TITLE: Units in group rings.
QUESTION [14 upvotes]: Let $G$ be a finite solvable group of order $n$, and let $g_1 ... g_n$ be an enumeration of its elements. Let $a_1 ... a_n$ be a sequence of integers, such that $\sum a_i$ is relatively prime to $n$. 
Consider $\mathbb{C}[G]$, the group ring of $G$ with complex coefficients. Does the element $\sum a_i g_i$ necessarily have to be a unit in the group ring? (I believe that the element does have to be a unit, and have a proof in the cyclic and abelian case, but was hoping for a reference in greater generality, at least in the case when $G$ is solvable.)

REPLY [9 votes]: David Speyer's example shows that the answer is "no", in general, even in the cyclic case. However, here are a few general remarks about the case of finite Abelian groups.
Each linear character (ie, group homomorphism $G \to \mathbb{C}^{\times}$), say $\lambda,$ extends by linearity to an algebra homomorphism from $\mathbb{C}G \to \mathbb{C}$ (which we call $\lambda^{+}$ here).
When $G$ is Abelian, an element $u \in \mathbb{C}G$ is a unit of $\mathbb{C}G$ if and only $\lambda^{+}(u) \neq 0$ for each linear character $\lambda$ of $G.$
In a positive direction, the answer to your question is positive when $G$ is a finite Abelian $p$-group. If $u = \sum_{i}a_{i}g_{i}$ with $p$ not dividing $\sum_{i}a_{i}$ (each $a_{i} \in \mathbb{Z}$), then 
for each linear character $\lambda$ of $G$, we have 
$\lambda^{+}(u) \in \mathbb{Z}[\eta]$, but $\lambda^{+}(u) \equiv \sum_{i} a_{i} \not \equiv 0$ (mod $\pi$), where $\pi$ is the unique prime ideal of $\mathbb{Z}[\eta]$ containing $p$ ( $\eta$ being a primitive $p^{e}$-th root of unity where $G$ has exponent $p^{e}$). Then certainly $\lambda^{+}(u) \neq 0.$