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I have a question about this well-known theorem about free groups by focusing on the proof stated by D. L. Johnson in his book "Presentations of groups ":

Theorem (Nielsen-Schreier): Let $F$ be a free group and $H$ a subgroup of $F$. Then $H $ is free.

In this book, he started the proof by:

Let $X$ be a set of free generators for $F$ and $U$ a Schreier transversal for $H$ (Lemma 2). The resulting set $A$ generates $H$ (Lemma 3), and thus, so does the subset $B$ (Lemma 4).

May I ask why "so does the subset $B$ "?

  • Lemma 2. Every subgroup $H$ of $F$ has a Schreier transversal.

  • Lemma 3. The elements of the set $A := \{ ux \overline{\mathbf{ux}}^{-1} \mid u \text{ is in } U \text{ and } x \text{ in }X^{+}\cup X^{-} \}$ generate $H$.

  • Lemma 4. We have $B := \{ ux \overline{\mathbf{ux}}^{-1} \mid u \text{ is in } U, x \text{ in } X \text{ and } ux \text{ does not belong to } U \}$ and $A \setminus \{e \}= B\cup B^{-1}$.

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I can't follow some of the notation, but if you know that $A = B \cup B^{-1}$ and that $\langle A\rangle = H$, then doesn't $\langle A\rangle \supset \langle B\rangle \supset A$ imply $\langle B\rangle = \langle A\rangle = H$? –  Dylan Moreland Oct 30 '11 at 17:58
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Yes, it follows immediately from Lemma 4 that $\langle A \rangle = \langle B \rangle$. –  Derek Holt Oct 30 '11 at 18:09
    
those $\overline{ux}$ shouldn't be bolded –  janmarqz Jan 8 at 4:19

1 Answer 1

I think these ideas are clarified by the use of groupoids, as giving more geometry to the algebra. See

Higgins, P.J. Notes on categories and groupoids, Mathematical Studies, Volume 32. \newblock Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of Categories, No. 7 (2005) pp 1--195.

A Schreier transversal is there seen as a maximal tree in a covering graph. This method also uses the important idea of a covering map (or morphism) of groupoids, which is closely related to actions of a group or groupoid on sets, and, of course, to the usual theory of covering maps of spaces.

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