Are permutation matrices over reals reductive?

Question

Let $H$ denote the subgroup of permutation matrices of $\operatorname{GL}(n,\mathbb{R})$. Is $H$ a reductive group?

My Work

I know that $H$ is a linear algebraic group being a finite subgroup of $\operatorname{GL}(n,\mathbb{R})$. And, I have found various facts saying that the permutation matrices as an algebraic subgroup of $\operatorname{GL}(n,\mathbb{C})$ is reductive, e.g.,

Symmetric subgroups of $\operatorname{GL}(n,\mathbb{C})$ are linearly reductive and being linearly reductive is equivalent to being reductive over algebraically closed fields of zero characteristic.

I have tried to apply the following facts to this problem but have not had success:

1. A linear algebraic group over a field of characteristic $0$ is reductive if and only if its Lie algebra is a reductive Lie algebra.
2. Let $G$ be a reductive group and $H$ a closed subgroup, then the $H$ is reductive if and only if the quotient space $G/H$ is affine.

I have primarily tried applying fact (2) above, since $\operatorname{GL}(n,\mathbb{R})$ is a reductive group and the subgroup of permutation matrices is a closed subgroup. But, I am unable to determine whether $G/H$ is affine.

Motivation

I am curious about this question because I have a real algebraic variety with a group action on it by real permutation matrices, and I am curious about applying techniques of Geometric Invariant Theory to this problem. I do not know a lot about GIT, but from I have read one gets nice statements statements when the group is algebraic and reductive, hence my question.

• As you say, $H$ is a finite group. – Lord Shark the Unknown Aug 28 '18 at 3:49
• @LordSharktheUnknown I could not find a reference for the statement that every finite group is reductive when I was reading about them. Is it just a simple exercise, or would you be able to point me towards a reference for that statement - even an exercise in a book would suffice? Thank you. – Aguila Aug 28 '18 at 3:55
• See your comment 1. – Lord Shark the Unknown Aug 28 '18 at 3:58
• Which definition would you like to use? Note that most nice results about algebraic group actions require it to be connected in order for the geometric structure to have much use. – Tobias Kildetoft Aug 28 '18 at 4:26
• Note that being reductive by definition is a statement about a maximal solvable connected normal subgroup. This makes it very easy to show that this is a reductive group. – Tobias Kildetoft Aug 28 '18 at 17:02