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I've seen written the following sentence:

Let $G_i$ be a collection of more than one non-trivial group. Prove that their free product is non-abelian.

Now if $\{G_i\}$ denotes the collection of more than one non-trivial group, must this collection be finite? I've only seen the definition for free products of two groups. I can see how this could repeatedly be applied to a finite number of groups. But what about a countable or an uncountable number of groups? Does here the author just imply that $\{G_i\}$ is finite?

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2 Answers 2

up vote 3 down vote accepted

It's certainly possible to define arbitrary free products; essentially, the free product of the collection $\{G_\alpha\}$ consists of all finite words in the elements of $G_\alpha$, with only the relations coming from the $G_\alpha$ themselves. PlanetMath has some more details.

In particular, though, the free product of $\{G_\alpha\}$ contains the free product of any two of the $G_\alpha$, so it would suffice to prove your claim for free products of two groups.

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All the author is saying is that the set $\{G_i\}$ is a collection of two groups or more which are assumed to be nontrivial; in particular the collection could be countable or even uncountable. The definition of the free product of an uncountable number of groups (say) is not that different from the definition of the free product of two groups; in the first case, the letters of the words of the group can come from uncountably many groups, although the length of the words must be finite, while in the second case the letters of the words can come from only two groups.

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