Let $G$ be a finite non-abelian group of order $n$. Given $g\in G$ we denote its order by $\mathrm{ord}(g)$.

Consider the group algebra $\mathbb{F}[G]$ for some field $\mathbb{F}$.

Given an ordering $g_1,\ldots,g_n$ of all the elements of $G$, consider the product $$ P=(1+g_1+g_1^2+\cdots+g_1^{\mathrm{ord}(g_1)-1})(1+g_2+g_2^2+\cdots+g_2^{\mathrm{ord}(g_2)-1})\cdots (1+g_n+g_n^2+\cdots+g_n^{\mathrm{ord}(g_n)-1}) \in \mathbb{F}[G].$$

Q1: Take $\mathbb{F}=\mathbb{C}$. Does there exist an ordering $g_1,\ldots,g_n$ such that $P$ is proportional to $g_1+g_2+\cdots+g_n$?

Q2: Let $p$ be a prime dividing $n$ and take $\mathbb{F}=\mathbb{F}_p$. Does there there exist an ordering $g_1,\ldots,g_n$ such that $P$ vanishes in $\mathbb{F}_p[G]$?

A few comments:

1. If the answer to Q1 is positive, then necessarily $P = m (g_1+\cdots+g_n)$ for $m=\prod_{i=1}^{n} \mathrm{ord}(g_i)/|G|$. In particular, by Sylow's Theorems, say, $p\mid n$ implies $p\mid m$. Thus, a positive solution to Q1 implies a positive solution to Q2.
2. My colleague B. Bedert, who introduced me to group algebras, suggested that the hardest case of these questions is when $G$ is a simple non-abelian group.
3. If $G$ is abelian, $P$ is independent of the ordering and the answer to both questions is easily seen to be positive. However, for $G=A_5$, I was able to find (computationally) some orderings for which $P$ is not proportional to $g_1+\ldots+g_{5!}$, which is the reason I merely ask for an existence of an ordering.
4. This question is motivated by an old question I asked.