# Homework Help: Understanding the Coproduct in Grp as a Universal Object

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1. Jul 29, 2015

### NihilTico

1. The problem statement, all variables and given/known data
Coproducts exist in Grp. This starts on page 71. of his Algebra.

2. Relevant equations

Allow me to present the proof in it's entirety, modified only where it's convenient or necessary for TeXing it. I've underlined areas where I have issues and bold bracketed off my snarky/expository asides.

Let $\{G_i\}_{i\in I}$ be a family of groups. We work in a category whose objects are families of group-homomorphisms $\{g_i\colon G_i\to G\}_{i\in I}$ and whose morphisms are the obvious ones. We must find a universal element in this category. For each index $i$, we let $S_i$ be the same set $G_i$ if $G_i$ is infinite and we let $S_i$ be denumerable if $G_i$ is finite. We let $S$ be a set having the same cardinality as the set-theoretic disjoint union of the sets $S_i$ (i.e. their coproduct in Set). We let $\Gamma$ be the set of group structures on $S$ , and for each $\gamma\in\Gamma$ we let $\Phi_{\gamma}$ be the set of all families of homomorphisms $\varphi=\{\varphi_i\colon G_i\to S_{\gamma}\}$ [Aside: I interpret this $\varphi$ with an indexing set $i\in{I}$ on $\{\varphi_i\colon G_i\to S_{\gamma}\}$ which I believe Lang dropped for convenience. Additionally, $S_{\gamma}$ is the group $S$ with group structure $\gamma$.]

Each pair $(S_{\gamma},\varphi)$ where $\varphi\in\Phi_{\gamma}$, is then a group, using $\varphi$ merely as an index. We let $F_0=\prod\limits_{\gamma\in\Gamma}\prod\limits_{\varphi\in\Phi_{\gamma}}(S_{\gamma},\varphi)$, and for each $i$, we define a homomorphism $f_i\colon G_i\to F_0$ by prescribing the component of $f_i$ on each factor $(S_{\gamma},\varphi)$ to be the same as that of $\varphi_i$.

Let now $g=\{g_i\colon G_i\to G\}$ be a family of homomorphisms. Replacing $G$ if necessary by the subgroup generated by the images of the $g_i$, we see that $\left|G\right|\le\left|S\right|$, because each element of $G$ is a finite product of elements in these images. Embedding $G$ as a factor in a product $G\times S_{\gamma}$ for some $\gamma$, we may assume that $\left|G\right|=\left|S\right|$. [Non sequitur a la Lang?]

There exists a homomorphisms $g_{*}\colon F_0\to G$ such that $g_{*}\circ f_i=g_i$ for all $i$. Indeed, we may assume without loss of generality that $G=S_{\gamma}$ for some $\gamma$ and that $g=\psi$ for some $\psi\in\Phi_{\gamma}$. We let $g_{*}$ be the projection of $F_0$ on the factor $(S_{\gamma},\psi)$.

Let $F$ be the subgroup of $F_0$ generated by the union of the images of the maps $f_i$ for all $i$. The restriction of $g_{*}$ to $F$ is the unique homomorphism satisfying $f_i\circ g_{*}=g_i$ for all $i$. We have thus constructed our universal object.

Here are my questions related to the chunks of underlines.

1.
Is there any motivation for choosing $S$ as the disjoint union of the $S_i$'s? Could someone help elucidate that?

2.
Is there any reason for replacing $G$ by the subgroup generated by the union of the images of the $g_i$'s -- that's how I interpret that phrase Lang has written there.

3.
What the hell is happening with embedding $G$ into $G\times S_{\gamma}$? For instance, take $G=\{0\}$; then it is clear that $S$ is never of the same cardinality as $G$, contradicting whatever Lang has said, as far as (more like 'if') I understand him.

4.
The 4th underline chunk is directly related to the 3rd.

5.
Again, any words of wisdom to offer about the thought process behind choosing the union of the images of the $f_i$'s to generate $F$?

3. The attempt at a solution

I generally bang my head against a wall whenever I attempt to read this section and there are so many notes in the margins of my textbook that I haven't attempted ameliorating this by constructing my own such object. I'm almost afraid to try as some of the worst universal objects can be a daunting task to come up with. However, I suspect the object Lang is going for should enjoy the property that any group of the right size should be able to embed in it, similar to the construction of his Proposition 12.1.

I apologize for everything that's going on in this question, but I haven't been able to figure it out all day and I'm generally displeased with the style of writing in this section.

2. Jul 30, 2015

### NihilTico

Mea culpa, by 'his Algebra' I mean Lang's.

3. Aug 20, 2015

### NihilTico

(Edit: I realize I am relying heavily on Lang's prior proof of the existence of the free group (apparently owed to J. Tits). Please let me know if I should fill it out in detail here if it is not readily accessible to you via the internet or by the text.)

