Q: A basic question on semi-direct product induced group action Let $G'=A\rtimes G$ be semi direct product of $A,G$ s.t. $G\to Aut(A)$ is given. Then by definition of semi-direct product $g,g'\in G,a,a'\in A, (a,g)\times (a',g')=(a (g\cdot a'),gg')$ where $g\cdot\in Aut(A)$. 
It is clear that $(1,g)(a,1)(1,g)^{-1}=(g\cdot a,1)$ defined by the semi-direct product. 
Now consider $G'=A\rtimes G\to G$ projection map by projecting out $A$ where $A$ is normal in $G'$ by construction. Pick any fiber of $g'\in G$ above $g\in G$ via projection map. Consider $(b,g')(a,1)(b,g')^{-1}=(b (g\cdot a)b^{-1},1)$. I do not see any good reason that $b(g\cdot a)b^{-1}=g\cdot a$ unless $[b,ga]=1$ which I doubt this is true in general.  
$\textbf{Q:}$ Did I get $(b,g')$ action computation wrong here? It seems that the book is implying that action of $G$ on $A$ can be identified by action of any preimage in $G'$ on $A$.(i.e. $g'\in G\to g\in G$, then $a\in A, g\cdot a=g'ag'^{-1}$.) This is related to looking at number of extensions of $1\to A\to G'\to G\to 1$ exact sequence with $A,G$ given s.t. action of $G$ on $A$ is given by $ ^{\sigma}(a)=\sigma'a\sigma'^{-1}$ where $\sigma\in G,\sigma'\in G'$ and $\sigma'$ is the pre-image of $\sigma$.
Ref. Cohomology of Number Fields by Neukirch, Chpt 1, Sec 2.
