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Suppose given a prime $p$.

Question: Do there exist finite groups $G$ and $H$ such that ${\bf F}_p G$ is isomorphic to ${\bf F}_p H$, but such that ${\bf Z}_p G$ is not isomorphic to ${\bf Z}_p H$ ?

Variants: Suppose given $s\geqslant 2$ and replace ${\bf Z}_p$ resp. ${\bf F}_p$ by ${\bf Z}/p^s$.

Variant: Suppose $G$ and $H$ to be $p$-groups. (It is unknown whether there are nonisomorphic $p$-groups with isomorphic group rings over ${\bf F}_p$ , but still, maybe someone knows an argument in favour of ${\bf F}_p G \simeq {\bf F}_p H$ $\Rightarrow$ ${\bf Z}_p G \simeq {\bf Z}_p H$ in this case?)

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    $\begingroup$ As for the last variant of your problem: It is known that for $p$-groups $G$ and $H$, an isomorphism $\mathbb Z_p G \cong \mathbb Z_p H$ implies $G\cong H$ (see Roggenkamp: Isomorphisms of p-adic group rings). Therefore a positive answer to the last variant of your question would imply the modular isomorphism problem (which I believe is still open). $\endgroup$
    – Florian Eisele
    Aug 17, 2012 at 20:05
  • $\begingroup$ Can someone explain why I'm confused because $\mathbb{F}_p\cong\mathbb{Z}_p$ ? $\endgroup$
    – Chris Gerig
    Aug 17, 2012 at 20:10
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    $\begingroup$ [I think the notation $\mathbb{Z}_p$ for $\mathbb{F}_p=\mathbb{Z}/p$ should be abandoned (even by topologists)! ] $\endgroup$
    – Qfwfq
    Aug 17, 2012 at 20:56
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    $\begingroup$ The following paper by Sehgal is also relevant: math.ualberta.ca/people/Faculty/Sehgal/publications/007.pdf $$ $$ It shows that for two $p'$-groups $G$ and $H$ we have $\mathbb Z_p G \cong \mathbb Z_p H$ iff $\mathbb F_p G \cong \mathbb F_p H$. In particular you'll find an example of the sort you're looking for in your question neither for $p$-groups nor for $p'$-groups. I suspect the answer to your question is "there are no such groups", but a proof is out of reach due to the connection with the modular isomorphism problem. $\endgroup$
    – Florian Eisele
    Aug 17, 2012 at 21:36
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    $\begingroup$ This isn't an answer to your question, but if you take $G=S_n$ to be the symmetric group then $\Bbb{F}_pS_n$ is isomorphic to a cyclotomic quiver Hecke algebra $R_p\otimes_{\Bbb{Z}}\Bbb{F}_p$ but $\Bbb{Z}_pS_n$ and $R_p\otimes_{\Bbb{Z}}\Bbb{Z}_p$ are NOT isomorphic if $p\le n$. $\endgroup$
    – Andrew
    Dec 18, 2013 at 7:08

1 Answer 1

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In a very short paper put on the arXiv recently, García, Margolis and del Río give examples of nonisomorphic finite $2$-groups $G$ and $H$ with $\mathbb{F}_2G\cong\mathbb{F}_2H$, thus solving the modular isomorphism problem. The smallest examples have order $512$.

Combined with Roggenkamp and Scott's theorem that $\mathbb{Z}_pG\cong\mathbb{Z}_pH\Rightarrow G\cong H$ for $p$-groups, this answers the question when $p=2$.

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    $\begingroup$ Just for completeness, their example have cardinal $2^n$ for any given $n\ge 9$ and are 3-step nilpotent, with derived subgroup a cyclic group of order 4. $\endgroup$
    – YCor
    Jun 19, 2021 at 8:53

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