ii) is similar as a Lemma of Cohomology.

Theorem

Hall subgroups are conjugate if $G$ is solvable.

\begin{proof} Let $G$ be a solvable group. Let $M⊲G$ such that $|G/M|$ is a prime $p$. The existence of $M$ is guaranteed by the composition series of $G$. Let $H,K$ be Hall $\pi$-subgroups of $G$. If $p\notin \pi$, then $H,K⩽M$ and $H,K$ are conjugation by induction. If $p\in \pi$, then $H\cap M$, $K\cap M$ are Hall $\pi$-subgroups of $M$. By induction hypothesis, there is a $x\in M$ such that $(H\cap M{)}^x=K\cap M$.

WLOG we may assume that Hall $\pi$-subgroups $H$ and $K$ satisfies $H\cap M=K\cap M$. It suffices to show $H$ and $K$ are conjugate. Let $L=⟨H,K⟩$. If $L<G$, then by induction hypothesis, $H$ and $K$ are Hall $\pi$-subgroups of $L$ and so they conjugate in $L$. Then we suppose that $L=G$, that is, $G=⟨H,K⟩$.

Let $N:=H\cap M$. Note that $N$ is a normal subgroup of $G$. Since $H/N$ and $K/N$ are Hall subgroups of $G/N$, by induction hypothesis, we have a $\overline{g}\in G/N$ such that $(H/N{)}^{\overline{g}}=K/N$. As $H/N\cong {\mathbb{Z}}_p$, there is a $h\in H$ such that $H=⟨N,h⟩$. Let $k=h^g$ for some preimage of $\overline{g}$ under $G\to G/N$. Then by $K/N\cong {\mathbb{Z}}_p$, we have $K=⟨N,k⟩$ and so $H^g=K$. Therefore, $H$ and $K$ are conjugate. \end{proof}

Theorem

Let $G$ be a finite solvable group and $\pi$ be any set of primes. Then any subgroup whose order is a product of primes in $\pi$ is contained in some Hall $\pi$-subgroup.

\begin{proof} Suppose $H⩽G$ is a $\pi$-subgroup of $G$ and $K$ is a Hall $\pi$-subgroup of $G$. Note that

|H|/|H\cap K|=|H|/(|H||K|/|HK|)=|HK|/|K|\mbox{ is a divisor of $|G|/|K|$},

we have that $\gcd (|H|/|H\cap K|,|H\cap K|)=1$. So if $H\cap K\ne 1$, then $H\cap K$ is a Hall $\pi$-subgroup of $H$. Since the Hall $\pi$-subgroup of $H$ is itself, there is $H\cap K=H$ and $K⩾H$. Therefore, if $H$ is not contained in any Hall $\pi$-subgroup, then $H\cap K=1$ for all Hall $\pi$-subgroup $K$.

Take a prime $p\in \pi$. Let $P$ be a Sylow $p$-subgroup of $H$, and let $P_K$ be a Sylow $p$-subgroup of Hall $\pi$-subgroup $K$. As $P_K$ is a Sylow $p$-subgroup of $G$, then there exists a $g\in G$ such that $P⩽P_K^g$. Then $K^g$ is a Hall $\pi$-subgroup and $K^g\cap H⩾P$, contradiction. \end{proof}

Schur-Zassenhaus

Let $N⊲G$ be a Hall subgroup of $G$. Then:

• $N$ has complement in $G$;
• if one of $N,G/N$ is solvable and $H,H_1$ are complements of $N$ in $G$, then there exists $u\in N$ such that $H^u=H_1$.

Remark. By Feit–Thompson theorem, one of $|N|,|G/N|$ is odd and so ii) always hold.