Powers in compact coset spaces 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 $V$.)
 A: It seems the following.
In general the answer is no, because compactness does not imply sequential compactness. Let $\Bbb T=\{z\in\Bbb C:|z|=1\}$ be the unit circle endowed with the standard topology. Put $G={\Bbb T}^{\Bbb T}$. By Tychonov Theorem, $G$ is a compact space. Let $K=\{e\}$ be the trivial subgroup of $G$. Select an element $g=(g_z)_{z\in\Bbb T}\in G$ such that $g_z=z$ for each $z\in\Bbb T$. Suppose that there exists an increasing sequence $\{i_n\}$ of positive integers such that the sequence $\{g^{i_n}\}$ converges to the identity of the group $G$. Let $U_0=\{z\in\Bbb T: \operatorname{Re} z\ge 0\}$ be a neighborhood of the identity of the group $\Bbb T$.  For each natural number $n$ put $T_n=\{z\in\Bbb T: i_mz\in U_0\mbox{ for each }m>n\}$.
The continuity of power on the group $\Bbb T$ implies that the set $T_n$ is closed for each natural number $n$. The assumption implies that $\Bbb T=\bigcup_{n\in\Bbb N} T_n$. By Baire Theorem, there exists a number $m$ such that a set $T_m$ has non-empty interior. Therefore there exists an open arc $U\subset T_m$ of the circle $\Bbb T$. Since the sequence $\{i_n\}$ is increasing, there exists a number $n>m$ such that $i_n>1/\mu(U)$, where $\mu$ is the standard measure on $\Bbb T$ such that $\mu(\Bbb T)=1$. But then $U_0\supset i_nT_m\supset i_n\overline U=\Bbb T$, a contradiction.
I hope you will be almost happy with this addendum. Let $G$ be a Hausdorff totally disconnected and locally compact topological group, $K$ be a cocompact normal subgroup of the group $G$ and $g\in G$. For each $n$ put $i_n=n!$. I claim that a sequence $\{g^{i_n}K\}$ converges to $K$ in the coset space $G/K$. Indeed, by [Pon, Theorem 16], the group $G$ has a base $\mathcal B$ at the identity consisting of its open compact subgroups. Let $H\in\mathcal B$ be an arbitrary group. Since the group $K$ is normal then $HK$ is a group. Since $\{hHK: h\in G\}$ is an open cover of the compact space $G/K$, there exists a finite subset $F$ of the group $G$ such that $G=\bigcup\{hHK: h\in G\}$. Then the pigeonhole principle implies that there exist natural numbers $k<l$ and an element $h\in F$ such that $g^k,g^l\in hHK$. Then $g^{l-k}\in K^{-1}H^{-1}h^{-1}hHK=KHK=HK$. Since the set $HK$ is a group, $g^{i_n}\in HK$ for each $n\ge l-k$.
[Pon] Lev S. Pontrjagin, Continuous groups, 2nd ed., M., (1954) (in Russian).
A: $\DeclareMathOperator{\eps}{\varepsilon}$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 $+\infty$.

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).
