# What's the Sylow group of the absolute Galois group of a henselian field?

Let $(K,w)$ be a henselian field such that the residue field $k=k(w)$ is an algebraic extension of a finite field $\mathbb{F}_p$. Let $\ell\neq p$ be a prime with $\mu_\ell\subset K$.

Now, in the paper I'm reading, it states:

Let $G_\ell$ be a pro-$\ell$ Sylow group of $G_K$. By the ramification theory for general valuations (see e.g. O. Endler, Valuation Theory, Springer, 1972, §20) we have $G_\ell\cong \mathbb{Z}_\ell(1)^r \rtimes G_k(\ell)$ where the action is defined via the cyclotomic character with $r=\dim_{\mathbb{F}_\ell}(\Gamma_w/\ell\Gamma_w)$ and $G_k(\ell)$ being the pro-$\ell$ Sylow group of $G_k$ hence either $\cong \mathbb{Z}_\ell$ or $\cong 0$.

I've looked into "Valuation Theory", §20 by O. Endler and other books on the subject ramification theory, but the only thing I've found is how the ramification group is the pro-$p$ Sylow group of the inertia group, which, of course, is something entirely different, right?

Anyway, I'm wondering why this assertion holds and if there are any better references?

Update: I've found the same assertion in the paper On Grothendieck's Conjecture of Birational Anabelian Geometry by Florian Pop, Ann. Math. 138 (1994), p.155, 1.6 (2). But there's no reference.

• Is it clear what $G_k(\ell)$ is? – Lubin Mar 9 '17 at 23:23
• @Lubin It's the pro-$\ell$ Sylow group of $G_k$, I've updated my question. – ryanblack Mar 10 '17 at 11:50

Let $T\subset G_K$ be the absolute inertia group and $V\subset T$ be the ramification group. $V\subset T$ is the unique $p$-Sylow group of $T$. Valued Fields by A.J. Engler, A. Prestel, Theorem 5.3.3 (iii) states that

$$T/V\cong\prod_{\ell\neq p}\mathbb{Z}_\ell^{r_\ell}$$

where $r_\ell=\dim_{\mathbb{F_\ell}}(\Gamma/\ell\Gamma)$. Thus the $\ell$-Sylow group of $T$ is just $\mathbb{Z}_\ell^{r_\ell}$. We have the short exact sequence:

$$1\longrightarrow T \longrightarrow G_K \longrightarrow G_k \longrightarrow 1$$

Now $G_k\subset G_{\mathbb{F}_p}\cong\hat{\mathbb{Z}}$, thus $G_k(\ell)\cong\mathbb{Z}_\ell$ or $\cong 0$. This sequence gives rise to an exact sequence of their $\ell$-Sylow groups:

$$1\longrightarrow \mathbb{Z}_\ell^{r_\ell}\longrightarrow G_\ell \longrightarrow G_k(\ell)\longrightarrow 1$$

which splits since $G_k(\ell)$ is a free pro-$\ell$ group. Therefore, we get an isomorphism:

$$G_\ell \cong \mathbb{Z}_\ell^{r_\ell}\rtimes G_k(\ell)$$

All there's left to do is to verify that $G_k(\ell)$ acts on $\mathbb{Z}_\ell^{r_\ell}$ via the cyclotomic character.