Let $n$ a positive integer and $k \in \mathbb{Z}^n$ such that for all integer $a \geq 2$ and $h \in \mathbb{Z}^n$ we have $k \neq ah$. Here $\cdot$ is the scalar product.

Is it true that $\{x \in \mathbb{Z}^n : k \cdot x = 0\} \cong \mathbb{Z}^{n-1}$ (group isomorphism)?

The answer is yes for $n=1,2$ but the general case seems difficult to me.

Thanks for any suggestions.

`$\{x \in \mathbb{Z}^n : k \cdot x = 0\} \cong \mathbb{Z}^{n-1}$`

:-( – user21706 Mar 8 '12 at 20:14`$h_{i,j} : \mathbb{Z}^n \to \mathbb{Z}^n : x \mapsto x + (x \cdot e_i) e_j$`

is an isomorphism. For any $k \in \mathbb{Z}$ we definite`$S_k = \{x \in \mathbb{Z}^n : k \cdot x = 0\}$`

. If $k = h_{i,j}(k^\prime)$ then $k \cdot x = 0$ iff $k^\prime \cdot x = -(k^\prime \cdot e_i)(x \cdot e_j)$ iff $k^\prime \cdot x^\prime = 0$ where $x^\prime = h_{j,i}(x)$, in conclusion $S_k \cong S_{k^\prime}$. – user21706 Mar 9 '12 at 9:53