Group Theory - 2024spring

NOTE

这是2024年春的“群论及其应用”笔记。

第三周缺课，第十周和第十一周的内容没有整理。

Week 1: Constructing groups (done)

Semi-direct product

Example.

$$Z_7:Z_6=\{Z_7\times Z_6,(Z_7:Z_2)\times Z_3,(Z_7:Z_3):Z_2,Z_7:Z_6\}.$$

We call $Z_p:Z_{p-1}$ the affine group when it has trivial center, denoted as $\mathrm{AGL}(1,p)$.

from finite field

It is mentioned in Singer Cycle.

Let $F=GF(p^f)$ and let $G=F^{+}\cong Z_p^f$ and $H=F^{\ast }\cong Z_{p^f-1}$. Define a group $X=\{gh|g\in G,h\in H\}$ where $g^h=g\cdot h$.

And the automorphism group of $X$ is $\mathrm{Aut}(X)=X:⟨\tau ⟩=(Z_p^f:⟨g⟩):⟨\tau ⟩$ where the order of $\tau$ is $f$.

$\tau$ acts on $⟨g⟩$ as Frobenius action.

example. Let $X=Z_3^2:Z_8$ and $\mathrm{Aut}(X)=(Z_3^2:Z_8):Z_2:=(Z_3:⟨g⟩):⟨\tau ⟩$. Let $a=g^2$ and $b=g\tau$. Then we can verify $a^b=a^{-1}$, $b^a=b^{-1}$ and $a^2=b^2$.

Let $Y=⟨a,b⟩$ where $|a|=|b|=4$. So $Y=Q_8$.

$\mathrm{Aut}(X)$ also can be written as $Z_3^2:(Z_8:Z_2)$ and it has a subgroup $Z_3^2:Q_8$.

exe. Prove that all elements of order $3$ in $X$ are conjugate.

exe. $|X|=72$, $Z(X)=1$. Show that $X$ is unique and $X=Z_3^2:Q$ where $Q\in \{Z_8,Q_8,D_8\}$.

(hint: consider the Sylow 3 group $H$, which is normal and cannot be cyclic by trivial center. Then by NC lemma, $G/C_G(H)\le \mathrm{Aut}(H)$.)

Week 2 The two groups whose elements with same order are conjugate (done)

Question. (1970s) If $G$ is such that elements of the same order are conjugate, then $G=S_2$ or $S_3$.

Solved by P. Fitzpetrich 1985, W.Feit-G. Seitz 1988, Jiping Zhang 1988.

Week 4: Burnside transfer (done)

Let $H<G$ and $\Omega =[G:H]=\{Hx|x\in G\}=\{H{x}_1,\cdots ,H{x}_n\}$. Define $\hat{g}:Hx\mapsto Hxg$ and ${\tau }_g\in \mathrm{Sym}(n)$ be the permutation induced from $\hat{g}$.

Suppose $H{x}_{i}g=H{x}_j$. Then $h_i(g)=x_{i}g{x}_j^{-1}=x_{i}g{x}_{i^{\tau (g)}}^{-1}\in H$.

Definition

Define

$$\begin{array}{rl}{V}_{G\to H}:G& \to H/H^{\prime },\\ g& \mapsto ({\Pi }_{i=1}^n{h}_i(g))H^{\prime }\end{array}$$

as a transfer from $G$ to $H/H^{\prime }$.

Lemma

• $V_{G\to H}$ is a homomorphism from $G$ to $H/H^{\prime }$;

• $V_{G\to H}$ does not depend on the choices of $x_1,\cdots ,x_n$;

• Writing $\hat{g}$ in cycle form,

V_{G\to H}(g)=\Pi_{i=1}^n(x_ig^{l_i}x_i^{-1})H,

where $l_i$ is defined as following. $\hat g$ partitions $\Omega$ i to $t$ orbits:

(Hx_1,Hx_1g,\cdots,Hx_1g^{l_1-1})(Hx_2,Hx_2g,\cdots,Hx_2g^{l_1-2})\cdots(Hx_t,\cdots,Hx_tg^{l_t-1}).

\begin{proof} (i) and (ii) are easy. For (iii), take $x_1,\cdots ,x_1{g}^{l_1-1},\cdots ,x_t,\cdots ,x_t{g}^{l_t-1}$ as representatives. \end{proof}

Definition

A group $G$ is called $p$-nilpotent if $G=NP$, where $N⊲G$ and $P$ is a Sylow $p$-subgroup, and $N\cap P=1$. (equivalent: If $G$ has a normal subgroup $N$ such that $|G/N|=|P|$, where $p$ is a prime divisor of $|G|$. )

Burnside

Let $G$ be a finite group, and $P$ a Sylow $p$-group. If $N_G(P)=C_G(P)$, then $G$ is $p$-nilpotent.

\begin{proof} Since $N_G(P)=C_G(P)$ is abelian, $P^{\prime }=1$. Let $g\in P\backslash \{1\}$. Then $V_{G\to P}(g)={\Pi }_{i=1}^t{x}_i{g}^{l_i}{x}_i^{-1}$. As $g^{l_i}$ and $x_i{g}^{l_i}{x}_i^{-1}\in P$ and $P$ abelian, $C_G(g{)}^{l_i}\ge P$ and $C_G(x_i{g}^{l_i}{x}_i^{-1})=x_i{C}_G(g^{l_i})x_i^{-1}\ge P$. Hence $C_G(g^{l_i})\ge P,x_i^{-1}P{x}_i$.

By Sylow theorem, there exists $u\in C_G(g^{l_i})$ such that $(x_{i}u{)}^{-1}P(x_{i}u)=P$.

Also $V_{G\to P}(g)={\Pi }_{i=1}^t{g}^{l_1+\cdots +l_t}=g^{|G/P|}$ is not trivial. So $V_{G\to P}(G)=P=P/P^{\prime }$ with kernel $K$. By isomorphism theorem, there is $G/K\cong P$. So $K⊲G$ such that $G=KP$ and so $G$ is $p$-nilpotent. \end{proof}

Theorem

Let $G$ be finite of which Sylow subgroups are all cyclic. Let $p$ be the smallest prime divisor of $G$ and let $P$ be a Sylow $p$-subgroup of $G$. Then:

• $G$ is $p$-nilpotent.
• $G$ is solvable.

\begin{proof} Since $P$ is cyclic, $N_G(P)/C_G(P)\lesssim \mathrm{Aut}(P)$. It follows that $N_G(P)/C_G(P)=1$. So $G$ has a normal subgroup $K$ with $|G/K|=|P|$. Then $K$ is $p_1$-nilpotent. Repeat this procedure and we get $G⊳K⊳K_1\cdots ⊳K_m$ where $K_m$ is a Sylow $q$-subgroup. So $G$ is solvable. \end{proof}

Corollary

If a Sylow $2$-group of $G$ is cyclic, then $G$ is $2$-nilpotent and by Feit-Thompson Theorem solvable $K⊲G$. Further $G$ is solvable.

