Automorphisms of free groups

Question

Suppose $U$ is a subgroup of finite index in the free group on $k$ generators $F_k$.

Suppose $\sigma$ is an automorphism of $F_k$ such that $\sigma|_U = \text{id}$, then must $\sigma = \text{id}$?

Answer 1

So...it turns out this is one of the situations where you read a sentence in a paper, and only after you spend forever thinking about why its true, that you notice the sentence ends in a colon, and the very next sentence has the proof. Basically, the proof is this:

Let $g_1,\ldots,g_k$ be the free generators of $F_k$. Then for any $i$ we can build a surjective homomorphism $F_k\rightarrow\mathbb{Z}$ sending any word in $F_k$ to the number of $g_i$'s appearing in that word. If no power of $g_i$ is in $U$, then $U$ is contained in the kernel of this homomorphism, and hence must be infinite index. Therefore, for every $i$, there is an $n$ such that $g_i^n\in U$. But that means $\sigma(g_i)^n = \sigma(g_i^n) = g_i^n$, and since $F_k$ is free, that means $\sigma(g_i) = g_i$. Since the $g_i$'s generate $F_k$, $\sigma = \text{id}$.