# Example III.3.2(1) of Baumslag's “Topics in Combinatorial Group Theory”: proving $F=\operatorname{gp}(1+\xi\mid \xi\in\Xi)$ is free.

This question is a little tricky (for me, at least), since in the textbook the proof of

Theorem 5: Every subgroup of a free group is free.

is not yet provided (even though I've seen such proofs in Magnus et al.'s "Combinatorial Group Theory [...]," in Lyndon & Schupp's "Combinatorial Group Theory," and in Johnson's "Presentation$$\color{red}{s}$$ of Groups").

## Here is the exercise:

Let $$\Xi$$ be a set of non-commuting variables. Consider the group of units of the ring of formal power series, with integer coefficients, in the non-commuting variables coming from $$\Xi$$. Prove that

$$F=\operatorname{gp}(1+\xi\mid \xi\in\Xi)$$

is free on $$\{1+\xi\mid \xi\in\Xi\}$$. (This is a theorem of W. Magnus.)

Before this exercise is the following

## Hint:

Before starting out on the proof of Theorem 5, we give some examples of some groups which turn out to be free. The proofs require the criterion (iv), of Theorem 1 of Chapter 1. (Emphasis added.)

Okay, so here is that criterion:

Theorem 1: (iv) Let $$G$$ be a group and suppose that $$X$$ generates $$G$$. If every reduced $$X$$-word is different from $$1$$, then $$G$$ is a free group, freely generated by $$X$$.

Notation: Let $$X\subseteq G$$. The subgroup of $$G$$ generated by $$X$$ is denoted $$\operatorname{gp}(X)$$.

This notation is so that such subgroups are not confused with presentations.

Thoughts:

This is a question I think I should be able to do myself, especially since it seems to be a matter of letting $$x_\xi:=1+\xi$$ - or something to that effect - as there are no restrictions on the elements of $$\Xi$$. But then I suspect I'd be using Theorem 5, which would result in an annoying anachronism in my understanding of the material in the book.

• I'm a little unclear on what you want. Are you asking how to prove "the exercise" using Theorem 1? – Lee Mosher Apr 15 at 11:43
• Yes, @LeeMosher. I'm sorry it's unclear. Please let me know how I can improve the question. – Shaun Apr 15 at 11:44
• Writing an actual question somewhere, with a question mark, would help. – Lee Mosher Apr 15 at 11:45
• Is that any better, @LeeMosher? – Shaun Apr 15 at 11:47
• There's still no question mark, but clear enough from the comments, I suppose. – Lee Mosher Apr 15 at 11:49

So to start, you should take a sequence of elements $$\xi_1,...,\xi_n \in \Xi$$ and a sequence of exponents $$\epsilon_1,...,\epsilon_n \in \{-1,+1\}$$, and you should assume that for all $$i=1,...,n-1$$ if $$\xi_i = \xi_{i+1}$$ then $$\epsilon_i = \epsilon_{i+1}$$. Using those assumptions, your job is then to prove that product $$(1 + \xi_1)^{\epsilon_1}\cdots(1 + \xi_n)^{\epsilon_n} \quad (*)$$ is not the identity element of the ring of formal power series. The assumptions were made in order to express that the product $$(*)$$ is a reduced word in the set of generators $$X = \{1 + \xi \mid \xi \in \Xi\}$$.