## The group ring isomorphism problem

Posted: June 6, 2010 in Group Algebras, Noncommutative Ring Theory Notes
Tags: , ,

Let $k$ be a field or $\mathbb{Z}$ and $G_1,G_2$ two groups. It is clear that if $G_1 \cong G_2,$ then $k[G_1] \cong k[G_2],$ as $k$ algebras. The group ring isomorphism problem is this question that whether or not $k[G_1] \cong k[G_2],$ as $k$ algebras, implies $G_1 \cong G_2$. Another version of the group ring isomorphism problem is this: given a group $G_1$ and a field $k$ find all groups $G_2$ such that $k[G_1] \cong k[G_2],$ as $k$ algebras. These questions have been answered in special cases only. For example, an old result due to Perlis and Walker states that if $G_1,G_2$ are finite abelian groups and $\mathbb{Q}[G_1] \cong \mathbb{Q}[G_2],$ as $\mathbb{Q}$ algebras, then $G_1 \cong G_2.$ In 2001 Hertweek found two non-isomorphic groups $G_1, G_2$ of order $2^{21}97^{28}$ such that $\mathbb{Z}[G_1] \cong \mathbb{Z}[G_2].$

For now, we’ll only show that it is possible to have $k[G_1] \cong k[G_2]$ and $G_1 \ncong G_2$:

Theorem. If $G_1, G_2$ are two finite abelian groups of order $n,$ then $\mathbb{C}[G_1] \cong \mathbb{C}[G_2] \cong \mathbb{C}^n$ as $\mathbb{C}$-algebras.

Proof. We have already seen in this post that $J(\mathbb{C}[G])=(0)$ for any group $G.$ So if $G$ is a finite abelian group of order $n,$ then $\mathbb{C}[G]$ is a commutative semisimple algebra. Thus, since $\mathbb{C}$ is algebraically closed, the Wedderburn-Artin theorem gives us $\mathbb{C}[G] \cong \mathbb{C}^n$ as $\mathbb{C}$-algebras. $\Box$

Example. Let $G_1$ be the Klein four-group and let $G_2$ be the cyclic group of order four. Then $G_1 \ncong G_2$ but, by the theorem, $\mathbb{C}[G_1] \cong \mathbb{C}[G_2] \cong \mathbb{C}^4$ as $\mathbb{C}$-algebras.