Suppose I have a finite group $G$, a non-trivial proper subgroup $H$, a field $k$ (restricting to $k=\mathbb C$ would be fine), and non-zero elements $a,u$ in the group algebra $kG$ satisfying the following:

  • $u$ is a unit.

  • $a$ is a zero divisor.

  • $a\in kH$, $u\not\in kH$.

  • $a$ is not in the augmentation ideal. In other words, if we write $a=\sum_{h\in H}c_h h$ with $c_h\in k$, then $\sum c_h\neq 0$ (so by rescaling we may suppose this sum is $1$).

  • $ua=au = a$.

Does it then follow that $u=1$?

If not, does anything change if we further impose $H\subseteq Z(G)$?

I'm not sure if the augmentation condition is actually necessary, by the way: I do not have a counterexample if I remove the assumption. The assumption that $a\in kH$ but $u\not\in kH$ is to prevent a trivial counterexample: $a=\sum_{h\in H}h$ and $u\in H$. Indeed, it follows from the assumptions that $u\not\in G$.

It may be worth pointing out that under these assumptions $kG$ is not an integral domain. Indeed, any group $G$ with torsion always has zero divisors in any group ring. A conjecture of Kaplansky, if true, would assert that torsionless groups do not have this problem: $kG$ would be an integral domain. But I am very specifically looking at a problem with finite groups.


1 Answer 1


Take $G=C_2\times C_2$, generated by elements $x,y$, and let $H$ be the subgroup generated by $x$.

Then you can take $u=1+y-xy$ and $a=1+x$.

  • $\begingroup$ Turns out this won't quite disprove what I'm looking at, but it does disprove what I posted. Thanks. $\endgroup$ Aug 14, 2015 at 22:09

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