I'm wondering if the next exercise in 'An introduction to homological algebra' by Weibel is correct:
Let $G$ be the profinite group $\widehat{\mathbb{Z}}_p$. Show that $H^i(G;\mathbb{Z}) = \begin{cases} \widehat{\mathbb{Z}}_p & \text{ if $i$ even}\newline 0 & \text{ if $i$ odd} \end{cases}$
Is it possible that $\widehat{\mathbb{Z}}_p$ should be replaced by $\mathbb{Z}(p^\infty)$ ? I've found on the website of Weibel that it should be for $H^0(G;\mathbb{Z})=\mathbb{Z}$, $H^2(G;\mathbb{Z})=\mathbb{Z}(p^\infty)$ and for all other $i$ $H^i(G;\mathbb{Z})=0$, but shouldn't that be for odd $i$ and the rest still $\mathbb{Z}(p^\infty)$ ? Since the cohomology groups $H^i(\mathbb{Z}/m\mathbb{Z};\mathbb{Z})$ are $\mathbb{Z}/m\mathbb{Z}$ for $i$ even ?