I don't know if the next thing is true, but I'm not able to find a counterexample: suppose you have a surjective group homomorphism of finite groups f:G \rightarrow G' and normal subgroups H \lhd G, H' \lhd G', such that the induced homomorphism f':G/H \rightarrow G'/H' is also surjective. Is it true that f\mid_H: H \rightarrow H' is also surjective ? In particular, is this true for decomposition groups and inertia groups in the case of Galois extensions $L/K/\mathbb{Q}_p$ ?
surjectivity of group homomorphisms
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0I'm not entirely clear on what your second question is. You want $\mathbf{Q}_p\subseteq K\subseteq L$ to be a tower of Galois extensions, and then you're asking if the inertia group of $L/\mathbf{Q}_p$ surjects onto the inertia group of $K/\mathbf{Q}_p$? Also, since you're working over $\mathbf{Q}_p$, "decomposition group" just means "Galois group." – 2011-12-09
1 Answers
This is not true; in fact, for any surjective group homomorphism f:G\to G', we can choose $H=\{e\}\triangleleft G$ and H'=G'\triangleleft G', so that $G/H\cong G$ and G'/H' is trivial, hence the induced homomorphism f':G/H\to G'/H' is certainly surjective, but f|_H:H\to H' is not surjective unless the group G' is itself trivial.
I think that if
- $L/\mathbb{Q}$ and $K/\mathbb{Q}$ are Galois extensions with $L\supseteq K$
- $p$ is a prime of $\mathbb{Q}$, $\mathcal{P}$ is a prime of $L$ lying over $p$, and $\mathfrak{p}=\mathcal{P}\cap K$ is the prime of $K$ beneath $\mathcal{P}$
$\begin{align*}G&=D(\mathcal{P}/p)=\{\sigma\in\operatorname{Gal}(L/\mathbb{Q})\mid \sigma(\mathcal{P})=\mathcal{P}\}\\ H&= I(\mathcal{P}/p)=\{\sigma\in D(\mathcal{P}/p)\mid \overline{\sigma}:\mathcal{O}_L/\mathcal{P}\to\mathcal{O}_L/\mathcal{P}\text{ is the identity} \}\end{align*}$
\begin{align*}G'&=D(\mathfrak{p}/p)=\{\tau\in\operatorname{Gal}(K/\mathbb{Q})\mid \tau(\mathfrak{p})=\mathfrak{p}\}\\ H'&= I(\mathfrak{p}/p)=\{\tau\in D(\mathfrak{p}/p)\mid \overline{\tau}:\mathcal{O}_K/\mathfrak{p}\to\mathcal{O}_K/\mathfrak{p}\text{ is the identity} \}\end{align*}
f:G\to G' is the restriction of the restriction map $r:\operatorname{Gal}(L/\mathbb{Q})\to\operatorname{Gal}(K/\mathbb{Q})$, which is well-defined because if $\tau(\mathfrak{p})=\mathfrak{q}\neq\mathfrak{p}$ then, for any lift of $\tau$ to a $\sigma\in \operatorname{Gal}(L/\mathbb{Q})$, we would have that $\sigma(\mathcal{P})=\mathcal{Q}$ for some prime $\mathcal{Q}$ lying over $\mathfrak{q}$, so $\sigma(\mathcal{P})\neq\mathcal{P}$, and hence $f$ is surjective because $r$ is
then in fact f':G/H\to G'/H' is always surjective, as it is equivalent to the restriction map of Galois groups $\operatorname{Gal}(\mathcal{O}_L/\mathcal{P}\;\;/\;\;\mathbb{Z}/(p) )\to\operatorname{Gal}(\mathcal{O}_K/\mathfrak{p}\;\;/\;\;\mathbb{Z}/(p))$.
So I think your question just becomes, is f|_H: H\to H' always surjective? I think the answer is again yes, intuitively because of the multiplicativity of ramification indices, but I am having trouble proving it. Hopefully someone will be able to sort this out.
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1Haha, I didn't even see the little ${}_p$ down there :) Oh well. By the way, you don't have to accept my answer; you can write up your solution and accept it, so that the whole web can get the benefit of the answer. This is explicitly encouraged by the SE network of sites; see [here](http://meta.stackexchange.com/questions/12513/should-i-not-answer-my-own-questions) and [here](http://blog.stackoverflow.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/). – 2011-12-09