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I am wondering (after reading the Skolem-Noether theorem) whether a group algebra $\Bbb{k}[G]$ can be central (ie. Its center be equal to $\Bbb{k}$). Obviously, the group would be centerless but apart from that?

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  • $\begingroup$ I believe the dimension of the center of $k[G]$ as a $k$-vector space is equal to the number of conjugacy classes of $G$. $\endgroup$ – sharding4 May 28 '17 at 15:52
  • $\begingroup$ @sharding4 right! So there is only one conjugacy class and the group is trivial. If you write it down I will mark it as an answer $\endgroup$ – user128787 May 28 '17 at 15:55
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    $\begingroup$ If $G$ is infinite, and all non-identity conjugacy classes are too, then the centre of $kG$ will be just $k$. $\endgroup$ – Lord Shark the Unknown May 28 '17 at 16:00
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If $G$ is a group and $a=\sum_g a_gg\in k[G]$, then $a$ is in the center of $k[G]$ iff the coefficients $a_g$ are constant on conjugacy classes. Indeed, given $h\in G$, the condition that $ha=ah$ says exactly that $a_{h^{-1}g}=a_{gh^{-1}}$ for all $g\in G$. Replacing $g$ with $hg$, this is equivalent to saying $a_g=a_{hgh^{-1}}$ for all $g\in G$.

So in particular, the center of $k[G]$ has a basis consisting of sums of the form $\sum_{g\in C} g$ where $C$ is a finite conjugacy class of $G$. So the dimension of the center is the number of finite conjugacy classes. In particular, the center is $k$ iff all conjugacy classes except the identity class are infinite. If $G$ is finite, this means the center can be $k$ only if $G$ is trivial.

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