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
$$ P=(1+g_1+g_1^2+\cdots+g_1^{\mathrm{ord}(g_1)-1})(1+g_2+\cdots+g_2^{\mathrm{ord}(g_2)-1})\cdots (1+g_n+\cdots+g_n^{\mathrm{ord}(g_n)-1}) \in \mathbb{F}[G].$$

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

Some comments:


1. If $G$ is abelian, $P$ is independent of the ordering and the answer 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.
2. My colleague  B. Bedert, who introduced me to group algebras, suggested that the hardest case is when $G$ is a simple non-abelian group.
3. This question is motivated by an [old question I asked][2]. 
4. [See here][3] for a variant of the question which was solved. The solution there suggests that even $n$ might be problematic, so I will also be happy with an answer that restricts to odd $n$.

  [1]: https://mathoverflow.net/a/477750/31469
  [2]: https://mathoverflow.net/questions/206961/proof-of-cauchys-theorem-from-group-theory-generalizable
  [3]: https://mathoverflow.net/q/477740/31469