# Any injective homomorphism $F_n\to F_n$ with image of finite index is bijective ($n\geq 2$)

$$\DeclareMathOperator{\im}{Im}$$ Let $$\Phi: F_n\to F_n$$ be an injective homomorphism between free groups with $$n\geq 2$$ and $$\im{\Phi}$$ having finite index. Prove that $$\Phi$$ is bijective, i.e. $$\im{\Phi}$$ has index 1.

Proof: Let $$\Phi: F_n \to F_n$$ be an injective homomorphism between free groups of rank $$n\geq 2$$ with $$\im{\Phi}$$ having finite index in $$F_n$$. Since $$\im{\Phi}$$ is a subgroup of $$F_n$$ and by the Nielson-Schreier Theorem any subgroup of a free group is free, let $$\im{\Phi}=F_m$$ for some $$m\geq 2$$. Moreover, since the index of $$F_m$$ is finite therefore $$F_m$$ is free on precisely $$k(n-1)+1$$ generators for some positive integer $$k$$. Now, since $$\Phi$$ is injective then $$\ker{\Phi}=1$$, so by the First Isomorphism Theorem we will have \begin{align*} F_n / \ker{\Phi} = F_n / 1 &\cong \im{\Phi}=F_m\\ F_n &\cong F_m\\ \end{align*} which means that $$F_n$$ and $$F_m$$ have the same rank, so $$\Phi$$ is an isomorphism, hence bijective. $$\Box$$

Is this correct?

As a background, I am allowed to know elementary group theory, as well as material from Chapter 1 of Algebraic Topology by Hatcher.

• Yes this is completely correct. One remark: there are two parts of the argument. One is to say that if $G \hookrightarrow H$ is injective map then $G$ can be identified with its image. The second is that a subgroup of a free group of rank $n$ and index $k$ is free of rank $k(n-1)+1$. Now of these two statements the first is elementary and the second is much more sophisticated. So it's a little jarring seeing someone use a sophisticated fact and then giving a detailed proof of the totally trivial part using the First Isomorphism Theorem. But that's a stylistic criticism not a mathematical one. Commented Mar 7, 2022 at 16:28

## 1 Answer

I cleaned up the reasoning in accordance with @user994373's comment above.

$$\DeclareMathOperator{\im}{Im}$$ Proof: Let $$\Phi$$ be an injective homomorphism from $$F_n$$ to itself for $$n\geq 2$$ and with $$\im{\Phi}$$ having finite index. Since $$\Phi$$ is injective, by the First Isomorphism Theorem we have $$\ker{\Phi}=1$$ so $$F_n \cong \im{\Phi}$$. Next, since $$\im{\Phi}$$ has finite index, by the Nielson-Schreier Theorem we have that $$\im{\Phi}\leq F_n$$ is a free subgroup of $$F_n$$, and in particular $$\im{\Phi}=F_m$$ where $$m=k(n-1)+1$$ for some positive integer $$k$$. Therefore we have $$F_n \cong F_m$$ and since isomorphic free groups have the same rank by the uniqueness property, therefore $$n=m$$, so $$\im{\Phi}=F_n$$ and $$\Phi$$ is a bijection. $$\Box$$