Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 9 / 1 / g / Solution

True. Given a compatible left-invariant group metric $d$, definePullback by the bijection $x\mapsto x^{-1}$ preserves all metric axioms, and because inversion is a homeomorphism, $d_R$ induces the same topology. For right translation by $a$, left invariance givesThus $d_R$ is the required compatible right-invariant group metric. No metrizability theorem is needed.