Proposition

Let $G$ be a finite nonabelian simple group. Let $p$ be a prime divisor of $|G|$.

• If $p$ is the smallest prime divisor, then $|G|$ is divisible by $p^3$ or $12$.
• If $|G|=pm$ with $\gcd (p,m)=1$, then for a Sylow $p$-subgroup $P$,
  • $C_G(P)⪇N_G(P)<G$;
  • $|N_G(P)/C_G(P)|$ is divided by $p-1$.

\begin{proof} (1) Suppose $|G|$ is not divisible by $p^3$, then $p^2||G|$. Otherwise, a Sylow $p$-subgroup $P=C_p$ and $G$ is $P$-nilpotent, contradiction by $G$ simple. Thus $P$ is cyclic or $P\cong C_p\times C_p$. The first case is impossible. Hence $P\cong C_p\times C_p$ and $N_G(P)\ne C_G(P)$. Further $N_G(P)/C_G(P)\lesssim \mathrm{Aut}(P)={\mathrm{GL}}_2(p)$ and $|{\mathrm{GL}}_2(p)|=p(p-1{)}^2(p+1)$. Since $p$ is the smallest one, $p=2$ and $p+1=3$. It yields that $12||G|$.

(2) (i) is easy. For (ii), $N_G(P)/C_G(P)\lesssim \mathrm{Aut}(P)\cong C_{p-1}$. \end{proof}

Exercise. $|G|=5\times 7\times 13$. Prove that $G$ is cyclic.

\begin{proof} $G$ is $13$-nilpotent. $G=G_{13}\times G_{\{5,7\}}$. \end{proof}

Week 5 (done)

Lemma

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

Theorem

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

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

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

and

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

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}

Theorem

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

• If $H<G$, the $H<N_G(H)$.
• If $H{<}_{\max }G$, then $H⊲G$ and $|G|/|H|=p$.
• $G$ is solvable.
• $Z(G)>1$.
• If $N$ is a normal subgroup of $G$ and $|N|=p$, then $N\le Z(G)$.

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

Corollary

A group of order $p^2$ with prime $p$ is abelian.

Theorem

Let $|G|=p^3$. Then either $G$ is abelian, or:

• for $p=2$, $G=D_8$ or $Q_8$.
• for $p\ge 3$, $G={\mathbb{Z}}_{p^2}:{\mathbb{Z}}_p$ or $G=⟨a,b|a^p=b^p=c^p=1,[a,b]=c⟩$ where the exponent of $G$ equals to $p$.

\begin{proof} Assume $G$ is non-abelian. First, suppose $G$ has an element $a$ of order $p^2$. Then there exists $b\in G\backslash ⟨a⟩$ and $G=⟨a,b⟩$. If $|b|=p$, then $G=⟨a⟩:⟨b⟩$ as in (i) for $p=2$ and as in (ii) for $p\ge 3$.

Suppose $|b|=p^2$. Then $b^p=a^{\lambda }$ for some integer $\lambda$. If $p=2$, then $a^b=a^3=a^{-1}$ and $b^a=b^{-1}$. It deduces that $G=Q_8$. If $p$ is odd, then $a^b=a^{1+kp}$ for some $k$ with $1\le k\le p-1$. Claim $(b{a}^{-l}{)}^p=1$. Let $b^{\prime }=b{a}^{-l}$. Since the order of $b^{\prime }$ is $p$, we get back to the last case.

Assume $G$ does not have elements of order $p^2$. Then each of elements of $G$ is of order $p$. If $p=2$, then $G$ is abelian. Since $G/Z(G)$ is abelian, $|G^{\prime }|\le |Z(G)|=p$. Further $G^{\prime }=⟨c⟩={\mathbb{Z}}_p$ and $G/G^{\prime }=⟨a{G}^{\prime },b{G}^{\prime }⟩={\mathbb{Z}}_p\times {\mathbb{Z}}_p$, where $c=[a,b]$. \end{proof}

Remark

Define $G={\mathbb{Z}}_{p^2}:{\mathbb{Z}}_p$ as $p_{-}^{1+2}$ and $G=⟨a,b|a^p=b^p=c^p=1,[a,b]=c⟩$ as $p_{+}^{1+2}$.

Exercise. Prove:

• $\mathrm{Aut}(D_8)\cong D_8$
• $\mathrm{Aut}(Q_8)\cong S_4$
• $\mathrm{Aut}({\mathbb{Z}}_{p^2}:{\mathbb{Z}}_p)\cong P:{\mathbb{Z}}_{p-1}$
• $\mathrm{Aut}(p_{+}^{1+2})\cong \mathrm{GL}(2,p)$

open question. (…closed by zyz) automorphism group acts on G has three orbits.

Week 6 triangle group (done)

Definition

Define the triangle group $T_{m,n,l}$ as

$$X=⟨x,y||x|=m,|y|=n,|xy|=l⟩.$$

Consider $A:=(\frac{1}{m}+\frac{1}{n}+\frac{1}{l})\pi$.

Assume that $m\le n\le l$ and $G$ is nontrivial.

Case 1. $A>\pi$.

If $m=1$, then $X\cong {\mathbb{Z}}_n$.

If $m=2$ and $n=2$, then $X\cong D_{2l}$.

If $m=2$ and $n\ge 3$, then $n=3$ and $l\in \{3,4,5\}$, which corresponding $A_4,S_4,A_5$, respectively.

• [ ]

Case 2. $A=\pi$

All possible solution: $(3,3,3),(2,3,6),(2,4,4)$. They correspond 正三角形，正六边形，正方形. They are the only polygons filling the whole plane.

Those groups are infinity and solvable.

Case 3. $A<\pi$

These groups are infinity and non-solvable.

Week 8 Quasiprimitive (done)

Transclude of Finite-quasiprimitive-graphs#^pn888w

Holomorph simple (HS)

This part is mentioned here.

Proposition

Let $M\times N⊲G\le \mathrm{Sym}(\Omega )$ where $M\cong N\cong T$ is simple. Then $M:\mathrm{Inn}(M)\le G\le M:\mathrm{Aut}(M)$.

\begin{proof} Note that $M\times N$ can be identified by

$$M\times N=\hat{M}\times \check{M}=\hat{M}:\tilde{M}=M:\mathrm{Inn}(M).$$

By NC lemma, there is $N_G(M)/C_G(M)=G/N\le \mathrm{Aut}(M)$.

\end{proof}

Holomorph compound (HC)

This part is mentioned here.

