Inertia group of $\operatorname{Gal}{(\overline{\mathbb{Q}}_p/{\mathbb{Q}}_p)}$ I want to find inertia group of $\text{Gal}{(\overline{\mathbb{Q}}_p/{\mathbb{Q}}_p)}$ .
It is well known that the inertia group is isomorphic to $\operatorname{Gal}{(\overline{\mathbb{Q}}_p/{\mathbb{Q}}_p^{nr})}$.
But is it impossible to clarify this group in a more explicit form ?
 A: I'm using the decomposition descirbed by Asvin in the comments. Let $G=\mathrm{Gal}(\overline{\Bbb Q_p}/\Bbb Q_p)$ and let $G_i$ be the higher ramification groups, in particular, $G_0$ is the inertia group.
$G_0/G_1$ is isomorphic to $\mathrm{Gal}(\Bbb Q_p^{tr}/\Bbb Q_p^{ur})$, i.e. the Galois group of the maximal tamely ramified extension over the maximal unramified extension. It's well known that this Galois group is just $\prod_{\ell \neq p}\Bbb Z_{\ell}$.
So we have an exact sequence
$1 \to G_1 \to G_0 \to \prod_{\ell \neq p}\Bbb Z_\ell \to 1$.
Now $G_1$ is a pro-p group, in fact it's actually a free pro-p group of countably infinite rank (see Neukirch-Schmidt-Wingberg (https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/NSW2.3.pdf) Proposition 7.5.1.)
Because $G_1$ is pro-p and $\prod_{\ell \neq p}\Bbb Z_\ell$ doesn't have any quotient which is a p-group, $G_1$ is a Hall subgroup of $G_0$, so the sequence splits by the profinite Schur-Zassenhaus theorem (see Theorem 2.3.15 in Ribes-Zalesskii), so we get that $G_0$ is a semidirect product $G_1 \rtimes\prod_{\ell \neq p} \Bbb Z_\ell$ where $G_1$ is a free pro-p group of countably infinite rank. I'm not sure how to describe the action of $\prod_{\ell \neq p}\Bbb Z_l$ on $G_1$, however.
