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Let $G$ be a finite group and let $\mathbb{Q}G=M_{n_1}(D_1)\times\cdots\times M_{n_k}(D_k)$ be the decomposition of $\mathbb{Q}G$ as a product of rings of matrices over divisions rings. Let $Z_i$ be the center of $D_i$. Then each $Z_i$ is an algebraic extension of $\mathbb{Q}$. The question is: how much is it known about these algebraic number fields?. More precisely, for which family of groups $G$ it is known that each $Z_i$ is a Galois extension of $\mathbb{Q}$ and in such a case what can we say about the corresponding Galois groups?.

Any comments and references will be strongly appreciated.

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In more concrete terms (without mention of Wedderburn's theorem), you're asking about the structure of the center of $\mathbf{Q}[G]$ (as a finite product of number fields). –  user30379 Jan 23 '13 at 15:24
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2 Answers

up vote 12 down vote accepted

The short answer is yes, the centres $Z_i$ are contained in the cyclotomic extension $E_n= {\mathbb Q}(e^{(2\pi i)/n})$ where $n$ is the order of the group. As pranavk says, we need only prove that the simple factors of the centre of ${\mathbb Q}[G]$ are contained on $E_n$. But the centre is spanned by the averages $ C_x =\sum gxg^{-1}$ where the sum is over all the elements of $G$. Given an absolutely irreducible representation $V$ of $G$ of dimension $r$ , each $C_x$ acts by a scalar $\lambda $ say. By taking traces, you see that $\lambda $ is a rational multiple of the trace of $x$; since $x$ has order dividing $n$, all its eigenvalues in $V$ are $n$-th roots of unity, and hence the trace of $x$ lies in $E_n$.

Now,in your notation, each $M_{n_i}(D_i)$ after tensoring (over $Z_i$) with the algebraic closure of ${\mathbb Q}$ is of the form $End (V)$ for some absolutely irreducible representation $V$ of $G$, hence each $Z_i$ (which as pranavk observed, lies in the centre of ${\mathbb Q}[G]$) lies in $E_n$.

I believe all this is worked out in Serre's book on finite groups (Springer notes) and is well known to experts (I am not one!).

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@Aakumadula: Just to be precise, in the 2nd paragraph you're tensoring against $\overline{\mathbf{Q}}$ over $Z_i$ (using some embedding) rather than over $\mathbf{Q}$. Also, though it is an irrelevant refinement, the argument shows that we can take $n$ to be the exponent of $G$ and not just the order (as is also mentioned in Serre's book, if I remember correctly). –  user30379 Jan 23 '13 at 16:03
    
I would like to add that sometimes, the fields could be smaller. For example, if $G=S_n$, then the theory of Young diagrams shows that all the fields $Z_i$ are ${\mathbb Q}$. –  Aakumadula Jan 23 '13 at 17:03
    
@pranavk: Thanks. you are quite correct. I am tensoring over $Z_i$. –  Aakumadula Jan 23 '13 at 17:46
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I think you can take $n$ to be the exponent of the group (instead of the order). –  Bruce Westbury Jan 23 '13 at 19:43
    
I saw this link at MO today. This is relevant to the OP's question mathoverflow.net/questions/32386/… –  Aakumadula Jan 26 '13 at 16:47
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The simplest case of your question is the case where $G$ is the cyclic group of order $n$, it is known that $\mathbb{Q}G\simeq\mathbb{Q}[X]/(X^n-1)$. As $X^n-1=\Pi_{d|n}\Phi_d(X)$ where $\Phi_d(X)$ is the $d$-th cyclotomic polynomial, it follows that $\mathbb{Q}G\simeq\Pi_{d|n}\mathbb{Q}(\zeta_d)$ where $\zeta_d\in\mathbb{C}$ is a primitive $d$-th root of unity.

Edit: This is also a very particular but concrete case of the idea of @Aakumadula.

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