If \(G = G_1 * G_2\), show that
\[ G/[G, G] \cong (G_1/[G_1, G_1]) \oplus (G_2/[G_2, G_2]). \]
[Hint: Use the extension condition for direct sums and free products to define homomorphisms
\[ G/[G, G] \to (G_1/[G_1, G_1]) \oplus (G_2/[G_2, G_2]) \]
that are inverse to each other.]
Solution
Goal: Prove that the abelianization of a free product of groups \(G = G_1 * G_2\) is canonically isomorphic to the direct sum (product) of the individual abelianizations: \(G_{\text{ab}} \cong G_1^{\text{ab}} \oplus G_2^{\text{ab}}\).
1. Universal property setup:Let \(\pi_i: G_i \to G_i^{\text{ab}} = G_i/[G_i, G_i]\) (\(i=1, 2\)) and \(\pi_G: G \to G_{\text{ab}} = G/[G, G]\) denote the canonical abelianization projections.Let \(\iota_i: G_i \to G_1 * G_2\) and \(j_i: G_i^{\text{ab}} \to G_1^{\text{ab}} \oplus G_2^{\text{ab}}\) denote the canonical inclusions.
2. Construction of the forward homomorphism \(\Phi: G_{\text{ab}} \to G_1^{\text{ab}} \oplus G_2^{\text{ab}}\):Proof:
2.1. For \(i = 1, 2\), the composite map \(\phi_i = j_i \circ \pi_i: G_i \to G_1^{\text{ab}} \oplus G_2^{\text{ab}}\) is a group homomorphism into the abelian group \(G_1^{\text{ab}} \oplus G_2^{\text{ab}}\).
2.2. By the universal property of the free product \(G = G_1 * G_2\), there exists a unique homomorphism \(\phi: G \to G_1^{\text{ab}} \oplus G_2^{\text{ab}}\) such that \(\phi \circ \iota_i = \phi_i\) for \(i = 1, 2\).
2.3. Since the target group \(G_1^{\text{ab}} \oplus G_2^{\text{ab}}\) is abelian, the commutator subgroup \([G, G] \subseteq \ker \phi\).
2.4. By the universal property of quotient groups (abelianization), \(\phi\) factors uniquely to a homomorphism:\[\Phi: G_{\text{ab}} \to G_1^{\text{ab}} \oplus G_2^{\text{ab}}, \quad \Phi(\pi_G(\iota_i(g_i))) = j_i(\pi_i(g_i)).\]
3. Construction of the reverse homomorphism \(\Psi: G_1^{\text{ab}} \oplus G_2^{\text{ab}} \to G_{\text{ab}}\):Proof:
3.1. For \(i = 1, 2\), the composite map \(\psi_i = \pi_G \circ \iota_i: G_i \to G_{\text{ab}}\) maps into the abelian group \(G_{\text{ab}}\).
3.2. By the universal property of abelianization of \(G_i\), each \(\psi_i\) factors uniquely through \(G_i^{\text{ab}}\) to yield a homomorphism \(\bar{\psi}_i: G_i^{\text{ab}} \to G_{\text{ab}}\) satisfying \(\bar{\psi}_i(\pi_i(g_i)) = \pi_G(\iota_i(g_i))\).
3.3. By the universal property of the direct sum of abelian groups, the pair \((\bar{\psi}_1, \bar{\psi}_2)\) induces a unique homomorphism:\[\Psi: G_1^{\text{ab}} \oplus G_2^{\text{ab}} \to G_{\text{ab}}, \quad \Psi(x_1, x_2) = \bar{\psi}_1(x_1) \cdot \bar{\psi}_2(x_2).\]
4. Verification that \(\Phi\) and \(\Psi\) are mutual inverses:Proof:
4.1. On \(G_{\text{ab}}\): Every element of \(G_{\text{ab}}\) is a product of cosets \(\pi_G(\iota_1(g_1))\) and \(\pi_G(\iota_2(g_2))\).\[(\Psi \circ \Phi)(\pi_G(\iota_1(g_1))) = \Psi(\pi_1(g_1), 0) = \bar{\psi}_1(\pi_1(g_1)) = \pi_G(\iota_1(g_1)).\]Similarly for elements from \(G_2\). Since these generators span \(G_{\text{ab}}\), \(\Psi \circ \Phi = \operatorname{id}_{G_{\text{ab}}}\).
4.2. On \(G_1^{\text{ab}} \oplus G_2^{\text{ab}}\): For any \((x_1, 0) \in G_1^{\text{ab}} \oplus \{0\}\), choose \(g_1 \in G_1\) with \(\pi_1(g_1) = x_1\).\[(\Phi \circ \Psi)(x_1, 0) = \Phi(\bar{\psi}_1(\pi_1(g_1))) = \Phi(\pi_G(\iota_1(g_1))) = (\pi_1(g_1), 0) = (x_1, 0).\]Similarly for \((0, x_2)\). Since these generate \(G_1^{\text{ab}} \oplus G_2^{\text{ab}}\), \(\Phi \circ \Psi = \operatorname{id}_{G_1^{\text{ab}} \oplus G_2^{\text{ab}}}\).
5. Conclusion:\(\Phi\) is an isomorphism, establishing \(G/[G, G] \cong (G_1/[G_1, G_1]) \oplus (G_2/[G_2, G_2])\). Q.E.D.