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The question in title has been considered for finite groups $G$ and $H$, but I do not know its situation, how far it is known whether $G$ and $H$ could be isomorphic. I have two simple questions regarding it.

Q.0 If $\mathbb{Z}[G]\cong \mathbb{Z}[H]$ then $|G|$ should be $|H|$; because, $G$ is a free basis for the free additive abelian group $\mathbb{Z}[G]$, am I right? (I am asking this because in Isaac's character theory, I saw something different argument, not too lengthy, but I was thinking for the above natural arguments.)

Q.1 Are there known examples of finite groups $G\ncong H$ with $\mathbb{Z}[G]\cong \mathbb{Z}[H]$? (In the book of character theory by Isaacs, he stated for metablelian groups $G,H$, $\mathbb{Z}[G]\cong \mathbb{Z}[H]$ implies $G\cong H$; it was proved by Whitcomb, in $1970$; but book has not been further revised, I don't known what is done after $1970$).

Beginner
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  • For Q. 0 the answer is yes for the reason you say. I am finding Qn 2 confusing. What did Isaacs say is true - that there are known examples? If so then the answer is obviously yes. – Derek Holt May 02 '17 at 12:53
  • @Derek: Thanks for noticing the error in question 1. I corrected it. – Beginner May 03 '17 at 04:05
  • @Derek: By any chance, do you know the corresponding statement for $\mathbb{C}[G]\cong\mathbb{C}[H]$? Is it easier to find a counter-example in this case? – Prism May 03 '17 at 04:16
  • @Prism: non-isomorphic abelian groups of same order, or extra-special p-groups of same order, dihedral and quaternion groups of order 8, etc. (Instead of fields, if we consider rings, then there are less counterexamples.) – Beginner May 03 '17 at 05:07
  • @Beginner: Thanks a lot! – Prism May 03 '17 at 05:08

2 Answers2

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You're right about the zeroth question.

For the first, a counterexample was constructed by Hertweck in 2001. There are two non-isomorphic groups $G$ and $H$ of order $2^{21}97^{28}$ with $\mathbb{Z}[G]\cong\mathbb{Z}[H]$.

Jeremy Rickard
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This isomorphism problem was stated by Higman in this thesis $1940$: $$ \mathbb{Z}[G]\cong\mathbb{Z}[H] \Longrightarrow G\cong H ? $$ It is true for nilpotent groups, for metabelian groups (Whitcomp $1968$), and was disproved by Hertweck in $2001$, see this question.

Dietrich Burde
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