Equivalence of finiteness of $spliG$ and periodicity isomorphisms being induced by cup product

Question

I am trying to prove that the following are equivalent for a group $G$ with periodic cohomology with period $q$ after $k$ steps:

$(i)\ spliG<\infty$ (where $spliG$ is the supremum of injective length of $\mathbb{Z}G$ - projective modules)

$(ii)$ There is an element $g\in H^q(G,\mathbb{Z})$ such that the cup (Yoneda) product $\ \_\bigcup g:H^i(G,\_)\to H^{i+q}(G,\_)$ is an isomorphism for every $i>k$

It was easy to prove that $(i)\Longrightarrow (ii)$, since $(i)$ is equivalent to the following:

There is a $q$-extension of the form $0\to\mathbb{Z}\to X\to P_{q-2}\to\ldots\to P_0\to\mathbb{Z}\to0$ where $P_i$ is a projective $\mathbb{Z}G$-module for every $i=0,1,\ldots,q-2$ and $pd_{\mathbb{Z}G}X<\infty$.

and $(ii)$ is equivalent to the following:

There is a $\mathbb{Z}$-split, $\mathbb{Z}G$-exact sequence $0\to\mathbb{Z}\to X$, where $X$ is $\mathbb{Z}$-free and $pd_{\mathbb{Z}G}X<\infty$.

However, I can't figure out how to prove $(ii)\Longrightarrow(i)$.

Any help or suggestions would be greatly appreciated.

Thank you.

Edit: There is a mistake in the original question. $(i)$ and $(ii)$ have been interchanged. Actually, the fact that the periodicity isomorphisms are induced by cup product with an element $g\in H^q(G,\mathbb{Z}$ is equivalent to the existence of a $q$-extension $0\to\mathbb{Z}\to X\to P_{q-2}\to\ldots\to P_0\to\mathbb{Z}\to0$ where $P_i$ is a projective $\mathbb{Z}G$-module for every $i=0,1,\ldots,q-2$ and $pd_{\mathbb{Z}G}X<\infty$

while

$spliG<\infty$ is equivalent to the existence of a $\mathbb{Z}$-split, $\mathbb{Z}G$-exact sequence $0\to\mathbb{Z}\to X$, where $X$ is $\mathbb{Z}$-free and $pd_{\mathbb{Z}G}X<\infty$.

So, the correct question is how to prove that:

If $G$ has periodic cohomology with period $q$ after $k$ steps and $spliG<\infty$ then there exists a $q$-extension of the form $0\to\mathbb{Z}\to X\to P_{q-2}\to\ldots\to P_0\to\mathbb{Z}\to0$ where $P_i$ is a projective $\mathbb{Z}G$-module for every $i=0,1,\ldots,q-2$ and $pd_{\mathbb{Z}G}X<\infty$.

I apologize for the mix-up.

Answer 1

Let $P_\bullet \to \mathbb{Z}$ be a projective resolution as a $\mathbb{Z}[G]$-module, $\Delta : P_\bullet \to P_\bullet \otimes P_\bullet$ an approximation of the diagonal, and $\phi : P_\bullet \to \mathbb{Z}[q]$ be a chain map representing $g$. Then the chain map $$f : P_\bullet \overset{\Delta}\to P_\bullet \otimes P_\bullet \overset{Id \otimes \phi}\to P_\bullet \otimes \mathbb{Z}[q] = P_\bullet[q]$$ represents capping with $g$. Let $C_\bullet$ be the desuspension of the mapping cone of $f$, fitting into a SES $$0 \to P_\bullet[q-1] \to C_\bullet \to P_\bullet \to 0.$$ Then $C_\bullet$ is a chain complex of projective $\mathbb{Z}[G]$-modules, and the homology of $C_\bullet$ is $\mathbb{Z}$ in degrees 0 and $q-1$.

Let $X := C_{q-1}/d_q(C_{q})$, so there is a chain complex $$X \to C_{q-2} \to C_{q-3} \to \cdots \to C_0$$ and a resolution $$\cdots \to C_q \to C_{q-1} \to X \to 0.$$

The first chain complex is exact except at the top and bottom, leaving homology $\mathbb{Z}$ in degrees 0 and $q-1$. This establishes the first property and it remains to show that $X$ has finite projective dimension.

We can use the resolution to compute $Ext^n_{\mathbb{Z}[G]}(X, Y) = H_{n+q-1}(Hom_{\mathbb{Z}[G]}(C_\bullet, Y))$ for $n > 0$. But the latter is the relative term in a LES whose other terms are $$- \smile g : H^i(G ; Y) \to H^{i+q}(G ; Y).$$ By assumption this map is an isomorphism for $i > k$ and all $Y$, so $Ext^n_{\mathbb{Z}[G]}(X, Y)=0$ for $n$ large enough and all $Y$. This means that $X$ has finite projective dimension.