GTM148 抄书笔记 Part II. (Chapter IV~V)

上一篇: GTM148 抄书笔记 Part I.

■■■■■■■■■■■■■■■■■■■■■■■■■

Contents

• Contents
• Chapter IV. The Sylow Theorems
  • $p$-Groups
  • The Sylow Theorems
• Chapter V. Normal Series
  • Some Galois Theory
  • The Jordan-Hölder Theorem
  • Solvable Groups
  • Two Theorems of P.Hall
  • Central Series and Nilpotent Groups
  • Frattini Subgroup

Chapter IV. The Sylow Theorems

$p$-Groups

Definition 4.1.1 If $p$ is a prime, then a $p$-group is a group in which every element has order a power of $p$.

Theorem 4.1.2 If $G$ is a finite abelian group whose order is divisible by a prime $p$, then $G$ contains an element of order $p$.

Proof

Write $|G|=pm$, where $m\ge 1$. We proceed by induction on $m$ after noting that the base step is clearly true. For the inductive step, choose $x\in G$ of order $t>1$. If $p|t$, then Proposition 2.2.14 shows that $x^{t/p}$ has order $p$, and the lemma is proved. We may, therefore, assume that the order of $x$ is not divisible by $p$. Since $G$ is abelian, $\langle x\rangle$ is a normal subgroup of $G$, and $G/\langle x\rangle$ is an abelian group of order $|G|/t=pm/t$. Since $p\nmid t$, we must have $m/t<m$ an integer. By induction, $G/\langle x\rangle$ contains an element $y^*$ of order $p$. But the natural map $\nu\!:G\rightarrow G/\langle x\rangle$ is a surjection, and so there is $y\in G$ with $\nu(y)=y^*$. Thus, the order of $y$ is a multiple of $p$, and we have returned to the first case.
■

Definition 4.1.3 Let $G$ be a group. Partition $G$ into its conjugacy classes and count (recall that $\mathrm Z(G)$ consists of all the elements of $G$ whose conjugacy class has just one element):

$$|G|=|\mathrm Z(G)|+\sum_{i}[G:\mathrm C_G(x_i)],$$

where one $x_i$ is selected from each conjugacy class with more than one element.
The equation above is called the class equation of the finite group $G$.

Theorem 4.1.4 (Cauchy) If $G$ is a finite group whose order is divisible by a prime $p$, then $G$ contains an element of order $p$.

Proof

If $x\in G$, then the number of conjugates of $x$ is $[G:\mathrm C_G(x)]$, where $\mathrm C_G(x)$ is the centralizer of $x$ in $G$. If $x\notin \mathrm Z(G)$, then its conjugacy class has more than one element, and so $|\mathrm C_G(x)|< |G|$. If $p\big||\mathrm C_G(x)|$ for such a noncentral $x$, we are done, by induction. Therefore, we may assume that $p\nmid|\mathrm C_G(x)|$ for all noncentral $x$ in $G$. Better, since $|G|=[G:\mathrm C_G(x)]|\mathrm C_G(x)|$, we may assume that $p\big|[G:\mathrm C_G(x)]$.

Consider the class equation $|G|=|\mathrm Z(G)|+\sum_{i}[G:\mathrm C_G(x_i)]$. Since $|G|$ and all $[G:\mathrm C_G(x_i)]$ are divisible by $p$, it follows that $|\mathrm Z(G)|$ is divisible by $p$. But $\mathrm Z(G)$ is abelian, and so it contains an element of order $p$, by Lemma 4.2.2.
■

Corollary 4.1.5 A finite group $G$ is a $p$-group if and only if $|G|$ is a power of $p$.

Proof

If $|G|=p^m$, then Lagrange's theorem shows that $G$ is a $p$-group. Conversely, assume that there is a prime $q\neq p$ which divides $|G|$. By Cauchy's theorem, $G$ contains an element of order $q$, and this contradicts $G$ being a $p$-group.
■

Theorem 4.1.6 If $G\neq \mathbf 1$ is a finite $p$-group, then its center $\mathrm Z(G)\neq \mathbf 1$.

Proof

Consider the class equation $|G|=|\mathrm Z(G)|+\sum_{i}[G:\mathrm C_G(x_i)]$. Each $\mathrm C_G(x_i)$ is a proper subgroup of $G$, for $x_i\notin\mathrm Z(G)$. By Corollary 4.1.5, $[G:\mathrm C_G(x_i)]$ is a power of $p$. Thus, $p$ divides each $[G:\mathrm C_G(x_i)]$, and so $p$ divides $|\mathrm Z(G)|$.
■

Corollary 4.1.7 If $p$ is a prime, then every group $G$ of order $p^2$ is abelian.

Proof

If $G$ is not abelian, then $\mathrm Z(G)<G$; since $\mathbf 1\neq\mathrm Z(G)$, we must have $|\mathrm Z(G)|=p$. The quotient group $G/\mathrm Z(G)$ is defined, since $\mathrm Z(G)\lhd G$, and it is cyclic, because $|G/\mathrm Z(G)|=p$; this contradicts Proposition 3.1.19.
■

Theorem 4.1.8 Let $G$ be a finite $p$-group.

1. If $H$ is a proper subgroup of $G$, then $H< \mathrm N_G(H)$.
2. Every maximal subgroup of $G$ is normal and has index $p$.

Proof

1. If $H\lhd G$, then $\mathrm N_G(H)=G$ and the theorem is true. If $X$ is the set of all the conjugates of $H$, then we may assume that $|X|=[G:\mathrm N_G(H)]\neq 1$. Now $G$ acts on $X$ by conjugation and, since $G$ is a $p$-group, every orbit of $X$ has size a power of $p$. As $\{H\}$ is an orbit of size $1$, there must be at least $p-1$ other orbits of size $1$. Thus there is at least one conjugate $gHg^{-1}\neq H$ with $\{gHg^{-1}\}$ also an orbit of size $1$. Now $agHg^{-1}a^{-1}=gHg^{-1}$ for all $a\in H$, and so $g^{-1}ag\in \mathrm N_G(H)$ for all $a\in H$. But $gHg^{-1}\neq H$ gives at least one $a\in H$ with $g^{-1}ag\notin H$, and so $H< \mathrm N_G(H)$.
2. If $H$ is a maximal subgroup of $G$, then $H<\mathrm N_G(H)$ implies that $\mathrm N_G(H)=H$; that is, $H\lhd G$. By Proposition 2.7.4, $[G:H]=p$.
   ■

Lemma 4.1.9 If $G$ is a finite $p$-group and $r_1$ is the number of subgroups of $G$ having order $p$, then $r_1\equiv 1\pmod p$.

Proof

Let us first count the number of elements of order $p$. Since $\mathrm Z(G)$ is abelian, all its elements of order $p$ together with $\mathbf 1$ form a subgroup $H$ whose order is a power of $p$; hence the number of central elements of order $p$ is $|H|-1\equiv -1\pmod p$. If $x\in G$ is of order $p$ and not central, then its conjugacy class $x^G$ consists of several elements of order $p$; that is, $|x^G|>1$ is an "honest" power of $p$. It follows that the number of elements in $G$ of order $p$ is congruent to $-1\!\!\mod p$; say, there are $mp-1$ such elements. Since the intersection of any distinct pair of subgroups of order $p$ is trivial, the number of elements of order $p$ is $r_1(p-1)$. But $r_1(p-1)=mp-1$ implies $r_1\equiv 1\pmod p$.
■

Theorem 4.1.10 If $G$ is a finite $p$-group and $r_s$ is the number of subgroups of $G$ having order $p^s$, then $r_s\equiv 1\pmod p$.

Proof

Let $H$ be a subgroup of order $p^s$, and let $K_1,\ldots,K_a$ be the subgroups of $G$ of order $p^{s+1}$ which contain it; we claim that $a\equiv 1\pmod p$. Every subgroup of $G$ which normalizes $H$ is contained in $N=\mathrm N_G(H)$; in particular, each $K_j$ lies in $N$, for Theorem 4.1.8(ii) shows that $H\lhd K$ for all $j$. By the Correspondence Theorem, the number of subgroups of order $p$ in $N/H$ is equal to the number of subgroups of $N$ containing $H$ which have order $p^{s+1}$. By Lemmma 4.1.9, $a\equiv 1\pmod p$.

Now let $K$ be a subgroup of order $p^{s+1}$, and let $H_1,\ldots,H_b$ be its subgroups of order $p^s$; we claim that $b\equiv 1\pmod p$. By Theorem 4.1.8(ii), $H_i\lhd K$ for all $i$. Since $H_1H_2=K$ (for the $H_i$ are maximal subgroups of $K$), the Product Formula (Theorem 2.4.2) gives $|D|=p^{s-1}$, where $D=H_1\cap H_2$, and $[K:D]=p^2$. By Corollary 4.1.7, the group $K/D$ is abelian; moreover, $K/D$ is generated by two subgroups of order $p$, namely, $H_i/D$ for $i=1,2$, and so it is not cyclic. Thus $K/D\cong \mathbb Z_p\times \mathbb Z_p$. Therefore, $K/D$ has $p^2-1$ elements of order $p$ and hence has $p+1=(p^2-1)/(p-1)$ subgroups of order $p$. The Correspondence Theorem gives $p+1$ subgroups of $K$ of order $p^s$ containing $D$. Suppose there is some $H_j$ with $D\nleq H_j$. Let $E=H_1\cap H_j$; as above, there is a new list of $p+1$ subgroups of $K$ of order $p^s$ containing $E$, one of which is $H_1$. Indeed, $H_1=ED$ is the only subgroup on both lists. Therefore, there are $p$ new subgroups and $1+2p$ subgroups counted so far. If some $H_l$ has not yet been listed, repeat this procedure beginning with $H_1\cap H_l$ to obtain $p$ new subgroups. Eventually, all the $H_i$ will be listed, and so the number of them is $b=1+mp$ for some $m$. Hence, $b\equiv 1\pmod p$.

Let $H_1,\ldots,H_{r_s}$ be all the subgroups of $G$ of order $p^s$, and let $K_1,\ldots,K_{r_{s+1}}$ be all the subgroups of order $p^{s+1}$. For each $H_i$, let there be $a_i$ subgroups of order $p^{s+1}$ containing $H_i$; for each $K_j$, let there be $b_j$ subgroups of order $p^s$ contained in $K_j$. Now

$$\sum_{i=1}^{r_s}a_i=\sum_{j=1}^{r_{s+1}}b_j,$$

for either sum counts each $K_j$ with multiplicity the number of $H$'s it contains. Since $a_i\equiv 1\pmod p$ for all $i$ and $b_j\equiv 1\pmod p$ for all $j$, it follows that $r_s\equiv r_{s+1}\pmod p$. Lemma 4.1.9 now gives the result, for $r_1\equiv 1\pmod p$.
■

Lemma 4.1.11 (Landau) Given $n>0$ and $q\in\mathbb Q$, there are only finitely many $n$-tuples $(i_1,\ldots,i_n)$ of positive integers such that $q=\sum_{j=1}^n(1/i_j)$.

Proof

We do an induction on $n$; the base step $n=1$ is obviously true. Since there are only $n!$ permutations of $n$ objects, it suffices to prove that there are only finitely many $n$-tuples $(i_1,\ldots,i_n)$ with $i_1\le i_2\le \cdots\le i_n$ which satisfy the equation $q=\sum_{j=1}^n(1/i_j)$. For any such $n$-tuple, we have $i_1\le n/q$, for

$$q=1/i_1+\cdots+1/i_n\le 1/i_1+\cdots+1/i_1=n/i_1,$$

But for each positive integer $k\le n/q$, induction gives only finitely many $(n-1)$-tuples $(i_2,\ldots,i_n)$ of positive integers with $q-(1/k)=\sum_{j=2}^n(1/i_j)$. This completes the proof, for there are only finitely many such $k$.
■

Theorem 4.1.12 For every $n\ge 1$, there are only finitely many finite groups having exactly $n$ conjugacy classes.