$G$ is of HC type if $M\times N⊲G\le \mathrm{Sym}(\Omega )$ where $M\cong N=T^l$ with $l\ge 2$ and $T$ is nonabelian simple.

Proposition

If $G$ is of HC type, then

$$M:\tilde{M}<G\le M:\mathrm{Aut}(M)$$

where $\mathrm{Aut}(M)=\mathrm{Aut}(T)\wr S_l$.

Classification of quasiprimitive groups

See here: Classification of Quasiprimitive Groups

This section is mentioned in Quasiprimitive Group of Type HS, HC and HA are Primitive.

Lemma

A quasiprimitive permutation group of type HS, HC or HA is primitive.

\begin{proof} i) Let $G$ be of type HS. Let $G=T:H$ with $\mathrm{Inn}(T)\le H\le \mathrm{Aut}(T)$. Then $H=G_{\omega }$ where $\omega =1\in T$. It remains to show $H$ is a maximal subgroup. Otherwise, there is a $K$ such that $H<K\le G$. Then an element $g\in K\backslash H$ has the form $g=(t,h)\in G$. Thus $(t,1)=(t,h)(1,h^{-1})\in K$. For any element $x\in H$, we have $(1,x^{-1})\in K$ and $(1,x^{-1})(t,1)=(t^x,x^{-1})\in K$. Hence $(t^x,1)=(t^x,x^{-1})(1,x)\in K$ and so $M$ is a subgroup of $K$, where

$$M=⟨(t^x,1):x\in H⟩\le T\times \{1\}.$$

Since $H\ge \mathrm{Inn}(T)$, the action of $T$ on $M$ by conjugation fixes $M$. So $M⊲T$ and $M=T$. Thus $G=⟨M,H⟩\le K$. So $H$ is the maximal subgroup and so $G$ is primitive.

iii) Let $G$ be of type HA. Then $G=M:G_{\omega }$, where $M$ is an abelian unique minimal normal subgroup.

Suppose $G$ is imprimitive on $\Omega$. By Equivalent Condition of Primitivity, there is a subgroup $H$ such that $G_{\omega }<H<G$ and $1\ne H\cap M=M_0<M$. Then $M_0⊲M,H$ and hence $M_0⊲⟨M,H⟩=G$. Contradiction.

(iii) is mentioned here.

\end{proof}

For example, let $G$ be a nonabelian simple group. Then $\hat{G}$ is a quasiprimitive group on $G$ and is not primitive, because any point stabilizer is trivial.

Week 9 Coset action (done)

Let $G$ be a group and $H$ be a subgroup of $G$. Then there is a group action $G$ acting on $\Omega :=[G:H]$. Note that $G_{(\Omega )}={\mathrm{Core}}_G(H)$, we have $G/{\mathrm{Core}}_G(H)\cong G^{\Omega }$. If $H$ is core-free in $G$, then $G$ is faithful.

Definition

Let $G$ acts on $\Omega$ and $\Sigma$ transitively and faithfully. Then $G^{\Omega }$ and $G^{\Sigma }$ are said to be equivalent if there is a $1$-$1$ map $\phi$ from $\Omega$ to $\Sigma$, and an automorphism $\Phi \in \mathrm{Aut}(G)$ such that the following diagram commutes.

$$\text{require}AMScd\begin{array}{ccc}\omega & \stackrel{\phi }{\to }& {\omega }^{\phi }\\ g\downarrow & & \downarrow g^{\Phi }\\ {\omega }^g& \underset{\phi }{\to }& ({\omega }^g{)}^{g^{\Phi }}\end{array}$$

Theorem

$G^{\Omega }$ and $G^{\Sigma }$ are equivalent iff $G_{\omega }$ and $G_{\sigma }$ are conjugate in $G$.

Week 10: Classification of 2-transitive group

Exercise 13. Characterize quasiprimitive permutation group $G\le \mathrm{Sym}(\Omega )$ such that $\mathrm{rank}(G)\le 5$. If $G$ is of holomorph simple, then $r\ge ?$

Burnside

A $2$-transitive permutation group is affine or almost simple.

Proposition

Let $G\le \mathrm{Sym}(\Omega )$ be a $2$-transitive group. Then:

• $G$ is primitive.
• $G_{\omega }$ is faithful on $\Omega$ and transitive on $\Omega \setminus \{1\}$.
• If $N⊲G$, then $N_{\omega }$ is $\frac{1}{2}$-transitive.

Lemma

Let $G\le \mathrm{Sym}(\Omega )$ be a $2$-transitive group. Let $N⊲G$ such that $N=T^k$ with $k\ge 2$ and $T$ non-abelian simple.

Then $N$ does not have a proper normal subgroup which is regular on $\Omega$. (i.e., $G$ is not HS, HC, SD, CD).

\begin{proof} Suppose $M⊲N$ is regular on $\Omega$. Then $|\Omega |=|M|$ and $N_{\omega }\ne 1$ is half-transitive on $\Omega \setminus \{1\}$. Let $m$ be the size of an orbits of $N_{\omega }$ on $\Omega \setminus \{1\}$. Then $m\mid (|\Omega |-1)$. Also $m\mid |N_{\omega }|$ yields $m\mid \gcd (|\Omega |-1,|N_{\omega }|)$ and $m\mid \gcd (|\Omega |-1,|N_{\omega }|)=(|T{|}^l-1,|T^{k-l}|)=1$, because $N=M:N_{\omega }$. Thus $m=1$, which is impossible. \end{proof}

Lemma

$N$ is not regular on $\Omega$. Furthermore, $G$ is not TW.

\begin{proof} Suppose $N$ is regular on $\Omega$. Then $G=N{G}_{\omega }$. By $N$ regular, $N\cap G_{\omega }=\{1\}$ and so $G=N:G_{\omega }$. Then $\Omega$ can be identified by $G_{\omega }$ acting $N$ by conjugation. Notice that elements in the same orbits of $N$ have the same order. Clearly, $|N|$ has at least $3$ prime divisor $r,s,t$. Let $x,y,z\in N$ be such that $|x|=r$, $|y|=s$ and $|z|=t$. Then $x^{G_{\omega }}$, $y^{G_{\omega }}$ and $z^{G_{\omega }}$ are three orbits and so $rank(G)\ge 4$. Contradiction. \end{proof}

Example. If $N=T=A_5$ and $G_{\omega }=S_5$, then $(r,s,t)=(2,3,5)$ and $\mathrm{rank}(G)=4$, where $G=A_5:S_5$.

Let $G$ be $2$-transitive and let $N{⊲}_{\min }G$ such that $N=T^k$ does not have a normal subgroup which is regular on $\Omega$, with $k\ge 2$ and $T$ nonabelian simple.

