I'm looking for a proof or counterexample to the following claim:
Claim: Let $(G,\cdot,^{-1},1)$ be a topological group with a (compatible) metric $d$ which is left-invariant, i.e. $$ d(x,y) = d(g\cdot x,g\cdot y) $$ Let $(x_n)$ be a Cauchy sequence. Then $(x_n^{-1})$ is a Cauchy sequence.
Without the assumption of left-invariantness, the group of positive reals $\mathbb{R}_+$ with ordinary multiplication and standard Euclidean metric works, as $(\frac1n)_{n=1}^\infty$ is a Cauchy sequence but $(n)_{n=1}^\infty$ is not. But multiplication doesn't exactly preserve length...
Background: I'm studying completions of metric groups. According to a book I'm reading, in order to complete a metric group , one need to change the metric from $d(x,y)$ to $D(x,y)=d(x,y)+d(x^{-1},y^{-1})$. This forces sequences $(x_n)$ to not be Cauchy, if $(x_n^{-1})$ isn't Cauchy, to so that completion has inverses and therefore is also a group. $d(x^{-1},y^{-1})$ induces the same topology, so $D(x,y)$ does too. But I cannot really come up with a counterexample which illustrates the need from passing from $d$ to $D$.