I had some time to think about this morning, so here is what I have come up with in the style of Lang's construction of the free group, but rejecting whatever his explanation is here.

I will section off my explanation and the actual meat of the construction.

Here are my thoughts:

As per usual when I construct universal objects, I find it helpful to collect data regarding the object I wish to construct. The coproduct of an arbitrary family of groups $\left\{G_i\right\}_{i\in I}$ where $i$ is an arbitrary indexing set (should it exist in Grp) is the initial object in the category $\mathfrak{C}$ whose objects are collections of morphisms from $G_i$ to some $G$. Certainly, for every group, there exists such a collection as there always exists the trivial homomorphism, so let $\left\{g_i\colon G_i\to G\right\}_{i\in I}$ be such an arbitrary collection. If the coproduct exists, then it is a collection $\left\{f_i\colon G_i\to S\right\}$ for some group $S$ such that there exists a unique homomorphism $h\colon S\to G$ for which the following diagram commutes for each $i\in{I}$ (Forgive the terrible TeX'ing):
$\begin{array}{cccc} & f_{i} & G_{i}\\ & \swarrow & \downarrow & g_{i}\\ S & \dashrightarrow & G\\ & h \end{array}$

Should it exist, there ought to be certain qualities it has that we can divine from the basic properties of groups. The prospective construction must work for any group $G$ and for our arbitrary family of groups, $G_i$.

It is easy to expunge the direct sum as an option here, as, clearly, if the construction were the direct sum, then as the order of multiplication matters for every group $G$, it is not true that the product of morphisms is unique. Indeed, if we defined $h(g_i)=\prod_{i\in I}f_i(p_i(g_i))$ where $p_i$ is the projection on the $i$th component as in Ab, then, for a non-abelian group, it is not necessarily the case that our product is unique as a possible permutation of the order of the product will change the result. Moreover, a homomorphism $h\colon S\to G$ can only pull in finitely many elements of $S$ into an operation. However, if it were to do so, we would run into the same issue of uniqueness that we had in using the direct sum.

Given the above observations, a reasonable inference is that $S$ ought to contain the image of any possible homomorphism of each $G_i$ as a subgroup in a certain way. Indeed, given $G_1$ and $G_2$, it is clear that the image of $G_1$ under $f_1\colon G_1\to S$ must have something to with the image of any homomorphism from $G_1$ to another group to work for all groups $G$. Additionally, we note that if $G_1$ and $G_2$ map into the same group, then it is a basic fact that the smallest subgroup containing their images is the subgroup generated by the union of their images.

Here is the essential construction:

This strongly suggests that we look at the free group on the set given by the set-theoretic disjoint union of the $G_i$'s. Let $G'=\biguplus_{i\in I}G_i$. As in Lang's prior construction, we 'prune' the tree $F_0$ into $F$ by looking at the subgroups that the image of $G'$ generates.

However, we now 'prune' this tree further by requiring that the restriction of the set map to each $G_i$ be a homomorphism. Thus, the object $S$ we are left with is one whose components are tuples of the form $(S'_{\gamma},\{\varphi_i\}_{i\in I})$ where $S'_{\gamma}$ is the restriction of some group structure on a subset of a 'large enough' set to the subgroup generated by the image of the set-map of $G'$ to it, and whose restriction to the $G_i$ are the homomorphisms $\varphi_i$ for each $i \in I$, respectively.

Now there is obviously a tuple $(S'_{\gamma},\{\varphi_i\}_{i\in I})$ that is isomorphic to the subgroup of $G$ generated by the union of the images of the $g_i$ $\varphi_i=g_i$ for each $i$. Just as in the case of the construction of the free group. We let $g'\colon G'\to S$ be such that $\bar{g}\in G_{j}\subseteq G'$ maps to every tuple member in the cartesian product—recalling that every tuple is of the form $(S'_{\gamma},\{\varphi_i\}_{i\in I})$ as described above—such that its image in each is the image of $\bar{g}$ under the $\varphi_j$ homomorphism in the tuple. Thus, we let $h\colon S\to {G}$ be the projection on the component satisfying $S'_{\gamma}=G$ and $g_i=\varphi_i$.

Uniqueness is simple. If $g'\colon G'\to S$ and $\beta\colon G'\to G$ (where $\beta$ is the map whose restriction to the $G_i$ in $G'$ are the $g_i$ maps give initially) are set maps into groups and the image of $G'$ generates $S$ (it does, by construction, in the construction I have just given above), then there exists at most one homomorphism $\kappa$ of $F$ into $G$ such that $\kappa\circ g'=\beta$. This follows as (should such a homomorphism exist), it is a trivial fact that the map of generators of a group uniquely determines a homomorphism.

Does that appear to be correct?

Last edited: Aug 20, 2015