# Q: Commutator subgroup of free group of infinite rank [duplicate]

I know that $\ F_n$/[$\ F_n$,$\ F_n$] is isomorphic to $\Bbb Z^n$, but I do not know what happens if the rank is infinite. In particular, if the rank is countable, is the resulting group isomorphic to the direct sum or product of countable many $\Bbb Z$s? And what happens if the rank is uncountable?

• Direct sum, for any rank. – Qiaochu Yuan Sep 3 '17 at 22:20

$F_n/[F_n,F_n]$ is the Abelianization of $F_n$. That is, it is the free Abelian group of rank $n$, where $n$ is allowed to be any cardinal. A free Abelian group of rank $n$ is the direct sums of $n$ copies of $\mathbb{Z}$. For instance, $\bigoplus_{i \in \mathbb{Z}} \mathbb{Z}$ is the free Abelian group of countably infinite rank and $\bigoplus_{i \in \mathbb{R}} \mathbb{Z}$ is an example of a free Abelian group of uncountably infinite rank.