You should look at my paper with Mikhail Ershov and Thomas Weigel about commensurators of profinite groups http://arxiv.org/PS_cache/arxiv/pdf/0810/0810.2060v1.pdf.
Edit: Let $G$ be a profinite group. There are two natural ways to associate a topology with $\rm{Comm}(G)$. With the Aut-topology you take the biggest topology such that the homomorphisms of the automorphism group an open subgroups of $G$ to $\rm{Comm}(G)$ are continues for all open subgroups. You do not always get a topological group and if you do, its properties are not always nice. But many times it does coincide with a natural topolgoy that exists. The other topology is the strong-topolgy in which you force the image of $G$ in $\rm{Comm}(G)$ to have the quotient topolgy and to be an open subgroup in $\rm{Comm}(G)$. Now, you always do get a locally compact topological groups, but it is not always $\sigma$-compact. (Look at sections 7 and 8 of the paper.)
Let me give two examples: 1. Take $G=\mathbb{Z}_p$, then $\rm{Comm}(G)=\mathbb{Q}^{*}_p$. The Aut-topology yields the natural topology while the strong topolgy yields the discrete topology. 2. Take $G=\rm{PSL}_n(\mathbb{F}_p[[t]])$, then $\rm{Comm}(G)=\rm{PSL}_n(\mathbb{F}_p((t))) \rtimes \rm{Aut}(\mathbb{F}_p((t)))$. In the Aut-topology $\rm{PSL}_n(\mathbb{F}_p[[t]]) \rtimes \rm{Aut}(\mathbb{F}_p((t)))$ is an open subgroup. However, in the strong-topology $\rm{PSL}_n(\mathbb{F}_p[[t]])$ is an open subgroup and as Colin mentioned below $\rm{Comm}(G)$ is not $\sigma$-compact.