Let $G$ be a topological group, let $K$ be a closed cocompact subgroup (i.e. the coset space $G/K$ is compact in the quotient topology) and let $g \in G$. Is there a sequence (edit: or net) of positive powers $g^{i\_n}$ of $g$ such that $g^{i\_n}K$ converges to $K$ in the coset space $G/K$?


If the answer is `no' in general, what if $G$ is totally disconnected and locally compact? (For the application, I'd be happy if I could at least get powers of $g$ to land in $UKV$ for any pair of identity neighbourhoods $U$ and $V$.)



Powers in compact coset spaces
$\DeclareMathOperator{\eps}{\varepsilon}$No. I'll write $P$ instead of $K$, as $K$ often denotes a compact subgroup. Write $X=G/P$.


For $x\_0\in X$ such that $g\mapsto gx\_0$ induces a homeomorphism $G/P\to X$, the question is whether $(g^nx\_0)\_{n\ge 1}$ always accumulates at $x\_0$. This fails in the most classical (typically non-normal) case: $G=\mathrm{PGL}\_2(\mathbf{K})$, where $\mathbf{K}$ is, say, $\mathbf{R}$ or $\mathbf{Q}\_p$, and $P$ is conjugate the upper triangular group. In this case we can identify $X$ to $\mathbb{P}^1(\mathbf{K})$ and choose to identify $x\_0$ to $1$ (so $P$ is a given conjugate of the upper triangular group). Then pick $g(x)=2x$. Then $g^n(x)$ converges to $+\infty$.




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Even the "$g^n\in UPV$" expectation fails. For this I haven't found a better argument than a brute computation. Here, in the real case, I take $P$ the lower triangular group, and $g=\begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix}$, so $g^n=\begin{pmatrix}2^n & 2^n-1\\ 0 & 1\end{pmatrix}$. Then if $u,v$ satisfy $\|u-I\|\_\infty,\|v-I\|\_\infty\le \eps=1/6$ then $ug^nv\notin P$. For this I just compute the upper-right entry of $ug^nv$: writing $u=\begin{pmatrix} u\_1 & e\_1\\ \* & \*\end{pmatrix}$ and $v=\begin{pmatrix} \* & e\_2\\ \* & v\_2\end{pmatrix}$ then $$(ug^nv)\_{12}=2^nu\_1e\_2+(2^n-1)u\_1v\_2+e\_1v\_2.$$
Hence the condition $ug^nv\in P$, that is, $(ug^nv)\_{12}=0$, yields $-(2^n-1)u\_1v\_2=2^nu\_1e\_2+e\_1v\_2$, hence $|(2^n-1)u\_1v\_2|\le |2^nu\_1e\_2+e\_1v\_2|$, which in turn yields $(1-\eps)^2|(2^n-1)|\le \eps(1+\eps)(2^n+1)$, that is, $\frac{1-2^n}{1+2^n}\le \frac{\eps(1+\eps)}{(1-\eps)^2}$. But for $n\ge 1$ $\frac{1-2^n}{1+2^n}\ge 1/3$ while for $\eps\le 1/6$, $\frac{\eps(1+\eps)}{(1-\eps)^2}\le 7/25<1/3$. 
The $p$-adic case should be checked similarly with $2$ replaced with $1/p$ (and always a bit easier by the ultrametric inequality).


