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Just a note, I'm using Dummit and Foote (specificially second edition Prentice Hall, section 6.3 "A word on free groups") as my reference material. I also found a similar question here, but this is more concerned with why one would show uniqueness of the isomorphism. Anyway, I'll get to it.

In the text, $F(S)$ is the group of all the reduced words constructed by elements in $S$ and their inverses with the group operation essentially concatenation of words. After the construction of this group, the universal property is stated and proven. All of this I am ok with.

My confusion stems from the corollary of this theorem, which states

$F(S)$ is unique up to a unique isomorphism which is the identity map on the set $S$.

I'm confused because as far as I can tell, $F(S)$ was contructed in an unambiguous way. It is the set of all reduced words constructible from $S\cup S^{-1}\cup\{1\}$ with concatenation as the group operation. Why do you then want to show $F(S)$ is unique? Where is the ambiguity in the construction that I am missing?

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Dana
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  • As usual: it is not easy to show that if both you and I begin with the construction defined there we'll get the same elements and thus with the same group. Thus, if we have two insce both contain $;S;$ and the universal property and etc... – DonAntonio May 21 '16 at 19:22
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    The point is that you can define the notion of a group $G$ being free over a set $S$ (the universal property about extending mappings to group homomorphisms). The construction of your $F(S)$ with reduced words and so on is just the construction of such a free group with its (important) mapping property whose existence is a priori not obvious. So logically one should start by defining a group as being free if it has this universal property and then construct such (an important) group with reduced words. – M.U. May 21 '16 at 20:21
  • M.U., that makes a whole lot more sense. I'm not sure why the text doesn't present it in that particular order. It spends a lot of space on the construction, and then goes back abruptly to the general concept of a free group. But I suppose you're not the person I got a bone to pick with. Thanks again. – Dana May 21 '16 at 21:47
  • The point D&F are trying to make is that the universal mapping property characterizes $F(S)$, but only up to a unique isomorphism. To get a feel for why this caveat is important, imagine $S = {x,y}$ and we "switch the generators". Which one is the "real" free group over $S$? – David Wheeler May 22 '16 at 02:06
  • Hi @DavidWheeler I think I understand where you are coming from. If you defined $F(S)=(1, x,x^{2}, \ldots, x^{n}, \ldots, y, y^{2}, \ldots, y^{n}, \ldots, xy, x^{2}y, \ldots, x^{i}y, \ldots, yx, y^{2}x, \ldots, y^{j}x, \ldots)$ and so on, then I could see how talking about "switching generators" yields an isomorphism. But the way D&F construct $F(S)$ is by writing just a general word ${s_{1}}^{\epsilon_{1}}{s_{2}}^{\epsilon_{2}}\cdots{s_{n}}^{\epsilon_{n}}$ where $s_{i}$ is in $S\cup{S^{-1}}\cup{1}$. So the notion of switching generators seems a little out of place in their construction. – Dana May 22 '16 at 14:07
  • Do you see how isomorphic sets induce isomorphic free groups? Mathematicians often talk of, for example "the" free group on two generators, but what they mean is that any two such free groups (on two generators) are "the same" up to a specific "re-labelling" of the generators. Algebraically, we don't care (they "act" just the same). Ontologically, it's a different story (our actual generators may be binary strings in a computer system, or loops in a certain topological space, which clearly aren't "the same things"). Note that if we "tag" the elements of $S$ by an indexing set.... – David Wheeler May 23 '16 at 00:37
  • The properties of the index (which might, for example, be derived from the natural order of the natural numbers) then say we have "different" inclusions into $F(S)$, which is important since $F(S)$ is, in general, non-abelian (the order of the $s_i$ in the word $s_1^{\epsilon_1}s_2^{\epsilon_2}\cdots s_n^{\epsilon_n}$ matters). – David Wheeler May 23 '16 at 00:40
  • Thanks DavidWheeler. I'm not quite sure I'm following you unfortunately. I do see how isomorphic sets induce isomorphic free groups. The free group of sets with the same cardinality are going to be the same algebraically. And I understand how reindexing $S$ makes a "different" $F(S)$, i.e. it is just a permutation of any other $F(S)$ generated by a different indexing. But I don't think this is what D&F are getting at here. I think they are saying any other construction of the free group that has the universal property is isomorphic to this construction, like @M.U. said. But I could be wrong. – Dana May 23 '16 at 01:11
  • Here is a "pithier" example: consider the free product (https://en.wikipedia.org/wiki/Free_product) of $\Bbb Z$ with itself: $\Bbb Z \ast \Bbb Z$ and compare this to $F({x,y})$. Contrast this with the free group on one generator (say $x$), and the integers. The idea is there's not a "unique" way to make a free group. – David Wheeler May 23 '16 at 01:12
  • Hi David, but in the text they are not talking about different sets. They are talking about one set $S$ and two different free groups generated from this one set $F(S)$ and $F'(S)$. So maybe we're talking about two different things?''' – Dana May 23 '16 at 01:22

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a group G contains S as a subset and any other group H containing S as subset, then there exists one and only one morphism f from G to H groups such the restriction f to S is the identity of S, then this G is said free group on S (universel proprety of free group on S). 1) F (S) is a built, hence the existence. 2) If G1 and G2 are two free groups on S then are both isomorphic (therefore to F (S) also). 3) Note: I replace the injection ensembilste by inclusion.

m.idaya
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