surjectivity of group homomorphisms 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 $L/K/\mathbb{Q}_p$ ?
 A: 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.
