---
tag: FYSQ4BP2JK
kind: proof
name: Proof of the universal property of the abelianisation
msc: [20F12]
proves: 0EJ4PC8NJ6
uses: [0H9K5D3RDV, 08SJ5K2CBG, 06RGT4V6Y6, 0RG9SPQSWM, 0HV9Y7WG5R, 08FKQD7NJ3, 071DDFVG4J, 0Z424HK8CQ, 0C9ZKYSPYR, 0GJ83E78A5]
sources:
  - url: https://en.wikipedia.org/wiki/Commutator_subgroup
    locator: Abelianization
---

## Proof

We show that $\varphi$ kills the commutator subgroup, and then apply the universal property of the quotient group.

1. 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]].
2. By the definition of the [[abelianisation|0H9K5D3RDV]], $G^{\mathrm{ab}} = G/[G, G]$, and $\pi(g) = g[G, G]$.
3. 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]].
4. Since $A$ is abelian, we may reorder the factors in step 3. This gives $\varphi([g, h]) = \varphi(g)^{-1} \varphi(g) \varphi(h)^{-1} \varphi(h) = e_A$.
5. So the set $C$ of all commutators lies in the [[kernel|0Z424HK8CQ]] $\ker \varphi$.
6. By [[the kernel is a normal subgroup|0C9ZKYSPYR]], $\ker \varphi$ is a subgroup of $G$ that contains $C$.
7. The [[commutator subgroup|08SJ5K2CBG]] $[G, G]$ is the [[subgroup generated by|0RG9SPQSWM]] $C$, the intersection of all subgroups that contain $C$. By step 6, $[G, G] \subseteq \ker \varphi$.
8. 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 $\bar{\varphi} \circ \pi = \varphi$.

## Idea

In 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.