What is the "no small subgroups" argument for $GL(n,\mathbb R)$? That is, how do we show that in $GL(n,\mathbb R)$ there exists a neighborhood of the identity which contains no subgroup other than the trivial one? I had some scribbling (for the $n=2$ case) but could not arrive at a clean proof.



"No Small Subgroups" Argument
Here is Asaf's agrument expanded a bit. It has the advantage of working for all Lie groups simultaneously.


Given a Lie group $G$ with Lie algebra $\mathfrak{g}$, consider the exponential map $\exp:\mathfrak{g}\rightarrow G$. It is known that it is a diffeomorphism on a small enough open set $U\subseteq\mathfrak{g}$.


Choosing an inner product on $\mathfrak{g}$, we may assume wlog that $U$ has the form $U = \{v\in \mathfrak{g} : \; |v| < \epsilon\}$ for some $\epsilon > 0$. Let $V\subseteq U$ with $V = \{v\in\mathfrak{g} : \; |v| < \epsilon/2\}$.


I claim that $\exp(V)$ contains no nontrivial subgroups. Indeed, suppose $H\subseteq \exp(V)$ is a subgroup and choose $g\in H$ so $g = \exp(v)$ for some $v\in V$. I claim that $2v \in V$ as well. To see this, notice that since $g^2 \in H\subseteq \exp(V)$, we must have $g^2 = \exp(w)$ for some $w\in V$. Then $\exp(w) = g^2 = \exp(v)^2 = \exp(2v)$ which implies $w=2v$ since $\exp|\_U$ is a diffeomorphism. Thus, $2v \in V$.


But now can iterate this argument showing $2^n v \in V$ for all $n$. Since $|2^n v| = 2^n |v|$, this implies $v=0$, i.e. that $g =e$ so $H$ is trivial.


