2
$\begingroup$

I'm reading the book "Combinatorial group Theory and applications to geometry" by D.j. Collins.

In Chapter 2.3 , definition 2.3.1 i show a definition about Νielsen αutomorphisms.

We denote with $F_n$ the free group with $n$ generators with basis $X=(x_1,x_2,\dots,x_n)$.

These are the automorphisms where they act on the tuple $X$ as elementary Nielsen transformations as explained in the book:

For an element $x_i \in X$ we define the automorphism which sends $x_i$ to $x_i^{-1}$ and for different elements $x_i,x_j \in X$ where $i \neq j$ we define the automorphism which sends $x_ix_j$.

This is the Definition 2.3.1:

For any permutation $π$ of $1,2,\dots,n$ and $\epsilon_i=\{1,-1\}$ let $\phi:(x_1,x_2,\dots,x_n) \rightarrow (x_{\pi(1)}^{\epsilon_1},\dots,x_{\pi(n)}^{\epsilon_n})$. This is a permutation automorphism.

I'm trying to see how an automorphism of free group $F_n$, may be represented by a composition of permutations automorphisms.

My first thought is that there is an injective and surjective map to the permutation group $S_n: \{1,2,\dots,n\} \rightarrow \{1,2,\dots,n\}$. So, if there is a matrix with all permutation from Nielsen automorphisms action in the tuple $(x_1,x_2,\dots,x_n)$ , I can write it as product of disjoint cycles and transpositions where each cycle can represent and a step of this action.

My problem is how exactly can i represent the first automorphism which sends $x_i$ to $x_i^{-1}$ as an element of the group $S_n$?

A short example on which I'm trying to apply my way: $(x_1,x_2) \rightarrow ((x_1x_2)^{-1},x_2)\rightarrow(x_2^{-1}x_1^{-1},x_2x_2^{-1}x_1^{-1})\rightarrow (x_2^{-1}x_1^{-1},(x_1^{-1})^{-1})\rightarrow (x_2^{-1}x_1^{-1}x_1,x_1)\rightarrow(x_2,x_1)$

Edit later: How a permutation automorphism can be expressible by a composition of nielsen automorphisms?Any example would be helpful.

$\endgroup$
  • $\begingroup$ Arbitrary automorphisms of $F_n$ are not always expressible as compositions of permutation automorphisms. In particular, a Nielsen automorphism that inverts a generator is not expressible in this way. Put another way, the permutation automorphisms generate a proper subgroup of $\operatorname{Aut} F_n$. $\endgroup$ – James Nov 4 '15 at 14:36
  • $\begingroup$ But how a permutation automorphism can be expressible by a composition of nielsen automorphisms? $\endgroup$ – mpouxa Nov 4 '15 at 15:06
  • 3
    $\begingroup$ You have shown how to do a transposition of two generators using Nielsen automorphisms, and every element of $S_n$ is a composite of transpositions, so every permutation of the generators can be so expressed. You can then replace some of them by inverses immediately, because there is a Nielsen automorphism that does that. $\endgroup$ – Derek Holt Nov 4 '15 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.