Is there a CAS that can solve a given system of equations in a finite group algebra $kG$?

MathOverflow Asked by Bernhard Boehmler on December 1, 2020

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 $kG$.


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

Any help is appreciated.

Thank you very much.

2 Answers

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:

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);

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


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


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

Answered by John Wiltshire-Gordon on December 1, 2020

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);
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.

Answered by Carl-Fredrik Nyberg Brodda on December 1, 2020

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