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Here is a theorem about characteristic property of the free group:

Theorem (Lee TM). Let $S$ be a set. For any group $H$ and any map $f:S\to H$, there exists a unique homomorphism $g:F(S)\to H$ extending $f$.

Here $F(S)$ is free group on $S$.

I know what it says but I don't know why it should be useful. i.e. What is the strategy of such theorems? How it can help to understand $F(S)$?

Can anyone enlighten it by a simple example?

C.F.G
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  • What does "stratergy of a theorem" mean? – markvs Sep 19 '20 at 18:52
  • How it can help us. – C.F.G Sep 19 '20 at 18:53
  • The answer depends on who are "us". Free groups are important. This is their definition. – markvs Sep 19 '20 at 18:55
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    It tells you that it is very easy to define homomorphisms from free groups to other groups since you only have to define a "normal" map on the generators. That is clearly a useful property. It is similar to defining linear maps on a basis of a vector space. You do not need to worry about the group homomorphism property but just get it for free. – Con Sep 19 '20 at 18:55
  • So this theorem does not help in understanding $F(S)$? i.e. $F(S)\simeq ?$ – C.F.G Sep 19 '20 at 18:57
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    Well, free groups are uniquely determined by their rank (up to isomorphism of course). I don't know what else you would like to have. – Con Sep 19 '20 at 18:58
  • By rank? any reference? – C.F.G Sep 19 '20 at 18:59
  • So what does this theorem says? – C.F.G Sep 19 '20 at 19:00
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    rank = cardinality of any set of generators. Should be in any book/notes that introduce free groups. – Con Sep 19 '20 at 19:00
  • I told you above what it says. It tells you how to define group homomorphisms from a free group to another group. It actually even gives a bijection between the maps from $S$ to a group $G$ and homomorphisms from $F(S)$ to $G$. – Con Sep 19 '20 at 19:00
  • sorry I read it partially. – C.F.G Sep 19 '20 at 19:02
  • It helps "us" prove that certain groups are free, that is statements like $F(S)\cong ...$ – markvs Sep 19 '20 at 19:03
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    This is the definition of a free group, as the free object in the category of groups (that is, it is the reason why they are specifically called free groups). Proofs which use this definition appear in many standard texts. Also, I find it weird that your statement is as a theorem attributed to someone who isn't a standard person in the history of free groups. So I want to ask: what text are you working from? – user1729 Sep 19 '20 at 19:04
  • I guess in OP's source free groups are defined via their construction, which I do not like, but this is sadly done sometimes. – Con Sep 19 '20 at 19:05
  • Lee Topological manifold. – C.F.G Sep 19 '20 at 19:05
  • Yeah, they are defined via their construction. Quote from Lee:" In some texts, a free product is defined as any group satisfying the characteristic property, or as the coproduct in the category of groups." – C.F.G Sep 19 '20 at 19:06
  • @TMO: Can I ask why you don't like? it is helpful to understand free groups. – C.F.G Sep 19 '20 at 19:08
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    Because the construction does not tell you anything at first. It does not motivate the concept. One should define free groups via this property so that one knows what they should do (or explain that more) and then show their existence by constructing them in the usual way. This is at least my opinion. – Con Sep 19 '20 at 19:10
  • Unlike you, I don't understand the concept without heuristic definition. – C.F.G Sep 19 '20 at 19:14

2 Answers2

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An analogy with vector spaces may be helpful. If $V$ is a vector space with basis $B$, then any set function on $B$ into another vector space $W$ can be extended to a linear map $V \to W$. So $V$ is a free vector space on the basis $B$. In the case of vector spaces this property does not characterise $V$ because every vector space has a basis. This is not true in the case of groups, even if they are abelian. Free groups are special in that they admit "basis" expansions, and group homomorphisms from a free group are determined completely by their action on "basis elements". In the same way that a vector space is determined (upto linear isomorphism) by its vector space dimension, which is the cardinality of any basis for that space, a free group is determined (upto group isomorphism) by the cardinality of its generating set, called its rank.

abhi01nat
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7

This characterisation emphasises that free groups are related to homomorphisms. Here is a simple proof which uses this characterisation:

Theorem. If $|X|\geq2$ then $F(X)$ is non-soluble.

Proof. The symmetric group $S_5$ can be generated by two elements. Therefore, by this characterisation of free groups, $F(X)$ subjects onto $S_5$ for all sets $X$ with at least two elements. As $S_5$ is non-soluble, as as solubility is preserved under homomorphic images, the result follows.

(Previously this answer used $S_3$ instead of $S_5$ to prove that free groups are non-nilpotent. The answer is now stronger, as non-soluble implies non-nilpotent.)

user1729
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