normal subgroup basis of $\mathrm{GL}(n, \mathbb{Q}_{p})$ Consider the group $G=\mathrm{GL}(n, \mathbb{Q}_p)$. This group is locally compact and totally disconnected, and we have a basis of open subgroups given by 
$$
K(p^m) = \left\{ \begin{pmatrix} a&b\\c&d\end{pmatrix} \equiv \begin{pmatrix}1&0\\0&1\end{pmatrix} mod \,p^m\right\} \subseteq \mathrm{GL}(n, \mathbb{Z}_{p})
$$
for $n\geq 0$. These are even normal subgroup of $\mathrm{GL}(n, \mathbb{Z}_{p})$, but not in $\mathrm{GL}(n, \mathbb{Q}_{p})$. Can we find any other basis of open subgroups that is even normal in $\mathrm{GL}(n, \mathbb{Q}_p)$? If not, how can we prove it? Such normality holds for any totally disconnected compact group, so non-compactness of $\mathrm{GL}(n, \mathbb{Q}_p)$ will be a problem. However, I can't prove this. 
 A: The answer is no.
Assuming you are actually asking about $GL(d, \Bbb Q_p)$ where $d \ge 2$, here is an idea (for $d=2$, but it should not be hard to generalise): Any open subgroup $U$ must contain some element of the form $\pmatrix{1&0 \\p^n &1}$ for high enough $n$. But assuming such a subgoup is normal, the conjugate of that element with $\pmatrix{0&p^{-n} \\ 1 &0}$, which is $\pmatrix{1&1 \\0 &1}$, must also be in $U$, so the intersection of all open normal subgroups contains $\pmatrix{1&1 \\0 &1}$, hence no set of such open normal subgroups can be a basis of neighbourhoods of the identity element $\pmatrix{1&0 \\0 &1}$.
A: $\DeclareMathOperator{GL}{GL}\DeclareMathOperator{SL}{SL}\DeclareMathOperator{PSL}{PSL}$Here is another, more abstract proof. It uses some more powerful facts, but on the upside, requires pretty much no computations (if you use these facts as black boxes).
Instead of showing that $\GL_n$ does not have a basis of open normal subgroups, we will show that $\SL_n$ does not have one (which is slightly stronger, at least a priori). It works for any non-discrete, Hausdorff topological field, not just for the $p$-adics (just like the other answer).
Suppose $K$ is a non-discrete topological field and $n>1$. We will show that every open normal subgroup of $\SL_n(K)$ has index at most $n$. It follows that there is a minimal open normal subgroup $N_0\unlhd \SL_n(K)$ (also of index at most $n$). Since $K$ is non-discrete, neither is $\SL_n(K)$, so $N_0$ is nontrivial, and so if $K$ is Hausdorff, it follows that there is a neighbourhood of the identity in $\SL_n(K)$ which does not contain $N_0$.
Let $N\unlhd \SL_n(K)$ be an open normal subgroup. Since $K$ is not discrete, it follows that $N$ is nontrivial and contains a non-central element (e.g. a non-scalar diagonal matrix).
It follows that $N/Z(\SL_n(K))$ is a nontrivial normal subgroup in $\PSL_n(K)$, which is a simple group. Thus, $N\cdot Z(\SL_n(K))=\SL_n(K)$. But the center of $\SL_n(K)$ has at most $n$ elements (corresponding to $n$-th roots of unity in $K$). The result follows.
