I study Free Product of Groups in [Munkres, Topology], and I have some concerns about it.

$\textbf{Basic definition}$ 1. Let $G$ be a group, and let $\{G_\alpha\}$ be a family of subgroups of $G$. Then the $n-$tuple $$(x_{\alpha_1},x_{\alpha_2}, ..., x_{\alpha_n}) \in G_{\alpha_1} \times G_{\alpha_2} \times ... \times G_{\alpha_n}$$ is a reduce word if for all $i$, $x_{\alpha_i} $ and $x_{\alpha_{i+1}}$ belong to different subgroups $G_\alpha$ and $x_{\alpha_j} \neq e$, $e$ is the indentity in $G$, for all $1 \leq j \leq n.$

  1. Let $G$ be a group, and let $\{G_\alpha\}$ be a family of subgroups of $G$. Suppose $G_\alpha \cap G_\beta = \{e\}$ for all $\alpha \neq \beta.$ Then $G$ is a $\textbf{free product of $G_\alpha$}$ if for each $x$ there is $\textbf{only one reduced word that represents $x$.}$

$\textbf{Lemma}$ Let $G$ be a group and $\{G_\alpha\}$ be a family of groups. Let $i_\alpha : G_\alpha \rightarrow G$ be a group homomorphism. If each $\alpha$, $i_\alpha$ is injective and $G$ is a free product of $i_\alpha(G_\alpha),$ then $G$ satisfies

$\textbf{(*)}$ Given a group $H$ and a family of homomorphism $h_\alpha : G_\alpha \rightarrow H,$ there exists a homomorphism $h : G \rightarrow H$ such that $h(i_\alpha(x)) = h_\alpha(x)$ for each $\alpha.$

Furthermore, $h$ is unique.

$\textbf{(Uniqueness of free product)}$ Let $\{G_\alpha\}_{\alpha \in J}$ be a family of groups. Suppose $G$ and $G'$ are groups and $i_\alpha : G_\alpha \rightarrow G$ and $i_\alpha^{'} : G_\alpha \rightarrow G'$ are families of monomorphisms, such that $\{i_\alpha(G_\alpha)\}$ and $\{i_\alpha^{'}(G_\alpha)\}$ generates $G$ and $G'$, respectively. If $G$ and $G'$ have the property $(*)$, then there exists a unique isomorphism $\phi : G \rightarrow G'$ such that $\phi(i_\alpha(x)) = i_\alpha^{'}(x)$ for all $\alpha.$

I suppose that the Theorem should utalize the Lemma, but in the Theorem, $G$ and $G'$ are "generated" by the image, but not the free product of the image. Also, I am not sure if the uniqueness part of $h$ in the lemma is included in $(*)$.

Any suggestion to begin the proof ? Where is the important point that the Lemma comes to help?

  • $\begingroup$ So, is the lemma clear, and you're interested in the proof of 'Uniqueness of free products'? $\endgroup$ – Berci Feb 18 '18 at 23:46
  • $\begingroup$ I would like to prove the uniqueness property, yes. I think the content of the Lemma is clear to me, except the point that is uniqueness of the homomorphism $h$ is part of $*$ or not. Particularly, in the uniqueness theorem, it says $G$ and $G'$ satisfy $(*)$, does they include the part where $h$ is unique or not ? $\endgroup$ – Both Htob Feb 19 '18 at 3:30
  • $\begingroup$ Yes, uniqueness of $h$ is an essential part of (*). $\endgroup$ – Berci Feb 19 '18 at 16:05

The lemma is about a characteristic property of the free product, and this uniqueness theorem basically proves that this property is indeed characteristic, i.e. determines the free product up to isomorphism.

Actually, for the proof of (this form of) the uniqueness theorem, we don't need to use the lemma. However, together with the lemma they yield a similar uniqueness for the described construction:

Let $G_\alpha$ be subgroups of a group $G$, $\ G_\alpha\cong G'_\alpha$ subgroups of $G'$, both satisfying the given definition of free product, then we have $G\cong G'$.

As for the proof, the property (*) has to be applied twice with $H=G'$ and twice with $H=G$.

As for the lemma, given $h_\alpha:G_\alpha\to H$, for each $x\in G$ there is a unique reduced word $\langle g_{\alpha_1},\dots,g_{\alpha_n}\rangle$ such that $g_{\alpha_1}\cdot \ldots \cdot g_{\alpha_n}=x$ in $G$, so by the constraints $h(g_{\alpha_k})=h_{\alpha_k}(g_{\alpha_k})$, $\ h$ must satisfy $$h(x)=h_{\alpha_1}(g_{\alpha_1})\cdot \ldots \cdot h_{\alpha_n}(g_{\alpha_n})$$ This proves uniqueness, and this formula also defines $h$. It's only left to prove that $h$ is a homomorphism.

  • $\begingroup$ I a bit confuse ? So the Lemma is not needed for proving the uniqueness Theorem ? $\endgroup$ – Both Htob Feb 19 '18 at 3:33
  • $\begingroup$ The Lemma is proved in the book, but it says that the uniquess theorem should follows immediately form the Lemma. So I try to apply the Lemma to prove the uniqueness theorem. However, like I mentioned above, there is a different in statement between Lemma and Uniquesness Theorem. Also, this theorem should concerns the uniqueness for free product of group, but its statement assume just the groups $G$ and $G'$ are just generated by the image, which means it might not be the free product of the image. I kinda confuse what connect between the Lemma and the Theorem. $\endgroup$ – Both Htob Feb 19 '18 at 3:37
  • $\begingroup$ You have to apply (*) - including uniqueness of $h$ - for $G_\alpha$'s with codomain $G$ (here $h=I'd_G$), with codomain $G'$ (that gives an $f:G\to G'$), then two other times for $G'_\alpha$'s (yielding a $g:G'\to G$ and $id_{G'} $), then uniqueness in (*) implies $fg=id_{G'} $ and $gf=I'd_G$. $\endgroup$ – Berci Feb 19 '18 at 17:05
  • $\begingroup$ so the uniqueness of $h$ in the Lemma still holds even if $G$ in the Lemma is a free product, but in the Theory, $G$ and $G'$ are just generated by the image which might not be free product ? $\endgroup$ – Both Htob Feb 19 '18 at 18:10

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