Consider an extension of groups $$1 \to N \to G \to Q \to 1$$ where $Q \simeq \mathbb{Z}$ is infinite cyclic. Such an extension necessarily splits as a semidirect product $G \simeq N \rtimes Q$. My question is about the cohomological Lyndon–Hochschild–Serre spectral sequence $$E_2^{p,q} = H^p(Q, H^q(N,\mathbb{K})) \implies H^{p+q}(G,\mathbb{K})$$ where $\mathbb{K}$ is some arbitrary field. I think one can deduce from this an equality of dimensions $$\dim_{\mathbb{K}} H^n(G,\mathbb{K}) = \dim_{\mathbb{K}} H^n(N,\mathbb{K})^Q + \dim_{\mathbb{K}} H^1(Q,H^{n-1}(N,\mathbb{K}))\,,$$ but I couldn't find a reference of this argument and I've heard this float around.
Since $Q$ has cohomological dimension $1$, the $E_2^{p,q}$ page has only two non-zero vertical columns:
| $E_2^{0,-}$ | $E_2^{1,-}$ | $E_2^{2,-}$ |
|---|---|---|
| $\vdots$ | $\vdots$ | $0$ |
| $H^3(N,\mathbb{K})^Q$ | $H^1(Q,H^3(N,\mathbb{K}))$ | $0$ |
| $H^2(N,\mathbb{K})^Q$ | $H^1(Q,H^2(N,\mathbb{K}))$ | $0$ |
| $H^1(N,\mathbb{K})^Q$ | $H^1(Q,H^1(N,\mathbb{K}))$ | $0$ |
| $\mathbb{K}$ | $H^1(Q,\mathbb{K})$ | $0$ |
| $0$ | $0$ | $0$ |
| $\vdots$ | $\vdots$ | $\vdots$ |
Since the differentials $d_2$ have degree $(2,-1)$, one immediately checks that they are all the zero map. Hence, the $E_3^{p,q}$ has the same entries as the $E_2^{p,q}$ page. Now the differentials have degree $(3,-2)$, and once again by inspection they are all the zero map and the fourth page is the same as the second page. Repeating this process, it shows that the sequence stabilizes at the second page, and hence $H^n(G,\mathbb{K})$ admits a filtration with quotients isomorphic to $H^{n-p}(Q,H^p(N,\mathbb{K}))$, from which the dimension claim follows.