Let $K$ be a finite extension of $\mathbb Q_p$ for some prime $p$. $I_K$ is the inertia group of $\bar K/K$.
Can we prove the inertia group is commutative?
Actually the case is: For some representation $V$ (finite-dimensional over $\mathbb Q_l$) of $I_K$, if $I_K$ acts unipotently(i.e. all elements only have $1$ as the eigenvalue), I need to prove that all elements of $I_K$ can be simultaneously triangularized.
If $I_K$ is commutative or its image in $\operatorname {Aut}_{\mathbb Q_l}(V)$ is commutative, then it's easy. How to get the conclusion?
Let $V_K$ be the ramification group, which is a closed subgroup of $I_K$. By Proposition 7.5.1 of Cohomology of Number fields (Neukirch-Schmidt-Wingberg) $V_K$ is a free pro-$p$ group of countably infinite rank, thus $V_K$ is not commutative. Then $I_K$ is not commutative a fortiori.