• Then $\Omega ={\Delta }^k$ such that $G$ acting on $\Omega$ by $({\delta }_1,\cdots ,{\delta }_k{)}^{({\tau }_1,\cdots ,{\tau }_k)\pi }=({\delta }_{1^{{\pi }^{-1}}}^{{\tau }_{1^{{\pi }^{-1}}}},\cdots ,sth)$. We say $G$ acts on $\Omega$ in product action.

Then $N_{\omega }\ne 1$, otherwise $N$ is regular. Write $N$ as $N=T_1\times \cdots \times T_k$ where $T_1\cong \cdots \cong T_k$. Let $H_i$ be the projection of $N_{\omega }$ into $T_i$. Then $H_i=N_{\omega }{M}_i/M_i\cong N_{\omega }/(N_{\omega }\cap M_i)$, where $M_i={\Pi }_{j\ne i}{T}_j$ and $N/M_i\cong T_i$.

Then $H_i\lesssim T_i$ and $H_1\cong \cdots \cong H_k$. $G_{\omega }$ is transitive on $\{T_1,\cdots ,T_k\}$ and normalize $N_{\omega }$. Then $G_{\omega }$ is transitive on $\{H_1,\cdots ,H_k\}$.

Then $N_{\omega }\le K:=H_1\times \cdots \times H_k<N$ and $K$ is normalized by $G_{\omega }$. Thus $G_{\omega }\le K{G}_{\omega }<G$. Since $G$ is primitive, $G_{\omega }$ is maximal and so $G_{\omega }=K{G}_{\omega }$ and $K=N_{\omega }\le G_{\omega }$.

Consider $N=T_1\times \cdots \times T_k$ acting on $\Omega$. Identifying $\Omega$ with $[N:N_{\omega }]$,

$$\Omega =[N:N_{\omega }]=[(T_1\times \cdots \times T_k):(H_1\times \cdots \times H_k)]=[T_1:H_1]\times \cdots \times [T_k:H_k]$$

with $N$ acting on $\Omega$ defined by ${\Delta }_1\times \cdots \times {\Delta }_k$.

i.e. if $\omega =({\delta }_1,\cdots ,{\delta }_k)$, then ${\omega }^{(t_1,\cdots ,t_k)}=({\delta }_1^{t_1},\cdots ,{\delta }_k^{t_k})$.

Example.

Problem.

Week 11: Classification of Quasiprimitive to Classification of Quasiprimitive Rank $3$

Theorem

If $\mathrm{rank}(G)=3$ and $G$ is quasiprimitive on $\Omega$, then $G$ is HA, AS or $N=T^2$ and $G$ acts on $\Omega =\Delta \times \Delta$ in product action, where $G$ induces a $2$-transitive action on $\Delta$.

\begin{proof} By the following lemmas. \end{proof}

Lemma

Let $M⊲G$ be regular on $\Omega$. Then $\mathrm{rank}(G)⩾\#\text{mbox}fusionclassesofelementsofM$. In particular, if $M$ is nonsolvable, then $\mathrm{rank}(G)⩾4$.

\begin{proof} Identify $\Omega$ with $M$ and let $\omega =1$. Then $G=M:G_{\omega }$ and $G_{\omega }$ acts on $M$ by conjugation. Now the action of $G_{\omega }⩽\mathrm{Aut}(M)$ on $M$ partitions into orbits, whose number $⩾$ the number of fusion classes of $M$. Suppose that $M$ is non-solvable. Then by Burnside’s Theorem, there exist $x$, $y$, $z\in G$ such that $|x|=r$, $|y|=s$ and $|z|=t$. Since $x$, $y$, $z$ are in different fusion classes, we have $\mathrm{rank}(G)⩾4$. \end{proof}

Corollary

With the lemma above, $G$ is not of type HS, HC, TW.

Lemma

Let $G$ be a quasiprimitive group of type SD. Then $\mathrm{rank}(G)⩾4$.

\begin{proof} Let N=T^k\lhd_\min G. Write $N=T_1\times \cdots \times T_n$ where $T_i$ are isomorphic. Then $N_{\omega }=\{(t,\cdots ,t):t\in T\}\cong T$, and $R=T_1\times \cdots \times T_{n-1}\times \{1\}$ is regular on $\Omega$. Note that $\Omega =[N:N_{\omega }]$ and $N_{\omega }(t_1,\cdots ,t_n)\in [N:N_{\omega }]$ with a representation $(t_k^{-1}{t}_1,\cdots ,t_k^{-1}{t}_{k-1},1)\in R$. Observe that $N_{\omega }(s_1,\cdots ,s_{n-1},1)=N_{\omega }(t_1,\cdots ,t_{n-1},1)$ iff $(s_1,\cdots ,s_{n-1},1)=(t_1,\cdots ,t_{n-1},1)$.

Note that $G=N:G_{\omega }=(T_1\times \cdots \times T_k):G_{\omega }$, $G_{\omega }$ normalizes $N$ and induces an action on $\{T_1,\cdots ,T_k\}$ transitively by N\lhd_\min G (need proof). The stabilizer $G_{\omega }⩽\mathrm{Aut}(T)\times S_k$ and $G_{\omega }$ acts on $\{T_1,\cdots ,T_n\}$ by conjugation transitively and faithful ($C_G(N)⊲G$ and $C_G(N)\cap N=1$ yields $C_G(N)=1$). Each element of $G_{\omega }$ has the form $\sigma \pi$ with $\sigma \in \mathrm{Aut}(T)$ and $\pi \in S_k$.

Pick two elements $x:=(s,1,\cdots ,1)$ and $y:=(t,1,\cdots ,1)$ are two nontrivial elements of $R$. Then $x,y$ lie in the same orbits of $G_{\omega }$ iff there exists $\sigma \pi \in G_{\omega }$ such that $(s,1,\cdots ,1)=(t,1,\cdots ,1{)}^{\sigma \pi }=(t^{\sigma },1,\cdots ,1{)}^{\pi }$.

• If $1^{\pi }=1$, then $s=t^{\sigma }$ and so they are conjugate in $\mathrm{Aut}(T)$.
• If $1^{\pi }=k$, the $(s,1,\cdots ,1)=(1,\cdots ,1,t^{\sigma })$ where the latter is conjugate to $\left((t^{\sigma }{)}^{-1},\cdots ,(t^{\sigma }{)}^{-1},1\right)$. Then $(s,1,\cdots ,1)=\left((t^{\sigma }{)}^{-1},\cdots ,(t^{\sigma }{)}^{-1},1\right)$ and so $k=2$, i.e., $(s,1)=(t^{{\sigma }^{-1}},1)$ and so $s,t^{{\sigma }^{-1}}$ are conjugate by $\sigma \in \mathrm{Aut}(T)$.
• If $1^{\pi }\notin \{1,k\}$, then $(s,1,\cdots ,1)=(1,\cdots ,1,t^{\sigma },1,\cdots ,1)$, which is impossible.

