So you have a free group $F_n$, freely generated by $\alpha_1 \cdots \alpha_n$. Pick any $n$ elements $g_1 \cdots g_n$ and define an endomorphism $\psi$ of $F_n$ by $\psi(\alpha_i) = g_i^{-1}\alpha_ig_i$ and extend as usual.
It looks very much to me that $\psi$ will always be injective, but I'm having a hard time proving this. Any ideas?

Cheers in advance.


  • 3
    $\begingroup$ @Seirios It is not clear to me. In fact, $S_4$ is generated by $(1,2)$ and $(1,2,3,4)$ but not by $(1,3) = (1,2)^{(1,2)(1,2,3,4)}$ and $(1,2,3,4)$, so in $F_2 = \langle\alpha_1,\alpha_2\rangle$, the endomorphism $\psi$ given by $g_1 = \alpha_1\alpha_2$ and $g_2 = 1$ is not an automorphism. $\endgroup$
    – Aurel
    Sep 25 '17 at 7:24
  • 1
    $\begingroup$ @aurel $g_1=\alpha_1\alpha_2$ gives the same result as $g_1=\alpha_2$. I think you mean $g_1= \alpha_2\alpha_1$. Also, if you are suing right actions, then $(1,2)^{(1,2)(1,2,3,4)} = (1,2)^{(1,2,3,4)}=(2,3)$. $\endgroup$
    – Derek Holt
    Sep 25 '17 at 8:04
  • $\begingroup$ @DerekHolt Indeed, I messed up my actions, thanks. What I meant was that $(1,2)(1,2,3,4)$ sends $1\mapsto 1$ and $2\mapsto 3$ and therefore conjugates $(1,2)$ to $(1,3)$ for a suitable normalisation of conjugation. $\endgroup$
    – Aurel
    Sep 25 '17 at 8:41

It's injective. Indeed, the image is free of rank $k\le n$, and is injective if and only if $k=n$ (as $F_n$ is Hopfian: is not a proper quotient of itself). Since the image surjects onto the abelianization $\mathbf{Z}^n$, we have $k=n$. More generally, any endomorphism of a free group that maps onto a finite index subgroup of the abelianization has to be injective.

As Aurel mentions, it can fail to be an automorphism: writing $F_2=\langle x,y\rangle$, the subgroup generated by $x$ and $(yx)y(yx)^{-1}$ is a proper subgroup (of infinite index).


Another argument:

As said above, it suffices to show that the subgroup $S=<\{g_i \alpha_i g_i^{-1}\}>$ has rank $n$. We make the subgroup graph $\Gamma_S$ of $S$ by identifying at a basepoint the roots of $n$ "lollipop" graphs, the $i$th of which has a stick directed away from the basepoint reading $g_i$ and a loop at the top reading $\alpha_i$. Now we reduce this graph using Stallings foldings. It is easy to see that the rank of this graph could only decrease by folding together two of the loops. Since they are all labeled differently, none would be identified. Since $\Gamma_S$ has rank $n$, $S$ has rank $n$.

As above, this also shows that $\psi$ need not be an automorphism. It will be an automorphism exactly when $\Gamma_S$ is the bouquet with 1 vertex and $n$ edges.


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