Let $\Lambda \subset \mathbb{Z}^n$ be a proper sublattice (so that $\Lambda \ne \mathbb{Z}^n$). We say that $\Lambda$ is cyclically generated if there exists a matrix $M \in \text{GL}\_n(\mathbb{Z})$ and an element $\mathbf{u} \in \mathbb{Z}^n$ such that $\Lambda$ is equal the $\mathbb{Z}$-span of $\{\mathbf{u}, M \mathbf{u}, M^2 \mathbf{u}, \cdots \}$. Is it true that all proper sublattices are cyclically generated? If so, how would one prove this, and if not, what's a counterexample?



Can all proper sublattices of $\mathbb{Z}^n$ be generated cyclically?
Some standard latticework allows us to write $\Lambda = \text{im}(A)$ for some $n$ by $n$ matrix $A$ (with nonzero determinant). Using Smith normal form, there are some $U, V \in GL\_n(\mathbb{Z})$ such that $UAV = D$, where $D$ is a diagonal matrix with diagonal elements $d\_1 | d\_2 | \dots | d\_n$.


Write $\vec{e}\_i$ for each of the standard basis vectors. Then let $\vec{u} := U^{-1} d\_1 \vec{e}\_1$, and let $M := U^{-1}BU$, where $B$ is the bidiagonal matrix with all $1$s on the diagonal, and $\frac{d\_{i + 1}}{d\_i}$ on the subdiagonal. Note that $det(B) = 1$, and so $B \in GL\_n(\mathbb{Z})$; correspondingly, so is $M$. I claim that $(\vec{u}, M)$ is a cyclic pair for $\Lambda$.


Proof: Write $\vec{v}\_i := U^{-1} d\_i \vec{e}\_i$ for each $i$. Note that $\vec{u} = \vec{v}\_1$. Then each $\vec{v}\_i \in \text{im}(U^{-1}D) = \text{im}(U^{-1}DV^{-1}) = \text{im}(M) = \Lambda$; further, it's not hard to see that they generate $\Lambda$ (by the same reasoning taken backwards).


We can check that $M \vec{v}\_i = U^{-1}BU U^{-1} d\_i \vec{e}\_i = U^{-1} B d\_i \vec{e}\_i = U^{-1} (d\_i \vec{e}\_i + d\_{i + 1} \vec{e}\_{i + 1}) = \vec{v}\_i + \vec{v}\_{i + 1}$ for $1 \leq i < n$. Noting the base case that $\vec{v}\_1 = \vec{u}$, we can use induction to see that $\vec{v}\_{i + 1} = M \vec{v}\_i - \vec{v}\_i \in \text{span}(\{M^j \vec{u}\}\_{j = 0}^i)$. As the $\{\vec{v}\_i\}$ are a generating set, $\text{span}(\{M^j \vec{u}\}\_{j = 0}^{n - 1}) = \Lambda$, and we are done.


