Is $\mathcal O^{\ast}/U^{(n)}\cong (\mathcal O/\mathfrak p)^{\ast}$ as topological groups? I have the following: $K$ is a field with discrete valuation $v$, $\mathcal O$ its valuation ring and $\mathfrak p$ the maximal ideal and $U^{(n)}=1+\mathfrak p^n$ the $n$-th unit group for $n\geq 1$. I know there is a group isomorphism $\mathcal O^{\ast}/U^{(n)}\cong (\mathcal O/\mathfrak p^n)^{\ast}$. 
My question is whether this is also a homeomorphism of topological groups.
Thank you! 

Yes, because your topological groups are discrete. The higher unit groups are open in $\mathscr{O}^\times$, and $\mathfrak{p}^n$ is open in $\mathscr{O}$. This makes both $\mathscr{O}^\times/U^{(n)}$ and $\mathscr{O}/\mathfrak{p}^n$ discrete (and discreteness of $(\mathscr{O}/\mathfrak{p}^n)^\times$ follows).
In general, if $G$ is a topological group and $H$ is an open subgroup, then the coset space $G/H$ is discrete with its natural quotient topology from $G$. 
