Let $G$ be a finite group and $k$ be a finite field of characteristic $p>0$ such that $p\mid |G|$.

Let $M$ be a $kG$-module which has an embedding $M\hookrightarrow kG^{reg}$ into the regular $kG$-module $kG^{reg}$.

Then $M$ corresponds to a right ideal of $kG$.


Is there a MAGMA command / procedure that has as input the $kG$-module $M$ and as output the corresponding right ideal $I$ in terms of giving me elements of the algebra $kG$ that generate $M$ as the right ideal $I$?

I only could find how to get a basis of the module $M$ in the manual, but not how to transform this into the elements of the algebra $kG$.

I would be very grateful for any help.

Thanks in advance.

EDIT (04.10.2020) :

I am in the following concrete situation:

Let $P$ be a non-trivial $p$-subgroup of $G$. Let $T$ be the trivial group.

Consider the $kP$-module $M:= k \uparrow_T^{P}\cong {kP}^{\text{reg}}$.

$M$ is a permutation module. Hence, $k_P$ is a submodule of $M$ (namely, $\{\lambda\cdot (1,1,1,\dots , 1)^t | \lambda\in k\} \cong k$, if $M$ is given as a representation via permutation matrices).

Since $kG$ is free as $kP$-module, the induction functor is exact.

Thus, via induction, we get the $kG$-module $N:=k\uparrow_P^G$ as a submodule of $M\uparrow_P^G\cong {kG}^{\text{reg}}$.

I would like to get $N$ (and all of its direct summands) as a right ideal of $kG$ and couldn't find out how to do this in MAGMA.

Cross-reference: https://mathoverflow.net/questions/373514/magma-question-concerning-the-transformation-of-a-kg-module-m-into-a-right

  • $\begingroup$ I don’t understand the question. You said $M$ is embedded. That tells you exactly what elements are the copy of $M$ inside the group ring. Of course you can’t have a method that takes a module and tells you the image because the image inside the group ring may not exist. You’d need to specify the map, too. And the map tells you right away what you’re looking for, does it not? $\endgroup$
    – rschwieb
    Oct 4, 2020 at 3:18
  • $\begingroup$ Thank you very much for the comment. You are right. My formulation of the question was very bad. I edited the question. $\endgroup$ Oct 4, 2020 at 14:08


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