Existence of a special ordering of the elements of a finite group

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: 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 $\mathbb{F}_p[G]$?

This question is motivated by an old question I asked.

Answer 1

The answer to question 2 is no, but for a stupid reason. It fails for the cyclic group of order two. But as long as a Sylow $p$ subgroup $S$ of $G$ has order at least three, it is true. Order the elements of the group so that the elements of $S$ come first. If $g$ is an element of order $p^r$ then $1+g+\dots+g^{p^r-1}=(1-g)^{p^r-1}$ is in the $(p^r-1)$st power of the radical $J(kS)$. So the product of all of them is at least in the $(p-1)(|S|-1)$th power of the radical. Since $J(kS)^{|S|}=0$, if $p$ is odd we are done. If $S$ is non-cyclic then $J(kS)^{|S|-1}=0$, and we are done. For a cyclic group with elements of order four, we get at least $J^3$ for one term, and we are done. So the only exception is when $S$ is cyclic of order two.

Answer 2

Even without restricting to nonabelian groups, the only counterexample is the cyclic group of order $2$. Below is an elementary argument.

Note that if this is true for a given prime $p$ and a group $G$ (abelian or not), then it is also true for any supergroup (just ensure that the elements of $G$ occur consecutively, in the "correct" order).

Now, if $G$ is abelian, then it does not depend on the ordering, and if it's cyclic of order $p>2$, then $P=(\sum_{g\in G}g)^{p-1}=p(\sum_{g\in G}g)^{p-2}$. (Because for any $g_0\in G$ we have $g_0(\sum_{g\in G}g)=\sum_{g\in G}g$.) So we may assume that $p=2$.

Write $P_g$ for the factor of $P$ corresponding to $g$. Suppose first that $g,h$ are distinct of order $2$. Arguing as before, $(1+gh)P_{gh}=0$. On the other hand, $(1+g)(1+h)=(1+g)(1+gh)$, and hence $(1+g)(1+h)P_{gh}=0$.

Thus, we may assume that there is a unique element of order $2$. Then it is necessarily central, so either $n=2$ (and this is the counterexample), or there is a cyclic group of order $2m$ for $m>1$. Then letting $g$ be of order $2$ and $h$ of order $2m$, we have that $P_gP_h=(1+g)P_h=2P_h=0$ (by the same trick, since $g=h^m$).