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In Topological manifolds (Book), Lee defined the free product then free group by construction. i.e. defining a word and making set of all words a group. After that he defined Free Abelian Group in different way. A reader expects this definition: "A Free abelian Group is a free group that is also abelian", but he first defined the following operator

A formal linear combination of elements of $S$ is a map from $S$ to $\Bbb Z$ that takes the value zero for all but finitely many $\sigma\in S$.

then

Under the operation of pointwise addition, the set of all such functions is an abelian group, denoted by $\Bbb ZS$ and called the free abelian group on $S$.

Why he used this definition instead of expected one? Can anyone explain the above definition for $S=\{a,b,c\}$?

Edited after comments: Are these two different things? "Free abelian group" and " abelian free group"?

C.F.G
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    There exists exactly one non-trivial, free group which is also abelian, the group of integers under addition. – Dan Rust Sep 20 '20 at 19:10
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    The reader's expectations are going to be dashed. Free abelian groups of rank $\ge 2$ are not free groups. Free groups of rank $\ge 2$ are not abelian. – Lee Mosher Sep 20 '20 at 19:12
  • So what is the meaning of term "abelian" here? – C.F.G Sep 20 '20 at 19:16
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    Are these two different things? "Free abelian group" and " abelian free group"? – C.F.G Sep 20 '20 at 19:17
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    @C.F.G Yes, there are exactly two abelian free groups (the trivial group and the integers), but there is one free abelian group $FS$ for every set $S$. – Jackozee Hakkiuz Sep 20 '20 at 19:20
  • Could you please explain what is the relation between "Free abelian groups" and "Free groups"? (Over a fixed set $S$, we have $F(S)$ and $FA(S)$ (free abelian). ) – C.F.G Sep 20 '20 at 19:26
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    The word "free" in mathematics had a categorical meaning. Whenever you see the term "a free X" then you should be thinking "a free object in the category X". So free groups are the free objects in the category of groups, while free abelian groups are the free objects in the category of abelian groups. You also get free nilpotent groups, etc. – user1729 Sep 20 '20 at 19:48
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    Oh, also, free products also have a categorical meaning: they are the coproduct in the category of groups. I'm not completely confident I can explain why it deserves the "free" label though. – user1729 Sep 20 '20 at 19:56
  • For the last question this is a related post – C.F.G Sep 20 '20 at 20:54

2 Answers2

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As in the comments: free (abelian group) is not the same as abelian (free group). The order/grouping of words cannot be interchanged, as it happens, (though in a different universe it might have been otherwise), because most free groups are not abelian (that is "abelian (free group)" refers to only one non-trivial thing, $\mathbb Z$), while there are many free (abelian groups).

In both cases the constructions given are typical, but do not clearly explain the function of the things constructed. A simple and standard categorical characterization succeeds in characterizing both: given a category $C$ (of either all groups, or just abelian groups, or possibly some other types of things), so that every object $X$ in $C$ has an underlying set ${\mathrm set}X$ (for example, $X$ is a set plus structure), a free object $F(S)$ in $C$ on a set $S$ is an object in $C$ such that every map of sets $f:S\to {\mathrm set}X$ gives a unique $C$-morphism $F(S)\to X$ (restricting to $f$ on $S$...) We can check that the two constructions you mention succeed in exhibiting free objects in the two categories.

The free group $F(S)$ on a set $S$ does map to the free abelian group $FA(S)$, induced from the identity map $f:S\to S$... so the free abelian group is a quotient of the free group. You can check that the kernel is generated by commutators, unsurprisingly.

EDIT: it might be useful to add another operation here, "abelianization" of a group $G$, which creates the largest quotient ${\mathrm ab}G$ of $G$, with a quotient map $G\to {\mathrm ab}G$. (No, it's not a-priori clear that there is a unique such, etc...) The map from free group on $S$ to free abelian group on $S$ is exactly the abelianization map.

paul garrett
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The word "free" in mathematics had a categorical meaning. Whenever you see the term "a free X" then you should be thinking "a free object in the category X".

Free groups. Your previous question gives the definition of a free group as the free object in the category of groups:

Let $S$ be a set. A free group on $S$ is a group such that for any group $G$ and any map $f:S\to G$, there exists a unique homomorphism $\varphi:F(S)\to G$ extending $f$.