Proof

Assume that $G$ is a finite group having exactly $n$ conjugacy classes. If $|\mathrm Z(G)|=m$, then the class equation is

$$|G|=|\mathrm Z(G)|+\sum_{j=m+1}^n[G:\mathrm C_G(x_j)].$$

If $i_j=|G|$ for $1\le j\le m$ and $i_j=|G|/[G:\mathrm C_G(x_j)]=|\mathrm C_G(x_j)|$ for $m+1\le j\le n$, then $1=\sum_{j=1}^n(1/i_j)$. By Lemma 4.1.11, there are only finitely many such $n$-tuples, and so there is a maximum value for all possible $i_j$'s occurring therein, say, $M$. It follows that a finite group $G$ having exactly $n$ conjugacy classes has order at most $M$. But there are only finitely many nonisomorphic groups of any given order.
■

Proposition 4.1.13 Let $G$ be a finite $p$-group, and $|G|=p^n$, where $n\in \mathbb N^*$.

1. For any $0\le k< n$, there exists a normal subgroup of order $p^k$ of $G$.
2. Let $H$ be a nontrivial normal subgroup of $G$, then $H\cap \mathrm Z(G)\neq\mathbf 1$; moreover, if $|H|=p$, then $H\le \mathrm Z(G)$.
3. Let $H$ be a proper subgroup of $G$. If $|H|=p^{s}$, then there is a subgroup of order $p^{s+1}$ containing $H$.

Proposition 4.1.14 The number of normal subgroups of order $p^s$ of a finite $p$-group $G$ is congruent to $1$ mod $p$.

Proposition 4.1.15 Let $p$ be a prime,

1. if $G$ is an abelian group of order $p^2$, then $G$ is either cyclic or isomorphic to $\mathbb Z_p\times \mathbb Z_p$.
2. if $G$ is a nonabelian group of order $p^3$, then $|\mathrm Z(G)|=p,\,G/\mathrm Z(G)\cong \mathbb Z_p\times \mathbb Z_p$, and $\mathrm Z(G)=G'$, the commutator subgroup.

Definition 4.1.16 If $p$ is a prime, then an elementary abelian $p$-group is a finite group $G$ isomorphic to $\mathbb Z_p\times\cdots\times\mathbb Z_p$.

Proposition 4.1.17

1. If $G$ is an abelian group of prime exponent $p$, then $G$ is a vector space over $\mathbb Z_p$, and every homomorphism $\varphi\!:G\rightarrow G$ is a linear transformation.
2. A finite abelian $p$-group $G$ is elementary if and only if it has exponent $p$.

The Sylow Theorems

Definition 4.2.1 If $p$ is a prime, then a Sylow $p$-subgroup $P$ of a group $G$ is a maximal $p$-subgroup.

Observe that every $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup; this is obvious when $G$ is finite, and it follows from Zorn's lemma when $G$ is infinite.

Lemma 4.2.2 Let $P$ be a Sylow $p$-subgroup of a finite group $G$.

1. $|\mathrm N_G(P)/P|$ is prime to $p$.
2. If $a\in G$ has order some power $p$ and $aPa^{-1}=P$, then $a\in P$.

Proof

1. If $p$ divides $|\mathrm N_G(P)/P|$, then Cauchy's theorem shows that $\mathrm N_G(P)/P$ contains some element $Pa$ of order $p$; hence, $S^*=\langle Pa\rangle$ has order $p$. By the Correspondence Theorem, there is a subgroup $S\le \mathrm N_G(P)\le G$ containing $P$ with $S/P\cong S^*$. Since both $P$ and $S^*$ are $p$-groups, then $S$ is a $p$-group, contradicting the maximality of $P$.
2. Replacing $a$ by a suitable power of $a$ if necessary, we may assume that $a$ has order $p$. Since $a$ normalizes $P$, we have $a\in \mathrm N_G(P)$. If $a\notin P$, then the coset $Pa\in \mathrm N_G(P)/P$ has order $p$, and this contradicts (i).
   ■

Theorem 4.2.3 (Sylow)

1. If $P$ is a Sylow $p$-subgroup of a finite group $G$, then all Sylow $p$-subgroups of $G$ are conjugate to $P$.
2. If there are $r$ Sylow $p$-subgroups, then $r$ is a divisor of $|G|$ and $r\equiv 1\pmod p$.

Proof

Let $X=\{P_1,\ldots,P_r\}$ be the family of all the conjugates of $P$, where we have denoted $P$ by $P_1$. In Theorem 3.4.7, we saw that $G$ acts on $X$ by conjugation: there is a homomorphism $\psi\!:G\rightarrow\mathrm S_X$ sending $a\mapsto \psi_a$, where $\psi_a(P_i)=aP_ia^{-1}$. Let $Q$ be a Sylow $p$-subgroup of $G$. Restricting $\psi$ to $Q$ shows that $Q$ acts on $X$, and we have that every orbit of $X$ under this action has size dividing $|Q|$; that is, every orbit has size some power of $p$. If one of these orbits has size $1$, then there would be an $i$ with $\psi_a(P_i)=P_i$ for all $a\in Q$; that is, $aP_ia^{-1}=P_i$ for all $a\in Q$. By Lemma 4.2.2(ii) if $a\in Q$, then $a\in P_i$; that is, $Q\le P_i$; since $Q$ is a Sylow $p$-subgroup, $Q=P_i$. If $Q=P=P_1$, we conclude that every $P$-orbit of $X$ has size an "honest" power of $p$ except $\{P_1\}$ which has size $1$. Therefore, $|X|=r\equiv 1\pmod p$.

Suppose there were a Sylow $p$-subgroup $Q$ that is not a conjugate of $P$; that is, $Q\notin X$. If $\{P_i\}$ is a $Q$-orbit of size $1$, then we have seen that $Q=P_i$, contradicting $Q\in X$. Thus, every $Q$-orbit of $X$ has size an honest power of $p$, and so $p$ divides $|X|$; that is, $r\equiv 0\pmod p$. The previous congruence is contradicted, and so no such subgroup $Q$ exists. Therefore, every Sylow $p$-subgroup $Q$ is conjugate to $P$.

Finally, the number $r$ of conjugates of $P$ is the index of its normalizer, and so it is a divisor of $|G|$.
■

Corollary 4.2.4 A finite group $G$ has a unique Sylow $p$-subgroup $P$, for some prime $p$, if and only if $P\lhd G$.

Theorem 4.2.5 If $G$ is a finite group of order $p^em$, where $(p,m)=1$, then every Sylow $p$-subgroup $P$ of $G$ has order $p^e$.

Proof

We claim that $[G:P]$ is prime to $p$. Now $[G:P]=[G:N][N:P]$, where $N=\mathrm N_G(P)$, and so it suffices to prove that each of the factors is prime to $p$. But $[G:N]=r$, the number of conjugates of $p$, so that $[G:N]\equiv 1\pmod p$, while $[N:P]=|N/P|$ is prime to $p$, by Lemma 4.2.2(i). Thus by Lagrange's theorem, we have $|P|=p^e$.
■

Corollary 4.2.6 Let $G$ be a finite group and let $p$ be a prime. If $p^k$ divides $|G|$, then $G$ contains a subgroup of order $p^k$.

Lemma 4.2.7 If $p$ is a prime not dividing an integer $m$, then for all $n\ge 1$, the binomial coefficient $\binom{p^nm}{p^n}$ is not divisible by $p$.

Proof

Write the binomial coefficient as follows:

$$\dfrac{p^nm(p^nm-1)\cdots(p^nm-i)\cdots(p^nm-p^n+1)}{p^n(p^n-1)\cdots(p^n-i)\cdots(p^n-p^n+1)}$$

Since $p$ is prime, each factor equal to $p$ of the numerator (or of the denominator) arises from a factor of $p^nm-i$ (or of $p^n-i$). If $i=0$, then the multiplicity of $p$ in $p^nm$ and in $p^n$ are the same because $p\nmid m$. If $1\le i\le p^n$, then $i=p^kj$, where $0\le k< n$ and $p\nmid j$. Now $p^k$ is the highest power of $p$ dividing $p^n-i$, for $p^n-i=p^n-p^kj=p^k(p^{n-k}-j)$ and $p\nmid p^{n-k}-j$. A similar argument shows that the highest power of $p$ divideing $p^nm-i$ is also $p^k$. Therefore, every factor of $p$ upstairs is canceled by a factor of $p$ downstairs, and hence the binomial coefficient has no factor equal to $p$.
■

Theorem 4.2.8 (Wielandt's Proof) If $G$ is a finite group of order $p^nm$, where $(p,m)=1$, then $G$ has a subgroup of order $p^n$.

Proof

If $X$ is the family of all subsets of $G$ of cardinal $p^n$, then $|X|$ is the binomial coefficient in the lemma, and so $p\nmid |X|$. Let $G$ act on $X$ by left translation: if $B$ is a subset of $G$ with $p^n$ elements, then for each $g\in G$, define $gB=\{gb|b\in B\}$. Now $p$ cannot divide the size of every orbit of $X$ lest $p\big| |X|$; therefore, there is some $B\in X$ with $|\mathcal O(B)|$ not divisible by $p$, where $\mathcal O(B)$ is the orbit of $B$. If $G_B$ is the stabilizer of $B$, then $|G|/|G_B|=[G:G_B]=|\mathcal O(B)|$ is prime to $p$. Hence, $|G_B|=p^nm'\ge p^n$ (for some $m'$ dividing $m$). On the other hand, if $b_0\in B$ and $g\in G_B$, then $gb_0\in gB=B$; moreover, if $g$ and $h$ are distinct elements of $G_B$, then $gb_0$ and $hb_0$ are distinct elements of $B$. Therefore, $|G_B|\le |B|=p^n$, and so $|G_B|=p^n$.
■

Theorem 4.2.9 (Frattini Argument) Let $K$ be a normal subgroup of a finite group $G$. If $P$ is a Sylow $p$-subgroup of $K$ (for some prime $p$), then

$$G=K\mathrm N_G(P).$$

Proof

If $g\in G$, then $gPg^{-1}\le gKg^{-1}=K$, because $K\lhd G$. It follows that $gPg^{-1}$ is a Sylow $p$-subgroup of $K$, and so there exists $k\in K$ with $kPk^{-1}=gPg^{-1}$. Hence, $P=(k^{-1}g)P(k^{-1}g)^{-1}$, so that $k^{-1}g\in \mathrm N_G(P)$. The required factorization is thus $g=k(k^{-1}g)$.
■

Proposition 4.2.10 If $Q$ is a normal $p$-subgroup of a finite group $G$, then $Q\le P$ for every Sylow $p$-subgroup $P$.

Proposition 4.2.11 Let $|G|=p^nm$, where $p\nmid m$. If $s\le n$ and $r_s$ is the number of subgroups of $G$ of order $p^s$, then $r_s\equiv 1\pmod p$.

Proposition 4.2.12 Let $X$ be a finite $G$-set, and let $H\le G$ act transitively on $X$. Then $G=HG_x$ for each $x\in X$.

Proposition 4.2.13 Let $P\le G$ be a Sylow subgroup. If $\mathrm N_G(P)\le H\le G$, then $H$ is equal to its own normalizer; that is, $H=\mathrm N_G(H)$.

Proposition 4.2.14 Let $G$ be a finite group and let $P\le G$ be a Sylow subgroup. If $H\lhd G$, then $H\cap P$ is a Sylow subgroup of $H$ and $HP/H$ is a Sylow subgroup of $G/H$.

Theorem 4.2.15 Let $|G|=pq$, where $p>q$ are primes. Then either $G$ is cyclic or $G=\langle a,b\rangle$, where $b^p=a^q=\mathbf 1,aba^{-1}=b^m$, and $m^q\equiv 1\pmod p$ but $m\not\equiv 1\pmod p$. If $q\nmid p-1$, then the second case cannot occur.

Proof

By Cauchy's theorem, $G$ contains an element $b$ of order $p$: let $S=\langle b\rangle$. Since $S$ has order $p$, it has index $q$. It follows from Proposition 3.4.15 that $S\lhd G$. Cauchy's theorem shows that $G$ contains an element $a$ of order $q$; let $T=\langle a\rangle$. Now $T$ is a Sylow $q$-subgroup of $G$, so that the number $c$ of its conjugates is $1+kq$ for some $k\ge 0$. As above, either $c=1$ or $c=p$. If $c=1$, then $T\lhd G$ and $G\cong S\times T$, and so $G\cong \mathbb Z_p\times \mathbb Z_q\cong \mathbb Z_{pq}$. In case $c=kq+1=p$, then $q|p-1$, and $T$ is not a normal subgroup of $G$. Since $S\lhd G$, $aba^{-1}=b^m$ for some $m$; furthermore, we may assume that $m\not\equiv 1\pmod p$ lest we return to the abelian case. It is easy to prove, by induction on $j$, that $a^{j}ba^{-j}=b^{m^j}$. In particular, if $j=q$, then $m^q\equiv 1\pmod p$.
■

Corollary 4.2.16 If $p>q$ are primes, then every group $G$ of order $pq$ contains a normal subgroup of order $p$. Moreover, if $q$ does not divide $p-1$, then $G$ must be cyclic.

