I am trying to prove the following:
If $g$ is a flat Riemannian metric on $\mathbb T^n$ (the $n$-dimensional torus), then $(\mathbb T^n, g)$ is isometric to a Riemannian quotient of the form $\mathbb R^n/\Lambda$, where $\Lambda$ is a complete lattice acting on $\mathbb R^n$ by vector addition.
By "complete lattice", we mean an additive subgroup $\Lambda \leq \mathbb R^n$ of the form $$ \Lambda = \{m_1 v_1 + \cdots + m_n v_n : m_1, \ldots, m_n \in \mathbb Z\} $$ for some fixed basis $\{v_1, \ldots, v_n\}$ of $\mathbb R^n$.
What I've tried: Since $\mathbb T^n$ is complete, connected, and has constant sectional curvature equal to $0$ in this flat metric, $\mathbb T^n$ is isometric to $\mathbb R^n/\Gamma$ for some discrete subgroup $\Gamma \leq \mathrm{E}(n)$ acting freely on $\mathbb R^n$. Here, $\mathrm{E}(n) = \mathbb R^n \rtimes \mathrm{O}(n)$ is the complete group of Euclidean isometries with group law $$ (b,A)(b',A') = (b+Ab', AA') $$ whose action on $\mathbb R^n$ is given by $(b,A) \cdot x = b + Ax$ for $x \in \mathbb R^n$, $(b,A) \in \mathrm E(n)$.
It suffices to show that for every $(b,A) \in \Gamma$, $A$ is the $n \times n$ identity matrix $I_n$. To this end, $\Gamma$ is a subgroup of the automorphism group of the covering $q : \mathbb R^n \to \mathbb R^n/\Gamma \approx \mathbb T^n$. Since this is the universal covering, $\mathrm{Aut}_q(\mathbb R^n) \cong \pi_1(\mathbb T^n) \cong \mathbb Z^n$. Thus, $\Gamma \cong \mathbb Z^k$ for some $0 \leq k \leq n$.
Also, if $(b,A) \in \mathrm E(n)$, we have $(b,A) \cdot x = b+Ax = x$ for some $x \in \mathbb R^n$ if and only if $b=(I_n-A)x$ for some $x \in \mathbb R^n$, so if $(b,A) \in \Gamma$, then $I_n - A$ is singular, so $1$ is an eigenvalue for $A$.
My question: Does commutativity of $\Gamma$ ensure that $A=I_n$ for every $(b,A) \in \Gamma$?
Obviously the group of matrices (without translation vectors) in the elements of $\Gamma$ must form a commutative subgroup of $\mathrm{O}(n)$, but this doesn't ensure $\Gamma$ itself is commutative; I have lots of low-dimensional counter-examples, but nothing sufficiently general. And referring to the group law, we have $b+Ab' = b' + A'b$ for every $(b,A), (b', A') \in \Gamma$, so $(A-I_n)b' = (I_n-A')b$, but I'm having trouble seeing why that helps. (Though if $A = A'$, we have that $b-b'$ is fixed by $A$.) Any thoughts?