Let $G$ be a connected matrix group and let $H$ be any subgroup of $G$ containing open nbhd $U$ of $e$. Then $H=G$.

Question

Let $G$ be a connected matrix group and let $H$ be any subgroup of $G$ containing open nbhd $U$ of $e$. Then $H=G$.

For proving this, in Matrix groups by Curtis, following set was considered $$W=U \cup U^2 \cup U^3 \cdots $$ $W$ is open as it is union of open sets. It is closed as if $x$ is limit point of $W$. Then $xU$ is nbhd of $x$ thus $xu=u_1u_2u_3 \cdots u_m$ for $u_i \in U$.
Thus $x=u_1u_2u_3 \cdots u_m u^{-1}$ but $u^{-1} $ need not be in $U$ or $W$(not completely sure here). So I dont understand how this argument works?
Also, why exactly do we need $W$ the argument seems to work equally well for just $U$?

Answer 1

Let $V$ be $U \cap U^{-1}$; $V$ is open and equal to $V^{-1}$. Now let $K=\cup_{i=1}^{+\infty} V^i$; obviously $K \subset H$, and $K$ is also open, as it is a union of open sets, and you can verify that it is also a subgroup (HINT: $V^n \cdot V^m \subset V^{n+m}$); now we want to prove that $K=G$. If it's not the case, then $G= K \coprod_{g \notin K} gK$. This is a disjoint union of open sets, but G is connected; so we have $G=K$.

So we have shown that a connected (Lie) group (or group of matrices in your case) is "generated" by a nbhd of the origin. If you take only "U" it is not a subgroup, so the argument used to say that $G$ is the union of left cosets of $K$ doesn't work.

Answer 2

As others have noted, we need to take a symmetric neighborhood of the identity to validate the original argument.

Another way to prove the claim (without taking a symmetric neighborhood) is to show that $H$ is both open and closed in $G$. Since $G$ is connected, that would imply $G=H$.

$H$ is open in $G$ because, for every $h\in H$, $hU$ is an open neighborhood of $h$ contained in $H$.

Let $x$ be a limit point of $H$. So every open neighborhood of $x$ intersects $H$ at some point other than $x$ itself. Consider the open neighborhood $xU$. Let $h$ be an element of $xU\cap H$. Then $h=xu$ for some $u\in U\subseteq H$. In other words, $x=hu^{-1}$ is an element of $H$. So $H$ contains all of its limit points and hence is closed.

Answer 3

I will add something to Giuseppe's answer to validate the argument given in my question.
As Guiseppe said take symmetric nbhd. So we have the inverse of every element in our nbhd.
From the hint given by Giuseppe, it is clear that $W$ is now a subgroup. Thus now it makes sense to say that $x=u_1u_2u_3 \cdots u_m u^{-1} \in W$.
Note that $W$ being subgroup is a crucial fact here.