Example Application of the Theorem

For example, let's see what we get from the exact sequence

$$0 \to \mathbf{Z}\xrightarrow{m} \mathbf{Z}\to \mathbf{Z}/m\mathbf{Z}\to 0,$$

where $m$ is a positive integer, and $\mathbf {Z}$ has the structure of trivial $G$ module. By definition we have $\H^0(G,\mathbf{Z}) = \mathbf{Z}$ and $\H^0(G,\mathbf{Z}/m\mathbf{Z})=\mathbf{Z}/m\mathbf{Z}$. The long exact sequence begins

$$0 \to \mathbf{Z}\xrightarrow{m} \mathbf{Z}\to \mathbf{Z}/m\mathbf... ...mathbf{Z})\to \H^2(G,\mathbf{Z}) \xrightarrow{m} \H^2(G,\mathbf{Z}) \to \cdots$$

From the first few terms of the sequence and the fact that $\mathbf {Z}$ surjects onto $\mathbf{Z}/m\mathbf{Z}$, we see that $[m]$ on $\H^1(G,\mathbf{Z})$ is injective. This is consistent with our observation above that $\H^1(G,\mathbf{Z})=0$. Using this vanishing and the right side of the exact sequence we obtain an isomorphism

$$\H^1(G,\mathbf{Z}/m\mathbf{Z}) \cong \H^2(G,\mathbf{Z})[m].$$

As we observed above, when a group acts trivially the $\H^1$ is $\Hom$, so

$$\H^2(G,\mathbf{Z})[m] \cong \Hom (G,\mathbf{Z}/m\mathbf{Z}).$$ (11.1)

One can prove that for any $n>0$ and any module $A$ that the group $\H^n(G,A)$ has exponent dividing $\char93 G$ (see Remark 11.3.4). Thus (11.2.1) allows us to understand $\H^2(G,\mathbf{Z})$, and this comprehension arose naturally from the properties that determine $\H^n$.