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I'm currently reading the section in Lang's 'Algebra' about free groups and their coproducts. I skipped it the first time around because I took a look at it and decided I'd be better off returning to it later, but it now feels like this is something I should complete to move foward.

My question is two-fold. I'll enumerate them because they are separate but sequential in the text.

(1)

Lang states and proves the following Lemma:

There exists a set $I$ and a family of groups $\{G_i\}_{i\in{I}}$ such that, if $g:S\to{G}$ is a map of $S$ into a group $G$, and $g$ generates $G$, then $G$ is isomorphic to some $G_i$.

proof. Let $T$ be a set which is infinite denumerable if $S$ is finite, and has the same cardinality as $S$ if $S$ is infinite. For each non-empty subset $H$ of $T$, let $\Gamma_H$ be the set of group structures on $H$. For each $\gamma\in\Gamma_H$, let $H_\gamma$ be the set $H$, together with the group structure $\gamma$. Then the family $\{H_\gamma\}$ for $\gamma\in\Gamma_H$ and $H$ ranging over subsets of $T$ is the desired family.

I'm not sure where the isomorphism is in this proof. Moreover, as I understand it, a group structure is just the set $H$ endowed with some group operation, correct? If so, what would that make $H_\gamma$? A tuple with one component the set $H$, and the other, the group $H$ (or $H_\gamma$)?

(2)

Lang uses this lemma by saying:

For each $i\in{I}$ we let $M_i$ be the set of mappings of $S$ into $G_i$. For each map $\varphi\in{M_i}$, we let $G_{i,\varphi}$ be the set-theoretic product of $G_i$ and the set with one element $\{\phi\}$, so that $G_{i,\varphi}$ is the 'same' group as $G_i$ indexed by $\varphi$. We let $$F_0=\prod_{i\in{I}}\prod_{\varphi\in{M_i}}G_{i,\varphi}$$ be the Cartesian product of the groups $G_{i,\varphi}$. We define a map $$f_0\colon S\to{F_0}$$ by sending $S$ on the factor $G_{i,\varphi}$ by means of $\varphi$ itself.

I'm not sure what kind of group ${G_{i,\varphi}}\ni(g,\varphi)$ is. Do we define operations by component-wise multiplication and let $\varphi$ act as an identity on its component?

As I understand it, the Cartesian product he describes yields a tuple where its components are of the form $(g,\varphi)$ where $g\in{G_{j}}$ and $\varphi\in{M_j}$ for some $j\in{I}$. Is that correct? I also believe that Lang explicitly avoids using the term 'direct product' here as he wants $F_0$ to be a set.

And I'm also uncomfortable with the way Lang presents $f_0$. Is he saying that we let some arbitrary component of the tuple produced by Cartesian product $F_0$, say $G_{j,\psi}$, be the codomain of the map $f_0$ and that it is 'mapped by $\psi$' onto the tuples in the $G_{j,\psi}$ component of $F_0$ consistent with how $\psi$ mapped $S$ in the case of $\psi\colon{S}\to{G_j}$?

Any input or clarification would be great. I'm sure I'm missing something very simple but I haven't been able to figure it out, sadly, so I've put this up on a 'bounty' in the hopes that someone will see it and have an insight and be able to explain it to me. Thanks!

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Regarding your question (1):

Note that there are several possible definitions of what a group is which are all "essentially the same" but differ in the formal presentation. For example, a group could be defined to be any of the following:

  • A 4-tuple $(H,\circ,e,inv)$, where $H$ is a set, $\circ$ a map $H \times H \to H$, $e \in H$ and $inv$ a map $H \to H$ such that the familiar group axioms hold (with $e$ being the neutral element and the $inv$ mapping an element to its inverse).
  • A 3-tuple $(H,\circ,e)$, where ...
  • A pair $(H,\circ)$, where ...

