Open neighborhood of the identity with no proper subgroup. I'm trying to prove that there exists an open neighborhood of the identity in the complex general linear group that contains no proper subgroup.  This is needed to show that a continuous representation of a profinite group has open kernel.
 A: Choose your neighborhood $U$ of the identity to consist of matrices $g$ with the property that the operator norm of $g-1$ is less than $1$. In particular, any element $g \in U$ has eigenvalues of distance (in the complex plane $\mathbf{C}$) less than $1$ from the number $1$. 
Now suppose you have a non-trivial subgroup contained in $U$ and let $g$ be an element of this subgroup. Then all the powers of $g$ are also contained in $U$; it now follows by our choice of $U$ that the eigenvalues of $g$ are all equal to $1$. So $g$ is unipotent; say $g=1+n$ with $n$ nilpotent. Suppose $n \neq 0$ and choose vectors $v_1 \neq 0$ and $v_2$ such that $v_1 \in \mathrm{ker}(n)$ and $n v_2=a v_1$ for some non-zero $a.$ 
It follows that for all integers $k$
$$g^kv_2=v_2+kav_1,$$ contradicting 
$$|g^k v_2-v_2 | < |v_2|.$$
This finishes the proof. It's motivated by thinking first about what happens for $\mathbf{C}^\times=\mathrm{GL}_1(\mathbf{C})$; the proof above is simply an adaptation of what I think is the most obvious route in that case. 
