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Given a finite group $G$ and a field $\mathbb{F}$ with $\operatorname{Char}\mathbb{F}\nmid |G|$, Mashke's Theorem tells us that the group algebra $\mathbb{F}G$ is semisimple. By the Artin-Wedderburn Theorem, then we have $\mathbb{F} G \cong M_{n_1}(D_1)\times \dots \times M_{n_r}(D_r)$ for some $n_1,\dots,n_r\in \mathbb{Z}^+$ and $D_1,\dots,D_r$ finite dimensional division rings over $\mathbb{F}$.

I have read that the integer $r$ above corresponds to the number of conjugacy classes of $G$, since $Z(\mathbb{F}G) = \langle \sum_{c\in C} c \mid C \text{ a conjugacy class of $G$}\rangle_{\mathbb{F}}$ has the same dimension over $\mathbb{F}$ as the number of conjugacy classes, and $Z(M_{n_1}(D_1)\times \dots \times M_{n_r}(D_r))$ is the product $M_{n_1}(Z(D_1))\times \dots \times M_{n_r}(Z(D_r))$. Then since $Z(D_i) = \mathbb{F}$ (I'm not sure if this is right), the dimension over $\mathbb{F}$ of $Z(M_{n_1}(D_1)\times \dots \times M_{n_r}(D_r))$ is $r$.

Given this fact, my first question is whether the following statement (and argument to show it) is true:

Given any finite abelian group $G$ and any field $\mathbb{F}$ with $\operatorname{Char}\mathbb{F}\nmid |G|$, we have $\mathbb{F}G \cong \prod_{n=1}^{|G|}\mathbb{F}$.

Proof: Since $G$ is abelian, the number of conjugacy classes is $|G|$, hence $\mathbb{F}G \cong M_{n_1}(D_1)\times \dots \times M_{n_r}(D_{|G|})$. Since $G$ is abelian, we must have $n_i = 1$ for all $i$, and each $D_i$ is a finite (field) extension of $\mathbb{F}$. By counting dimensions, $D_i = \mathbb{F}$ for each $i$.

I think the result holds for cyclic $G$ where $\mathbb{F}$ contains all $|G|$-th roots of unity, since then $\mathbb{F}G \cong \mathbb{F}[x]/(x^{|G|}-1)$ and (at least as $\mathbb{F}[x]$-modules) $\mathbb{F}[x]/(x^{|G|}-1)\cong \bigoplus_{i=1}^{|G|} \mathbb{F}[x]/(x-a_i) \cong \bigoplus_{i=1}^{|G|} \mathbb{F}$. If $\mathbb{F}$ doesn't have all $|G|$-th roots of unity, then the result above seems suspect.

My second question is:

For an arbitrary non-abelian group $G$, does the center $Z(G)$ correspond to some standard form (i.e. maybe a string of $|Z(G)|$ $1\times 1 $ matrices) in the decomposition $\mathbb{F}G\cong M_{n_1}(D_1)\times \dots \times M_{n_r}(D_r)$?

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Hint: (Perlis Walker) Let $G$ be a finite abelian group of order $n$, and $K$ be a field such that $char(K)\nmid n.$ Then $$KG\cong\oplus_{d/n}a_{d}K(\zeta_{d})$$ where $\zeta_{d}$ denotes a primitive root of unity of order $d$ and $a_{d}=\frac{n_{d}}{[K(\zeta_{d}):K]}$, $n_{d}$ denotes the number of elements of order $d$.

$D_i$ may not equal to $\mathbb F$

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