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I cannot find a finite group $G$ such that $\exists x\in \mathbb{Q}[G]$ with $x^2=2e$, where $\mathbb{Q}[G]$ is the group algebra of $G$ over $\mathbb{Q}$.

I also could not prove it does not exist. Can you give me an example or is it impossible?

In the same spirit, for $f$ the extension by linearity of the trivial morphism (ie. the sum of the coefficients), $\ker(f)$ has an $\mathbb{Q}$-algebra structure with unit $u:=e-\frac{1}{n}\sum_{g \in G}{g}$.

Is there a solution to $x^2=2u$ for some finite group $G$?

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    $\begingroup$ If you consider the one dimensional part of the direct sum coming from the trivial representation you should see $x^2=2e$ is impossible without a $\sqrt2$ coefficient. $\endgroup$
    – JP McCarthy
    Commented Aug 7, 2022 at 18:10
  • $\begingroup$ Indeed, but I do not manage to use it for the "$ker(f)$" case. When I tried to do it somehow explicitely, I got a system of quadrics in wich I found no obvious obstructions. $\endgroup$
    – Hugo MTV
    Commented Aug 7, 2022 at 20:27

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