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Let $G$ be a finite group. Then the group algebra $\mathbb{Q}G$ trivially contains $\mathbb{Q}$.

But when (i.e. for which $G$) does the augmentation ideal $I_G=\{\sum_{g\in G} r_g\,g \mid \sum_{g\in G} r_g=0,\ g \in G,\ r_g \in \mathbb{Q}\}$ of $\mathbb{Q}G$ contain a subring isomorphic to $\mathbb{Q}$.

I know that this is true for $\mathbb{Q}S_3$ with the subring $\{\frac{2}{3}r\ 1_G-\frac{1}{3}r\ t-\frac{1}{3}r\ t^2\mid t^3=1_G,r\in\mathbb{Q}\}$.

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Always.

Since $\Bbb QG$ is a semisimple $\Bbb Q$ algebra, $\Bbb QG=I_G\oplus J$ for another ideal $J$, and as rings $I_G$ and $J$ are both $\Bbb Q$ algebras, each having an idempotent acting as a multiplicative identity.

You just take the idempotent $e\in I_G$ which is the identity, and $e\Bbb Q\cong \Bbb Q$.

That is exactly what happened in your example, btw, since $\frac{2}{3}\ 1_G-\frac{1}{3}\ t-\frac{1}{3}\ t^2$ is the idempotent generator of $I_G$ in that ring.


Of course, you can make the exact same case for any semisimple group algebra $F[G]$ that the augmentation ideal contains a copy of $F$ and all of its subfields.

I couldn't immediately draw a conclusion about all algebras with nontrivial radicals, but there do exist counterexamples there. For example, $F_2[C_2]$ for the cyclic group of order 2 $C_2$ has a nilpotent augmentation ideal, so it cannot contain a copy of $F_2$.

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  • $\begingroup$ Ok, thanks, that was helpful. :) $\endgroup$ – vuur Jul 23 '14 at 12:50
  • $\begingroup$ @vuur No problem: glad you liked it! $\endgroup$ – rschwieb Jul 23 '14 at 13:07
  • $\begingroup$ I thought a little bit more about what you wrote. The idempotent $e_J$ associated to $J$ is $\frac{1}{\lvert G \rvert}\sum_{g\in G}\ g$. Then the idempotent of $I_G$ is $e_{I_G}=1_G-e_J$ which in the case of $\mathbb{Q}S_3$ gives $\frac{5}{6}$ for the identity and $-\frac{1}{6}$ for the other $5$ elements as coeffcients. My example from above then comes from the same construction applied to the subgroup $C_3$. $\endgroup$ – vuur Jul 23 '14 at 14:30

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