5
$\begingroup$

For $A$ a finite-dimensional algebra over a field $K$
Does there exist a finite group $G$, such that $A$ is a sub-algebra of $K[G]$ ?
Where $K[G]$ denotes the group-algebra of $G$ over $K$.

In case that the answer is no, would there be a way to "detect" when it is the case ?

I would not mind an answer under some "nice" conditions such as commutativity, associativity, etc.

I tried to do few low-dimensionnal examples through the use of the structure coefficients but it became quickly untractable by hand.

One the one hand, I have a feeling that possibility of taking $G$ as large as one wants would give some trivial construction but did not find any.
On the other hand I feel like it would imply quite a lot and that some algebraic structure such as hypergroups would loose a bit of interest.

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ If you're fine with a non-unital subalgebra, can't you take a group with an irrep of dimension at least that of $A$, and identify $A$ with a matrix representation? $\endgroup$
    – Kevin
    Commented Aug 2, 2022 at 15:02

1 Answer 1

Reset to default
11
$\begingroup$

Assuming that by "sub-algebra" you mean "unital sub-algebra":

Every group algebra has a one-dimensional module (the trivial module), so any subalgebra has a one-dimensional module.

But many finite-dimensional algebras (e.g., $M_n(K)$ for $k>1$) have no one-dimensional modules.

But there may be more interesting restrictions.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ I may not understand what you mean by "having a module" for that I am not that versed in algebra but isn't $K Id_n$ a one-dimensonal module of $M_n(K)$ ? $\endgroup$
    – Hugo MTV
    Commented Aug 2, 2022 at 14:25
  • $\begingroup$ @HugoMTV No, multiplying an element of $K\text{Id}_n$ by an element of $M_n(K)$ doesn't usually give an element of $K\text{Id}_n$. $\endgroup$
    – Jeremy Rickard
    Commented Aug 2, 2022 at 14:33
  • $\begingroup$ Nevermind, I thought that "having a module" meant having a sub-module when it meant "having a module over oneself. Thanks $\endgroup$
    – Hugo MTV
    Commented Aug 2, 2022 at 14:39
  • 1
    $\begingroup$ Nice answer! The same argument works for any augmented algebra as well. $\endgroup$
    – Salvatore Siciliano
    Commented Aug 2, 2022 at 18:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .