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  "body": "\n## Proof\n\nWe show that $\\varphi$ kills the commutator subgroup, and then apply the universal property of the quotient group.\n\n1. Let $G$ be a group, let $A$ be an [[abelian group|071DDFVG4J]] with identity $e_A$, and let $\\varphi \\colon G \\to A$ be a [[group homomorphism|0HV9Y7WG5R]].\n2. By the definition of the [[abelianisation|0H9K5D3RDV]], $G^{\\mathrm{ab}} = G/[G, G]$, and $\\pi(g) = g[G, G]$.\n3. Let $g, h \\in G$. Since [[homomorphisms preserve inverses|08FKQD7NJ3]], $\\varphi([g, h]) = \\varphi(g)^{-1} \\varphi(h)^{-1} \\varphi(g) \\varphi(h)$, where $[g, h] = g^{-1}h^{-1}gh$ is the [[commutator|06RGT4V6Y6]].\n4. Since $A$ is abelian, we may reorder the factors in step 3. This gives $\\varphi([g, h]) = \\varphi(g)^{-1} \\varphi(g) \\varphi(h)^{-1} \\varphi(h) = e_A$.\n5. So the set $C$ of all commutators lies in the [[kernel|0Z424HK8CQ]] $\\ker \\varphi$.\n6. By [[the kernel is a normal subgroup|0C9ZKYSPYR]], $\\ker \\varphi$ is a subgroup of $G$ that contains $C$.\n7. The [[commutator subgroup|08SJ5K2CBG]] $[G, G]$ is the [[subgroup generated by|0RG9SPQSWM]] $C$, the intersection of all subgroups that contain $C$. By step 6, $[G, G] \\subseteq \\ker \\varphi$.\n8. By step 7 and the [[universal property of the quotient group|0GJ83E78A5]] with $N = [G, G]$, there is exactly one homomorphism $\\bar{\\varphi} \\colon G^{\\mathrm{ab}} \\to A$ with $\\bar{\\varphi} \\circ \\pi = \\varphi$.\n\n## Idea\n\nIn an abelian group every commutator is trivial, so a homomorphism into an abelian group sends each commutator of $G$ to the identity. Hence its kernel contains the commutator subgroup. The universal property of the quotient then factors the homomorphism through $G/[G, G]$ in exactly one way.\n",
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