1
$\begingroup$

The abelianization of a group $G$ is an abelian group $A$ and a homomorphism $\varphi: G \to A$ such that if $B$ is any abelian group, and $\phi: G \to B$ is any homomorphism, there is a unique homomorphism $\psi: A \to B$ (which might depend on $\phi$) such that $\psi \varphi = \phi$.

Question. If $A$ exists, is it unique in the sense that for any other $\varphi': G \to A'$ with the properties above, there is an isomorphism $\rho: A' \to A$ such that $\rho \varphi' = \varphi$?

$\endgroup$
1
$\begingroup$

Yes, it is unique precisely in the sense you give.

Indeed, if we have $\varphi':G\to A'$ with these properties, by the universal property of $A'$ applied to $\varphi:G\to A$ we must have exactly one group homomorphism $\rho :A'\to A$ such that $\varphi =\rho \varphi'$. But similarly you can also find a unique $\rho':A\to A'$ such that $\varphi'=\rho'\varphi$. Now you have $$\rho\rho'\varphi=\rho\varphi'=\varphi=id_A\varphi,$$ and thus $\rho\rho'$ and $id_A$ are two factorization of $\varphi$ through itself; by uniqueness of such a factorization, $\rho\rho'=id_A$. The same argument shows that $\rho'\rho =id_{A'}$, and thus $\rho$ is an isomorphism with inverse $\rho'$.

Note that the universal property implies not only that there is an isomorphism such that $\rho\varphi'=\varphi$, but that there is only one such isomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy