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Let $k$ be a finite field with char$(k)=p>0$. Let $G$ be a finite group.

Consider the group algebra $kG$.

I would like to solve a given system of equations in $kG$.

Question:

Is there a computer algebra system that can solve a given system of equations in $kG$?

Any help is appreciated.

Thank you very much.

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    $\begingroup$ Is there a particular small example that you might like to see? $\endgroup$ Sep 18, 2020 at 22:33
  • $\begingroup$ Hm...for example finding all idempotent elements of the group algebra $\endgroup$ Sep 18, 2020 at 23:32

2 Answers 2

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As you say you have a given system of equations, you can do this by brute-force in GAP. The following constructs a group algebra $kG$ in GAP -- this one constructs $kG$ when $k = GF(4)$ and $G = C_2$.

G := CyclicGroup(2); 
k := FiniteField(4);
kG := GroupRing(k,G);

The output is then

gap> kG;
<algebra-with-one of dimension 2 over GF(2^2)>

and, as expected, one can run commands such as

gap> Size(kG);
16
gap> Elements(kG);
[ <zero> of ..., (Z(2)^0)*<identity> of ..., (Z(2)^0)*<identity> of ...+(Z(2)^0)*f1,
  (Z(2)^0)*<identity> of ...+(Z(2^2))*f1, (Z(2)^0)*<identity> of ...+(Z(2^2)^2)*f1, (Z(2^2))*<identity> of ...,
  (Z(2^2))*<identity> of ...+(Z(2)^0)*f1, (Z(2^2))*<identity> of ...+(Z(2^2))*f1, (Z(2^2))*<identity> of ...+(Z(2^2)^
    2)*f1, (Z(2^2)^2)*<identity> of ..., (Z(2^2)^2)*<identity> of ...+(Z(2)^0)*f1,
  (Z(2^2)^2)*<identity> of ...+(Z(2^2))*f1, (Z(2^2)^2)*<identity> of ...+(Z(2^2)^2)*f1, (Z(2)^0)*f1, (Z(2^2))*f1,
  (Z(2^2)^2)*f1 ]

Now to solve your given system of equations is easy by brute force.

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    $\begingroup$ Thank you. Unfortunately, a problem is that this approach might take years, if the groups are slightly bigger. Therefore, I was wondering if there is some CAS or GAP - package which has already implemented some short cuts / fast algorithms. $\endgroup$ Sep 18, 2020 at 23:31
  • $\begingroup$ @BernhardBoehmler Maybe you can work in a concise matrix representation by switching to FaithfulModule(kG) - for example in this particular case it provides a faithful representation with 2x2 matrices $\endgroup$ Sep 19, 2020 at 6:38
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A partial answer in a very special case, but still possibly useful to record.

The search for idempotents in a group algebra can be simplified by using GAP's MeatAxe: https://www.gap-system.org/Manuals/doc/ref/chap69.html

For example, if $G$ is the dihedral group with eight elements, and $F$ the field with eight elements, the following code verifies that $kG$ is indecomposable as a $G$-module, which means that the only idempotents are zero and one.

G := DihedralGroup(8)
k := FiniteField(8);
cayley := Action(G, AsList(G), OnRight);
reg := PermutationGModule(cayley, k);
MTX.IsIndecomposable(reg);

More generally, if $M$ is a MeatAxe module, you can compute a direct sum decomposition using

MTX.Indecomposition(M)

Of course, this doesn't solve for all the idempotents---it only returns a few. You can also use

MTX.HomogeneousComponents(M)

to break the search into smaller algebras before using a more brute-force method.

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  • $\begingroup$ As a representation theorist, I'm embarrassed to have to ask, but I'm only used to thinking of representations over fields of characteristic $0$. Why does indecomposability imply that the only idempotents are trivial? $\endgroup$
    – LSpice
    Sep 19, 2020 at 3:29
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    $\begingroup$ @LSpice Any idempotent in the group algebra induces an idempotent endomorphism of the regular representation (multiplication on the side opposite the action). The image and kernel of this endomorphism would then be a direct sum decomposition. $\endgroup$ Sep 19, 2020 at 3:40
  • $\begingroup$ Got it, thanks. I was trying to think in terms of eigenspaces, which doesn't work in characteristic 2, and it didn't occur to me to substitute image and kernel. $\endgroup$
    – LSpice
    Sep 19, 2020 at 11:04

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