- Show that any finite abelian group is isomorphic to the direct product of its Sylow subgroups
Solution
Goal: Prove that every finite abelian group \(G\) is isomorphic to the direct product of its Sylow subgroups.
1. Definition of the Sylow subgroups:Proof:
1.1. 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 \(a_i \ge 1\).
1.2. For each \(i \in \{1, \dots, k\}\), define\[P_i = \left\{ x \in G : x^{p_i^{a_i}} = e \right\}.\]
1.3. 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 \(G\).
1.4. 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 \(G\).
1.5. Since \(G\) is abelian, \(P_i \trianglelefteq G\) for all \(i\).
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 \(j \in \{1, \dots, k\}\).
3. Generation of \(G\) via Bézout’s identity:Proof:
3.1. For each \(i \in \{1, \dots, k\}\), let \(m_i = |G| / p_i^{a_i}\).
3.2. Since \(p_1, \dots, p_k\) are distinct primes, \(\gcd(m_1, m_2, \dots, m_k) = 1\).
3.3. By Bézout’s identity in \(\mathbb{Z}\), there exist integers \(u_1, u_2, \dots, u_k \in \mathbb{Z}\) such that\[\sum_{i=1}^k u_i m_i = 1.\]
3.4. For any \(g \in G\):\[g = g^1 = g^{\sum_{i=1}^k u_i m_i} = \prod_{i=1}^k g^{u_i m_i}.\]
3.5. For each \(i\), let \(g_i = g^{u_i m_i}\). Then \(g_i^{p_i^{a_i}} = g^{u_i m_i p_i^{a_i}} = g^{u_i |G|} = (g^{|G|})^{u_i} = e^{u_i} = e\).
3.6. Thus \(g_i \in P_i\) for each \(i\), which proves that \(g \in P_1 P_2 \cdots P_k\).
3.7. Therefore \(G = P_1 P_2 \cdots P_k\).
4. Construction of the direct product isomorphism:Proof:
4.1. Define the map \(\Phi: P_1 \times P_2 \times \cdots \times P_k \to G\) by\[\Phi(g_1, g_2, \dots, g_k) = g_1 g_2 \cdots g_k.\]
4.2. Since \(G\) is abelian, \(\Phi\) is a group homomorphism.
4.3. By 3, \(\Phi\) is surjective.
4.4. If \(\Phi(g_1, \dots, g_k) = e\), then for each \(j\), \(g_j^{-1} = \prod_{i \neq j} g_i \in P_j \cap Q_j\).
4.5. By 2, \(P_j \cap Q_j = \{e\}\), so \(g_j = e\) for all \(j\).
4.6. Thus \(\ker \Phi = \{(e, \dots, e)\}\), so \(\Phi\) is injective.
4.7. Therefore \(\Phi\) is an isomorphism.
5. Conclusion:Proof:The finite abelian group \(G\) is isomorphic to the direct product of its Sylow subgroups \(P_1 \times P_2 \times \cdots \times P_k\).