This question already has an answer here:

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 $1970$).


marked as duplicate by Dietrich Burde, Derek Holt finite-groups May 2 '17 at 15:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – Derek Holt May 2 '17 at 12:53
  • $\begingroup$ @Derek: Thanks for noticing the error in question 1. I corrected it. $\endgroup$ – Beginner May 3 '17 at 4:05
  • $\begingroup$ @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? $\endgroup$ – Prism May 3 '17 at 4:16
  • $\begingroup$ @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.) $\endgroup$ – Beginner May 3 '17 at 5:07
  • $\begingroup$ @Beginner: Thanks a lot! $\endgroup$ – Prism May 3 '17 at 5:08

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 $\mathbb{Z}[G]\cong\mathbb{Z}[H]$.

  • 1
    $\begingroup$ Hertweek is the cousin of Hertweck ... $\endgroup$ – GEdgar May 2 '17 at 13:02
  • $\begingroup$ @GEdgar Whoops! Fixed now. $\endgroup$ – Jeremy Rickard May 2 '17 at 13:38

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.

  • $\begingroup$ +1. By the way, would you have any idea concerning this related question on MO? $\endgroup$ – Watson May 3 '17 at 16:54
  • 1
    $\begingroup$ @Watson Thank you! The MO question looks interesting, and is perhaps not so easy. $\endgroup$ – Dietrich Burde May 3 '17 at 18:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.