Proposition 4.2.17 If $p$ and $q$ are primes, then there is no simple group of order $p^2q$.

Chapter V. Normal Series

Some Galois Theory

We are going to assume in this exposition that every field $F$ is a subfield of an algebraically closed field $C$. Thus, if $f(x)\in F[x]$, the ring of all polynomials with coefficients in $F$, and if $f(x)$ has degree $n\ge 1$, then there are (not necessarily distinct) elements $\alpha_1,\ldots,\alpha_n$ in $C$ (the roots of $f(x)$) and nonzero $a\in F$ so that

$$f(x)=a(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)$$

in $C[x]$. The intersection of any family of subfields of a field is itself a subfield; define the smallest subfield of $C$ containing a given subset $X$ as the intersection of all those subfields of $C$ containing $X$. For example, if $\alpha\in C$, the smallest subfield of $C$ containing $X=F\cup \{\alpha\}$ is

$$F(\alpha)=\{f(\alpha)/g(\alpha):f(x),g(x)\in F[x],g(\alpha)\neq 0\};$$

$F(\alpha)$ is called the subfield obtained from $F$ by adjoining $\alpha$. Similarly, one can define $F(\alpha_1,\ldots,\alpha_n)$, the subfield obtained from $F$ by adjoining $\alpha_1,\ldots,\alpha_n$. In particular, if $f(x)\in F[x]$ and $f(x)=(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)\in C[x]$, then $F(\alpha_1,\ldots,\alpha_n)$, the subfield obtained from $F$ by adjoining all the roots of $f(x)$, is called the splitting field of $f(x)$ over $F$. Notice that the splitting field of $f(x)$ does depend on $F$.

Definition 5.1.1 Let $f(x)\in F[x]$ have splitting field $E$ over $F$. Then $f(x)$ is solvable by radicals if there is a chain of subfields

$$F=K_0\subset K_1\subset\cdots\subset K_t,$$

in which $E\subset K_t$ and each $K_{i+1}$ is obtained from $K_i$ by adjoining a root of an element of $K_i$; that is, $K_{i+1}=K_i(\beta_{i+1})$, where $\beta_{i+1}\in K_{i+1}$ and some power of $\beta_{i+1}$ lies in $K_i$.

Remark It can be shown that, for an arbitrary polynomial $f(x)$, the concept of "$f(x)$ is solvable by radicals" is equivalent to "there is a formula for the roots of $f(x)$".

Definition 5.1.2 If $E$ and $E'$ are fields, then a homomorphism is a function $\sigma\!:E\rightarrow E'$ such that, for all $\alpha,\beta\in E$,

$$\begin{aligned} \sigma(\mathbf 1)&=\mathbf 1,\\ \sigma(\alpha+\beta)&=\sigma(\alpha)+\sigma(\beta),\\ \sigma(\alpha\beta)&=\sigma(\alpha)\sigma(\beta). \end{aligned}$$

if $\sigma$ is a bijection, then $\sigma$ is an isomorphism; an isomorphism $\sigma\!:E\rightarrow E$ is called an automorphism.

Lemma 5.1.3 Let $f(x)\in F[x]$, let $E$ be its splitting field over $F$, and let $\sigma\!:E\rightarrow E$ be an automorphism fixing $F$ (i.e., $\sigma(a)=a$ for all $a\in F$). If $\alpha\in E$ is a root of $f(x)$, then $\sigma(a)$ is also a root of $f(x)$.

Proof

If $f(x)=\sum a_ix^i$, then $\mathbf 0=\sigma(f(\alpha))=\sigma(\sum a_i\alpha^i)=\sum \sigma(a_i)\sigma(\alpha)^i=\sum a_i\sigma(\alpha)^i$, and so $\sigma(\alpha)$ is a root of $f(x)$.
■

Lemma 5.1.4 Let $F$ be a subfield of $K$, let $\{\alpha_1,\ldots,\alpha_n\}\subset K$, and let $E=F(\alpha_1,\ldots,\alpha_n)$. If $K'$ is a field containing $F$ as a subfield, and if $\sigma\!:E\rightarrow K'$ is a homomorphism fixing $F$ with $\sigma(\alpha_i)=\alpha_i$ for all $i$, then $\sigma$ is the identity.

Proof

The proof is by induction on $n\ge 1$. If $n=1$, then $E$ consists of all $g(\alpha_1)/h(\alpha_1)$, where $g(x),h(x)\in F[x]$ and $h(\alpha_1)\neq 0$; clearly $\sigma$ fixes each such element. The inductive step is clear once we realizes that $F(\alpha_1,\ldots,\alpha_n)=F^*(\alpha_n)$, where $F^*=F(\alpha_1,\ldots,\alpha_{n-1})$.
■

Definition 5.1.5 If $F$ is a subfield of $E$, then the Galois group, denoted by $\mathrm{Gal}(E/F)$, is the group under composition of all those automorphisms of $E$ which fix $F$. If $f(x)\in F[x]$ and $E=F(\alpha_1,\ldots,\alpha_n)$ is the splitting field of $f(x)$ over $F$, then the Galois group of $f(x)$ is $\mathrm{Gal}(E/F)$.

Theorem 5.1.6 Let $f(x)\in F[x]$ and let $X=\{\alpha_1,\ldots,\alpha_n\}$ be the set of its distinct roots (in its splitting field $E=F(\alpha_1,\ldots,\alpha_n)$ over $F$). Then the function $\varphi\!:\mathrm{Gal}(E/F)\rightarrow\mathrm S_X\cong \mathrm S_n$, given by $\varphi(\sigma)=\sigma\big|_X$, is an imbedding; that is, $\varphi$ is completely determined by its action on $X$.

Proof

If $\sigma\in\mathrm{Gal}(E/F)$, then Lemma 5.1.3 shows that $\sigma(X)\subset X$; $\sigma\big|_X$ is a bijection because $\sigma$ is an injection and $X$ is finite. It is easy to see that $\varphi$ is a homomorphism; it is an injection, by Lemma 5.1.4.
■

Not every permutation of the roots of a polynomial $f(x)$ need arise from some $\sigma\in\mathrm{Gal}(E/F)$. For example, let $f(x)=(x^2-2)(x^2-3)\in\mathbb Q[x]$. Then $E=\mathbb Q(\sqrt 2,\sqrt 3)$ and there is no $\sigma\in\mathrm{Gal}(E/\mathbb Q)$ with $\sigma(\sqrt 2)=\sqrt 3$.

Definition 5.1.7 If $F$ is a subfield of a field $E$, then $E$ is a vector space over $F$ (if $a\in F$ and $\alpha\in E$, define scalar multiplication to be the given product $a\alpha$ of two elements of $E$). The degree of $E$ over $F$, denoted by $[E:F]$, is the dimension of $E$.

Proposition 5.1.8 Let $p(x)\in F[x]$ be an irreducible polynomial of degree $n$. If $\alpha$ is a root of $p(x)$ (in a splitting field), then $\{\mathbf 1,\alpha,\alpha^2,\ldots,\alpha^{n-1}\}$ is a basis of $F(\alpha)$ (viewed as a vector space over $F$), and so we have that $[F(\alpha):F]=n$.

Proposition 5.1.9 Let $F\subset E\subset K$ be fields, where $[K:E]$ and $[E:F]$ are finite, then we have $[K:F]=[K:E][E:F]$.

Proposition 5.1.10 Let $E$ be a splitting field over $F$ of some $f(x)\in F[x]$, and let $K$ be a splitting field over $E$ of some $g(x)\in E[x]$. If $\sigma\in\mathrm{Gal}(K/F)$, then $\sigma\big|_E\in\mathrm{Gal}(E/F)$.

Lemma 5.1.11 Let $p(x)\in F[x]$ be irreducible, and let $\alpha$ and $\beta$ be roots of $p(x)$ in a splitting field of $p(x)$ over $F$. Then there exists an isomorphism $\lambda^*\!:F(\alpha)\rightarrow F(\beta)$ which fixes $F$ and with $\lambda^*(\alpha)=\beta$.

Proof

By Proposition 5.1.8, every element of $F(\alpha)$ has a unique expression of the form $a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}$. Define $\lambda^*$ by

$$\lambda^*(a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1})=a_0+a_1\beta+\cdots+a_{n-1}\beta^{n-1}.$$

It is easy to see that $\lambda^*$ is a field homomorphism; it is an isomorphism because its inverse can be constructed in the same manner.
■

Remark There is a generalization of this lemma having the same proof. Let $\lambda\!:F\rightarrow F'$ be an isomorphism of fields, let $p(x)=a_0+a_1x+\cdots+a_nx^n\in F[x]$ be an irreducible polynomial, and let $p'(x)=\lambda(a_0)+\lambda(a_1)x+\cdots+\lambda(a_n)x^{n}\in F'[x]$. Finally, let $\alpha$ be a root of $p(x)$ and let $\beta$ be a root of $p'(x)$ (in appropriate splitting fields). Then there is an isomorphism $\lambda^*\!:F(\alpha)\rightarrow F'(\beta)$ with $\lambda^*(\alpha)=\beta$ and with $\lambda^*\big|_F=\lambda$.

Lemma 5.1.12 Let $f(x)\in F[x]$, and let $E$ be its splitting field over $F$. If $K$ is an "intermediate field", that is, $F\subset K\subset E$, and if $\lambda\!:K\rightarrow K$ is an automorphism fixing $F$, then there is an automorphism $\lambda^*\!:E\rightarrow E$ with $\lambda^*\big|_K=\lambda$.

Proof

The proof is by induction on $d=[E:F]$. If $d=1$, then $E=K$, every root $\alpha_1,\ldots,\alpha_n$ of $f(x)$ lies in $K$, and we may take $\lambda^*=\lambda$. If $d>1$, then $E\neq K$ and there is some root $\alpha$ of $f(x)$ not lying in $K$. Now $\alpha$ is a root of some irreducible factor $p(x)$ of $f(x)$; since $\alpha\notin K$, degree $p(x)=k>1$. By the generalized version of Lemma 5.1.11, there is $\beta\in E$ and an isomorphism $\lambda_1\!:K(\alpha)\rightarrow K(\beta)$ which extends $\lambda$ and with $\lambda_1(\alpha)=\beta$. By Proposition 5.1.9, $[E:K(\alpha)]=d/k<d$. Now $E$ is the splitting field of $f(x)$ over $K(\alpha)$, for it arises from $K(\alpha)$ by adjoining all the roots of $f(x)$. Since all the inductive hypotheses have been verified, we conclude that $\lambda_1$, and hence $\lambda$, can be extended to an automorphism of $E$.
■

Remark As with the previous lemma, Lemma 5.1.12 has a more general version having the same proof. It says that if $f(x)\in F[x]$, then any two (abstract) splitting fields of $f(x)$ over $F$ are isomorphic.

Theorem 5.1.13 Let $p$ be a prime, let $F$ be a field containing a primitive $p$th root of unity, say, $\omega$, and let $f(x)=x^{p}-a\in F[x]$.

1. If $\alpha$ is a root of $f(x)$ (in some splitting field), then $f(x)$ is irreducible if and only if $\alpha\notin F$.
2. The splitting field $E$ of $f(x)$ over $F$ is $F(\alpha)$.
3. If $f(x)$ is irreducible, then $\mathrm{Gal}(E/F)\cong \mathbb Z_p$.

