Algebra qual prep week 3: Sylow theory

Write \[ V=(\ZZ/p)^n\cong \FF_p^n. \] Every group endomorphism of the elementary abelian \(p\)-group \(V\) is automatically \(\FF_p\)-linear, because for \(a\in\FF_p\) and \(v\in V\), \[ \phi(av)=\phi(v+\cdots+v)=a\phi(v). \] Hence \[ \Endo_{\Grp}(V)=\Endo_{\FF_p}(V)\cong M_n(\FF_p), \] and the automorphisms are exactly the invertible linear maps: \[ \Aut_{\Grp}(V)\cong \GL_n(\FF_p). \]

To count them, choose the columns of an invertible matrix successively. The first column can be any nonzero vector, giving \(p^n-1\) choices. Once \(j\) linearly independent columns have been chosen, their span has \(p^j\) elements, so the next column has \(p^n-p^j\) choices. Therefore \[ \left|\Aut_{\Grp}\bigl((\ZZ/p)^n\bigr)\right| =|\GL_n(\FF_p)| =\prod_{j=0}^{n-1}(p^n-p^j). \]

Let \(G\) be a finite group and let \[ |G|=p^n m, \qquad p\nmid m. \] A subgroup of order \(p^n\) is called a Sylow \(p\)-subgroup.

The Sylow theorems are:

<1>1. Existence. There exists a subgroup \[ P\le G \] of order \(p^n\).

<1>2. Conjugacy and containment. Every \(p\)-subgroup of \(G\) is contained in a Sylow \(p\)-subgroup, and any two Sylow \(p\)-subgroups are conjugate in \(G\).

<1>3. Number of Sylow subgroups. If \(n_p\) denotes the number of Sylow \(p\)-subgroups, then \[ n_p\equiv1\pmod p \] and \[ n_p\mid m. \]

In particular, \[ n_p=1 \iff \text{the Sylow $p$-subgroup is normal in $G$}. \]

Let \(n_p\) be the number of Sylow \(p\)-subgroups. Sylow’s theorem gives \[ n_p\mid q,\qquad n_p\equiv1\pmod p. \] Since \(q\) is prime, \(n_p\) is either \(1\) or \(q\). But \(p>q\), so \(q\not\equiv1\pmod p\). Hence \(n_p=1\).

Thus the Sylow \(p\)-subgroup is unique and therefore normal. Its order is \(p\), so it is nontrivial and proper in a group of order \(pq\).

Goal: Prove that every finite abelian group \(G\) is isomorphic to the direct product of its Sylow subgroups.

<1>1. Definition of the Sylow subgroups: Proof: <2>1. Let \(G\) be a finite abelian group of order \(|G| = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}\), where \(p_1, \dots, p_k\) are distinct primes and \(a_i \ge 1\). <2>2. For each \(i \in \{1, \dots, k\}\), define \[P_i = \left\{ x \in G : x^{p_i^{a_i}} = e \right\}.\] <2>3. Because \(G\) is abelian, \((xy)^{p_i^{a_i}} = x^{p_i^{a_i}} y^{p_i^{a_i}} = e\) for all \(x, y \in P_i\), and \((x^{-1})^{p_i^{a_i}} = e\), so \(P_i\) is a subgroup of \(G\). <2>4. The order of every element in \(P_i\) is a power of \(p_i\), so \(P_i\) is a \(p_i\)-subgroup. By Lagrange’s Theorem and Sylow’s Theorems (or Cauchy’s Theorem), \(|P_i| = p_i^{a_i}\), so \(P_i\) is the unique Sylow \(p_i\)-subgroup of \(G\). <2>5. Since \(G\) is abelian, \(P_i \trianglelefteq G\) for all \(i\).

