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I have problem to understand the three final lines of the proof of the proposition 4.1 in page 23. The proposition is a generalization of the fact that $Aut(F_{n})$ is f.g. It says that if $F$ is a free group with basis X and $Aut_{f}(F)$ is the group that is generated by the elementary Nielsen transformations $a_{x} $ and $b_{xy}$, then for every $ u_{1}, .... u_{n} \in F $ and $a \in Aut_{f}(F)$ there is a $b \in Aut_{f}(F)$ such that $ u_{j} a = u_{j} b $ for every j. The proof starts supposing that X is infinite $X = (x_{1}, x_{2},..)$ finds a $Y= (x_{1}, ..., x_{p})$ finite such that $u_{1}, .... u_{n} \in Gp Y$ and a Z such that $Y \subseteq Gp Za^{-1}$ and we suppose that $Y \subseteq Z = (x_{1},..., x_{q}), p \le q $.

If $U = Za^{-1} = (u_{1}, ..., u_{n})$ and then there is a Nielsen transformation b such that $bU = V$ which is Nielsen reduced. Using a lemma can suppose that $V = (x_{1}, ... x_{p}, v_{p+1}, ..., v_{q})$ for some $v_{j}$. So we have that $V = bU = bZa^{-1}$.

Then uses that Z is initial segment of X and concludes that V is initial component of length q of $bXa^{-1}$. Then says that the fact that X is the matrix of the identity automorphism, implies the equation below.

I can't understand the equation $bXa^{-1}$ = $ba^{-1}$. I have a problem with notation because the book uses a notation of matrices for endomorphisms of free groups and I am not sure this means and how it is implied by the fact that $ X$ is the matrix of the identity automorphism.

About the notation: If we have an endomorphism f of the free group $F$ with basis X then we associate as matrix the ordered set $U = (u_{1}, u_{2} ..., ), u_{i}= x_{i}f $ viewed as words in $X$

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I don't have the book. :-( – Babak S. May 3 '13 at 9:28
Please include the content of the proof which you are confused about. It would be unwise to assume that anyone here has the same book (and edition) as you do. – Alexander Gruber May 3 '13 at 9:47
Thanks for your advice – user75691 May 3 '13 at 10:26
He explains this carefully in the second complete paragraph of page 22, so you really need to understand that. The point is that, if you apply the automorphism $\alpha^{-1}$ to the basis $X$ and then apply the Nielsen transformation $\beta$ to the resulting words, then the effect on $X$ is the same as the composite automorphism $\beta \alpha^{-1}$ (where $\beta$ comes first). I would advise ignoring the references to matrices. They are just put in to provide an analogy with elementary operations in linear algebra, which may or may not be helpful. – Derek Holt May 3 '13 at 11:52
@DerekHolt Can you make that an answer? – Alexander Gruber May 3 '13 at 23:14

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