A finite subgroup of $GL_n(\mathbb{R})$ such that $A^2=I_n$ has at most $2^n$ elements.

Question

I want to prove the following theorem:

Let $G$ be a finite subgroup of $GL_n(\mathbb{R})$ such that $A^2=I_n$ for all $A\in G$. Then $|G|\leq 2^n$.

I don't know even how to start. If someone has a tip, I would be very grateful!

Answer 1

A first step can be to solve the following general exercise : if $G$ is a group with $g^2 = e$ for all $g\in G$, then $G$ is abelian.

Applying this to your $G \subset \mathrm{GL}_n(\mathbb{R})$ yields co-diagonalizability of all the elements of $G$ (they're diagonalizable because they're cancelled by $X^2-1 = (X-1)(X+1)$, and all in the same basis because they commute; that's another classical result of linear algebra that you should try to prove).

Now, as in the comments, what can be the eigenvalues of the elements of $G$ and thus what can they be, up to a common conjugation ? This should yield the answer