Quotient group is complete, so is the group Let $G$ be a topological metrizable group and $K$ a normal subgroup of $G.$ Consider the homogeneous space $G/K$ and assume that both $K$ and $G/K$ are complete. I need to prove $G$ is complete.
More specifically, assume there is a right-invariant metric on $G$ such that the restriction of said metric to $K$ makes $K$ complete (hence closed in $G$) and the metric $\dot d(\dot x, \dot y) = d(xK, yK)$ on the homogeneous space $G/K$ makes it a complete topological space. How to prove $G$ is also complete?
I am not sure what I am not getting, I have seen some posts here and else where mentioning this result (but never proving it) and the few books I have read about topological groups always give this as an exercise. Now, my try so far goes as follows.
Consider a fundamental sequence $(x_n)$ in $G,$ since $K$ contains the neutral element, $\dot d(\dot x, \dot y) \leq d(x, y)$ and so the projection sends $x_n$ to $\dot x_n$ which is fundamental in the homoegenous space, making it to converge to some element $\dot x.$ Now, if $x$ belongs to $\dot x,$ one can show that $x_n x x_n^{-1}$ converges (possibly via a subsequence in $K$). This is where I am stuck, if $x_n x x_n^{-1} \to k,$ how to conclude $k = fxf^{-1}$ for a suitable $f$? Any help is appreciated.

I can finish your proof as follows. For each $n$ pick an element $x’_n\in \dot x$ such that $d(x_n, x’_n)< d(x_n, \dot x)+1/n$. Since the sequence $(\dot x_n)$ converges to $\dot x$ and is fundamental,  the sequence $(x’_n)$ is fundamental too. Since the space $\dot x\supset (x’_n)$ is complete in the induced metric, the sequence $(x’_n)$ converges to some point $x’\in\dot x$. Then the sequence $(x_n)$ converges to $x’ $ too.