Proof

1. If $\alpha\in F$, then $f(x)$ is not irreducible, for it has $x-\alpha$ as a factor. Conversely, assume that $f(x)=g(x)h(x)$, where degree $g(x)=k<p$. Since the roots of $f(x)$ are $\alpha,\omega\alpha,\omega^2\alpha,\ldots,\omega^{p-1}\alpha$, every root of $g(x)$ has the form $\omega^i\alpha$ for some $i$. If the constant term of $g(x)$ is $c$, then $c=\pm\omega^r\alpha^k$ for some $r$. As both $c$ and $\omega$ lie in $F$, it follows that $\alpha^k\in F$. But $(k,p)=1$, because $p$ is prime, and so $1=ks+tp$ for some integers $s$ and $t$. Thus

$$\alpha=\alpha^{ks+tp}=(\alpha^k)^s(\alpha^p)^t\in F.$$

2. Immediate from the observation that the roots of $f(x)$ are of the form $\omega^i\alpha$.
3. If $\sigma\in\mathrm{Gal}(E/F)$, then $\sigma(\alpha)=\omega^i\alpha$ for some $i$, by Lemma 5.1.3. Define $\varphi\!:\mathrm{Gal}(E/F)\rightarrow \mathbb Z_p$ by $\varphi(\sigma)=[i]$, the congruence class of $i$ mod $p$. It is easy to check that $\varphi$ is a homomorphism; it is an injection, by Lemma 5.1.4. Since $f(x)$ is irreducible, by hypotheses, Lemma 5.1.11 shows that $\mathrm{Gal}(E/F)\neq \mathbf 1$. Therefore, $\varphi$ is a surjection, for $\mathbb Z_p$ has no proper subgroups.
   ■

Theorem 5.1.14 Let $f(x)\in F[x]$, let $E$ be the splitting field of $f(x)$ over $F$, and assume that $f(x)$ has no repeated roots in $E$ (i.e., $f(x)$ has no factor of the form $(x-\alpha)^2$ in $E[x]$). Then $f(x)$ is irreducible if and only if $\mathrm{Gal}(E/F)$ acts transitively on the set $X$ of all the roots of $f(x)$.

Remark It can be shown that if $F$ has characteristic $0$ or if $F$ is finite, then every irreducible polynomial in $F[x]$ has no repeated roots.

Proof

Note first that Lemma 5.1.3 shows that $\mathrm{Gal}(E/F)$ does act on $X$. If $f(x)$ is irreducible, then Lemma 5.1.11 shows that $\mathrm{Gal(E/F)}$ acts transitively on $X$. Conversely, assume that there is a factorization $f(x)=g(x)h(x)$ in $F[x]$. In $E[x]$, $g(x)=\prod(x-\alpha_i)$ and $h(x)=\prod(x-\beta_j)$; since $f(x)$ has no repeated roots, $\alpha_i\neq\beta_j$ for all $i,j$. But $\mathrm{Gal}(E/F)$ acts transitively on the roots of $f(x)$, so there exists $\sigma\in\mathrm{Gal}(E/F)$ with $\sigma(\alpha_1)=\beta_1$, and this contradicts Lemma 5.1.3.
■

It is easy to see that if $\alpha_1$ is a root of $f(x)$, then the stabilizer of $\alpha_1$ is $\mathrm{Gal}(E/F(\alpha_1))\le \mathrm{Gal}(E/F)$, and $\mathrm{Gal(E/F(\alpha_1))}$ is the Galois group of $f(x)/(x-\alpha_1)$ over $F(\alpha_1)$. Thus, $f(x)/(x-\alpha_1)$ is irreducible (over $F(\alpha_1)$) if and only if $\mathrm{Gal}(E/F(\alpha_1))$ acts transitively on the remaining roots.

Theorem 5.1.15 Let $F\subset K\subset E$ be fields, where $K$ and $E$ are splitting fields of polynomials over $F$. Then $\mathrm{Gal}(E/K)\lhd\mathrm{Gal}(E/F)$ and

$$\mathrm{Gal}(E/F)/\mathrm{Gal}(E/K)\cong\mathrm{Gal}(K/F).$$

Proof

The functioni $\Phi\!:\mathrm{Gal}(E/F)\rightarrow\mathrm{Gal}(K/F)$, given by $\Phi(\sigma)=\sigma\big|_K$, is well defined (by Proposition 5.1.10, for $K$ is a splitting field), and it is easily seen to be a homomorphism. The kernel of $\Phi$ consists of all those automorphisms which fix $K$; that is, $\mathrm{ker}\Phi=\mathrm{Gal}(E/K)$, and so this subgroup is normal. We claim that $\Phi$ is a surjection. If $\lambda\in\mathrm{Gal}(K/F)$, that is, $\lambda$ is an automorphism of $K$ which fixes $F$, then $\lambda$ can be extended to an automorphism $\lambda^*$ of $E$ (by Lemma 5.1.12, for $E$ is a splitting field). Therefore, $\lambda^*\in\mathrm{Gal}(E/F)$ and $\Phi(\lambda^*)=\lambda^*\big|_K=\lambda$. The first isomorphism theorem completes the proof.
■

Proposition 5.1.16 Let $f(x)=x^n-a\in F[x]$, let $E$ be the splitting field of $f(x)$ over $F$, and let $\alpha\in E$ be an $n$th root of $a$. Then there are subfields

$$F=K_0\subset K_1\subset \cdots\subset K_t=F(\alpha)$$

with $K_{i+1}=K_i(\beta_{i+1}),\beta_{i+1}^{p(i)}\in K_i$, and $p(i)$ prime for all $i$.

Theorem 5.1.17 (Galois) Let $f(x)\in F[x]$ be a polynomial of degree $n$, let $F$ contain all $p$th roots of unity for every prime $p$ dividing $n!$, and let $E$ be the splitting field of $f(x)$ over $F$. If $f(x)$ is solvable by radicals, then there exist subgroups $G_i\le G=\mathrm{Gal}(E/F)$ such that:

1. $G=G_0\ge G_1\ge \cdots\ge G_t=\mathbf 1$;
2. $G_{i+1}\lhd G_i$ for all $i$; and
3. $G_i/G_{i+1}$ is cyclic of prime order for all $i$.

Remark

1. The hypothesis that $F$ contains various roots of unity can be eliminated.
2. If $F$ has characteristic $0$, then the converse of this theorem is also true.

Definition 5.1.18 A normal series of a group $G$ is a sequence of subgroups

$$G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1.$$

The factor groups of this normal series are the groups $G_i/G_{i+1}$ for all $i$; the length of the normal series is the number of strict inclusions; that is, the length is the number of nontrivial factor groups.

Definition 5.1.19 A finite group $G$ is solvable if it has a normal series whose factor groups are cyclic of prime order.

In this terminology, Theorem 5.1.17 and its converse say that a polynomial is solvable by radicals if and only if its Galois group is a solvable group.

The Jordan-Hölder Theorem

Definition 5.2.1 A normal series

$$G=H_0\rhd H_1\rhd\cdots\rhd H_m=\mathbf 1$$

is a refinement of a normal series

$$G=G_0\rhd G_1\rhd\cdots \rhd G_n=\mathbf 1$$

if $G_0,G_1,\ldots,G_n$ is a subsequence of $H_0,H_1,\cdots,H_m$.

Definition 5.2.2 A composition series is a normal series

$$G=G_0\rhd G_1\rhd\cdots \rhd G_n=\mathbf 1$$

in which, for all $i$, either $G_{i+1}$ is a maximal normal subgroup of $G_i$ or $G_{i+1}=G_i$.

Every refinement of a composition series is also a composition series; it can only repeat some of the original terms.

Proposition 5.2.3 A normal series is a composition series if and only if its factor groups are either simple or trivial.

Proposition 5.2.4

1. Every finite group has a composition series.
2. An abelian group has a composition series if and only if it is finite.

Proposition 5.2.5 If $G$ is a finite group having a normal series with factor groups $H_0,H_1,\ldots,H_n$, then $|G|=\prod|H_i|$.

Definition 5.2.6 Two normal series of a group $G$ are equivalent if there is a bijection between their nontrivial factor groups such that corresponding factor groups are isomorphic.

Lemma 5.2.7 (Zassenhaus Lemma) Let $A\lhd A^*$ and $B\lhd B^*$ be four subgroups of a group $G$. Then

$$\begin{aligned} A(A^*\cap B)&\lhd A(A^*\cap B^*),\\ B(B^*\cap A)&\lhd B(B^*\cap A^*), \end{aligned}$$

and there is an isomorphism

$$\dfrac{A(A^*\cap B^*)}{A(A^*\cap B)}\cong \dfrac{B(B^*\cap A^*)}{B(B^*\cap A)}.$$

Proof

Both $A$ and $A^*\cap B^*$ are subgroups of $A^*$, and $A\lhd A^*$. By the second isomorphism theorem, $A\cap B^*=A\cap(A^*\cap B^*)\lhd A^*\cap B^*$; similarly, $A^*\cap B\lhd A^*\cap B^*$. It follows from Lemma 2.6.3 and Proposition 2.4.17 that $D=(A^*\cap B)(A\cap B^*)$ is a normal subgroup of $A^*\cap B^*$.

If $x\in B(B^*\cap A^*)$, then $x=bc$ for $b\in B$ and $c\in B^*\cap A^*$. Define $f\!:B(B^*\cap A)\rightarrow (A^*\cap B^*)/D$ by $f(x)=f(bc)=cD$. To see that $f$ is well defined, assume that $x=bc=b'c'$, where $b'\in B$ and $c'\in B^*\cap A$; then $c'c^{-1}=b'^{-1}b\in (B^*\cap A^*)\cap B=B\cap A^*\le D$. It is routine to check that $f$ is a surjective homomorphism with kernel $B(B^*\cap A)$. The first isomorphism theorem gives $B(B^*\cap A)\lhd B(B^*\cap A^*)$ and

$$\dfrac{B(B^*\cap A^*)}{B(B^*\cap A)}\cong \dfrac{B^*\cap A^*}{D}.$$

Transposing the symbols $A$ and $B$ gives $A(A^*\cap B)\lhd A(A^*\cap B^*)$ and an isomorphism of the corresponding quotient group to $(B^*\cap A^*)/D$. It follows that the two quotient groups in the statement are isomorphic.
■

Theorem 5.2.8 (Schreier Refinement Theorem) Every two normal series of an arbitrary group $G$ have refinements that are equivalent.

Proof

Let

$$G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1$$

and

$$G=H_0\rhd H_1\rhd\cdots\rhd H_m=\mathbf 1$$

be normal series. Insert a "copy" of the second series between each $G_i$ and $G_{i+1}$. More precisely, define $G_{i,j}=G_{i+1}(G_i\cap H_j)$ for all $0\le j\le m$. Thus

$$G_{i,j}=G_{i+1}(G_i\cap H_j)\ge G_{i+1}(G_i\cap H_{j+1})=G_{i,j+1}.$$

Notice that $G_{i,0}=G_i$ because $H_0=G$ and that $G_{i,m}=G_{i+1}$ because $H_m=\mathbf 1$. Moreoever, setting $A=G_{i+1}, A^*=G_i,B=H_{j+1}$, and $B^*=H_j$ in the Zassenhaus lemma shows that $G_{i,j+1}\lhd G_{i,j}$. It follows that the sequence

$$G_{0,0}\rhd G_{0,1}\rhd \cdots\rhd G_{0,m}\rhd G_{1,0}\rhd\cdots\rhd G_{n-1,0}\rhd\cdots\rhd G_{n-1,m}=\mathbf 1$$

is a refinement of the first normal series (with $mn$ terms). Similarly, if $H_{i,j}$ is defined to be $H_{j+1}(H_j\cap G_i)$, then $H_{i,j}\ge H_{i+1,j}$ and

$$H_{0,0}\rhd H_{1,0}\rhd\cdots\rhd H_{n,0}\rhd H_{0,1}\rhd\cdots\rhd H_{0,m-1}\rhd\cdots\rhd H_{n,m-1}=\mathbf 1$$

is a refinement of the second normal series (with $mn$ terms). Finally, the function pairing $G_{i,j}/G_{i,j+1}$ with $H_{i,j}/H_{i+1,j}$ is a bijection, and the Zassenhaus lemma (with $A=G_{i+1},A^*=G_i,B=G_{j+1}$, and $B^*=H_j$) shows that corresponding factor groups are isomorphic. Therefore, the two refinements are equivalent.
■

Theorem 5.2.9 (Jordan-Hölder) Every two composition series of a group $G$ are equivalent.

