Inverse Sequence of Group Extensions?

Question

Let $Q$ and $K$ be finitely presented groups with $H^2(Q)$ finitely generated and $H_1(K) = H_2(K) =0$ and $Z(K) \ne 0$ but fg. If we always use the trivial outer action, is it possible to have an inverse sequence $Q \leftarrow G_2 \leftarrow G_3 \leftarrow \ldots$, where each $G_i$ is a group extension $1 \to K \to G_i \to G_{i-1} \to 1$, that does not stabilize as a sequence of direct products? Every time one does a non-direct product extension, it appears to me one takes a quotient of $H^2(Q)$, and, with no $H^2(K)$ to "replenish" it, it appears to me one will "exhaust" in a finite number of steps $H^2(G_i)$, leaving only direct product possible for group extensions after a finite number of steps.

Answer 1

If one considers the $K(G,1)$'s for the various groups involved (say $F$ is a $K(K,1)$, $B$ is a $K(Q,1)$, and $E_i$ is a $K(G_i,1)$), then each group extension corresponds to a fiber bundle; in particular, we have $F \hookrightarrow E_2 \twoheadrightarrow B$ for the first fiber bundle. There is a corresponding long exact sequence in cohomology for this fiber bundle

$$ 0 \rightarrow H^1(B) \rightarrow H^1(E_2) \rightarrow H^1(F) \rightarrow H^2(B) \rightarrow H^2(E_2) \rightarrow H^2(F) \rightarrow H^3(B) \rightarrow H^3(E_2) \rightarrow\ldots $$

which, as $K$ is superperfect, becomes

$$ 0 \rightarrow H^1(B) \overset{\cong}\rightarrow H^1(E_2) \rightarrow 0 \rightarrow H^2(B) \overset{\cong}\rightarrow H^2(E_2) \rightarrow 0 \rightarrow H^3(B) \rightarrow H^3(E_2) \rightarrow\ldots $$

(Recall (1) by Universal Coefficients, $H^1(F) \cong F_1(F)$ and $H^2(F) \cong F_2(F) \oplus T_1(F)$, where $F_i(F)$ is the free part of $H_i(F)$ and $T_i(F)$ is the torsion part of $H_i(F)$, and (2) $H_1(F)$ and $H_2(F)$ are both 0 as $K$ is superperfect, so $F_1(F)$, $T_1(F)$, and $F_2(F)$ are all 0, so $H^1(F)$ and $H^2(F)$ are both 0.)

Hence, $H^2(G_i)$ must be isomorphic to $H^2(Q)$ for all $i$, so there is always non-zero $H^2[G_{i-1}; Z(K)]$ with which to do a non-trivial group extension, and the inverse sequence need not stabilize.

(Special thanks to JW for allowing me to state the problem to him and therefore come up with the solution to my own problem on the walk from his office to the elevator .)