Therefore, $(s,1,\cdots ,1)$ and $(t,1,\cdots ,1)$ lie in the same $G_{\omega }$-orbits iff

• $s=t^{\sigma }$, or
• $k=2$ and $s=(t^{-1}{)}^{\sigma }$, and we say $s$ and $k$ are inverse conjugate in $\mathrm{Aut}(T)$.

Thus $\mathrm{rank}(G)⩾\#\text{mbox}orbitsof\mathrm{Aut}(T)\text{mbox}on\left\{\{t,t^{-1}\}:t\in T\right\}⩾4$.

\end{proof}

Example. $G=(A_5\times A_5).(2\times S_2)$ and $G_{\omega }=S_5\times S_2=\mathrm{Aut}(T)\times S_2$. Then $\mathrm{rank}(G)=4$.

Let $T⩽\mathrm{Sym}(\Delta )$, where $\Delta =\{0,1,\cdots ,m-1\}$. For $k⩾2$, define $G⩽T\wr S_k$ and $\Omega ={\Delta }^k$ such that $G$ acts on $\Omega$ naturally. We say $T\wr S_k$ acting on $\Omega$ is a blow-up of $T$ on $\Omega$.

Lemma

Using the notation above, we have

• $\mathrm{rank}(G)⩾k+1$;
• $\mathrm{rank}(G)=3$ iff $k=2$ and $T$ is $2$-transitive on $\Delta$.

\begin{proof} (i) Fix ${\omega }_0=(0,\cdots ,0)$. For a point $\omega =({\delta }_1,\cdots ,{\delta }_k)\in \Omega$, define $\mathrm{weight}(\omega )=\#i\text{mbox}with{\delta }_i\ne 0$. Then if $\mathrm{weight}({\omega }_1)\ne \mathrm{weight}({\omega }_2)$, then ${\omega }_1^g\ne {\omega }_2$ for any $g\in G_{{\omega }_0}$. Since $g$ preserves the weight of each point $\omega$ and the weight function ranges $\{0,1,\cdots ,k\}$, group $G_{{\omega }_0}$ has at least $k+1$ orbits and $\mathrm{rank}(G)⩾k+1$.

(ii) Assume $\mathrm{rank}(G)=3$, then $k=2$ and $\Omega =\Delta \times \Delta$. The set of all points with weight $1$ forms an orbits of $G_{{\omega }_0}$. Thus $T_0$ is transitive on $\Delta \setminus \{0\}$ and so $T$ is $2$-transitive on $\Delta$. Conversely, assume that $k=2$ and $T$ is $2$-transitive on $\Delta$. Since $T_0$ is transitive on $\Delta \setminus \{0\}$, then the set of all points with weight $1$ is an orbit and so for weight $2$. \end{proof}

Corollary

CD is not rank $3$.

\begin{proof} Since CD is a blow up of SD, and SD is not $2$-transitive. So CD is not rank $3$ by the lemma above. \end{proof}

week 12: Examples of $2$-transitive Groups (done)

Almost Simple Type

Theorem

Let $V={\mathbb{F}}^d$ with $d>1$, and let $G=\mathrm{GL}(V)$, then $G$ acting on $V\setminus \{0\}$ is transitive.

• If $\mathbb{F}={\mathbb{F}}_2$, then the action is $2$-transitive.
• If $|F|>2$, then the action is imprimitive with a block system $\mathcal{B}=\left\{⟨v⟩:v\in V\setminus \{0\}\right\}$.
• If $|F|>2$, then the kernel of the action on $\mathcal{B}$ is $Z(G)$.
• $G/Z(G)\cong \mathrm{PGL}(V)$ is $2$-transitive on $\mathcal{B}$.
• $\mathrm{PGL}(V)$ on $\mathcal{B}$ is not $3$-transitive iff $d>2$.

(v) is mentioned here.

Remark. Suppose $|F|>2$. Then $\mathrm{GL}(V)$ acting on $V\setminus \{0\}$ is transitive and imprimitive, of rank $|F|$:

• $\{(v,\lambda v):\lambda \in F^{\times }\}$.
• $\{(v,w):v,w\text{mbox}arelinearlyindependent\}$.

Affine Type

Theorem

Affine group is $2$-transitive. If $G={\mathrm{AGL}}_d(p)$ is $3$-transitive, then $p=2$.

Sharply $2$-transitive

Definition

$G$ is said to be sharply $n$-transitive on $\Omega$ if for any $({\omega }_1,\cdots ,{\omega }_k)$ and $({\omega }_1^{\prime },\cdots ,{\omega }_k^{\prime })$, there is the unique $g\in G$ such that $({\omega }_1,\cdots ,{\omega }_k{)}^g=({\omega }_1^{\prime },\cdots ,{\omega }_k^{\prime })$.

Proposition

$G:=F^{+}:F^{\times }$ is sharply $2$-transitive on $F^{+}$.

week 13: Hall subgroups I (done)

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 $\gcd (|H|,|G/H|)=1$.

Examples.

• 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 $G={\mathrm{PSL}}_2(11)$. $G$ has two subgroups $H\cong A_4$ and $K\cong D_{12}$, which are Hall $\{2,3\}$-subgroups of $G$. Note that they are 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 assume that $|\pi |=t⩾3$, and assume inductively that Hall theorem holds for groups of which the order has at most $t-1$ prime divisors.

i) First, assume that $G$ has Hall subgroups for all $\pi \subseteq \pi (G)$. Let $H_{p_i^{\prime }}:=H_i$ be a Hall $\{p_i{\}}^{\prime }$-subgroup of $G$, for $1⩽i⩽t$. By induction hypothesis, $H_{p_i^{\prime }}$ is solvable. Let $p$ be a prime divisor of $|H_{p_1^{\prime }}|$ such that $O_p(H_{p_1^{\prime }})\ne 1$. The existence is guaranteed by its minimal normal subgroup is elementary abelian. Then $p$ divides $|H_2|$ or $|H_3|$, WLOG we 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, $O_p(H_1)⩽P^g⩽H_2^g$ for some $g\in G$ by Sylow Theorems. 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 $N=⟨O_p(H_1{)}^x:x\in G⟩$, the normal closure of $O_p(H_1)$ in $N$. Then $N⊲G$ and $N⩽H_2^g$ as $O_p(H_1{)}^x=O_p(H_1{)}^{x_1{x}_2}=O_p(H_1{)}^{x_2}⩽H_2^g$, where $x=x_1{x}_2$ with $x_1\in H$ and $x_2\in H_2^g$, i.e. $N$ is the proper normal subgroup of $G$. For a Hall $\pi$-subgroup of $H$ of $G$, we have $N\cap H$ 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, $N$ and $G/N$ are solvable, which yields that $G$ is solvable.

ii) Conversely, let $G$ be a solvable group. Let $M{⊲}_{\min }G$. Then $M\cong {\mathbb{Z}}_p^d$ for some prime $p$ and some positive integer $d$. Note that $\overline{G}:=G/M$ has order properly dividing $|G|$. Thus by induction hypothesis, $\overline{G}$ has Hall $\pi$-subgroups for all $\sigma \subseteq \pi (\overline{G})$. Take a subset $\pi \subseteq \pi (G)$. 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 $M.\overline{H}$, which is a Hall $\pi$-subgroup of $G$.

