I am reading Charlap's book on crystallographic groups and have trouble understanding the proof of the following proposition:
Proposition. Let $\Phi$ be a subgroup of $\mathrm{GL} (n, \mathbb Z)$. Then there exists a symmetric positive-definite inner product $\rho$ on $\mathbb Z^n$ such that $\Phi$ is a subgroup of $\mathrm{Aut} (\rho) = \{ U \in \mathrm{GL} (n, \mathbb Z) \mid \rho (Ux, Uy) = \rho (x, y) \text{ for all } x, y \in \mathbb Z^n \}$ if and only if $\Phi$ is finite.
It is clear that if such a $\rho$ exists, then $\Phi \subset \mathrm{Aut} (\rho)$ must be finite, since $\mathrm{Aut} (\rho)$ itself is finite, as after conjugation with an element in $\mathrm{GL} (n, \mathbb R)$ it is discrete and contained in $\mathrm{O} (n)$, which is compact.
However, to prove the converse, the proof suggests to define $\rho$ by $$ \rho (x, y) = \sum_{U \in \Phi} (Ux)\cdot(Uy) $$ where $\cdot$ is the standard inner product.
However, this definition does not necessarily define an inner product on $\mathbb Z^n$, because the definition for $\mathbb Z^n$ also requires that
Given any homomorphism $f \colon \mathbb Z^n \to \mathbb Z$, there exists a unique $x_0 \in \mathbb Z^n$ such that $f (x) = \rho (x, x_0)$ for all $x \in \mathbb Z^n$.
(This condition implies that the matrix associated to the bilinear form should be in $\mathrm{GL} (n, \mathbb Z)$.)
One can for example check that if one considers the subgroup $$ \Phi = \left\langle \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix} \right\rangle \subset \mathrm{GL} (n, \mathbb Z) $$ then $\rho$ defined as above does not satisfy this property*. ($\Phi$ is a subgroup of the symmetry group of the hexagonal lattice.)
What is going wrong here?
*Consider $\mathbb Z^2$ with basis $\alpha = (1, 0)$ and $\beta =$ the rotation of $\alpha$ through 60°. Written in this basis $\Phi$ preserves the lattice generated by $\alpha$ and $\beta$ (the generator of $\Phi$ maps $\alpha \mapsto \beta$ and $\beta \mapsto \beta - \alpha$). Using the above definition for $\rho (x, y)$ gives that the quadratic form associated to $\rho$ is given by the symmetric matrix $$ \begin{pmatrix} 8 & 4 \\ 4 & 8 \end{pmatrix} $$ which is not an element of $\mathrm{GL} (2, \mathbb Z)$ und thus not an inner product $\mathbb Z^n$ as per the extra condition.