irreducible unitary representations of $(\mathbb{Z}^n,+)$ Being abelian, $(\mathbb{Z}^n,+)\equiv G$ necessarily has only 1-dimensional unitary irreducible representations. Any group element's action is then a simple scaling by complex number of unit length:
$$
\rho(g) = e^{i\lambda(g)}
$$
$\rho$ here is an irreducible unitary representation, $g \in G$, and $\lambda: G \to \mathbb{R}$.
The condition that $\rho$ be a representation places the following restriction on $\lambda$:
$$
\lambda(g) + \lambda(h) = \lambda(g+h) + 2 \pi N_{gh}
$$
where $N: G^2 \to \mathbb{Z}$.
From the structure of $G$ can it be shown that $N$ must necessarily be zero for all pairs $(g,h)$? It is clear that $N$ is symmetric, and $N_{0g} = N_{0h} = \lambda(0)$ for any $g,h \in G$. 
 A: Given $\rho:\mathbb{Z}^n\to S^1$, there are many different functions $\lambda:\mathbb{Z}^n\to\mathbb{R}$ satisfying the property
$$\rho(g)=\exp\big(i\lambda(g)\big).$$
Indeed, for every single $g\in\mathbb{Z}^n$, the phase $\theta$ in $\rho(g)=e^{i\theta}$ is only determined up to integer multiples of $2\pi $, and we can choose which of those $\theta$ values to make $\lambda(g)$... for every single $g$. That's a lot of independent choices that can be made to create $\lambda$.
The $\lambda$s that are additive (i.e. $\lambda(g+h)=\lambda(g)+\lambda(h)$) can be constructed as follows. First, pick a basis for $\mathbb{Z}^n$ (might as well use the coordinate basis vectors $e_j$), then pick the values $\lambda(e_j)$ at whim (consistent with $\rho(e_j)=\exp\big(i\lambda(e_j)\big)$), then "extend linearly," i.e. define
$$ \lambda(a_1e_1+\cdots+a_ne_n)=a_1\lambda(e_1)+\cdots+a_n\lambda(e_n).$$
All possible additive $\lambda$s arise in this way.
There are nonadditive $\lambda$s too. If $\lambda$ is additive, define $\overline{\lambda}$ to be the same function as $\lambda$ except change one of the outputs by an integer multiple of $2\pi$. Then $\overline{\lambda}$ will not be additive. Or, for the $n=1$ case, if $\rho(g)=1$ is a constant function, pick any nonadditive function $f:\mathbb{Z}\to\mathbb{Z}$, then define the function $\lambda(g)=2\pi f(g)$. This works but will not be additive.
