I saw this question on an old qualifying exam:
Let $G$ be a group of order $n\ge2$. Is such a group always isomorphic to a subgroup of $GL_{n-1}(\mathbb{Z})$?
A simpler problem would be to show that $G$ is isomorphic to a subgroup of $GL_{n}(\mathbb{Z})$. This seems to follow from that $G$ is isomorphic to a subgroup $H$ of $S_{n}$ and each element of $h$ is associated to a permutation matrix $h' \in GL_{n}(\mathbb{Z})$. We can consider that the subgroup $H$ which $G$ is isomorphic to is the set of maps $f_{g}: G \rightarrow G$ that multiply on the left by $g$. We then have that every element of $H$ fixes the identity element of $G$ and hence can be viewed as a permutation of $n-1$ elements.
Is there anything more to show from here?
You found an injection from $G$ to $GL_n(\mathbb Z)$. Notice that every element in the image preserves the subgroup $H$ of $\mathbb Z^n$ generated by $(1,\dots,1)$, so in this way we get an action of $G$ on $\mathbb Z^n/H$, which is isomorphic to $\mathbb Z^{n-1}$. This action is also injective.