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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?

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

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 $F(1)=\Bbb Z$.

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