Prove $\exists$ neighborhood of $I \in Gl(n,\mathbb{C})$ containing no nontrivial subgroup. Prove that there exists a neighborhood of the identity $I \in Gl(n,\mathbb{C})$ that contains no subgroup other than $\left\{ I \right\}$.
Thanks!
 A: We can prove there exists a neighborhood $V$ of $I$ in $\mathrm{GL}_{n}(\mathbb{C})$ such that the only subgroup of $\mathrm{GL}_{n}(\mathbb{C})$ in $V$ is $\lbrace I \rbrace$.
Consider the exponential map $\exp \, : \, \mathcal{M}_{n}(\mathbb{C}) \, \rightarrow \, \mathrm{GL}_{n}(\mathbb{C})$. For every $M \in \mathcal{M}_{n}(\mathbb{C})$, we have : 
$$ \exp(M) = I + M + o(M^{2}) $$
and $\exp(0)=I$. This proves that $\mathrm{D}_{0}\exp = \mathrm{Id}$, where $\mathrm{D}_{0}\exp$ denotes the differential of $\exp$ at $0$. Then, there exists a neighborhood $U$ of $0$ in $\mathcal{M}_{n}(\mathbb{C})$ and a neighborhood $V$ of $I$ in $\mathrm{GL}_{n}(\mathbb{C})$ such that $\exp$ is a diffeomorphism from $U$ to $V$. 
Set $W = \frac{U}{2}$ and $W' = \exp(W)$. Since $W'$ is open, it is a neighborhood of $I$. Let $M$ in $W'$. We can write $M = \exp(A)$ where $A \in W$. Then, there exists an integer $k \in \mathbb{N}$ such that $kA \in U \setminus W$. So, we have $M^{k} = \exp(kA) \in V \setminus W'$. 
Finally, $M^{k} \notin W'$. This ends the proof.
A: Select $U\subseteq \mathfrak{gl}(n,\mathbb C)$ bounded and small enough that the exponential map $\exp\colon \mathfrak{gl}(n,\mathbb C)\to GL(n,\mathbb C)$ is injective when restricted to $U$. 
Let $W=\bigcup_{0<t<1} t U$ and let $V=\exp(\frac12W)$.
For any $g\in V\setminus\{I\}$, we find a unique $h\in \frac12W\setminus\{0\}$ with $\exp(h)=g$. There exists a minimal $n\in\mathbb N$ with $nh\notin \frac12W$. But the still have $nh\in W$ because $\frac n2h\in\frac12W$. Consequently, $\exp(nh)=g^n$ is not in $V$ i.e. a group containing $g$ cannot be completely in $V$.
A: This uses the properties of exponential map heavily.
Let $U$ be a neighborhood of 0 in $M_{n}(\mathbb{C})$ such that the exponential map restricts to a diffeomorphism on $U$. We may assume (by shrinking if necessary) that $U$ is convex and bounded. We set $U_{1} := \frac{1}{2}U$. Let $ G \subseteq V := \exp{U_{1}}$ be a subgroup of $GL(n,\mathbb{C})$ and $A \in G$. Then we write $A = \exp{X}$ with $X \in U_{1}$ and assume that $X \neq 0$. Let $k \in \mathbb{N}$ be maximal with $kX \in U_{1}$ (the existence of $k$ follows as $U$ is bounded). Then
$$ A^{k+1} = \exp{(k+1)X} \in G \subseteq V $$
implies the existence of $Y \in U_{1}$ with $\exp{(k+1)X} = \exp{Y}$. Since $(k+1)X \in 2U_{1} = U$ follows from $\frac{k+1}{2}x \in [0,k]x \subseteq U_{1}$, and $\exp\vert_{U}$ is injective, we obtain $(k+1)X = Y \in U_{1}$, contradicting the maximality of $k$. Therefore, $A = I$.
This has been taken from "Structure and Geometry of Lie groups" by Hilgert and Neeb.
