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Let $G$ be a finite group and $N\le G$ be a subgroup. Consider the group algebra $kN$ as a subalgebra of $kG$ over an algebraically closed field $k$ of positive characteristic. What can we deduce from the simple relation $J(kN)=J(kG)$ (where $J()$ is the Jacobson radical)? Are there any conclusions concerning $G,N$? Especially the case where the Jacobson radical is not 0 is of interest.

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up vote 6 down vote accepted

In the case that $J(kG) \neq \{0\},$ (which is equivalent to $p | |G|,$ where $p$ is the characteristic of $k ),$ we may conclude that $N = G$ under the assumptions of the question. This is because $J(kG)$ is a two-sided ideal of $kG.$ Otherwise, for any group element $t \in G \backslash N,$ we would have $tJ(kN) \subseteq J(kG) \subseteq J(kN) \subseteq kN.$ On the other hand, $tJ(kN) \subseteq tkN,$ so any element of $tJ(kN)$ is a $k$-linear combination of elements of the right coset $tN.$ Since $J(kN) \neq \{0\},$ this is a contradiction.

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