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I need help with understanding the proof of the following Lemma(68.1, page 414) in Munkres' Topology, 2nd edition. I will be grateful if someone could take me through the proof or provide a different proof to what's found in the book.


Let $G$ be a group; Let $\{G_\alpha\}$ be a family of subgroups of $G$. If $G$ is the free product of the groups $G_\alpha$, then $G$ satisfies the following condition:

Given any group $H$ and any family of homomorphisms $h_\alpha : G_\alpha \to H$, there exists a homomorphism $h:G\to H$ whose restriction to $G_\alpha$ equals $h_\alpha$, for each $\alpha$. Furthermore, $h$ is unique.

Thanks.

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    $\begingroup$ What is the first part don't you understand? We're trying to prove existence and uniqueness of a map, so (as usual in these types of proofs) we show uniqueness first. This tells exactly what the map has to look like. It only remains to check that, if we define $h$ in that manner, that it has the desired properties. In this case, we have to show it is a well-defined homomorphism. He puts all the tough bits in that embedded lemma. $\endgroup$
    – dls
    Commented Mar 4, 2012 at 5:32
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    $\begingroup$ @dls: Hmmm... I usually show existence first, and then show uniqueness... $\endgroup$ Commented Mar 4, 2012 at 6:00
  • $\begingroup$ What part of the proof is giving you trouble? Do you understand what the free product of the $G_{\alpha}$ "looks like", or is that part of the problem? $\endgroup$ Commented Mar 4, 2012 at 6:15
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    $\begingroup$ @ArturoMagidin But don't you agree it's a common pattern, as in this proof, to assume existence, apply the fact that some diagram commutes and then deduce the form of the map (hence it must be unique)? Now you have a map in hand, so just show it has all the requested properties. $\endgroup$
    – dls
    Commented Mar 4, 2012 at 6:34
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    $\begingroup$ @dls: That's my point: that is not my experience. In my experience, the "common pattern" is (i) define a map explicitly; (ii) show it works; (iii) then show that any map that "works" must agree with the one we defined. In my experience, this is the usual pattern in any "there exists a unique blah" proof, not just with maps of universal properties: E.g., we prove unique factorization by first proving there is at least one factorization into primes, then showing any two factorizations are equal; not by first proving any two would have to agree and then showing there is one. Same with maps. $\endgroup$ Commented Mar 4, 2012 at 6:37

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The free product of the $G_{\alpha}$ consists of all words of the form $$\mathbf{g}=g_{1\alpha_1}g_{2\alpha_2}\cdots g_{m\alpha_m}$$ where $g_{i\alpha_i}\in G_{\alpha_i}-\{e\}$, $\alpha_i\neq\alpha_{i+1}$ for $i=1,\ldots,m-1$, $m\geq0$. Multiplication is by concatenation if the last index of $\mathbf{g}$ is different from the first index of $\mathbf{h}$; if they are equal, then we multiply $g_{m\alpha_m}$ by $h_{1\alpha_1}$, reduce as necessary, and continue.

The hardest part in proving that this is a group is proving associativity of the operation; trying to do it directly results in having to consider several different cases depending on how much cancellation there is.

Luckily, there is a better way of proving that the product is associative, known as "van der Waerden's trick." We embed this set into something that we already know is a group, and show that the product we have defined agrees with the operation on the known group.

Let $\mathbf{x}$ be a symbol that is not an element of any $G_{\alpha}$, and let $X$ consist of all words of the form $$g_{1\alpha_1}g_{2\alpha_2}\cdots g_{m\alpha_m}\mathbf{x},$$ with $g_{1\alpha_1}\cdots g_{m\alpha_m}$ as above. Let $S_X$ be the group of all permutations of $X$.

Let $G$ be the free product of the $G_{\alpha}$ with multiplication described as above. Given $\alpha$ and $h\in G_{\alpha}$, we define $\sigma_{h,\alpha}$ to be the element of $S_X$ given by $$\sigma_{h,\alpha}(\mathbf{g}\mathbf{x}) = \left\{\begin{array}{ll} hg_{1\alpha_1}\cdots g_{m\alpha_m}\mathbf{x} &\text{if }\alpha\neq\alpha_1\text{ or }(\alpha=\alpha_1\text{ and }h\neq g_{1\alpha_1}^{-1})\\ g_{2\alpha_2}\cdots g_{m\alpha_{m}}\mathbf{x} &\text{if }\alpha=\alpha_1\text{ and }h=g_{1\alpha_1}^{-1}\\ h\mathbf{x} & \text{if }m=0. \end{array}\right.$$ Given $\mathbf{h}=h_{1\beta_1}\cdots h_{m\beta_m}\in G$, we define $$\sigma_{\mathbf{h}} = \sigma_{h_{1\beta_1},\beta_1}\circ\cdots\circ \sigma_{h_{m\beta_m},\beta_m}.$$ (The role of $\mathbf{x}$ is as "punctuation", to ensure that there is actually an object in $X$ to which we are mapping when there is "full cancellation").

Note that the map $\mathbf{g}\mapsto\sigma_{\mathbf{g}}$ is one-to-one, since $\sigma_{\mathbf{g}} = \sigma_{\mathbf{h}}$ implies $\mathbf{g}=\sigma_{\mathbf{g}}(\mathbf{x})=\sigma_{\mathbf{h}}(\mathbf{x}) = \mathbf{h}$.