That is, there exists a unique homomorphism $\varphi$ such that the following diagram (image from wikipedia) commutes:

free group diagram

Then existence and uniqueness can be proven, so we can talk about the free group on $S$, and in fact it turns out that it is the cardinality of the set which matters, rather than the set itself (this is kinda obvious if you are used to thinking of groups as being "up to isomorphism"). As there are two-generated groups which are non-abelian, it is an easy exercise to use the above definition to prove that if $|S|>1$ then $F(S)$ is non-abelian. Slightly trickier is to prove that if $|S|=1$ then $F(S)\cong\mathbb{Z}$, but this yields the following:

A free group $F(S)$ is abelian if and only if $|S|=1$.

Free abelian groups. Okay, so that is the categorical view of a free group. Lets now switch categories to the category of abelian groups:

Let $S$ be a set. A free abelian group on $S$ is a group such that for any abelian group $G$ and any map $f:S\to G$, there exists a unique homomorphism $\varphi:FA(S)\to G$ extending $f$.

I made the word "abelian" bold as the two definitions are identical, except for this word (which has changed the category). The wikipedia article on free abelian groups does not have a diagram I can borrow, but if it did it would be identical to the free group one, apart from the label "$F_S$" which would be replaced with the label "$FA(S)$". It turns out that there is a nice description of free abelian groups on finite sets: if $S$ is a finite set then $FA(S)\cong \mathbb{Z}^{|S|}$.

We can similarly define the free nilpotent group on $S$ of class $n$, the free soluble group on $S$ of class $n$, free semigroups, free monoids, free abelian monoids, free lattices, and so on.

Free products. Free objects have a completely categorical definition (they are the "free object in the category X"). Free products of groups don't quite fit into the "free object" framework, but they sort of do (they are kind of the "free object in the category of products of groups", but I have no idea if the "category of products of groups" makes sense!). Anyway, the free product can be defined completely categorically, as it is the coproduct in the category of groups:

The free product of the groups $G$ and $H$ is a group $G\ast H$ with homomorphisms $\iota_G:G\rightarrow G\ast H$ and $\iota_H:H\rightarrow G\ast H$ such that given any other group $X$ with homomorphisms $f_G:G\rightarrow X$ and $f_H:H\rightarrow X$ there is a unique homomorphism $f:G\ast H\rightarrow X$ where $f_G=f\circ\iota_G$ and $f_H=f\circ\iota_H$.

It just so happens that the explicit constructions free groups and of free products are similar, and it is these definitions most people use on a day-to-day basis. However, it is always helpful to keep these categorical considerations in mind, especially as they link to other notions and ideas.

Coproducts, like free objects, are defined completely categorically, so we can ask if the category of abelian groups has a coproduct. It does: it is simply the direct product (the finitely-based Cartesian product).

user1729
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  • +1. Thanks for answer. I want to know is it necessary to learn about free abelian groups for computing fundamental group? because finally they are isomorphic to $\Bbb Z^n$ and these are not new groups (I think). – C.F.G Sep 21 '20 at 09:20
  • @C.F.G I would say free groups are more important in that setting, as free groups underlie presentations, and presentations crop up when computing fundamental groups. However, learning about free abelian groups is not a bad thing. Recently I was trying to show that a non-finitely generated abelian group was not free abelian, and this would have taken less time if I had previously taken the time to think about free abelian groups. – user1729 Sep 21 '20 at 09:26
  • Am I right that they are not new groups? – C.F.G Sep 21 '20 at 09:30
  • @C.F.G As I said in the answer, if $S$ is finite then yes, $FA(S)\cong \mathbb{Z}^{|S|}$ and you will be familiar with these groups already. However, if $S$ is infinite then they are more complicated. – user1729 Sep 21 '20 at 09:37
  • @C.F.G What do you mean by new? If you mean that you are more likely to have encountered free abelian groups before than free groups, you are right (because the former are simply powers of $\mathbb{Z}$ if the set you start with is finite). – Eike Schulte Sep 21 '20 at 09:38
  • I think we need not more that finite rank in computing fundamental group. – C.F.G Sep 21 '20 at 09:39
  • @C.F.G For simple examples, yes. However, many topological spaces do not have finitely generated fundamental group; the Hawaiian earring is a good example of an easily-defined space with uncountable fundamental group. – user1729 Sep 21 '20 at 09:42