# Homeomorphisms on a finite connected graph $X$ with $H_1(X; \Bbb Z)$ free abelian

For context, this is exercise 2.2.42 in Hatcher's Algebraic Topology.

Let $X$ be a finite connected graph having no vertex that is the endpoint of just one edge, and suppose that $H_1(X; \Bbb Z)$ is free abelian of rank $n>1$, so the group of automorphisms of $H_1(X; \Bbb Z)$ is $GL_n(\Bbb Z)$. Show that if $G$ is a finite group of homeomorphisms of $X$, then the homomorphism $\phi:G \to GL_n(\Bbb Z)$ assigning to $g:X \to X$ the induced homomorphism $g_*:H_1(X;\Bbb Z) \to H_1(X;\Bbb Z)$ is injective. Show the same result holds if the coefficient group $\Bbb Z$ is replaces by $\Bbb Z_m$ with $m>2$. What goes wrong when $m=2$?

My initial attempts:

To show that $\phi$ is injective I want to show that its kernel is trivial, i. e. that the only homeomorphism in $G$ that gets mapped to the identity matrix in $GL_n(\Bbb Z)$ is the identity homeomorphism.

A graph is a 1-dimensional CW-complex, so $H_1(X; \Bbb Z) = ker(d_1)$ where $d_1$ is the boundary map on the cellular chain complex. So I think that $H_1(X; \Bbb Z)$ is generated by the set of cycles of edges in X (and there are at least 2 of these cycles since $H_1(X; \Bbb Z)$ has rank $n>1$). The identity in $GL_n(\Bbb Z)$ leaves all generators of $H_1(X; \Bbb Z)$ fixed, so a map that induces the identity on $H_1(X;\Bbb Z)$ must leave each cycle in $X$ fixed. I think that this must mean that such a map is at least homotopic to the identity map on $X$, but I am not sure where to go from here.

I would appreciate any comments on my work so far and any suggestions on how to solve this problem.