Group Cohomology: Question $2$
I am learning group cohomology. In Wikipedia, I need to understand the section $\text{Higher cohomology groups are torsion}$.
The discussion says the cohomology groups $H^n(G,M)$ for finite groups $G$ are all torsion group. This is clear to me.
But it says that the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$ yields an isomorphism $$\text{Hom}(G, \mathbb{Q}/\mathbb{Z})=H^1(G, \mathbb{Q}/\mathbb{Z}) \cong H^2(G, \mathbb{Z}) $$ where $G$ acts trivially on each groups in the short exact sequence.
I understand from a previous discussion that if $G$ acts trivially on $M$, then $\text{Hom}(G, M)=H^1(G, M) $.
But I didn't understand why $H^1(G, \mathbb{Q}/\mathbb{Z}) \cong H^2(G, \mathbb{Z})$ ?
An explanation is given in the following para in the above section:
$\text{If the order of G is invertible in a $G$-module $M$ (for example, if $M$ is a $ \mathbb {Q}$ -vector space),}$ $\text{the transfer map can be used to show that $H^{n}(G,M)=0$ for $n ⩾ 1$}$. $\text{A typical application of this fact is as follows: the long exact cohomology sequence}$ $\text{ of the short exact sequence (where all three groups have a trivial $G$-action) }$
So according to the above suggestion we have the long exact cohomology sequence $$0 \to \mathbb{Z}^G \to \mathbb{Q}^G \to (\mathbb{Q}/\mathbb{Z})^G \xrightarrow{\delta_0} H^1(G,\mathbb{Z}) \to H^1(G,\mathbb{Q}) \to H^1(G, \mathbb{Q}/\mathbb{Z}) \xrightarrow{\delta_1} H^2(G, \mathbb{Z}) \to \cdots $$
We have to show $\delta_1$ is an isomorphism.
Edit: The order of $G$ is invertible in $\mathbb{Q}$, and so we have that $H^n(G,\mathbb{Q})=0$ for $n≥1$, why is so ?
Any help please
This is a direct consequence of Maschke's theorem (In fact, it is equivalent to it) : if the order of the characteristic of the field $\mathbb k$ does not divide the order of the group $G$, then every $\mathbb k G$-module is projective (that is, every short exact sequence of modules splits). This means that projective resolutions have length $0$, and that there is no cohomology (in fact, $H^n(G, V)$ vanishes for every representation $V$ and every $n \geq 1$).