# Free groups as free product of infinite cyclic groups

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?

• Yes, since $\mathbb{Z}$ is also the free group on 1 generator. 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 $$F(1)=\Bbb Z$$.