Existence of a special ordering of the elements of a finite group (II)

Question

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 $g_1+g_2+\cdots+g_n$?

Here "proportional" means it is equal to $c(g_1+g_2+\cdots+g_n)$ for a scalar $c\in \mathbb{C}$. If the answer is positive then $c$ is determined uniquely , as $\prod_{g\in G}\mathrm{ord}(g)/n$.

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 $g_1+\ldots+g_{60}$, 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.
4. See here 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 $n$.

Answer 1

Solvable groups have this property (hence in particular, it holds for $n=|G|$ odd), arguing as follows.

For a subgroup $H\subseteq G$ set $\sigma(H)=\sum_{h\in H}h.$ For ease of notation, for $g\in G$ write $\sigma(g)$ for $\sigma(\langle g\rangle)$. We say $g,h\in G$ are equivalent if $\langle g\rangle =\langle h\rangle$.

We actually deal with the following stronger property of $G$: We say $G$ has a cyclic factorization if there are pairwise inequivalent $g_1,\dots,g_r\in G$ with $g_i\neq 1$, for some $r$, such that $$\sigma(g_1)\dotsm \sigma(g_r)=k\sigma(G)$$ for some $k$ (which clearly must be a positive integer).

Since $\sigma(G)=\sigma(G)g$ for all $g\in G$, if $G$ has a cyclic factorization as just defined, then $G$ has a product of the required form in the question, which can be obtained by multiplying the cyclic factorization by the $\sigma(g)$ for the “unused” $g\in G\setminus\{ g_1,\dots,g_r\}$ in any order.

We claim that if $N\vartriangleleft G$ and $N$ and $G/N$ have cyclic factorizations, then so does $G$. Indeed let $\pi:G\rightarrow G/N$ be the quotient map and let $\sigma(\pi(g_1))\dotsm\sigma(\pi(g_s))$ be a cyclic factorization for $G/N$, where the $g_i$ are suitable elements of $G$. Note that the $g_i\notin N$ and are inequivalent. Since $\pi(\sigma(g))$ is a multiple of $\sigma(\pi(g))$, $\pi(\sigma(g_1)\dotsm\sigma(g_s))$ is a multiple of $\sigma(G/N)$; that is, in $\sigma(g_1)\dotsm\sigma(g_s)$ the total number of elements in each coset of $N$, counting multiplicity, is the same for each coset. Hence $$\sigma(g_1)\dotsm\sigma(g_s)\sigma(N)$$ is proportional to $\sigma(G)$. Finally, by hypothesis $N$ has a cyclic factorization so we can replace $\sigma(N)$ with a corresponding product $\sigma(n_1)\dotsm\sigma(n_l)$. Since the $g_i\notin N$, all the elements $g_1,\dots,g_s,n_1,\dots n_l$ are pairwise inequivalent as required.

The result for solvable $G$ now follows because abelian groups have cyclic factorizations (because they are direct products of cyclic groups).

It's perhaps worth noting that if $G$ need not be solvable but has an exact factorization $G=H_1\dotsm H_t$ (I mean $|G|=\lvert H_1\rvert\dotsm\lvert H_t\rvert$, so that every $g\in G$ can be written uniquely as $g=h_1\dotsm h_t$ with $h_i\in H_i$) and the $H_i$ all have cyclic factorizations, then so does $G$ by the same argument as for $G$ abelian, namely using $\sigma(G)=\sigma(H_1)\cdots\sigma(H_k)$. This might help establish the result for some groups that are not solvable. I have asked if such a factorization exists for all $G$, see Factorizing groups into a product of solvable subgroups.