<1>2. Linear independence / trivial intersections of Sylow subgroups: Proof: <2>1. For each \(j \in \{1, \dots, k\}\), define \(Q_j = P_1 P_2 \cdots P_{j-1} P_{j+1} \cdots P_k\). <2>2. Every element \(y \in Q_j\) has order dividing \(\prod_{i \neq j} p_i^{a_i} = |G| / p_j^{a_j}\), which is relatively prime to \(p_j\). <2>3. Every element \(x \in P_j\) has order dividing \(p_j^{a_j}\). <2>4. If \(z \in P_j \cap Q_j\), then the order \(o(z)\) divides both \(p_j^{a_j}\) and \(|G|/p_j^{a_j}\). <2>5. Since \(\gcd(p_j^{a_j}, |G|/p_j^{a_j}) = 1\), \(o(z) = 1\), which forces \(z = e\). <2>6. Thus \(P_j \cap Q_j = \{e\}\) for all \(j \in \{1, \dots, k\}\).

<1>3. Generation of \(G\) via Bézout’s identity: Proof: <2>1. For each \(i \in \{1, \dots, k\}\), let \(m_i = |G| / p_i^{a_i}\). <2>2. Since \(p_1, \dots, p_k\) are distinct primes, \(\gcd(m_1, m_2, \dots, m_k) = 1\). <2>3. By Bézout’s identity in \(\mathbb{Z}\), there exist integers \(u_1, u_2, \dots, u_k \in \mathbb{Z}\) such that \[\sum_{i=1}^k u_i m_i = 1.\] <2>4. For any \(g \in G\): \[g = g^1 = g^{\sum_{i=1}^k u_i m_i} = \prod_{i=1}^k g^{u_i m_i}.\] <2>5. For each \(i\), let \(g_i = g^{u_i m_i}\). Then \(g_i^{p_i^{a_i}} = g^{u_i m_i p_i^{a_i}} = g^{u_i |G|} = (g^{|G|})^{u_i} = e^{u_i} = e\). <2>6. Thus \(g_i \in P_i\) for each \(i\), which proves that \(g \in P_1 P_2 \cdots P_k\). <2>7. Therefore \(G = P_1 P_2 \cdots P_k\).

<1>4. Construction of the direct product isomorphism: Proof: <2>1. Define the map \(\Phi: P_1 \times P_2 \times \cdots \times P_k \to G\) by \[\Phi(g_1, g_2, \dots, g_k) = g_1 g_2 \cdots g_k.\] <2>2. Since \(G\) is abelian, \(\Phi\) is a group homomorphism. <2>3. By <1>3, \(\Phi\) is surjective. <2>4. If \(\Phi(g_1, \dots, g_k) = e\), then for each \(j\), \(g_j^{-1} = \prod_{i \neq j} g_i \in P_j \cap Q_j\). <2>5. By <1>2, \(P_j \cap Q_j = \{e\}\), so \(g_j = e\) for all \(j\). <2>6. Thus \(\ker \Phi = \{(e, \dots, e)\}\), so \(\Phi\) is injective. <2>7. Therefore \(\Phi\) is an isomorphism.

<1>5. Conclusion: Proof: The finite abelian group \(G\) is isomorphic to the direct product of its Sylow subgroups \(P_1 \times P_2 \times \cdots \times P_k\).

<1>1. The Sylow \(7\)-subgroup is normal.

<1>2. The Sylow \(11\)-subgroup is unique.

<1>3. The subgroup \(Q\) is central.

Let \[ G=\operatorname{Gal}(L/k). \] Since \(L/k\) is abelian Galois and \([L:k]=540\), \(G\) is a finite abelian group of order \[ 540=2^2 3^3 5. \]

<1>1. The possible primary components are \[ G_{(2)}\cong C_4\quad\text{or}\quad C_2\oplus C_2, \] \[ G_{(3)}\cong C_{27},\quad C_9\oplus C_3,\quad\text{or}\quad C_3\oplus C_3\oplus C_3, \] and \[ G_{(5)}\cong C_5. \]

<1>2. Hence there are exactly six possible Galois groups.

<1>3. An intermediate field \(E\) satisfies \([L:E]=2\) exactly when the corresponding subgroup \(\operatorname{Gal}(L/E)\le G\) has order \(2\).

<1>4. Such intermediate fields always exist.

<1>5. The number of such fields is determined by the \(2\)-primary part.