 Solution

ID: 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

d_{R} (x, y) = d (x^{- 1}, y^{- 1}) .

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 gives

d_{R} (x a, y a) = d (a^{- 1} x^{- 1}, a^{- 1} y^{- 1}) = d (x^{- 1}, y^{- 1}) = d_{R} (x, y) .

Thus

d_{R}

is the required compatible right-invariant group metric. No metrizability theorem is needed.

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