2

Question: We want to find all subgroups of $\mathrm{GL}(2,\mathbf{R})$ of index $2$.

Here is my first attempt: We already know that for a group $G$, if a subgroup $H$ satisfies $(G\colon H)=2$, then $H$ is a normal subgroup of $G$. So we first try to construct some normal subgroups of $\mathrm{GL}(2,\mathbf{R})$. Note that $$S\,\colon\!=\left\{A\in \mathrm{GL}(2,\mathbf{R})\,|\,\det(A)>0\right\}$$ is a normal subgroup of $\mathrm{GL}(2,\mathbf{R})$. And $\left|\mathrm{GL}(2,\mathbf{R})/S\right|=2$. So $S$ is one subgroup meeting our requirement.

But I am not sure whether there are other subgroups of $\mathrm{GL}(2,\mathbf{R})$ have index $2$.

Sunny. Y
  • 123
  • 6
  • 1
    Here's the general idea for finding subgroups of index 2. Let $G$ be any group. If $H \leq G$ has index 2, then $H$ is normal and $G/H$ is a group of order 2. Any group of order $2$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, so every subgroup of index 2 in $G$ is the kernel of a surjective homomorphism to $\mathbb{Z}/2\mathbb{Z}$. For your situation: using the fact that $GL(2,\mathbb{R})$ is generated by elementary matrices, see what you can discover about such homomorphisms! – diracdeltafunk Mar 10 '22 at 04:21
  • @diracdeltafunk Thank you for your help! That's a good idea! – Sunny. Y Mar 10 '22 at 17:46

1 Answers1

1

As the comment given by @diracdeltafunk suggests, since $H$ is a subgroup of index 2, it suffices to classify all group homomorphism $f:\mathrm{GL}_2(\mathbb{R})\to \mathbb{Z}/2\mathbb{Z}$. Since $\mathbb{Z}/2\mathbb{Z}$ is commutative and the commutator of $\mathrm{GL}_2$ is $\mathrm{SL}_2$,see Commutator Group of $\operatorname{GL}_2(\mathbb{R})$ is $\operatorname{SL}_2(\mathbb{R})$

the homomorphism $f$ factors through $\mathbb{R}^\times$. It should be very easy to classify all group homomorphism $\overline {f}: \mathbb{R}^\times\to \mathbb{Z}/2\mathbb{Z}$.

Q-Zhang
  • 1,608