Wedderburn's theorem for $\mathbb{Q}G$ - MathOverflow

Wedderburn's theorem for $\mathbb{Q}G$

2013-01-23T14:01:42Z

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.

Answer by Aakumadula for Wedderburn's theorem for $\mathbb{Q}G$

2013-01-23T15:54:44Z

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!).

Answer by Yazdegerd III for Wedderburn's theorem for $\mathbb{Q}G$

2013-01-23T16:10:25Z

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.