# $F_2\ltimes F_2^{2n-4}$ is a subgroup of $\mathrm{Aut}(F_n)$.

In a paper I read that $$F_2\ltimes F_2^{2n-4}$$ is a subgroup of $$\mathrm{Aut}(F_n)$$. The proof of this fact is as follows:

Choose $$F_2\leq \mathrm{Aut}(F_2)$$ and let it act diagonally on $$F_2^{2n-4}$$, i.e. the action is $$\varphi:F_2\times F_2^{2n-4}\to F_2^{2n-4}$$ given by $$\varphi(u,(v_1,\dots,v_{2n-4}))=(uv_1,\dots,uv_{2n-4})$$. This induces a semidirect product $$F_2^{2n-4}\rtimes_{\varphi}\ F_2$$. Now, to prove that this is a subgroup of $$\mathrm{Aut}(F_n)$$ we define the following map: \begin{align*} F_2^{2n-4}\rtimes_{\varphi}\ F_2&\rightarrow \mathrm{Aut}(F_n)\\ \mathbb{u}=((v_3,w_3,\dots,v_n,w_n),\alpha)&\mapsto \phi_\mathbb{u} \end{align*} where, if $$F_n=\langle x_1,\dots,x_n\rangle$$ we have $$\phi_\mathbb{u}(x_i)=\alpha(x_i)$$ for $$i=1,2$$ and $$\phi_\mathbb{u}(x_i)=w_ix_iv_i^{-1}$$ for $$i\geq 3$$, were we are assuming that $$\alpha$$ is an automorphism of $$F_2$$. We have to show that this is an injective map and then we will be done, but this is where I'm stuck.

Let $$\mathbb{u}_1=((v_{1,3},w_{1,3},\dots,v_{1,n},w_{1,n}),\alpha_1)$$ and $$\mathbb{u}_2=((v_{2,3},w_{2,3},\dots,v_{2,n},w_{2,n}),\alpha_2)$$ and suppose that $$\phi_{\mathbb{u}_1}=\phi_{\mathbb{u}_2}$$. Then $$\alpha_1(x_i)=\alpha_2(x_i)$$ for $$i=1,2$$ and so $$\alpha_1=\alpha_2$$. However, I don't know how to proceed with the second part, i.e. if $$w_{1,i}x_iv_{1,i}^{-1}=w_{2,i}x_iv_{2,i}^{-1}$$ for $$i\geq 3$$ I'm not sure how to prove that $$w_{1,i}=w_{2,i}$$ and $$v_{1,i}=v_{2,i}$$.

• What you wrote at the end can't be right: you have a tautology, of the form $a=a$. At least I don't think it can. Feb 5 at 11:58
• what is $F_n$ ? Feb 5 at 14:07
• @FabioLucchini $F_n$ is the free group generated by $n$ elements. Feb 5 at 14:33
• The problem is that you haven't said what $w_ix_iv_i^{-1}$ means, because $w_i$ and $v_i$ are elements of $F_2$, whereas $x_i$ is an element of $F_n$. Are we somehow regarding $w_i$ and $v_i$ as words in the generators $x_1$ and $x_2$ of $F_n$? If so, then it is clear that $w_{1,i}x_iv_{1,i}^{-1}=w_{2,i}x_iv_{2,i}^{-1}$ implies $w_{1,i}=w_{2,i}$ and $v_{1,i}=v_{2,i}$ when $i>2$. Feb 7 at 16:31