Let $S$ be an arbitrary set (countable or uncountable). It is clear that the free abelian group generated by $S$ is isomorphic to the direct sum

$$\bigoplus_{s\in S}\mathbb{Z}.$$

Is the free group generated by $S$ then isomorphic to

$$\ast_{s\in S}\mathbb{Z}$$ where $\ast$ denotes the free product of groups?

  • 2
    $\begingroup$ Yes, since $\mathbb{Z}$ is also the free group on 1 generator. $\endgroup$ Dec 28 '20 at 14:15


It's a general pattern based on the free-forgetful adjunction: the free functor preserves colimits, in particular coproducts, so since $S=\coprod_{s\in S}1$ where $1$ is a set with one element, then $$F(S)=\coprod_{s\in S}F(1)\,,$$ and the coproduct in the category of groups is free product and $F(1)=\Bbb Z$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.