If $p\notin \pi$, then $\gcd (p,|\overline{H}|)=1$. Let $\overline{Q}=O_q(\overline{G})⊲\overline{G}$ where $q\in \pi (\overline{G})$ such that $\overline{Q}\ne \{1\}$. Then the full preimage of $\overline{Q}$ has the form $M.\overline{Q}$. Let $Q$ be the Sylow $q$-subgroup of $M.\overline{Q}$, then $M.\overline{Q}=M:Q⊲G$. Consider $N_G(Q)⩽G$.

Suppose $x\in N_G(Q)\cap M\ne 1$. For any $y\in Q$, $y^x\in Q$ and so $x^{-1}yx{y}^{-1}\in Q$. Also $x^{-1}(yx{y}^{-1})\in M$. Thus $x^{-1}yx{y}^{-1}\in Q\cap M=\{1\}$, and so $x\in C_M(Q)$. Thus $N_M(Q)=C_M(Q)$. Since C_M(Q)\leqslant M\lhd_\min G, we obtain that $M=C_M(Q)$ and hence $M:Q=M\times Q⊲G$. As $Q\mathrm{char}M\times Q⊲G$, we have $Q⊲G$. Let $X=G/Q$. Then $q\in \pi$ and $X$ has a Hall $\pi$-subgroup $\overline{K}$. So the full preimage of $\overline{K}$ under $G\to X=G/Q$ is a Hall $p^{\prime }$-subgroup of $G$, so for $\pi \subseteq p^{\prime }$.

Now suppose $N_G(Q)\cap M=1$. Claim that $N_G(Q)\cong \overline{G}\cong N_{\overline{G}}(\overline{Q})$. Let $\overline{g}\in \overline{G}$. Then a preimage $g\in G$ of $\overline{g}$ under $G\to G/M$ is such that $g^{-1}MQg=MQ=M(g^{-1}Qg)$. Thus $g^{-1}Qg\subseteq MQ$, namely, $g^{-1}Qg$ is a Sylow $q$-subgroup of $MQ$. So there exists $x\in M$ such that $x^{-1}(g^{-1}Qg)x=Q$ by Sylow Theorems. Thus $gx\in N_G(Q)$. Suppose $g{x}^{\prime }\in N_G(Q)$ with $x^{\prime }\in M$. Then $x^{-1}{x}^{\prime }\in M$. Also $x^{-1}{x}^{\prime }=(gx{)}^{-1}(g{x}^{\prime })\in N_G(Q)$ and it yields $x=x^{\prime }$. Define

$$\begin{array}{rl}\phi :N_G(Q)& \to N_{\overline{G}}(\overline{Q})=Q,\\ gx& \mapsto \overline{g}.\end{array}$$

It is easy to verify $\phi$ is a homomorphism and well-defined: Let ${\overline{g}}_1\ne {\overline{g}}_2\in \overline{G}$. Suppose $g_1{x}_1=g_2{x}_2$. Then $g_2^{-1}{g}_1=x_2{x}_1^{-1}\in M$ and hence ${\overline{g}}_2^{-1}{\overline{g}}_1=\overline{x_2{x}_1^{-1}}=\overline{1}$.

To show $\phi$ is an isomorphism, suppose $g_1{x}_1\ne g_2{x}_2$ and ${\overline{g}}_1={\overline{g}}_2$. Then $g_{1}M=g_{2}M$ and $g_1=g_{2}x$ for $x\in M$. Hope: $g_1{x}_1\in N_G(Q)$.

• to be continued. (from week 13)

---

week 14: Hall subgroups II (done)

Assume $G$ is solvable. Let $\pi \subseteq \pi (G)$ such that $|\pi |=|\pi (G)|-1$. Let $M{⊲}_{\min }G$. Then $M={\mathbb{Z}}_p^d$ with $p$ prime, 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 $H$ of $\overline{H}$ under $G\to \overline{G}$ is a Hall $\pi$-subgroup of $G$.

If $p\notin \pi$, then $p\nmid |\overline{H}|$. Define $X:=M.\overline{H}$ as the full preimage of $\overline{H}$ under $G\to \overline{G}$. We aim to find a subgroup of order $|\overline{H}|$.

Remark

If $X<G$, then by induction, $X$ has a Hall $\pi$-subgroup which is isomorphic to $\overline{H}$, and is a Hall $\pi$-subgroup of $G$.

Assume $X=G$. Let $\overline{Q}=O_q(\overline{H})\ne 1$. Then $M.\overline{Q}=M:Q⊲G$. Suppose $C_M(Q)\ne 1$. Then $C_M(Q)=Z(M:Q):=P\mathrm{char}M:Q⊲G$ by claim as follows. Then $C_M(Q)=M$ yields $M:Q=M\times Q$.

We claim that $C_M(Q)=Z(M:Q)$.(*) If $C_M(Q)<Z(M:Q)$, then suppose $Z(M:Q)=P\times Q_1$. Since $P\mathrm{char}(P\times Q_1)=Z(M:Q)\mathrm{char}M:Q⊲G$. Thus $P⊲G$. As P\leqslant M\lhd_\min G, then $P=M$ and so $M:Q=M\times Q$. Now $Q\mathrm{char}M\times Q=M:Q⊲G$ and so $Q⊲G$.

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 pf $G$.

Definition

Define $O_p(G)$ as the largest normal $p$-subgroup of $G$. Note that $O_p(G)$ is the intersection of all Sylow $p$-groups of $G$. Then:

• $O_p(G){⊲}_{\mathrm{char}}G$ and $\overline{G}:=G/O_p(G)$ is such that $O_p(\overline{G})=1$.
• If $q\ne p$, then $⟨O_p(G),O_q(G)⟩=O_p(G)\times O_q(G)$ is a characteristic subgroup of $G$
• If $\pi (G)=\{p_1,\cdots ,p_k\}$, then Fitting subgroup is defined by $O_{p_1}(G)\times \cdots \times O_{p_k}(G)$.

