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.)



Units in group rings.
This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.


Generalizing this example, the statement is false for the cyclic group of order $pq$, with $p$ and $q$ two different primes. Let $g$ generate this group and let $\chi: G \to \mathbb{C}^{\ast}$ be a character with $\chi(g)$ a primitive $pq$-th root of unity. Let $\Phi\_{pq}(x) = \sum c\_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\Phi\_{pq}(g)= \sum c\_k g^k$ in $\mathbb{Z}[G]$ acts by $\Phi\_{pq}(\chi(g)) =0$ on the representation $\chi$. Thus $\sum c\_k g^k$ is not a unit. On the other hand, $\sum c\_k = \Phi\_{pq}(1)= 1$. (To compute the last, note that $\Phi\_{pq}(x) = \frac{(x^{pq}-1)(x-1)}{(x^q-1)(x^p-1)}$ and take the limit as $x \to 1$.)


I claim further that, if $G$ has any element of non-prime-power order, then $G$ fails to have this condition. Let $g$ be an element of order $pq$ and let $\chi: \langle g \rangle \to \mathbb{C}^{\ast}$ be an injective character. Let $V = \mathrm{Ind}\_{\langle g \rangle}^G \chi$. Then $V$ restricted to $\langle g \rangle$ has $\chi$ as a summand, and this summand is in the kernel of $\Phi\_{pq}(g)$ acting on $V$. So $\Phi\_{pq}(g)$ acts non-injectively on a representation of $G$, and thus is not a unit.


So the only groups for which this might be right are groups where every element has prime power order. These were classified by [Higman](https://doi.org/10.1112/jlms/s1-32.3.335), so you can dig into his paper if you care enough.




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On the positive side, the statement is true whenever $G$ is a $p$-group. Let $\alpha = \sum c\_g g \in \mathbb{Z}[G]$. I will show that the determinant of $\alpha$ acting on $\mathbb{Z}[G]$ is $\left( \sum c\_g \right)^{|G|} \bmod p$, and hence is not $0$ if $\sum c\_g \not \equiv 0 \bmod p$. Reducing $\mathbb{Z}[G]$ modulo $p$, we get an action of $\alpha$ on $\mathbb{F}\_p[G]$. More generally, I claim that $\alpha$ acts on any $G$-representation $V$ over $\mathbb{F}\_p$ by $\left( \sum c\_g \right)^{\dim V}$. This is simple: $V$ has a filtration whose associated graded is a $\dim V$-dimensional trivial representation. Passing to the associated graded doesn't change determinant, and $\alpha$ acts on the $1$-dimensional trivial representation by $\sum c\_g$.


