subgroups of $\mathbb{Z}_N^k$ Let $S$ be a subgroup of $\mathbb{Z}_N^k$, and $S^{\perp} = \{t | t \in \mathbb{Z}_N^k,\,\mbox{and}\,\forall s \in S,\, t\cdot s = 0\}$, here the dot product is defined as $t\cdot s = \sum_{i=1}^k t_is_i \mod N$. Is it possible to prove that for $y \notin S^\perp$, $$\sum_{s\in S}e^{2\pi i s\cdot y/N} = 0 $$
Here is my proof for the case of $k=1$. Since $S$ is a cyclic group, let $g$ be a generator of $S$, then $S = \{0, g, 2g, 3g, ... (|S|-1)g\}$, then the summation becomes the sum of a geometric sequence and leads to $0$ easily. But for $k > 1$, $S$ is not always cyclic, then my logic won't work any more.  Any ideas?
 A: Yes. This is a standard fact.
Consider the function
$$
\chi_y:S\to\Bbb{C}^*, s\mapsto e^{2\pi i\,s\cdot y/N}.
$$
Because $(s+s')\cdot y= s\cdot y+s'\cdot t$ (as elements of $\Bbb{Z}_N$) for all $s,s'\in S$ we see that $\chi_y$ is a homomorphism of groups: $\chi_y(s+s')=\chi_y(s)\chi_y(s')$.
The assumption that $y\notin S^\perp$ implies that there exists an element $s_0\in S$ such that $\chi_y(s_0)\neq1$. This is because $\chi_y(s)=1$ if and only if $s\cdot y\equiv0\pmod N$. Let us fix such an element $s_0\in S$.
Next we can consider your sum
$$
\Sigma(y):=\sum_{s\in S}\chi_y(s).
$$
Because $S$ is a group $s+s_0$ ranges over the elements of $S$ while $s$ does. 
In other words, if we write $s=s'+s_0$ and let $s'$ range over $S$, we get all the elements $s\in S$ exactly once. Utilizing this
$$
\begin{aligned}
\Sigma(y)&=\sum_{s\in S}\chi_y(s)\\
&=\sum_{s'\in S}\chi_y(s'+s_0)\\
&=\sum_{s'\in S}\chi_y(s')\chi_y(s_0)\\
&=\chi_y(s_0)\left(\sum_{s'\in S}\chi_y(s')\right)\\
&=\chi_y(s_0)\Sigma(y).
\end{aligned}
$$
So
$$
(\chi_y(s_0)-1)\Sigma(y)=0.
$$
Here $\chi_y(s_0)-1\neq0$, so we can conclude that $\Sigma(y)=0$.