Let $q\in \pi (|\overline{H}|)$ such that $\overline{Q}=O_q(\overline{H})\ne 1$. The existence is implied as $\overline{H}$ is solvable (hint: consider minimal normal subgroup). Then $M.\overline{Q}=M:Q$, where $Q\cong \overline{Q}$ is a Sylow $q$-subgroup of $M.\overline{Q}$. And $M:Q⊲X=M.\overline{H}$ as $\overline{Q}⊲\overline{H}$.

Further, claim that $N_M(Q)=C_M(Q)$. Let $x\in N_M(Q)$. Then for any $y\in Q$, we have $xy{x}^{-1}{y}^{-1}\in M\cap Q=1$ and so $N_M(Q)⩽C_M(Q)$. Now we finish the proof.

Also claim that $N_X(Q)\cong N_{\overline{H}}(\overline{Q})=\overline{H}$ (as $\overline{Q}⊲\overline{H}$). Let $\overline{g}\in \overline{H}=N_{\overline{H}}(\overline{Q})$. Then there is a unique $x\in M$ such that $gx\in N_X(Q)$. Define

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

Suppose $g_1{x}_1=g_2{x}_2$. Then $\overline{g_2^{-1}{g}_1}=\overline{x_2{x}_1^{-1}}=\overline{1}$, so ${\overline{g}}_2={\overline{g}}_1$, i.e., $\phi$ is injective. Thus $|N_{\overline{H}}(\overline{Q})|⩽|N_X(Q)|$. And either $|\overline{H}|=|N_{\overline{H}}(\overline{Q})|=|N_X(Q)|$, or $|N_{\overline{H}}(\overline{Q})|<|N_X(Q)|$. For the former, $N_X(Q)$ is a Hall $\pi$-subgroup of $G$. Suppose now that $|N_{\overline{H}}(\overline{Q})|<|N_X(Q)|$. Then $N_M(Q)\ne 1$. Otherwise, $N_X(Q)\cap M=1$ and then

$$N_X(Q)=N_X(Q)/(N_X(Q)\cap M)\cong M{N}_X(Q)/M⩽X/M=\overline{H},$$

which is a contradiction. Therefore, $C_M(Q)=N_M(Q)\ne 1$ and by (*).

\end{proof}

Proposition

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 $p$. Let $H,K$ be Hall $\pi$-subgroups of $G$. If $p\notin \pi$, then $H,K⩽M$ and $H,K$ are conjugation by induction.

Assume $p\in \pi$. Then $H/(H\cap M)\cong Z_p\cong K/(K\cap M)$ and $H\cap M$, $K\cap M$ are Hall $\pi$-subgroups of $M$. By induction, there is a $x\in M$ such that $(H\cap M{)}^x=K\cap M$. Now $H^x$, $K$ are Hall $\pi$-subgroups of $G$ such that $H^x\cap K⩾K\cap M$. Thus $K\cap M⊲⟨H^x,K⟩$.

As $H^x/(H\cap M)\cong {\mathbb{Z}}_p$, then $H^x=⟨H\cap M,g⟩$ such that $g^p\in H\cap M<K$. So $g^p\in K$. Similarly, $K=⟨H\cap M,h⟩$ such that $h^p\in H\cap M⩽H^x$. Thus $K\cap M{⊲}_p{H}^x{⊲}_{1\text{mbox}orp}⟨H^x,K⟩=⟨H^x,H\cap M,h⟩=⟨H^x,h⟩{⊲}_{1\text{mbox}orp}{H}^x$.

If $H^x{⊲}_p⟨H^x,K⟩$, then $|⟨H^x,K⟩|=|H^x|p=|K\cap M|p^x=|K|p$, not possible as $K$ is a Hall $\pi$-subgroup of $G$ and $p\in \pi$. Thus $H^x=⟨H^x,K⟩$ and $K⩽H^x$, so $K=H^x$.

\end{proof}

week 15 (done)

Nilpotent Subgroup

Definition

A group $G$ is called nilpotent if it has a chain of subgroups, which is called a central series,

$$G=K_1>K_2>\cdots >K_{s+1}=1$$

such that $K_i⊲G$ and $K_i/K_{i+1}⩽Z(G/K_{i+1})$.

Example. $A_4$ and $S_4$ are not nilpotent.

Theorem

The followings are equivalent:

• $G$ is nilpotent.
• if $H<G$, then $N_G(H)>H$.
• each maximal subgroup of $G$ is normal and of prime index.
• $G$ is a direct product of its Sylow subgroups.

\begin{proof} Assume (i) holds. Then $G$ has a central series

$$G=K_1>K_2>\cdots >K_{s+1}=1.$$

Let $H<G$. Then there exists $i$ such that $H⩾K_{i+1}$ and $K\ngeqq K_i$. Then $H/K_{i+1}$ and $Z(G/K_{i+1})$ centralizes each other. As $K_i/K_{i+1}⩽Z(G/K_{i+1})$, for any $h\in H$ and $y\in K_i$, there is $(h{K}_{i+1})(y{K}_{i+1})=(y{K}_{i+1})(h{K}_{i+1})$ and so $hy=(y{c}_1)(h{c}_2)$ for some $c_1,c_2\in K_{i+1}$. We have

$$(y{c}_1)(h{c}_2)=yh{c}_1^h{c}_2=yh{c}_1^{\prime }{c}_2\in yh{K}_{i+1}.$$

So $y^{-1}hy=h{c}_1^{\prime }{c}_2\in H{K}_{i+1}=H$, i.e. $y\in N_G(H)$ and $K_i⩽N_G(H)$. Thus $N_G(H)⩾H{K}_i>H$, as in (ii).

Assume (ii) holds. Let $M$ be a maximal subgroup of $G$. Then $M⊲N_G(M)⩽G$ and so $M⊲G$. The factor group $G/M$ does not have a proper subgroup as $M$ is maximal. So $|G/M|$ is a prime, as in (iii).

Assume (iii) holds. Let $P\in {\mathrm{Syl}}_p(G)$. If $P$ is not normal in $G$, then $P<G$ and $P<N_G(P)<G$. Let $M$ be a maximal subgroup of $G$ such that $N_G(P)⩽M<G$. Then $M⊲G$. By Frattini’s argument, $G=N_G(P)M=M$, a contradiction.

