Structure theorem for locally compact abelian groups

Remarks

Statement 2 follows from 1. A compact neighbourhood $V$ of the identity generates an open subgroup $U$, which is compactly generated, so $U \cong \R^n \times \Z^m \times K$. Inside it, $\R^n \times \set{0} \times K$ is open, because $\Z^m$ is discrete, and it is open in $G$ as well.

Examples: the $p$-adic numbers $\Q_p$ have the compact open subgroup $\Z_p$, so $n = 0$ in statement 3. The direct product $\R \times \Z \times \mathbb{T}$, with $\mathbb{T}$ the circle group, is compactly generated, as in 1.

The hypothesis in 1 is needed. The group $\Q$ with the discrete topology is not of the form $\R^n \times \Z^m \times K$. Since $\Q$ is discrete, such a decomposition would have $n = 0$. Its compact subgroups are finite, hence trivial, since $\Q$ is torsion free, so it would be $\Z^m$, but $\Q$ is not finitely generated.

Going deeper

For a student

The theorem reduces many questions about locally compact abelian groups to Euclidean spaces, discrete groups and compact groups, where Fourier analysis is classical. Statement 3 splits off the Euclidean part, so the remaining factor has a compact open subgroup, as the $p$-adic numbers and the discrete groups do. The proof and the theorem are closely tied to Pontryagin duality: the dual of $\R^n \times \Z^m \times K$ is $\R^n \times \mathbb{T}^m \times \widehat{K}$, with $\widehat{K}$ discrete.