Example Application of the Theorem

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

$\displaystyle 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

$\displaystyle 0 \to \mathbf{Z}\xrightarrow{m} \mathbf{Z}\to \mathbf{Z}/m\mathbf...
\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

$\displaystyle \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

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

William Stein 2012-09-24