# A question about Nielsen–Schreier theorem

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
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 '14 at 4:19