Assume (iv) holds. Let $L_1=Z(G)$. Then $L_1\ne 1$, and ${\overline{G}}_1=G/L_1$ satisfies (iv). Let ${\overline{L}}_2=Z(\overline{G})$, and let $L_2$ be the full preimage of ${\overline{L}}_2$ under $G\to G/L_1$. Then $L_2⊲G$ by ${\overline{L}}_2⊲\overline{G}$, and $L_2/L_1={\overline{L}}_2=Z(G/L_1)$. Let ${\overline{L}}_3=Z(G/L_2)$, and let $L_3=L_2.{\overline{L}}_3$ be the full preimage of ${\overline{L}}_3$ under $G\to G/L_2$. Then $L_3⊲G$ and $L_3/L_2={\overline{L}}_3=Z(G/L_2)$. Repeat the procedure, we get the central series, as in (i). \end{proof}

Frattini Subgroup

It is mentioned here.

Definition

Let $G$ be a finite group. The Frattini subgroup of $G$, denoted by $\Phi (G)$, is the intersection of all maximal subgroups of $G$.

Observation.

• $\Phi (G){⊲}_{\mathrm{char}}G$
• If $N⊲G$, then $\Phi (N)⩽\Phi (G)$.

Definition

An element $x\in G$ is called a non-generator if $⟨x,S⟩=G$, then $⟨S⟩=G$ for any $S\subseteq G$.

Example.

• Let $G=C_{p^2}$. Then for any $a\in G$, $a^p$ is a non-generator.
• $Q_8$ has non-generator s of order $1$ or $2$.
• The set of non-generators of $D_{2n}$ is $\{a^m,a^{m}b:(m,n)\ne 1\}$.
• Each non-identity element of a non-abelian simple group is not a non-generator element. In fact, each non-abelian simple group can be generated by “1.5” elements. That is, for a non-abelian simple group $G$ and any $1\ne g\in G$, there exists a $h\in G$ such that $G=⟨g,h⟩$.

Lemma

$\Phi (G)$ is the set of non-generators of $G$.

Link to original

Proof is omitted.

Gaschutz

Let $D,N⊲G$ such that $D⩽\Phi (G)\cap N$. Then $N/D$ is nilpotent iff $N$ is nilpotent.

\begin{proof} Assume that $N/D$ is nilpotent. Let $P$ be a Sylow $p$-subgroup of $N$. Then $PD/D$ is a Sylow $p$-subgroup of $N/D$, and

$$PD/D{⊲}_{\mathrm{char}}N/D⊲G/D.$$

Hence, $PD/D⊲G/D$ and $PD⊲G$. By Frattini’s argument, we have $N=PD{N}_N(P)=P{N}_N(P)=N_N(P)$, so $P⊲N$. By ^65aef6 (iv), $N$ is nilpotent. \end{proof}

Corollary

• $\Phi (G)$ is nilpotent.
• $G$ is nilpotent iff $G/\Phi (G)$ is nilpotent.
• (Wielandt) $G$ is nilpotent iff $G^{\prime }⩽\Phi (G)$.
• If $G$ is a $p$-group, then $G/\Phi (G)$ is elementary abelian.

\begin{proof} (i) Taking $N=D=\Phi (G)$. Then by ^wigmsn $N$ is nilpotent.

(ii) Let $N=G$ and $D=\Phi (G)$. Then $N/D=G/\Phi (G)$ is nilpotent iff $N=G$ is nilpotent, by ^wigmsn.

(iii) Suppose $G$ is nilpotent. Let $M$ be the maximal subgroup of $G$. Then $M⊲G$ and $G/M\cong {\mathbb{Z}}_p$ is abelian. Thus $M⩾G^{\prime }$. By the arbitrary of $M$, we have $G^{\prime }⩽\Phi (G)$. Conversely, assume that $G^{\prime }⩽\Phi (G)$. Then $G/\Phi (G)$ is abelian and so is nilpotent. Then by ^a15hbm (ii) $G$ is nilpotent.

(iv) Since $G$ is a $p$-group, then $G$ is nilpotent. Thus $\Phi (G)⩾G^{\prime }$ and so $G/\Phi (G)$ is abelian. If there is a $\overline{g}\in G/\Phi (G)$ with $|\overline{g}|$ is not a prime, then ${\overline{g}}^a$ is a non-generator with $|\overline{g}|=ab$, which is impossible. Thus $G/\Phi (G)$ is elementary abelian. \end{proof}

Exercise. There is a easy exercise. Pasted image 20240529212723.png

Fitting Subgroup

It is mentioned here.

Definition

The product of all normal nilpotent subgroups of $G$ is called the Fitting subgroup of $G$, denoted by $\mathrm{F}(G)$.

Note that $\Phi (G)$ is a normal nilpotent subgroup. So $\Phi (G)⩽\mathrm{F}(G)$.

Theorem

Let $G$ be a solvable group, and let $F=\mathrm{F}(G)$. Then:

• $F=O_{p_1}(G)\times \cdots \times O_{p_k}(G)$.
• $C_G(F)⩽F$, i.e., $F$ is self-centralizing. Moreover, we have $F/C_G(F)⩽\mathrm{Aut}(F)$.

\begin{proof} (i) Let $N$ be a nilpotent normal subgroup of $G$. Then $N=Q_1\times \cdots \times Q_t$ where $Q_i$ are Sylow subgroups. Now $Q_i{⊲}_{\mathrm{char}}N⊲G$ and so $Q_i⊲G$. So $Q_i⩽O_{q_i}(G)$ and so $N⩽O_{p_1}(G)\times \cdots \times O_{p_k}(G)$.

(ii) Let $C=C_G(F)$. Then $C\cap F=Z(F)$. Since $F$, $C$, $CF$ and $C\cap F$ are characteristic subgroups of $G$. Let $W=C\cap F$ and $\overline{G}=G/W$. Suppose $\overline{C}\ne 1$, then there exists $p$ such that $O_p(\overline{C})\ne 1$, and $1\ne O_p(\overline{C}){⊲}_{\mathrm{char}}\overline{G}$. Hence the full preimage $Y:=W.O_p(\overline{C}){⊲}_{\mathrm{char}}G$. Let $P$ be a Sylow subgroup of $Y$. Then $W_p<P⩽C$. By definition, $P$ centralizes $w_{p^{\prime }}$. So $Y=P\times W_{p^{\prime }}$. Then $P{⊲}_{\mathrm{char}}Y{⊲}_{\mathrm{char}}G$ and $P⊲G$. So that $w_p<P⩽F$

\end{proof}

Example. Let $G$ be a group such that each Sylow subgroup is cyclic. Then $G$ is solvable. Further, $G$ is metacyclic, i.e., $G/C$ is cyclic and $C$ is cyclic.

\begin{proof} Let $F=\mathrm{F}(G)$. Then $F$ is cyclic, and $G/C_G(F)\lesssim \mathrm{Aut}(F)$ is abelian. Since $F$ is abelian, $C_G(F)=F$ and $G/F$ is abelian and cyclic. \end{proof}

HW: Pasted image 20240529220050.png