An Exercise in Allen Hatcher's book on Spectral Sequences anyone knows how to solve Exercise 3 of Chapter 1 of Allen Hatcher's book on Spectral Sequences? The question is as follows:
For a fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$ associated to a short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$ show that the associated action of $\pi_1K(C,1)=C$ on $H_*(K(A,1);G)$ is trivial if $A$, regarded as a subgroup of $B$, lies in the center of $B$.
Any help will be greatly appreciated.
 A: If $A$, $B$, $C$ are finite groups that fit into an exact sequence $0 \to A \hookrightarrow B \xrightarrow{\phi} C \to 0$ with $A$ abelian, we will show that there is a well defined action of $\pi_1 K(C,1)$ on $\pi_1(K(A,1))$, and that it is given by the following purely algebraic description:
$c \cdot a= \phi^{-1}(c) a (\phi^{-1}(c))^{-1}$.  Here $\phi^{-1}(c)$ refers to a single lift of $c$ to $B$.  This will follow from
A General Fact:
Let $F_p \hookrightarrow E \xrightarrow{\pi} (B,p)$ be a fibration  with the induced map $\pi_1(F) \to \pi_1(E)$ injective, and $\pi_1(F)$ abelian.  (Thus we can view $\pi_1(F) \subset \pi_1(E)$. Then there is a well defined (monodromy) action of $\pi_1(B)$ on $\pi_1(F)$.  It has the same algebraic description as above: 


*

*$\phi_*$ is surjective(we will show it) and thus the exact sequence of homotopy groups breaks as $0 \to \pi_1(F) \subset \pi_1(E) \to \pi_1(B) \to 0$. 

*For $[\alpha] \in \pi_1(B,p)$, and $[\epsilon] \in \pi_1(F,p)$, we have $[\alpha] \cdot 
 [\epsilon]=\phi_*^{-1}([\alpha]) [\epsilon] (\phi_*^{-1}([\alpha]))^{-1}$.   


Proof:
Recall that $\alpha$ gives a map $\widetilde \alpha=\{\widetilde \alpha_t(p)\}_{p,t}: F \times [0,1] \to E$ with $F \times {0}$ mapping as the inclusion into $E$, and $\pi \widetilde \alpha(p)=\alpha$.  A standard fact is that $\alpha_1$ is a homotopy equivalence of $F$.  Let $\gamma$ be a path in $F$ connecting the basepoint $p \in F$ and $\widetilde \alpha_1(p)$.  Define $[\alpha] \cdot [\epsilon]:=[\gamma \widetilde{\alpha_1}(\epsilon) \gamma^{-1}] \in \pi_1(F,p)$.  This is an automorphism that does not depend on the choice of $\gamma$ because the inner automorphisms of $\pi_1(F,p)$ are 0.
Observe that in $\pi_1(E,p)$, $[\widetilde{\alpha}(p)\gamma^{-1}][ \gamma \widetilde{\alpha_1}( \epsilon) \gamma^{-1}][ \gamma \widetilde{\alpha(p)}^{-1}]=[\epsilon]$
Therefore $[\alpha] \cdot [\epsilon]=[\gamma \widetilde{\alpha_1}(\epsilon) \gamma^{-1}]=[\gamma \widetilde{\alpha}(p)^{-1}][ \epsilon] [\gamma \widetilde{\alpha}(p)^{-1}]^{-1}$
Note that $\gamma \widetilde{\alpha}(p)^{-1}$ is a loop in $E$ that projects to $\text{(constant path)} \,\,\alpha$.  This implies the surjectivity of $\pi_*$ and that $[\gamma \widetilde{\alpha}(p)^{-1}]$ is in $\pi_*^{-1}[\alpha]$. This implies part (2) of the proposition.

Now you know the algebraic description of the action.  Because your extension is a central extension, this action is trivial.  This gives the action on $\pi_*(K(A,1))$ since the action is trivial on the higher $pi_i$.  By naturality of the hurewicz map, $\pi_1(K(C,1))$ acts trivially on $H_*(K(A,1))$.