It is a well-known fact that, if $(H,\circ)$ is a group, $e$ and the map $inv$ are uniquely determined by the axioms imposed on them; in this sense, the above definitions are "essentially the same". With this in mind, a group structure $\gamma$ on a given set $H$ could be any of the following:

  • A 3-tuple $(\circ,e,inv)$, where $H$ is a set, $\circ$ a map $H \times H \to H$, $e \in H$ and $inv$ a map $H \to H$ such that ...
  • A pair $(\circ,e)$, where ...
  • A map $\circ\colon H \times H \to H$ such that ...

To simplify notation, let us settle for the last variant.

That being said, where is the isomorphism in Lang's proof? The following facts should help you find it:

  • $|\operatorname{im}(g)| \leq |S|$, where $|-|$ denotes the cardinality of a set and $\operatorname{im}(-)$ the image of a map.
  • If a group $G$ is generated by a subset $M \subset G$, then
    • If $M$ is finite, $G$ is at most countable.
    • If $M$ is infinite, then $|G| = |M|$.

Indeed, because of these facts, we have (using the notation of Lang's proof) $|G| \leq |T|$. We can therefore pick a subset $H \subset T$ and a bijection $f\colon G \to H$. We can thus transport $G$'s group structure to $H$ via $f$, i.e.: Let $\circ$ denote the group structure of $G$ (remember this just means that $\circ\colon G \times G \to G$ is the group operation of $G$). Then $$ \gamma\colon H \times H \to H, (h,h') \mapsto f(f^{-1}(h) \circ f^{-1}(h')) $$ is a group structure on $H$. Therefore, $(H,\gamma) = H_\gamma$ is in the constructed family and $f$ "is" a group isomorphism $G \cong H_\gamma$ by construction.

Regarding your question (2):

Your description of the group operation of $G_{i,\varphi}$ is correct. Note that there is an obvious isomorphism $\Psi_{i,\varphi}\colon G \overset{\sim}{\to} G_{i,\varphi}$, which we shall use later (how does it look?).
Regarding $F_0$: One element of $F_0$ is a tuple with components of the form you stated.
Regarding $f_0$: I am not sure whether you understood this part correctly. The codomain of $f_0$ is $F_0$ - not just one of its factors. But I can see where confusion might arise here. To understand what Lang does here, you should convince yourself that the Cartesian product together with the usual projection maps is the categorical product in the category of sets. This means roughly the following: A map into a Cartesian product is determined by its composites with all the projections; and, conversely, a family consisting of one map from one fixed set into each factor of a Cartesian product gives rise to a map of the fixed set into the Cartesian product. (Try to make this precise!)
Therefore, Lang can define a map $f_0\colon S \to F_0$ by giving, for each factor $G_{i,\varphi}$, a map $(f_0)_{i,\varphi}\colon S \to G_{i,\varphi}$. To this end, let $i \in I$ and $\varphi \in M_i$; then $\varphi$ is by definition a map $S \to G_i$ and we set $(f_0)_{i,\varphi} = \Psi_{i,\varphi} \circ \varphi$ (remember that $\Psi_{i,\varphi}\colon G_i \to G_{i,\varphi}$ is just the obvious group isomorphism). By the above discussion, the family of maps $((f_0)_{i,\varphi})$, where $i$ runs through $I$ and $\varphi$ through $M_i$ ($i$ being fixed by the first component!), determines a map $f_0\colon S \to F_0$. To be precise, $f_0$ is determined by $$ \forall i \in I \forall \varphi \in M_i : p_{i,\varphi} \circ f_0 = (f_0)_{i,\varphi} \quad , $$ where $p_{i,\varphi}\colon F_0 \to G_{i,\varphi}$ is the canonical projection.

While my explanation might be overly formal, I hope that is still helpful for you.

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  • $\begingroup$ I'd like to apologize, first, for getting back to you on this so late! Something came up and it's just been a crazy week for me. So the first part is rather obvious, I see. The isomorphism $\Psi_{i,\varphi}$ should map elements to the component containing that element. I believe I or Lang have proven that the product in Set is as you have described. So this makes it also fairly obvious (perhaps 'clear' is a better word here). Seeing it explained makes me feel like this is a big "D'oh!" moment for me! Haha, thank you very much! $\endgroup$ – Nobody Mar 14 '15 at 23:04

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