A counterexample to the isomorphism problem for integral group rings


Let $X$ be a finite group, and denote its integral group ring by $\mathbb{Z}X$. A group basis of $\mathbb{Z}X$ is a subgroup $Y$ of the group of units of $\mathbb{Z}X$ of augmentation $1$ such that $\mathbb{Z}X = \mathbb{Z}Y$ and $|X| = |Y|$. An example of a finite group $X$ is given such that $\mathbb{Z}X$ has a group basis which is not isomorphic to $X$. A main ingredient is the existence of a subgroup $G$ of $X$ which possesses a non-inner automorphism which becomes inner in the integral group ring $\mathbb{Z}G$.



Martin Hertweck