Proof

Composition series are normal series, so that every two composition series of $G$ have equivalent refinements. But a composition series is a normal series of maximal length; a refinement of it merely repeats several of its terms, and so its new factor groups have order $1$. Therefore, two composition series of $G$ are equivalent.
■

Definition 5.2.10 If $G$ has a composition series, then the factor groups of this series are called the composition factors of $G$.

Remark The Jordan-Hölder Theorem can be regarded as a unique factorization theorem.

Proposition 5.2.11 Let $G$ and $H$ be finite groups. If there are normal series of $G$ and of $H$ having the same set of factor groups, then $G$ and $H$ have the same composition factors.

Proposition 5.2.12 If $n\ge 5$, then $\mathrm S_n$ is not solvable.

Proposition 5.2.13 Assume that $G=H_1\times \cdots \times H_n=K_1\times \cdots \times K_m$, where each $H_i$ and $K_j$ is simple. Then we have that $m=n$ and there is a permutation $\pi$ of $\{1,2,\ldots,n\}$ with $K_{\pi(i)}\cong H_i$ for all $i$.

Solvable Groups

Definition 5.3.1 A solvable series of a group $G$ is a normal series all of whose factor groups are abelian. A group $G$ is solvable if it has a solvable series.

Remark Previously we have defined a "solvability" in Definition 5.1.18. The equivalence of these two definitions can be easily seen since we can always find in any abelian group $G$ a cyclic subgroup of some prime order, which is also a normal subgroup of $G$.

Theroem 5.3.2 Every subgroup $H$ of a solvable group $G$ is itself solvable.

Proof

If $G=G_0\rhd G_1\rhd \cdots\rhd G_n=\mathbf 1$ is a solvable series, consider the series $H=H_0\ge (H\cap G_1)\ge\cdots\ge(H\cap G_n)=\mathbf 1$. This is a normal series of $H$, for the second isomorphism theorem gives $H\cap G_{i+1}=(H\cap G_i)\cap G_{i+1}\lhd H\cap G_i$ for all $i$. Now $(H\cap G_i)/(H\cap G_{i+1})\cong G_{i+1}(H\cap G_i)/G_{i+1}\le G_i/G_{i+1}$; as $G_i/G_{i+1}$ is abelian, so is any of its subgroups. Therefore, $H$ has a solvable series.
■

Theroem 5.3.3 Every quotient of a solvable group is solvable.

Proof

It suffices to prove that if $G$ is a solvable group and $f\!:G\rightarrow H$ is a surjection, then $H$ is a solvable group. If

$$G=G_0\rhd G_1\rhd \cdots\rhd G_t=\mathbf 1$$

is a solvable series, then

$$H=f(G_0)\rhd f(G_1)\rhd\cdots\rhd f(G_t)=\mathbf 1$$

is a normal series for $H$: if $f(x_{i+1})\in f(G_{i+1})$ and $f(x_i)\in f(G_i)$, then $f(x_i)f(x_{i+1})f(x_i)^{-1}=f(x_ix_{i+1}x_i^{-1})\in f(G_{i+1})$, and so $f(G_{i+1})\lhd f(G_i)$ for every $i$. The map $\varphi\!:f(G_i)/f(G_{i+1})$, defined by $x_i\mapsto f(x_i)/f(G_{i+1})$, is a surjection, for it is the composite of the surjections $G_{i}\rightarrow f(G_i)$ and the natural map $f(G_i)\rightarrow f(G_i)/f(G_{i+1})$. Since $G_{i+1}\le \mathrm{ker}\varphi$, this map $\varphi$ includes a surjection $G_i/G_{i+1}\rightarrow f(G_i)/f(G_{i+1})$, namely, $x_iG_{i+1}\mapsto f(x_i)f(G_{i+1})$. Now $f(G_i)/f(G_{i+1})$ is a quotient of the abelian group $G_i/G_{i+1}$, and so it is abelian. Therefore, $H$ has a solvable series, and hence it is a solvable group.
■

Theorem 5.3.4 If $H\lhd G$ and if both $H$ and $G/H$ are solvable, then $G$ is solvable.

Proof

Let

$$G/H\rhd K_1^*\rhd K_2^*\rhd\cdots\rhd K_n^*=\mathbf 1$$

be a solvable series. By the correspondence theorem, we can construct a sequence looking like the beginning of a solvable series from $G$ to $H$; that is, there are subgroups $K_i$ (with $H\le K_i$ and $K_i/H\cong K_i^*$) such that

$$G\rhd K_1\rhd K_2\rhd\cdots\rhd K_n=H,$$

and $K_i/K_{i+1}(\cong K_i^*/K_{i+1}^*)$ is abelian. Since $H$ is solvable, it has a solvable series; if we splice these two sequences together at $H$, we obtain a solvable series for $G$.
■

Corollary 5.3.5 If $H$ and $K$ are solvable groups, then $H\times K$ is solvable.

Theorem 5.3.6 Every finite $p$-group $G$ is solvable.

Proof

The proof is by induction on $|G|$. By Theorem 4.1.6, $|\mathrm Z(G)|\neq 1$. Therefore, $G/\mathrm Z(G)$ is a $p$-group of order $<|G|$, and hence it is solvable, by induction. Since every abelian group is solvable, $\mathrm Z(G)$ is solvable. Therefore, $G$ is solvable, by Theorem 5.3.4.
■

Definition 5.3.7 The higher commutator subgroups of $G$ are defined inductively:

$$G^{(0)}=G;\qquad G^{(i+1)}=G^{(i)\prime};$$

that is, $G^{(i+1)}$ is the commutator subgroup of $G^{(i)}$. The series

$$G=G^{(0)}\rhd G^{(1)}\rhd G^{(2)}\rhd\cdots$$

is called the derived series of $G$.

Definition 5.3.8 An automorphism of a group $G$ is an isomorphism $\varphi\!:G\rightarrow G$. A subgroup $H$ of $G$ is called characteristic in $G$, denoted by $H\text{ char } G$, if $\varphi(H)=H$ for every automorphism $\varphi$ of $G$.

If $\varphi(H)\le H$ for every automorphism $\varphi$, then $H\text{ char }G$. For each $a\in G$, conjugation by $a$ (i.e., $x\mapsto axa^{-1}$) is an automorphism of $G$; it follows at once that every characteristic subgroup is a normal subgroup.

Lemma 5.3.9

1. If $H\text{ char }K$ and $K\text{ char }G$, then $H\text{ char }G$.
2. If $H\text{ char }K$ and $K\lhd G$, then $H\lhd G$.

Proof

1. If $\varphi$ is an automorphism of $G$, then $\varphi(K)=K$, and so the restriction $\varphi\big|_K\!:K\rightarrow K$ is an automorphism of $K$; since $H\text{ char }K$, it follows that $\varphi(H)=\left(\varphi\big|_K\right)(H)=H$.
2. Let $a\in G$ and let $\varphi\!:G\rightarrow G$ be conjugation by $a$. Since $K\lhd G$, $\varphi\big|_K$ is an automorphism of $K$; since $H\text{ char }K$, $\left(\varphi\big|_K\right)(H)\le H$. This says that if $h\in H$, then $aha^{-1}=\varphi(h)\in H$.
   ■

Theorem 5.3.10 For every group $G$, the higher commutator subgroups are characteristic, hence normal subgroups.

Proof

The proof is by induction on $i\ge 1$. Recall that the commutator subgroup $G'=G^{(1)}$ is generated by all commutators; that is, by all elements of the form $aba^{-1}b^{-1}$. If $\varphi$ is an automorphism of $G$, then $\varphi(aba^{-1}b^{-1})=\varphi(a)\varphi(b)\varphi(a)^{-1}\varphi(b)^{-1}$ is also a commutator, and so $\varphi(G')\le G'$. For the inductive step, we have just shown that $G^{(i+1)}\text{ char }G^{(i)}$; since $G^{(i)}\text{ char }G$, by induction, Lemma 5.3.9(i) shows that $G^{(i+1)}$ is characteristic in $G$.
■

Lemma 5.3.11 If $G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1$ is a solvable series, then $G_i\ge G^{(i)}$ for all $i$.

Proof

The proof is by induction on $i\ge 0$. If $i=0$, then $G_0=G=G^{(0)}$. For the inductive step, Theorem 2.5.4 gives $G_{i+1}\ge G_i'$, since $G_i/G_{i+1}$ is abelian. The inductive hypothesis gives $G_i\ge G^{(i)}$, so that $G_i'\ge G^{(i)\prime}=G^{(i+1)}$. Therefore, $G_{i+1}\ge G^{(i+1)}$, as desired.
■

Theorem 5.3.12 A group $G$ is solvable if and only if $G^{(n)}=\mathbf 1$ for some $n$.

Definition 5.3.13 A normal subgroup $H$ of a group $G$ is a minimal normal subgroup if $H\neq \mathbf 1$ and there is no normal subgroup $K$ of $G$ with $\mathbf 1< K< H$.

Theorem 5.3.14 If $G$ is a finite solvable group, then every minimal normal subgroup is elementary abelian.

Proof

Let $V$ be a minimal normal subgroup of $G$. Now Lemma 5.3.9(ii) says that if $H\text{ char }V$, then $H\lhd G$; since $V$ is minimal, either $H=\mathbf 1$ or $H=V$. In particular, $V'\text{ char }V$, so that either $V'=\mathbf 1$ or $V'=V$. Since $G$ is solvable, so is its subgroup $V$. It follows from Theorem 5.3.12 that $V'=V$ cannot occur here, so that $V'=\mathbf 1$ and so $V$ is abelian. Since $V$ is abelian, a Sylow $p$-subgroup $P$ of $V$, for any prime $p$, is characteristic in $V$; therefore, $V$ is an abelian $p$-group. But $\{x\in V:x^{p}=\mathbf 1\}\text{ char }V$, and so $V$ is elementary abelian.
■

Theorem 5.3.15 A finite group $G$ with no characteristic subgroups other than $G$ and $\mathbf 1$ is either simple or a direct product of isomorphic simple groups.

Proof

Choose a minimal normal subgroup $H$ of $G$ whose order is minimal among all nontrivial normal subgroups. Write $H=H_1$, and consider all subgroups of $G$ of the form $H_1\times H_2\times \cdots\times H_n$, where $n\ge 1$, $H_i\lhd G$, and $H_i\cong H$. Let $M$ be such a subgroup of largest possible order. We know that $M=G$ by showing that $M\text{ char }G$; to see this, it suffices to show that $\varphi(H_i)\le M$ for every $i$ and every automorphism $\varphi$ of $G$. Of course, $\varphi(H_i)\cong H=H_1$. We show that $\varphi(H_i)\lhd G$. If $a\in G$, then $a=\varphi(b)$ for some $b\in G$, and $a\varphi(H_i)a^{-1}=\varphi(b)\varphi(H_i)\varphi(b)^{-1}=\varphi(bH_ib^{-1})\le \varphi(H_i)$, because $H_i\lhd G$. If $\varphi(H_i)\nleq M$, then $\varphi(H_i)\cap M\nleq \varphi(H_i)$ and $|\varphi(H_i)\cap M|<|\varphi(H_i)|=|H|$. But $\varphi(H_i)\cap M\lhd G$, and so the minimality of $|H|$ shows that $\varphi(H_i)\cap M=\mathbf 1$. The subgroup $\langle M,\varphi(H_i)\rangle=M\times \varphi(H_i)$ is a subgroup of the same type as $M$ but of largere order, a contradiction. We conclude that $M\text{ char }G$, and so $M=G$. Finally, $H=H_1$ must be simple: if $N$ is a nontrivial normal subgroup of $H$, then $N$ is a normal subgroup of $M=H_1\times H_2\times\cdots\times H_n=G$, and this contradicts the minimal choice of $H$.
■

Corollary 5.3.16 A minimal normal subgroup $H$ of a finite group $G$ is either simple or a direct product of isomorphic simple subgroups.

Proposition 5.3.17 If $S$ and $T$ are solvable subgroups of $G$ with $S\lhd G$, then $ST$ is solvable.

Proposition 5.3.18 Every finite group $G$ has a unique maximal normal solvable subgroup $\mathscr S(G)$; moreover, $G/\mathscr S(G)$ has no nontrivial normal solvable subgroups.

