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Let $A$ be a set. We wish to construct the free group $F(A)$. It seems that this (invariably?) starts out like this:

Let $A'$ be a copy of $A$, and let $\mathscr A=A\cup A'$.

Let $\mathscr L$ be the set of finite strings of elements of $\mathscr A$.

Then it seems the usual approach does this:

Show that each element of $\mathscr L$ can be reduced to a unique irreducible form. Show that concatenation followed by reduction is a group operation on the set of irreducible forms, forming a group $F(A)$. Prove that this satisfies the universal property of a free group.

Possible alternative?

Having defined $\mathscr L$, define an equivalence relation $\sim$ by letting $s\sim t$ iff there is some finite sequence of reductions and/or expansions that transform $s$ into $t$. Let $F(A)=\mathscr L/\sim$.

This appears to make it really easy to prove that $F(A)$ is a group, and I believe it is still easy to see it is the free group, but I don't have a clue how to prove that each element has a unique irreducible representative, which would, I believe, make this a nice clean drop-in replacement for what appears to be the standard approach.

Any tips?

dfeuer
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    Chapter 11 in Rotman's book works with van der Waerden's trick, using permutations and stuff, which makes things much simpler. The use of the very equivalence relation you mention is used in Robinson's book, in chapter 2. – DonAntonio Dec 12 '13 at 17:49
  • @DonAntonio, could you give me the titles of the books and maybe also the authors' full names? Do you like them? – dfeuer Dec 12 '13 at 17:59
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    The two above are super-classical books: Joseph Rotman's "An Introduction to the theory of Groups", and Derek Robinson's "A Course in the Theory of Groups". They both are great, great books for newbies in this wonderful subject. – DonAntonio Dec 12 '13 at 18:11
  • @DonAntonio, thanks. I'll get them. – dfeuer Dec 12 '13 at 20:45

1 Answers1

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A topologist would say: Take the fundamental group of the wedge of circles of the given cardinality (rank of the group you want to get). However, this assumes that you know what fundamental groups are (and van Kampen's theorem).

Edit: If you are looking for purely algebraic approaches, take a look also at the answer

Slicker construction of the free product of groups

since the free product of $c$ copies of infinite cyclic groups satisfies is a free group of rank $c$ (one can see this via the universal property for both). I do not know how this answer compares with DonAntonio's.

Moishe Kohan
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  • Is it easy from this definition to show that such groups are free (w.r.t the universal property of free groups)? – Dan Rust Jan 22 '14 at 14:43
  • @DanielRust: Yes, if you know van Kampen's theorem. – Moishe Kohan Jan 22 '14 at 14:44
  • Don't you run into a problem then with having to define the free product of groups which is pretty close to having to define a free group in the first place? – Dan Rust Jan 22 '14 at 14:48
  • @DanielRust: No, you do not, you just verify that the fundamental group of the wedge satisfies the universal property (this is how I learned it from Massey's book on algebraic topology). – Moishe Kohan Jan 22 '14 at 14:51