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Let $\{G_\alpha\}$ be a collection of groups and $G$ be a group.

Can we interpret $(\oplus_\alpha G_\alpha)\ast G$ in terms of $G_\alpha \ast G$ and $G_\alpha \oplus G$?

Edit : At first glance, I thought that $(\mathbb{Z}\oplus\mathbb{Z})\ast\mathbb{Z}=\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$.

But, I think that it is not true.

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1 Answer 1

Your observation, that $(\mathbb{Z}\oplus \mathbb{Z})*\mathbb{Z}$ is isomorphic to $\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$, is incorrect if by $*$ you mean the free product of groups. (If you mean the coproduct of abelian groups, then that is the same as the direct sum, so you aren't saying much).

To see this, note that in $H=\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$ is the free group of rank $3$ (it is the coproduct of three copies of the free group of rank $1$), and therefore if $w,w'\in H$ commute, then they are both powers of the same element. But in $G=(\mathbb{Z}\oplus\mathbb{Z})*\mathbb{Z}$, the element $(1,0)$ and the element $(0,1)$, considered as elements of the free product, commute, even though there is no element of which they are both powers. So $G$ cannot be isomorphic to $H$.

In general, $\Bigl(\oplus_{\alpha} G_{\alpha}\Bigl)*G$ is a quotient of $\Bigl( *_{\alpha}G_{\alpha}\Bigr)*G$, which is, in turn, a quotient of $*_{\alpha}(G_{\alpha}*G)$. But I think that's about it. The free product of groups does not distribute over the direct sum, nor conversely.

Added. By the way: you had "direct product" in your title, but in your post you had the direct sum. Remember that when the index set is infinite, the direct sum is a proper subgroup of the direct product (and we don't usually use the $\oplus$ symbol except for abelian groups anyway); for arbitrary groups, not necessarily abelian, the subgroup of the direct product consisting of the elements with finite support is called the weak or restricted direct product.

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