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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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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). – Florian Eisele Aug 17 '12 at 20:05
Can someone explain why I'm confused because $\mathbb{F}_p\cong\mathbb{Z}_p$ ? – Chris Gerig Aug 17 '12 at 20:10
@Chris: I presume $\mathbb Z_p$ denotes the ring of $p$-adic integers, $\mathbb F_p$ the field with $p$ elements. – Florian Eisele Aug 17 '12 at 20:11
[I think the notation $\mathbb{Z}_p$ for $\mathbb{F}_p=\mathbb{Z}/p$ should be abandoned (even by topologists)! ] – Qfwfq Aug 17 '12 at 20:56
The following paper by Sehgal is also relevant: $$ $$ 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. – Florian Eisele Aug 17 '12 at 21:36

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