TITLE: "No Small Subgroups" Argument
QUESTION [10 upvotes]: 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.

REPLY [21 votes]: 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.