Finite groups with periodic cohomology
I'm trying to understand Chapter 12, Section 11 in Cartan + Eilenberg's Homological Algebra, which concerns finite groups with periodic cohomology. Unfortunately I am jumping right to this section in the book (I've been working in Serre's Local Fields, and I'm doing the exercise at the end of Chapter 8, Section 4, which refers to the above section in Cartan + Eilenberg), so I'm a little disconcerted with the notation change (and things like using $(\Pi:1)$ for the order of the group $\Pi$ - what is up with that?)
I suppose I have two main questions.
How do we know that, given a finite group $G$ with periodic cohomology, say with $${\widehat{H}}{}^n(G,A)\cong\!\!\!{\widehat{H}}{}^{n+q}(G,A)$$ for all $n\in\mathbb{Z}$ for some $q\in\mathbb{N}$, these isomorphisms must be given by cup-producting with a fixed element $g\in\widehat{H}{}^q(G,\mathbb{Z})$? This seems to be an implicit assumption in their investigation, and while I can very well believe that it's true (the cup-product satisfies some universal property, if I understand correctly), I don't see what's barring the isomorphisms from being "accidental".
The fact that the period $q$ is necessarily even (unless $G$ is trivial in which case $q=1$) seems very mysterious to me. Of course, this is the key property for the exercise I'm doing (defining a generalization of the Herbrand quotient), so I would like to have a firm grasp of why it's true. I can more or less follow the reasoning in Cartan + Eilenberg for why this is true, but it's just a proof by contradiction by making a computation using the cup product, and using the fact that $\mathbb{Z}/2\mathbb{Z}$ has periodic cohomology with even period. Furthermore, I again am not seeing why cup-producting with an element of $\widehat{H}{}^q(G,\mathbb{Z})$ is necessarily involved. So, are there any more intuitive explanations of why group cohomology, if it is periodic, has even period? Are there any references other than Cartan + Eilenberg I can look at for this fact?
As stated, neither 1 nor 2 is true. For example, (for $p>2$) $H^{\bullet}(\mathbb F_p;\mathbb F_p)=\mathbb F_p[\varepsilon,t]/\varepsilon^2$ ($\deg\varepsilon=1$, $\deg t=2$) — so (corresponding Tate) cohomology are 1-periodic, but this periodicity is not given by cup-product (of course, these groups are also 2-periodic, and 2-periodicity is given by cup-product).
P.S. One actually needn't compute ring structure in $H^{\bullet}(\mathbb F_p;\mathbb F_p)$ to show that the periodicity is not given by cup-product: indeed, it can't be given by an element of $H(G;\mathbb Z)$ because $H^1(\mathbb F_p;\mathbb Z)=0$; it can't be even given by cup-product with an element of $H^1(\mathbb F_p;\mathbb F_p)$ — because (by supercommutativity) square of such element must have order 2 and there are no such elements in $H^2$ (which is not very surprising for a vector space over a field of $char\ne2$).