It is now easy to verify that the multiplication in $G$ satisfies $\sigma_{\mathbf{gh}} = \sigma_{\mathbf{g}}\circ\sigma_{\mathbf{h}}$; so the multiplication we have defined on $G$ is associative, since it corresponds to composition in $S_X$ and we know $S_X$ is a group. Thus, this product makes $G$ into a group.

Moreover, there is a natural embedding of each $G_{\alpha}$ into $G$, mapping $g\in G_{\alpha}$ to the word that consists only of $g$. Call this $\iota_{\alpha}\colon G_{\alpha}\to G$.

So, taking for granted that this is a group, we prove that it has the desired universal property: let $H$ be an arbitrary group, and let $h_{\alpha}\colon G_{\alpha}\to H$ be group homomorphisms from each $G_{\alpha}$. We need to show that there exists a unique homomorphism $h\colon G\to H$ such that $h_{\alpha} = h\circ \iota_{\alpha}$ for every $\alpha$.

Existence. Given $\mathbf{g}=g_{1\alpha_1}\cdots g_{m\alpha_m}\in G$ as above, we define $$h(\mathbf{g}) = h_{\alpha_1}(g_{1\alpha_1})\cdots h_{\alpha_m}(g_m\alpha_m).$$ This makes sense, because $g_{i\alpha_i} \in G_{\alpha_i}$ for each $i$, hence $h_{\alpha_i}(g_{i\alpha_i})$ is defined and an element of $H$; and thus the right hand side is a product of finitely many elements of $H$, hence an element of $H$.

We need to show that $h\circ\iota_{\alpha}=h_{\alpha}$ and that $h$ is a group homomorphism. The first is easy: if $g_{\alpha}\in G_{\alpha}$, then $h\circ\iota_{\alpha}(g_{\alpha}) = h(g_{\alpha}) = h_{\alpha}(g_{\alpha})$; thus, $h\circ\iota_{\alpha}$ agrees with $h_{\alpha}$ at every $g_{\alpha}\in G_{\alpha}$, hence the two are equal.

For the second, we need to show that $h$ is multiplicative. By induction, it suffices to look at products $g_{1\alpha_1}g_{2\alpha_2}$. If $\alpha_1=\alpha_2$, then we have $$h(g_{1\alpha_1}g_{2\alpha_1}) = h_{\alpha_1}(g_{1\alpha_1}g_{2\alpha_1}) = h_{\alpha_1}(g_{1\alpha_1})h_{\alpha_1}(g_{2\alpha_1}) = h(g_{1\alpha_1})h(g_{2\alpha_1}),$$ since $h_{\alpha_1}$ is assumed to be a group homomorphism. If $\alpha_1\neq\alpha_2$, then $$h(g_{1\alpha_1}g_{2\alpha_2}) = h_{\alpha_1}(g_{1\alpha_1})h_{\alpha_2}(g_{2\alpha_2}) = h(g_{1\alpha_1})h(g_{2\alpha_2}).$$ From this, one can easily show by induction on the number of letters in the product that $h$ is multiplicative. Thus, a homomorphism exists.

To prove uniqueness, suppose that $k\colon G\to H$ is a homomorphism such that $k\circ\iota_{\alpha} = h_{\alpha}$ for every $\alpha$. We need to show that $k=h$.

Since $G$ is generated by the words $g_{\alpha} = \iota_{\alpha}(g_{\alpha})$, it suffices to show that $k$ and $h$ agree on those elements. But this is now immediate: we have: $$k(g_{\alpha}) = k(\iota_{\alpha}(g_{\alpha})) = h_{\alpha}(g_{\alpha}) = h(g_{\alpha}).$$ Since $k$ and $h$ are both group homomorphisms and they agree on a generating set for $G$, then $k=h$. Thus, the map is unique.

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  • $\begingroup$ Thanks very much for your response. Can you please explain the paragraph beginning "Moreover, there is a natural embedding...." why do we need it? Thanks. $\endgroup$
    – Nana
    Commented Mar 5, 2012 at 0:56
  • $\begingroup$ @Nana: In order to describe the universal property; otherwise, the groups $G_{\alpha}$ are not, formally, subgroups of $G$. The embeddings allow us to identify certain subgroups of $G$ which are canonically isomorphic to the $G_{\alpha}$, which is what allows us to talk about "restrictions [of $h$] to each $G_{\alpha}$". In fact, we are talking about the composition of $h$ with these canonical embeddings. $\endgroup$ Commented Mar 5, 2012 at 0:59
  • $\begingroup$ So the restriction of $h$ to $G_\alpha$ is given by the map $\iota_\alpha$? $\endgroup$
    – Nana
    Commented Mar 5, 2012 at 1:41
  • $\begingroup$ @Nana: Yes: if we identify $G_{\alpha}$ with its image under $\iota_{\alpha}$, then we can talk about "the restriction of $h$ to [the image of] $G_{\alpha}$". This amounts to talking about the composition $h\circ\iota_{\alpha}$. $\endgroup$ Commented Mar 5, 2012 at 1:49
  • $\begingroup$ Thanks very much for your help. $\endgroup$
    – Nana
    Commented Mar 5, 2012 at 2:08

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