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What are the groups $G$ and fields $\Bbb K$ for which $\Bbb K[G]\cong\Bbb K^{|G|}$ holds?

For example $\Bbb R[\Bbb F_2^n]\cong\Bbb R^{2^n}$ holds.

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As the group ring is identified with the set ${\mathbb K}^{(G)}$ which is the set oif maps from $G$ to $\mathbb K$ with finite support, the answer is: this is the case if and only if the group is finite. – Anton Deitmar Jan 25 at 7:50
4  
@Anton: if you want an isomorphism as algebras this is not enough. – Qiaochu Yuan Jan 25 at 7:50
    
@Qiaochu Yuan: What is the algebra structure on ${\mathbb K}^G$ when $G$ is not finite? – Anton Deitmar Jan 25 at 8:46
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@Anton: pointwise multiplication. – Qiaochu Yuan Jan 25 at 9:01
up vote 9 down vote accepted

This is true iff $G$ is finite and abelian, the characteristic of $K$ does not divide $G$, and $K$ has all $n^{th}$ roots of unity whenever $G$ has an element of order $n$. Hopefully it is clear why $G$ must be finite and abelian. The characteristic and root of unity conditions follow from writing $G$ as a product of cyclic groups and applying the Chinese remainder theorem.

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Yes sorry I missed about all roots of unity. So is this good enough for algebra isomorphism? – Student. Jan 25 at 8:02
    
@Arul: yes, that's what I'm claiming. – Qiaochu Yuan Jan 25 at 8:05
    
Oops yes I didn't say what I meant. I'll rewrite the comment, It' only a minor point anyway, but might as well say it right. – Geoff Robinson Jan 25 at 9:51
    
(Corrected): I suppose you could simplify the statement a bit: by stating that the field should have n distinct roots of unity whenever $G$ has an element of order $n$, you could omit the statement about the characteristic ( which would be implied by the other conditions). – Geoff Robinson Jan 25 at 9:51
    
Sure, but might as well make the characteristic hypothesis explicit for readability. – Qiaochu Yuan Jan 25 at 17:06

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