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Let $K$ be field with $\mathrm{char}(K) = p > 0$ and $G$ a finite group such that $KG$ is soluble. Then the $p$-Sylow-subgroup of $G$ is normal and contains the derived group of $G$ and every $p'$-Hall-subgroup $H$ is abelian. The radical $J=\mathrm{rad}(KG)$ of $KG$ is $\mathrm{Aug}(KP)KG$ and $KH$ is a radical complement.

I investigated the following: Take a unital subalgebra $T$ of $KH$ and determine $C_J(T)$ and afterwards $C_{KH}(C_J(T))$. Then $T$ is a subalgebra of $C_{KH}(C_J(T))$.

I could proove that starting this process again with $C_{KH}(C_J(T))$ then its stable. I describe for $T=KU$ and $U$ a subgroup of $H$ the subalgebras $C_J(T)$ and $C_{KH}(C_J(T))$ using group operation, too. In particular its possible to answer when $KU$ is stable under this construction, when not and when it reaches $KH$ or $Z(KG)\cap KH$.

My question is whether there are any ideas for a general unital subalgebra of $KH$ (maybe under the restriction that $K$ is a splitting field of $KH$. In this case the subalgebras are known by using the primitive othogonal idempotents in $KH$ (see Scheja & Storch). They can be contructed by the orthogonal idempotents using partitions. Since $T\le C_{KH}(C_J(T))$ the partition used for $C_{KH}(C_J(T))$ is a refinement of that used for $T$. The question is how the partion for $C_{KH}(C_J(T))$ can be decribed.)`

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  • $\begingroup$ I'm not familiar with the term soluble, applied to group algebras, or indeed to finite dimensional $K$-algebras. Or do you mean that $G$ is a soluble group? $\endgroup$ Commented Sep 9, 2014 at 9:46
  • $\begingroup$ For an associative K-algebra A soluble means that A/rad(A) is commutative. This is equivalent to that there is a descending chain of ideals beginning at A and ending at the zero-subalgebra with commutative factors. If $char(K)≠2$ than soluble is equivalent to the fact that the associative Lie-algebra $A^{\circ}$ of $A$ is soluble. So not all soluble groups leads to soluble group algebras. These are described in the intro of my question. $\endgroup$ Commented Sep 9, 2014 at 17:48
  • $\begingroup$ we should close the topic. $\endgroup$ Commented Dec 16, 2020 at 20:00

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