Let $S_3$ be the third symmetric group and let it act trivially on $\mathbb{Z}$. Consider the following theorem on page $154$ of Brown Group Cohomology
(9.1) Theorem The following conditions are equivalent:
(i) $G$ has periodic cohomology,
(ii) There exist $n,d$ with $d\neq 0$ such that $\hat{H}^n(G,M)\simeq \hat{H}^{n+d}(G,M)$ for all $G$-modules $M$,
(iii) For some $d\neq 0$, $\hat{H}^d(G,\mathbb{Z})\simeq|G|^{-1}\mathbb{Z}/\mathbb{Z},$
(iv) For some $d\neq 0$, $\hat{H}^d(G,\mathbb{Z})$ contains an element of order $|G|$.
I was interested in investigating whether or not $S_3$ has periodic cohomology. To this end, I calculated $H^2(S_3,\mathbb{Z})$. Note that by an explicit description of $H^1(G,M)$, we have $H^1(A_3,\mathbb{Z})=0$. Since $H^1(A_3,\mathbb{Z})$ vanishes, we have the following inflation-restriction exact sequence (see proposition 1.6.7 of Cohomology of Number Fields.)
$$0\to H^2(S_3/A_3,\mathbb{Z})\to H^2(S_3,\mathbb{Z})\to H^2(A_3,\mathbb{Z})\to H^3(S_3/A_3,\mathbb{Z})$$
By periodicity of cyclic cohomology, we have $H^2(S_3/A_3,\mathbb{Z}=\mathbb{Z}/2\mathbb{Z}$, $H^2(A_3,\mathbb{Z})=\mathbb{Z}/3\mathbb{Z}$, and $H^3(A_3,\mathbb{Z})=0$, so we have
$$0\to \mathbb{Z}/2\mathbb{Z}\to H^2(S_3,\mathbb{Z})\to \mathbb{Z}/3\mathbb{Z}\to 0$$
which implies by classification of finitely generated abelian groups yields $H^2(S_3,\mathbb{Z})=\mathbb{Z}/6\mathbb{Z}$. This means that $S_3$ satisfies (iii) in theorem 9.1 of Brown, and so $S_3$ should have periodic cohomology.
However, $\hat{H}^{-2}(S_3,\mathbb{Z})=Ab(S_3)=\mathbb{Z}/2\mathbb{Z}$, where $Ab(G)$ is the abelianization of $G$, but as we've shown $\hat{H}^2(S_3,\mathbb{Z})\simeq \mathbb{Z}/6\mathbb{Z}\not\simeq \mathbb{Z}/2\mathbb{Z}\simeq \hat{H}^{-2}(S_3,\mathbb{Z})$, so the cohomology cannot be $2$-periodic. What's going wrong here?
It's not true that the action of $S_3$ on $H^2(A_3)$ is trivial: the action is induced by conjugation. If we conjugate with a $2$-cycle, we get the inversion map on $A_3$. From this, via an explicit description of the conjugation map on cocycles, it's not difficult to figure out that the action is nontrivial. Hence the invariants are zero and we get that $H^2(S_3) \cong \Bbb Z/2\Bbb Z$.