My question is about a proposition that I found on Munkres book: Topology (page 419). My definitions are based on this book.

I give you the definition of free product that he uses.

$\textbf {Definition}$ Let $G$ be a group, let $\{G_\alpha\}_{\alpha\in J}$ a family of subgroups of $G$ that generates $G$. Suppose that $G_\alpha\cap G_\beta$ consists of the identity element alone whenever $\alpha\not=\beta$. We say that $G$ is the $\textbf{free product}$ of the groups $G_\alpha$ if for each $x\in G$, there is only one reduced word in the groups $G_\alpha$ that represents $x$. In this case, we write: $G=\prod_{\alpha \in J}^*G_\alpha$.

The following is the proposition that I don't understand:

$ \textbf {Proposition}$ Let $G=G_1*G_2$, where $G_1$ is the free product of the subgroups $\{H_\alpha\}_{\alpha\in J}$ and $G_2$ is the free product of the subgroups $\{H_\beta\}_{\beta\in K}$. $\textbf{If the index sets $J$ and $K$ are disjoint}$, then $G$ is the free product of the subgroups $\{H_\gamma\}_{\gamma\in J\cup K}$

So we have a group $G$ and subgroups $G_1$ and $G_2$ with the properties in the definition. Moreover, $G_1$ is the free product of some its subgroups and by transitivity these $H_\alpha$ are subgroups of $G$ too. Analogous for $G_2$.

I don't understand why he requires the sets $J$ and $K$ have to be disjoint. I mean, for me they $\textbf{always}$ have this property. Indeed, suppose they have a common index, say $c$. Then we have that $H_c$ is a subgroup of $G_1$ and at the same time $H_c$ is a subgroup of $G_2$ too, so their intersection cannot consist of the identity element alone and so they cannot be factors of the free product.

He makes the same request also to prove the analogous proposition about direct sums.

Where is my mistake?

  • 1
    $\begingroup$ I agree with you. With the given definitions, the disjointness of $J$ and $K$ follows from the assumption that $G = G_1 * G_2$. But a redundant assumption is not a mistake. $\endgroup$ – Derek Holt Jun 24 '16 at 15:28
  • $\begingroup$ Perhaps the intention of the author was to stress that for $\;G_1,\,G_2\;$ to be part of a free product then...etc. $\endgroup$ – DonAntonio Jun 24 '16 at 15:31
  • $\begingroup$ @DerekHolt I was wondering if the situation could change if, with an abuse of notation, he meant $G_1*G_2$ to be the external product of two groups. However it seems strange to me since in the analogous case with direct sums he defines external direct sums after the similar proposition. $\endgroup$ – Richard Jun 24 '16 at 15:40
  • $\begingroup$ These definitions are strange. It's simpler to define the "free product" (i.e. the coproduct) externally by a mapping property, then your result is straightforward with the indexing set given by $J\coprod K$ the coproduct of sets (aka the disjoint union). The exercise is then to prove that the internal definition given above defines a coproduct. $\endgroup$ – Justin Young Jun 24 '16 at 17:14
  • $\begingroup$ @JustinYoung I don't agree that the definition above is strange. At the same time, could you indicate to me some good references that use your definition of free product? $\endgroup$ – Richard Jun 24 '16 at 17:46

If $\;x\in I\cap J\;$ , then $\;G_x\;$ appears as free factor in both $\;G_a,\,G_2\;$ , so it won't be true that $\;G_r\cap G_s=1\;$ always, as one of the possible intersections will be $\;G_x\cap G_x=G_x\;$ .

The basic assumption is that the free products considered here are non-trivial, meaning: $\;G_i\neq 1\;$ forall the groups considered.

Yes, as commented, if it is given $\;G_1*G_2\;$ is already a free group then the condition seems to be superfluous.


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