TITLE: Free groups. Help me to understand this theorem.
QUESTION [0 upvotes]: This is the Theorem 2.9 from book Lyndon, Schupp "Combinatorial group theory".
Let $F$ - is free group, $G$ - is subgroup of $F$, $\forall g\in G$ define $G_g=Gp(\{h: h<g, h\in G\})$. So $A=\{g:g\in G, g\notin G_g\}$ is a basis of $G$.
I can't find mistake in the prove, but let $F=G=\mathbb{Z}_+$, then $G_g=G$ and $A$ is empty. So $A$ isn't basis.
What I do incorrectly?

REPLY [2 votes]: I don't know which particular ordering is being used here, but in CGT there are a few different orders that are commonly used: length order (which simply says that $v\lt w$ iff $|v|\lt|w|$, given that the words are already reduced), shortlex order (which says that $v\lt w$ iff $|v|\lt|w|$ or $|v|=|w|$ and $v$ is lexicographically before $w$), or dictionary order (which is just full lexicographic order). For all of these, canon is to take the inverse of a generator as being after the generator itself in the lexicographic order. In the case of the one-generator free group (which is isomorphic to $\langle\mathbb{Z},+\rangle$ as you note, but which I think it's more helpful here to write explicitly using a generator, say $a$, so that $F=\{a^m\}, m\in\mathbb{Z}$), the definitions are not equivalent to the usual order on $\mathbb{Z}$. Instead, for the first two $a^m\lt a^n$ if $|m|\lt|n|$, or $|m|=|n|$ and $m\gt n$ (since as mentioned above we have $a\lt a^{-1}$). Written out explicitly, the order is $e\lt a\lt a^{-1}$ $\lt a^2\lt a^{-2}\lt\ldots$. Full lexicographic order is slightly more complicated than this, but the explicit version is that $e\lt a\lt a^2\lt\ldots$ $\lt a^{-1}\lt a^{-2}\lt\ldots$. Now, consider taking $g=a$, the generator. What's the set $\{h: h\lt g\}$, under any of these orderings? What is the group $G_g$? Is $g$ a member of this group?
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