The group ring isomorphism problem

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


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