Generalize the result of Exercise 1 to arbitrary free products.
Solution
Goal: Prove that for an arbitrary indexed family of groups \(\{G_\alpha\}_{\alpha \in J}\), the abelianization of their free product is isomorphic to the direct sum of their individual abelianizations: \[\left( \bigast_{\alpha \in J} G_\alpha \right)_{\text{ab}} \cong \bigoplus_{\alpha \in J} (G_\alpha)_{\text{ab}}.\]
1. Notation and universal maps: Proof:
1.1. Let \(G = \bigast_{\alpha \in J} G_\alpha\), and let \(i_\alpha: G_\alpha \hookrightarrow G\) be the canonical inclusion homomorphisms into the free product.
1.2. For each \(\alpha \in J\), let \(\pi_\alpha: G_\alpha \to (G_\alpha)_{\text{ab}} = G_\alpha / [G_\alpha, G_\alpha]\) be the canonical quotient projection onto the abelianization.
1.3. Let \(A = \bigoplus_{\alpha \in J} (G_\alpha)_{\text{ab}}\), and let \(j_\alpha: (G_\alpha)_{\text{ab}} \hookrightarrow A\) be the canonical injection into the direct sum.
1.4. Let \(\pi: G \to G_{\text{ab}} = G / [G, G]\) be the quotient homomorphism onto the abelianization of \(G\).
2. Construction of the homomorphism \(\Phi: G_{\text{ab}} \to \bigoplus_{\alpha \in J} (G_\alpha)_{\text{ab}}\): Proof:
2.1. For each \(\alpha \in J\), define the composite homomorphism \(\phi_\alpha = j_\alpha \circ \pi_\alpha: G_\alpha \to A\).
2.2. By the universal property of the free product, there exists a unique group homomorphism \(\phi: G \to A\) such that \(\phi \circ i_\alpha = \phi_\alpha\) for all \(\alpha \in J\).
2.3. Because the codomain \(A\) is abelian, the commutator subgroup \([G, G] \subseteq \ker(\phi)\).
2.4. By the universal property of quotient groups (abelianization), \(\phi\) induces a unique homomorphism \(\Phi: G_{\text{ab}} \to A\) satisfying \(\Phi(\pi(g)) = \phi(g)\) for all \(g \in G\).
3. Construction of the inverse homomorphism \(\Psi: \bigoplus_{\alpha \in J} (G_\alpha)_{\text{ab}} \to G_{\text{ab}}\): Proof:
3.1. For each \(\alpha \in J\), consider the composite homomorphism \(\psi_\alpha = \pi \circ i_\alpha: G_\alpha \to G_{\text{ab}}\).
3.2. Since \(G_{\text{ab}}\) is abelian, \([G_\alpha, G_\alpha] \subseteq \ker(\psi_\alpha)\), so \(\psi_\alpha\) factors uniquely through \((G_\alpha)_{\text{ab}}\) to yield a homomorphism \(\bar{\psi}_\alpha: (G_\alpha)_{\text{ab}} \to G_{\text{ab}}\) satisfying \(\bar{\psi}_\alpha \circ \pi_\alpha = \psi_\alpha\).
3.3. By the universal property of the direct sum of abelian groups (coproduct in \(\mathbf{Ab}\)), the family \(\{\bar{\psi}_\alpha\}_{\alpha \in J}\) induces a unique homomorphism \(\Psi: A \to G_{\text{ab}}\) such that \(\Psi \circ j_\alpha = \bar{\psi}_\alpha\) for all \(\alpha \in J\).
4. Verification that \(\Phi\) and \(\Psi\) are mutually inverse: Proof:
4.1. \(\Phi \circ \Psi = \operatorname{id}_A\): For each \(\alpha \in J\) and each \(x \in (G_\alpha)_{\text{ab}}\), choose \(g \in G_\alpha\) with \(\pi_\alpha(g) = x\).Then: \[\Phi(\Psi(j_\alpha(x))) = \Phi(\bar{\psi}_\alpha(\pi_\alpha(g))) = \Phi(\pi(i_\alpha(g))) = \phi(i_\alpha(g)) = j_\alpha(\pi_\alpha(g)) = j_\alpha(x).\] Since the images \(j_\alpha((G_\alpha)_{\text{ab}})\) generate \(A\), \(\Phi \circ \Psi = \operatorname{id}_A\).
4.2. \(\Psi \circ \Phi = \operatorname{id}_{G_{\text{ab}}}\): For each generator \(\pi(i_\alpha(g)) \in G_{\text{ab}}\) (\(g \in G_\alpha\)): \[\Psi(\Phi(\pi(i_\alpha(g)))) = \Psi(j_\alpha(\pi_\alpha(g))) = \bar{\psi}_\alpha(\pi_\alpha(g)) = \pi(i_\alpha(g)).\] Since the elements \(\pi(i_\alpha(G_\alpha))\) generate \(G_{\text{ab}}\), \(\Psi \circ \Phi = \operatorname{id}_{G_{\text{ab}}}\).
5. Conclusion: \(\Phi\) is an isomorphism, so \(\left( \bigast_{\alpha \in J} G_\alpha \right)_{\text{ab}} \cong \bigoplus_{\alpha \in J} (G_\alpha)_{\text{ab}}\).Q.E.D.