1
$\begingroup$

Denote the symmetric group of order $n!$ by $S_n$. Let $H:=S_p$ for an odd prime $p$.

Every finite field $k$ is a splitting field $(^*)$ for $kH$, in particular $k:=\mathbb{F}_p$.

Questions:

Is $k:=\mathbb{F}_p$ also a splitting field for $kG$ where

a) $G:=H\times H$ ?

b) $G:=H \wr C_2$ ?

I would be interested in references appearing in the literature which deal with these (similar) questions.

Thank you in advance for the help.

Remark 1: $(^*)$ $k$ is a splitting field of $S_n$ if the $k$-algebra $kS_n$ splits over $k$, i.e. if for every simple $kS_n$-module $M$, we have End$_{kSn}(M)\cong k$. (cf. https://math.stackexchange.com/questions/422979/splitting-fields-of-symmetric-groups)

Remark 2: I looked at https://ncatlab.org/nlab/show/direct+product+group and due to remark 2.2 there my questions might not have an affirmative answer for arbitrary groups, but I was interested, if the statement is nevertheless true in these special cases.

Remark 3: I posted the same question at math.stackexchange.com (see https://math.stackexchange.com/questions/3800244/reference-request-concerning-splitting-fields-for-groups-that-are-related-to-spe) more then one week ago, but I did not receive an answer. I hope that it is ok to ask it here now.

Thanks in advance for the help.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ If k splits groups G,H then it splits $G\times H$. More generally the tensor product of split finite dimensional algebras is split. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 3, 2020 at 1:46
  • 1
    $\begingroup$ The result mentioned in the answer to this question mathoverflow.net/questions/369932/… certainly covers the case of HxH. The proof given there will handle the general case. It all boils down to the tensor product over k of matrix algebras over k if a matrix algebra over k. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 3, 2020 at 2:56

0

Reset to default

You must log in to answer this question.