GTM148 抄书笔记 Part II. [不定期更新]

Contents

• Contents
• Chapter IV. The Sylow Theorems
  • \(p\)-Groups

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.
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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):

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

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 \(|\mathrm Z(G)|\).
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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.
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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, \([G:H]=p\).
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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 \(r_1\equiv 1\pmod p\).
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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 \(r_1\equiv 1\pmod p\).
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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 \(k\).
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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.
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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 \(\mathbb Z_p\times\cdots\times\mathbb Z_p\).