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.