Proposition 5.3.19 Let $G$ be a finite group of order $>1$. If $G$ is solvable, then $G$ contains a nontrivial normal abelian subgroup; if $G$ is not solvable, then $G$ contains a nontrivial normal subgroup $H$ with $H=H'$.

Theorem 5.3.20 (Burnside's Theorem) If $p$ and $q$ are primes, then every group of order $p^mq^n$ is solvable.

Proposition 5.3.21

1. If $H\lhd G$ and $(|H|,[G:H])=1$, then $H\text{ char }G$.
2. If $H\text{ char }G$ and $H\le K\le G$, then $K/H\text{ char }G/H$ implies $K\text{ char }G$.

Proposition 5.3.22 Let $G$ be a group, then $\mathrm Z(G)\text{ char }G$.

Two Theorems of P.Hall

Theorem 5.4.1 (P.Hall) If $G$ is a solvable group of order $ab$, where $(a,b)=1$, then $G$ contains a subgroup of order $a$. Moreover, any two subgroups of order $a$ are conjugate.

Proof

The proof is by induction on $|G|$; the base step is trivially true.

Case 1. $G$ contains a normal subgroup $H$ of order $a'b'$, where $a'|a,b'|b$, and $b'<b$.

Existence. In this case, $G/H$ is a solvable group of order $(a/a')(b/b')$, which is strictly less than $ab$; by induction, $G/H$ has a subgroup $A/H$ of order $a/a'$. Now $A$ has order $(a/a')|H|=ab'<ab$; since $A$ is solvable, it has a subgroup of order $a$, as desired.

Conjugacy. Let $A$ and $A'$ be subgroups of $G$ of order $a$, and let $k=|AH|$. Since $AH\le G$, Lagrange's theorem gives $|AH|=\alpha\beta$, where $\alpha\big|a$ and $\beta\big|b$, we have $a'b'\big|\alpha\beta$, so that $b'\big|\beta$. But the product formula gives $k|aa'b'$, so that $\beta|b'$. We conclude that $|AH|=k=ab'$. A similar calculation shows that $|A'H|=ab'$ as well. Thus, $AH/H$ and $A'H/H$ are subgroups of $G/H$ of order $a/a'$. As $|G/H|=(a/a')(b/b')$, these subgroups are conjugate, by induction, say by $xH\in G/H$. It is quickly checked that $xAHx^{-1}=A'H$. Therefore, $xAx^{-1}$ and $A'$ are subgroups of $A'H$ of order $a$, and so they are conjugate, by induction.

If there is some proper normal subgroup of $G$ whose order is not divisible by $b$, then the theorem has been proved. We may, therefore, assume that $b\big||H|$ for every proper normal subgroup $H$. If $H$ is a minimal normal subgroup, however, then Theorem 5.3.14 says that $H$ is an (elementary) abelian $p$-group for some prime $p$. It may thus be assumed that $b=p^m$, so that $H$ is a Sylow $p$-subgroup of $G$. Normality of $H$ forces $H$ to be unique (for all Sylow $p$-subgroups are conjugate). The problem has now been reduced to the following case.

Case 2. $|G|=ap^m$, where $p\nmid a$, and $G$ has a normal abelian Sylow $p$-subgroup $H$, which is the unique minimal normal subgroup in $G$.

Existence. The group $G/H$ is a solvable group of order $a$. If $K/H$ is a minimal normal subgroup of $G/H$, then $|K/H|=q^n$ for some prime $q\neq p$, and so $|K|=p^mq^n$; if $Q$ is a Sylow $q$-subgroup of $K$, then $K=HQ$. Let $N^*=\mathrm N_G(Q)$ and let $N=N^*\cap K=\mathrm N_K(Q)$. We claim that $|N^*|=a$.

The Fratiini argument gives $G=KN^*$. Since $G/K=KN^*/K\cong N^*/(N^*\cap K)=N^*/N$, we have $|N^*|=|G||N|/|K|$. But $K=HQ$ and $Q\le N\le K$ gives $K=HN$; hence $|K|=|HN|=|H||N|/|H\cap N|$, so that

$$\begin{aligned} |N^*|=|G||N|/|K|&=|G||N||H\cap N|/|H||N|\\&=(|G|/|H|)|H\cap N|=a|H\cap N|. \end{aligned}$$

Hence $|N^*|=a$ if $H\cap N=\mathbf 1$. We show that $H\cap N=\mathbf 1$ in two stages: (i) $H\cap N=\mathrm Z(K)$; and (ii) $\mathrm Z(K)=\mathbf 1$.

(i). Let $x\in H\cap N$. Every $k\in K=HQ$ has the form $k=hs$ for $h\in H$ and $s\in Q$. Now $x$ commutes with $h$, for $H$ is abelian, and so it suffices to show that $x$ commutes with $s$. But $(xsx^{-1})s^{-1}\in Q$, because $x$ normalizes $Q$, and $x(sx^{-1}s^{-1})\in H$, because $H$ is normal; therefore, $xsx^{-1}s^{-1}\in Q\cap H=\mathbf 1$.

(ii). By Lemma 5.3.9(ii), $\mathrm Z(K)\lhd G$. If $\mathrm Z(K)\neq \mathbf 1$, then it contains a minimal subgroup which must be a minimal normal subgroup of $G$. Hence $H\le\mathrm Z(K)$, for $H$ is the unique minimal normal subgroup of $G$. But since $K=HQ$, it follows that $Q\text{ char }K$. Thus $Q\lhd G$, by Lemma 5.3.9(ii), and so $H\le Q$, a contradiction. Therefore, $\mathrm Z(K)=\mathbf 1,H\cap N=\mathbf 1$, and $|N^*|=a$.

Conjugacy. We keep the notation of the existence proof. Recall that $N^*=\mathrm N_G(Q)$ has order $a$; let $A$ be another subgroup of $G$ of order $a$. Since $|AK|$ is divisible by $a$ and by $|K|=p^mq^n$, it follows that $|AK|=ab=|G|, AK=G$, and $G/K=AK/K\cong A/(A\cap K)$, and $|A\cap K|=q^n$. By the Sylow theorem, $A\cap K$ is conjugate to $Q$. As conjugate subgroups have conjugate normalizers, by Proposition 3.1.9(i), $N^*=\mathrm N_G(Q)$ is conjugate to $\mathrm N_G(A\cap K)$, and so $a=|N^*|=|\mathrm N_G(A\cap K)|$. Since $A\cap K\lhd A$, we have $A\le \mathrm N_G(A\cap K)$ and so $A=\mathrm N_G(A\cap K)$ since they both have order $a$. Therefore, $A$ is conjugate to $N^*$.
■

Definition 5.4.2 If $G$ is a finite group, then a Hall subgroup $H$ of $G$ is a subgroup whose order and index are relatively prime; that is, $(|H|, [G:H])=1$.

Definition 5.4.3 If $\pi$ is a set of primes, then a $\pi$-number is an integer $n$ all of whose prime factors lie in $\pi$; the complement of $\pi$ is denoted by $\pi'$, and so a $\pi'$-number is an integer $n$ none of whose prime factors lies in $\pi$.

Definition 5.4.4 If $\pi$ is a set of primes, then a group is a $\pi$-group if the order of each of its elements is a $\pi$-number; a group is a $\pi'$-group if the order of each of its elements is a $\pi'$-number.

Remark It is obvious that every Sylow $p$-subgroup in a finite group is a Hall $p$-subgroup, and Hall's theorem says that Hall $\pi$-subgroups always exists in finite solvable groups. However, Hall $\pi$-subgroups with $|\pi|\ge 2$ of a group $G$ need not exist.

Definition 5.4.5 If $p$ is a prime, and $G$ is a finite group of order $ap^n$, where $a$ is a $p'$-number, then a $p$-complement of $G$ is a subgroup of order $a$.

Hall's theorem implies that a finite solvable group has a $p$-complement for every prime $p$.

Theorem 5.4.6 (P.Hall) If $G$ is a finite group having a $p$-complement for every prime $p$, then $G$ is solvable.

Proof

We proceed by induction on $|G|$; assume, on the contrary, that there are nonsolvable groups satisfying the hypotheses, and choose one such, say, $G$, of smallest order. If $G$ has nontrivial normal subgroup $N$, and if $H$ is any Hall $p'$-subgroup of $G$, then checking orders shows that $H\cap N$ is a Hall $p'$-subgroup of $N$ and $HN/N$ is a Hall $p'$-subgroup of $G/N$. Since both $N$ and $G/N$ have order smaller than $|G|$, it follows that both $N$ and $G/N$ are solvable. But Theorem 5.3.4 now shows that $G$ is solvable, a contradiction.

We may assume, therefore, that $G$ is simple, Let $|G|=p_1^{e_1}\ldots p_n^{e_n}$, where the $p_i$ are distinct primes and $e_i>0$ for all $i$. For each $i$, let $H_i$ be a Hall $p_i'$-subgroup of $G$, so that $[G:H_i]=p_i^{e_i}$, and thus $|H_i|=\prod_{j\neq i}p_j^{e_j}$. If $D=H_3\cap\cdots\cap H_n$, then $[G:D]=\prod_{i=3}^np_i^{e_i}$, by Proposition 3.4.12, and so $|D|=p_1^{e_1}p_2^{e_2}$. Now $D$ is a solvable group, by Burnside's theorem. If $N$ is a minimal normal subgroup of $D$, then Theorem 5.3.14 says that $N$ is elementary abelian; for notation, assume that $N$ is a $p_1$-group, then Proposition 3.4.12 shows that $[G:D\cap H_2]=\prod_{i=2}^np_i^{e_i}$, so that $|D\cap H_2|=p_1^{e_1}$ and $D\cap H_2$ is a Sylow $p_1$-subgroup of $D$. By Proposition 4.2.10, $N\le D\cap H_2$ and so $N\le H_2$. But, as above, $|D\cap H_1|=p_2^{e_2}$, and comparison of orders gives $G=H_2(D\cap H_1)$. If $g\in G$, then $g=hd$, where $h\in H_2$ and $d\in D\cap H_1$; if $x\in N$, then $gxg^{-1}=hdxd^{-1}h^{-1}=hyh^{-1}$ (where $y=dxd^{-1}\in N$, because $N\lhd D$) and $hyh^{-1}\in H_2$ (because $N\le H_2$). Therefore, $N^G\le H_2$, where $N^G$ is the normal subgroup of $G$ generated by $N$. Since $H_2< G$, $N^G\neq\mathbf 1$ is a proper normal subgroup of $G$, and this contradicts the assumption that $G$ is simple.
■

Proposition 5.4.7 If $G$ is a finite group and $H$ is a normal Hall subgroup, then $H\text{ char }G$.

Definition 5.4.8 Let $G$ be a finite group. If $\pi$ is a set of primes, define $\mathrm O_\pi(G)$ to be the subgroup of $G$ generated by all the normal $\pi$-subgroups of $G$.

By definition we know that $\mathrm O_\pi(G)$ is a characteristic subgroup of $G$.

Proposition 5.4.9 Let $G$ be a finite group, then $\mathrm O_\pi(G)$ is the intersection of all the maximal $\pi$-subgroups of $G$.

Central Series and Nilpotent Groups

Definition 5.5.1 If $H,K\le G$, then

$$[H,K]=\langle[h,k]:h\in H\text{ and }k\in K\rangle,$$

where $[h,k]$ is the commutator $hkh^{-1}k^{-1}$.

It is easily observed that $[H,K]\le G$, and $G^{(i+1)}=[G^{(i)},G^{(i)}]$.

We say that a subgroup $K$ normalizes $H$ if $K\le \mathrm N_G(H)$. It is easy to see that $K$ normalizes $H$ if and only if $[H,K]\le H$.

We say that a subgroup $K$ centralizes $H$ if $K\le \mathrm C_G(H)$. It is easy to see that $K$ centralizes $H$ if and only if $[H,K]=\mathbf 1$.

If $x,y\in G$ and $[x,y]\in K$, where $K\lhd G$, then $x$ and $y$ "commute mod $K$"; that is, $xKyK=yKxK$ in $G/K$.

Lemma 5.5.2

1. If $K\lhd G$ and $K\le H\le G$, then $[H,G]\le K$ if and only if $H/K\le \mathrm Z(G/K)$.
2. If $H,K\le G$ and $f\!:G\rightarrow L$ is a homomorphism, then $f([H,K])=[f(H),f(K)]$.

Proof

1. If $h\in H$ and $g\in G$, then $hKgK=gKhK$ if and only if $[h,g]K=K$ if and only if $[h,g]\in K$.
2. Both sides are generated by all $f([h,k])=[f(h),f(k)]$.
   ■

Definition 5.5.3 Define characteristic subgroups $\mathrm C_i(G)$ of $G$ by induction:

$$\mathrm C_0(G)=G; \qquad\mathrm C_{i+1}(G)=[\mathrm C_i(G), G].$$

It is easy to check that $\mathrm C_1(G)=G'=G^{(1)}$ and $\mathrm C_{i+1}(G)\le \mathrm C_i(G)$. Moreover, Lemma 5.5.2(i) shows that $[\mathrm C_i(G), G]=\mathrm C_{i+1}(G)$ gives $\mathrm C_i(G)/\mathrm C_{i+1}(G)\le \mathrm Z(G/\mathrm C_{i+1}(G))$.

Definition 5.5.4 The lower central series (or descending central series) of $G$ is the series

$$G=\mathrm C_0(G)\ge \mathrm C_1(G)\ge\cdots$$

(this need not be a normal series because it may not reach $\mathbf 1$).

Definition 5.5.5 The higher centers $\mathrm C^i(G)$ are the characteristic subgroups of $G$ defined by induction:

$$\mathrm C^0(G)=\mathbf 1;\qquad\mathrm C^{i+1}(G)/\mathrm C^{i}(G)=\mathrm Z(G/\mathrm C^i(G));$$

that is, if $\nu_i\!:G\rightarrow G/\mathrm C^i(G)$ is the natural map, then $\mathrm C^{i+1}(G)$ is the inverse image of the center.

It is obvious that $\mathrm C^1(G)=\mathrm Z(G)$.

Definition 5.5.6 The upper central series (or ascending central series) of $G$ is

$$\mathbf 1=\mathrm C^0(G)\le\mathrm C^1(G)\le\mathrm C^2(G)\le\cdots.$$

Theorem 5.5.7 If $G$ is a group, then there is an integer $n$ with $\mathrm C^n(G)=G$ if and only if $\mathrm C_n(G)=\mathbf 1$. Moreover, in this case, $\mathrm C_{i}(G)\le \mathrm C^{n-i}(G)$ for all $i$.

Proof

Assume that $\mathrm C^n(G)=G$, and let us prove that the inclusion holds by induction on $i$. If $i=0$, then $\mathrm C_0(G)=\mathrm C^n(G)$. If $\mathrm C_{i}(G)\le \mathrm C_{n-i}(G)$, then

$$\mathrm C_{i+1}(G)=[\mathrm C_i(G),G]\le [\mathrm C^{n-i}(G),G]\le \mathrm C^{n-i-1}(G)$$

the last inclusion follwing from Lemma 5.5.2. In particular, if $i=n$, then $\mathrm C_n(G)\le \mathrm C^0(G)=\mathbf 1$.

Assume that $\mathrm C_{n}(G)=\mathbf 1$, and let us prove that $\mathrm C_{n-j}(G)\le \mathrm C^j(G)$ by induction on $j$. If $j=0$, then $\mathrm C_n(G)=\mathbf 1=\mathrm C^0(G)$. If $\mathrm C_{n-j}(G)\le\mathrm C^j(G)$, then the third isomorphism theorem gives a surjective homomorphism $G/\mathrm C_{n-j}(G)\rightarrow G/\mathrm C^j(G)$. Now $[\mathrm C_{n-j-1}(G),G]=\mathrm C_{n-j}(G)$, so that Lemma 5.5.2 gives $\mathrm C_{n-j-1}(G)/\mathrm C_{n-j}(G)\le\mathrm Z(G/\mathrm C_{n-j}(G))$. So we have

$$\mathrm C_{n-j-1}(G)\mathrm C^j(G)/\mathrm C^j(G)\le\mathrm Z(G/\mathrm C^j(G))=\mathrm C^{j+1}(G)/\mathrm C^j(G).$$

Therefore, $\mathrm C_{n-j-1}(G)\le\mathrm C_{n-j-1}(G)\mathrm C^j(G)\le\mathrm C^{j+1}(G)$, as desired. In particular, if $j=c$, then $G=\mathrm C_0(G)\le \mathrm C^n(G)$.
■

Theorem 5.5.8 (Schur) If $G$ is a group with $G/\mathrm Z(G)$ finite, then $G'$ is also finite.

Proof

Let $g_1,\ldots,g_n$ be representatives of the cosets of $\mathrm Z(G)$ in $G$; that is, each $x\in G$ has the form $x=g_iz$ for some $i$ and some $z\in \mathrm Z(G)$. For all $x,y\in G$, we have $[x,y]=[g_iz,g_jz']=[g_i,g_j]$. Hence, every commutator has the form $[g_i,g_j]$ for some $i,j$, so that $G'$ has a finite number $(<n^2)$ of generators.

Each element $g'\in G'$ can be written as a word $c_1\cdots c_t$, where each $c_i$ is a commutator. It suffices to prove that if a factorization of $g'$ is chosen so that $t=t(g')$ is minimal, then $t(g')<n^3$ for all $g'\in G'$.

We prove first, by induction on $r\ge 1$, that if $a,b\in G$, then $[a,b]^r=(aba^{-1}b^{-1})^r=(ab)^r(a^{-1}b^{-1})^ru$, where $u$ is a product of $r-1$ commutators. This is obvious when $r=1$. Note, for the inductive step, that if $x,y\in G$, then $xy=yxx^{-1}y^{-1}xy=yx[x^{-1},y^{-1}]$; that is, $xy=yxc$ for some commutator $c$. Thus, if $r>1$, then

$$\begin{aligned} (aba^{-1}b^{-1})^{r+1}&=aba^{-1}b^{-1}(aba^{-1}b^{-1})^r\\ &=ab(a^{-1}b^{-1})((ab)^r(a^{-1}b^{-1})^r)u\\ &=ab((ab)^r(a^{-1}b^{-1})^r)(a^{-1}b^{-1})cu \end{aligned}$$

for some commutator $c$, as desired.

Since $yx=x^{-1}(xy)x$, we have $(yx)^n=x^{-1}(xy)^nx=(xy)^n$, because $[G:\mathrm Z(G)]=n$ implies $(ab)^n\in \mathrm Z(G)$. Therefore, $(a^{-1}b^{-1})^n=((ba)^{-1})^n=((ba)^n)^{-1}=((ab)^n)^{-1}$. It follows that $[a,b]^n$ is a product of $n-1$ commutators.

Now $xyx=(xyx^{-1})x^2$, so that two $x$s can be brought together at the expense of replacing $y$ by a conjugate of $y$. Take an expression of an element $g'\in G'$ as a product of commutators $c_1,\ldots,c_t$, where $t$ is minimal. If $t>n^3$, then there is some commutator $c$ occuring $m$ times, where $m>n$ (for there are fewer than $n^2$ distinct commutators). But all such factors can be brought together to $c^m$ at the harmless expense of replacing commutators by conjugates (which are still commutators); that is, the number of commutator factors in the expression is unchanged. And we have proven that the length of the minimal expression for $g'$ is shortened, and this is a contradiction. Therefore, $t<n^3$, and so $G'$ is finite.
■

Definition 5.5.9 A group $G$ is nilpotent if there is an integer $n$ such that $\mathrm C_{n}(G)=\mathbf 1$; the least such $n$ is called the class of the nilpotent group $G$.

Theorem 5.5.7 shows, for nilpotent groups, that the lower and upper central series are normal series of the same length.

A group is nilpotent of class $1$ if and only if it is abelian. By Theorem 5.5.7, a nilpotent group $G$ of class $2$ is described by $\mathrm C_1(G)=G'\le \mathrm Z(G)=\mathrm C^1(G)$. Every nonabelian group of order $p^3$ is nilpotent of class $2$, by Proposition 4.1.15.

Theorem 5.5.10 Every finite $p$-group is nilpotent.

Proof

By Theorem 4.1.6, every finite $p$-group has a nontrivial center. If, for some $i$, we have $\mathrm C^i(G)<G$, then $\mathrm Z(G/\mathrm C^i(G))\neq\mathbf 1$ and so $\mathrm C^i(G)<\mathrm C^{i+1}(G)$. Since $G$ is finite, there must be an integer $i$ with $\mathrm C^i(G)=G$; that is, $G$ is nilpotent.
■

Theorem 5.5.11

1. Every nilpotent group $G$ is solvable.
2. If $G\neq\mathbf 1$ is nilpotent, then $\mathrm Z(G)\neq \mathbf 1$.

Proof

1. An easy induction shows that $G^{(i)}\le \mathrm C_{i-1}(G)$ for all $i$. It follows that if $\mathrm C_{n}(G)=\mathbf 1$, then $G^{(n+1)}=\mathbf 1$; that is, if $G$ is nilpotent (of class $\le n$), then $G$ is solvable (with derived length $\le n+1$).
2. Assume that $G\neq\mathbf 1$ is nilpotent of class $c$, so that $\mathrm C_n(G)=\mathbf 1$ and $\mathrm C_{n-1}(G)\neq\mathbf 1$. By Theorem 5.5.7, $\mathbf 1\neq\mathrm C_{n-1}(G)\le\mathrm C^1(G)=\mathrm Z(G)$.
   ■

Theorem 5.5.12 Every subgroup $H$ of a nilpotent group $G$ is nilpotent. Moreoever, if $G$ is nilpotent of class $n$, then $H$ is nilpotent of class $\le n$.

Proof

It is easily proved by induction that $H\le G$ implies $\mathrm C_i(H)\le\mathrm C_i(G)$ for all $i$. Therefore, $\mathrm C_n(G)=\mathbf 1$ forces $\mathrm C_n(H)=\mathbf 1$.
■

Theorem 5.5.13 If $G$ is nilpotent of class $n$ and $H\lhd G$, then $G/H$ is nilpotent of class $\le n$.

Proof

If $f\!:G\rightarrow L$ is a surjective homomorphism, then Lemma 5.5.2 gives $\mathrm C_i(L)\le f(\mathrm C_i(G))$ for all $i$. Therefore, $\mathrm C_n(G)=\mathbf 1$ forces $\mathrm C_n(L)=\mathbf 1$. The theorem follows by taking $f$ to be the natural map.
■

Remark If $H\lhd G$ and both $H$ and $G/H$ are nilpotent, then $G$ need not be nilpotent. For example, $\mathrm S_3$ is not nilpotent, but both $\mathrm A_3\cong \mathbb Z_3$ and $\mathrm S_3/\mathrm A_3\cong \mathbb Z_2$ are abelian, hence nilpotent.

Theorem 5.5.14 If $H$ and $K$ are nilpotent, then their direct product $H\times K$ is nilpotent.

Proof

An easy induction shows that $\mathrm C_i(H\times K)\le\mathrm C_i(H)\times \mathrm C_i(K)$ for all $i$. If $M=\max\{n,m\}$, where $\mathrm C_{n}(H)=\mathbf 1=\mathrm C_{m}(K)$, then $\mathrm C_{M}(H\times K)=\mathbf 1$ and $H\times K$ is nilpotent.
■

Theorem 5.5.15 If $G$ is nilpotent, then it satisfies the normalizer condition: if $H< G$, then $H<\mathrm N_G(H)$.

Proof

There exists an integer $i$ with $\mathrm C_{i+1}(G)\le H$ and $\mathrm C_i(G)\nleq H$. Now $[\mathrm C_i,H]\le[\mathrm C_i,G]=\mathrm C_{i+1}\le H$, so that $\mathrm C_i$ normalizes $H$; that is, $\mathrm C_i\le\mathrm N_G(H)$ Therefore, $H$ is a proper subgroup of $\mathrm N_G(H)$.
■

Proposition 5.5.16 The following conditions on a finite group $G$ are equivalent:

1. $G$ is nilpotent;
2. $G$ satisfies the normalizer condition;
3. Every maximal subgroup of $G$ is normal.
4. Every Sylow $p$-subgroup of some prime $p$ is normal.

