# Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?

Due to the first (and very helpful) answer I received, I've reformulated the question a little: $G$ and $H$ are now assumed to be $p$-groups.

Let $\{p}$ be a prime, and let $\mathbb{F}_p$ be the field of $\{p}$ elements. Let $G,H$ be finite $p$-groups, and let $\mathbb{k}[G]$ denote the group algebra.

Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?

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No, take $G= Z/4Z$ and $H=(Z/2Z)^2$. If the field k contains a 4th root of 1 then both group algebras are isomorphic to k+k+k+k (I am assuming the the charachterisitc is not 2). But these group algebras are not isomorphic if k does not contian $\sqrt{-1}$, which is the case for F_p if p = 3,5 (mod 8)
$\mathbb{F}_p$ contains $\sqrt{-1}$ precisely when $p \equiv 1 \mod 4$, right? –  Maurizio Monge Apr 5 '11 at 18:11
Sorry, somehow I was thinking about $\sqrt{2}$ –  kassabov Apr 7 '11 at 9:08