Let $F_n$ be the free group generated by $x_1,\cdots,x_n$ and let $K_n$ be the reduced free group, which is defined in the following two equivalent ways:
- $K_n=F_n/I$, where $I$ is the normal subgroup generated by all elements of the form $[x_i,x_i^g]$, where $[a,b]=a^{-1}b^{-1}ab$ and $a^b=b^{-1}ab$, $i\in\{1,\cdots,n\}$ and $g\in F_n$.
- $K_n=F_n/J$, where $J$ is the normal subgroup generated by all iterated commutators $[\cdots[x_{i_1},x_{i_2}],\cdots,x_{i_t}]$ with $i_1,\cdots,i_t\in\{1,\cdots,n\}$ and $i_{p}=i_{q}$ for some $1\leq p<q\leq t$.
It is a (non-trivial) fact that $I=J$ and thus the two definitions are equivalent.
Now I would like to define two homomorphisms: (Here $\mathrm{Aut}(K_n)$ is the automorphism group of $K_n$.)
- $\alpha:\mathrm{Aut}(K_n)\to\mathrm{Aut}(K_{n+1})$ by setting $(\alpha(f))(z)=f(z)$ for $z\in K_n\subseteq K_{n+1}$ and $((\alpha(f))(x_{n+1})=x_{n+1}$ for $f\in\mathrm{Aut}(K_n)$;
- $\beta:K_n\to\mathrm{Aut}(K_{n+1})$ by setting $(\beta(g))(z)=z$ for $z\in K_n\subseteq K_{n+1}$ and $((\beta(g))(x_{n+1})=g x_{n+1}g^{-1}$ for $g\in G$.
Question: Are $\alpha$ and $\beta$ well-defined?
The answer is "yes", if we replace $K_{n+1}$ by $K_n*\mathbb{Z}$ (the free product); that is, define $\alpha:\mathrm{Aut}(K_n)\to\mathrm{Aut}(K_n*\mathbb{Z})$ and $\beta:K_n\to\mathrm{Aut}(K_n*\mathbb{Z})$ in the same way. But I am not sure how to deal with the case $K_{n+1}$.