Proposition 5.5.17

1. If $G$ is nilpotent of class $n$, then $G/\mathrm Z(G)$ is nilpotent of class $n-1$.
2. If $H\le\mathrm Z(G)$ and if $G/H$ is nilpotent, then $G$ is nilpotent.
3. If $G$ is a nilpotent group and $H$ is a minimal normal subgroup of $G$, then $H\le \mathrm Z(G)$.

Theorem 5.5.18 A finite group $G$ is nilpotent if and only if it is the direct product of its Sylow subgroups.

Proof

If $G$ is the direct product of its Sylow subgroups, then it is nilpotent, by Theorems 5.5.10 and 5.5.14.

For the converse, let $P$ be a Sylow $p$-subgroup of $G$ for some prime $p$. By Proposition 4.2.13, $\mathrm N_G(P)$ is equal to its own normalizer. On the other hand, if $\mathrm N_G(P)<G$, then Theorem 5.5.15 shows that $\mathrm N_G(P)$ is a proper subgroup of its own normalizer. Therefore, $\mathrm N_G(P)=G$ and $P\lhd G$, which shows that $G$ have a unique Sylow $p$-subgroup for every prime $p\big||G|$. Since the intersection of any Sylow $p$-subgroup and Sylow $q$-subgroup is always trivial, we know that $G$ equals to the direct product of its Sylow subgroups.
■

Corollary 5.5.19 Let $G$ be a finite nilpotent group of order $n$. If $m|n$, then $G$ has a subgroup of order $m$.

Theorem 5.5.20 If $G$ is a nilpotent group, then every maximal subgroup $H$ is normal and has prime index.

Proof

By Theorem 5.5.15, $H< \mathrm N_G(H)$; since $H$ is maximal, $\mathrm N_G(H)=G$, and so $H\lhd G$. Proposition 2.7.4 now shows that $G/H$ has prime order.
■

Theorem 5.5.21 Let $G$ be a nilpotent group.

1. If $H$ is a nontrivial normal subgroup, then $H\cap \mathrm Z(G)\neq\mathbf 1$.
2. If $A$ is a maximal abelian normal subgroup of $G$, then $A=\mathrm C_G(A)$.

Proof

1. Since $\mathrm C^0(G)=\mathbf 1$ and $G=\mathrm C^n(G)$ for some $n$, there is an integer $i$ for which $H\cap \mathrm C^i(G)\neq\mathbf 1$; let $m$ be the minimal such $i$. Now $[H\cap \mathrm C^m(G),G]\le H\cap [\mathrm C^{m}(G),G]\le H\cap \mathrm C^{m-1}(G)=\mathbf 1$, because $H\lhd G$, and this says that $\mathbf 1\neq H\cap\mathrm C^m(G)\le H\cap \mathrm Z(G)$.
2. Since $A$ is abelian, $A\le \mathrm C_G(A)$. For the reverse inclusion, assume that $g\in\mathrm C_G(A)$ and $g\notin A$. It is easy to see, for any subgroup $H$ (of any group $G$) and for all $g\in G$, that $g\mathrm C_G(H)g^{-1}=\mathrm C_G(g^{-1}Hg)$. Since $A\lhd G$, it follows that $g\mathrm C_G(A)g^{-1}=\mathrm C_G(A)$ for all $g\in G$, and so $\mathrm C_G(A)\lhd G$. Therefore, $\mathrm C_G(A)/A$ is a nontrivial normal subgroup of the nilpotent group $G/A$; by (i), there is $x\notin A$ with $Ax\in(\mathrm C_G(A)/A)\cap \mathrm Z(G/A)$. The correspondence theorem gives $\langle A,x\rangle$ a normal abelian subgroup of $G$ strictly containing $A$, and this contradicts the maximality of $A$.
   ■

Definition 5.5.22 A normal series

$$G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1$$

with each $G_i\lhd G$ and $G_i/G_{i+1}\le \mathrm Z(G/G_{i+1})$ is called a central series.

Proposition 5.5.23

1. If $G$ is nilpotent, then both the upper and the lower central series of $G$ are central series.
2. A group $G$ is nilpotent if and only if it has a central series $G=G_0\rhd G_1\rhd \cdots\rhd G_n=\mathbf 1$. Moreover, if $G$ is nilpotent of class $n$, then $\mathrm C_{i}(G)\le G_i\le \mathrm C^{n-i}(G)$ for all $i$.

Proposition 5.5.24 If $H$ and $K$ are normal nilpotent subgroups of a finite group $G$, then $HK$ is a normal nilpotent subgroup.

Proposition 5.5.25 Let $G$ be a finite group, then $G$ has a unique maximal normal nilpotent subgroup $\mathscr F(G)$, which is called the Fitting subgroup of $G$, and we have $\mathscr F(G)\text{ char }G$ when $G$ is finite.

Proposition 5.5.26 Let $G$ be a group, then $[\mathrm C_i(G), \mathrm C_j(G)]\le \mathrm C_{i+j+1}(G)$ for all $i,j$.

Frattini Subgroup

Definition 5.6.1 If $G$ is a group, its Frattini subgroup $\Phi(G)$ is defined as the intersection of all the maximal subgroups of $G$. If $G$ is an infinite group with no maximal subgroups, we define $\Phi(G)=G$.

It is clear that $\Phi(G)\text{ char }G$, and so $\Phi(G)\lhd G$.

Definition 5.6.2 An element $x\in G$ is called a nongenerator if it can be omitted from any generating set: if $G=\langle x,Y\rangle$, then $G=\langle Y\rangle$.

Theorem 5.6.3 For every group $G$, the Frattini subgroup $\Phi(G)$ is the set of all nongenerators.

Proof

Let $x$ be a nongenerator of $G$, and let $M$ be a maximal subgroup of $G$. If $x\notin M$, then $G=\langle x,M\rangle=M$, a contradiction. Therefore, $x\in M$, for all $M$, and so $x\in\Phi(G)$. Conversely, if $z\in\Phi(G)$, assume that $G=\langle z,Y\rangle$. If $\langle Y\rangle\neq G$, then there exists a maximal subgroup $M$ with $\langle Y\rangle\le M$. But $z\in M$, and so $G=\langle z,Y\rangle\le M$, a contradiction. Therefore, $z$ is a nongenerator.
■

Theorem 5.6.4 Let $G$ be a finite group. Then,

1. $\Phi(G)$ is nilpotent;
2. if $G$ is a finite $p$-group, then $\Phi(G)=G'G^{p}$, where $G^p$ is the subgroup of $G$ generated by all $p$-th powers;
3. if $G$ is a finite $p$-group, then $G/\Phi(G)$ is a vector space over $\mathbb Z_p$.
4. if $G$ is a finite $p$-group, then $G$ is cyclic if and only if $G/\Phi(G)$ is cyclic.

Proof

1. Let $P$ be a Sylow $p$-subgroup of $\Phi(G)$ for some $p$. Since $\Phi(G)\lhd G$, the Frattini argument gives $G=\Phi(G)\mathrm N_G(P)$. But $\Phi(G)$ consists of nongenerators, and so $G=\mathrm N_G(P)$; that is, $P\lhd G$ and hence $P\lhd \Phi(G)$. Therefore, by Proposition 5.5.16, $\Phi(G)$ is nilpotent.
2. If $M$ is a maximal subgroup of $G$, where $G$ is now a $p$-group, then Theorem 5.5.20 gives $M\lhd G$ and $[G:M]=p$. Thus, $G/M$ is abelian, so that $G'\le M$; moreover, $G/M$ has exponent $p$, so that $x^p\in M$ for all $x\in G$. Therefore, $G'G^p\le \Phi(G)$. For the reverse inclusion, observe that $G/G'G^p$ is an abelian group of exponent $p$, hence is elementary abelian, and hence is a vector space over $\mathbb Z_p$. Clearly $\Phi(G/G'G^p)=\mathbf 1$. If $H\lhd G$ and $H\le\Phi(G)$, then it is easy to check that $\Phi(G)$ is the inverse image of $\Phi(G/H)$ under the natural map (for maximal subgroups correspond). It follows that $\Phi(G)=G'G^p$.
3. Since $G'G^p=\Phi(G)$, the quotient group $G/\Phi(G)$ is an abelian group of exponent $p$; that is, it is a vector space over $\mathbb Z_p$.
4. Let $|G|=p^n$. If $G\cong \mathbb Z_{p^{n}}$, then $G$ has only one maximal subgroup and $\Phi(G)\cong \mathbb Z_{p^{n-1}}$, which gives that $G/\Phi(G)\cong \mathbb Z_p$. Conversely, If $G/\Phi(G)$ is cyclic, then $G/\Phi(G)\cong \mathbb Z_p$ since $G/\Phi(G)$ has exponent $p$. So we have $|\Phi(G)|=p^{n-1}$. But any maximal subgroup $M$ of $G$ has order $p^{n-1}$, which gives that $G$ has only one maximal subgroup. By Proposition 2.3.4, $G$ is cyclic of order $p^n$.
   ■

Theorem 5.6.5 For every (possible infinite) group $G$, one has $G'\cap\mathrm Z(G)\le\Phi(G)$.

Proof

Denote $G'\cap\mathrm Z(G)$ by $D$. If $D\nleq \Phi(G)$, there is a maximal subgroup $M$ of $G$ with $D\nleq M$. Therefore, $G=MD$, so that each $g\in G$ has a factorization $g=md$ with $m\in M$ and $d\in D$. Since $d\in\mathrm Z(G)$, $gMg^{-1}=mdMd^{-1}m^{-1}=mMm^{-1}=M$, and so $M\lhd G$. By Proposition 2.7.4, $G/M$ has prime order, hence is abelian. Therefore, $G'\le M$. But $D\le G'\le M$, contradicting $D\nleq M$.
■

Proposition 5.6.6 (Wielandt) A finite group $G$ is nilpotent if and only if $G'\le\Phi(G)$.

Proposition 5.6.7

1. A finite group $G$ is nilpotent if and only if $G/\Phi(G)$ is nilpotent.
2. Let $N$ be a normal subgroup of a finite group $G$. If $G/\Phi(N)$ is nilpotent, then $G$ is nilpotent.
3. Let $G$ be a finite group. If $\mathrm Z(G)\le\Phi(G)$, and $G/\mathrm Z(G)$ is nilpotent, then $G$ is nilpotent.

Definition 5.6.8 A minimal generating set of a group $G$ is a generating set $X$ such that no proper subset of $X$ is a generating set of $G$.

Theorem 5.6.9 (Burnside Basis Theorem) If $G$ is a finite $p$-group, then any two minimal generating sets have the same cardinality, namely, $\dim G/\Phi(G)$. Moreover, every $x\notin \Phi(G)$ belongs to some minimal generating set of $G$.

Proof

If $\{x_1,\ldots,x_n\}$ is a minimal generating set, then the family of cosets $\{\bar x_1,\ldots,\bar x_n\}$ spans $G/\Phi(G)$ (where $\bar x$ denotes the coset $x\Phi(G)$). If this family is dependent, then one of them, say $\bar x_1$, lies in $\langle\bar x_2,\ldots,\bar x_n\rangle$. There is thus $y\in\langle x_2,\ldots, x_n\rangle\le G$ with $x_1y^{-1}\in\Phi(G)$. Clearly, $\{x_1y^{-1},x_2,\ldots,x_n\}$ generates $G$, so that $G=\langle x_2,\ldots,x_n\rangle$, by Theorem 5.6.3, and this contradicts minimality. Therefore, $n=\dim G/\Phi(G)$, and all minimal generating sets have the same cardinality.

If $x\notin\Phi(G)$, then $\bar x\neq\mathbf 0$ in the vector space $G/\Phi(G)$, and so it is part of a basis $\{\bar x,\bar x_2,\ldots,\bar x_n\}$. If $x_i$ represents the coset $\bar x_i$, for $i\ge 2$, then $G=\langle \Phi(G),x,x_2,\ldots,x_n\rangle=\langle x,x_2,\ldots,x_n\rangle$. Moreover, $\{x,x_2,\ldots,x_n\}$ is a minimal generating set, for the cosets of a proper subset do not generate $G/\Phi(G)$.
■

■■■■■■■■■■■■■■■■■■■■■■■■■

下一篇: GTM148 抄书笔记 Part III.