16.104 (2006)

If $G$ is a finite group, then every element $a$ of the rational group algebra $\mathbb{Q}[G]$ has a unique Jordan decomposition $a = a_s + a_n$, where $a_n \in \mathbb{Q}[G]$ is nilpotent, $a_s \in \mathbb{Q}[G]$ is semisimple over $\mathbb{Q}$, and $a_s a_n = a_n a_s$. The integral group ring $\mathbb{Z}[G]$ is said to have the additive Jordan decomposition property (AJD) if $a_s, a_n \in \mathbb{Z}[G]$ for every $a \in \mathbb{Z}[G]$. If $a \in \mathbb{Q}[G]$ is invertible, then $a_s$ is also invertible and so $a = a_s a_u$ with $a_u = 1 + a_s^{-1}a_n$ unipotent and $a_s a_u = a_u a_s$. Such a decomposition is again unique. We say that $\mathbb{Z}[G]$ has multiplicative Jordan decomposition property (MJD) if $a_s, a_u \in \mathbb{Z}[G]$ for every invertible $a \in \mathbb{Z}[G]$. See the survey (A. W. Hales, I. B. S. Passi, in: Algebra, Some Recent Advances, Birkhäuser, Basel, 1999, 75–87).

Is it true that there are only finitely many isomorphism classes of finite 2-groups $G$ such that $\mathbb{Z}[G]$ has MJD but not AJD?