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A Hausdorff topological group admits a compatible left-invariant group metric exactly when it has a countable neighbourhood basis at the identity element. Choose symmetric neighbourhoods of the identity element $U_n$ with $U_{n+1}^3\subset U_n$ and trivial intersection. Assign a step in $U_n$ cost $2^{-n}$, and take the infimum of total costs of chains joining two points. The dyadic product estimate for group neighbourhoods makes this a genuine compatible metric. The Hausdorff assumption separates distinct points; without it the same construction can give only a pseudometric.

Assume first that the topological group is Hausdorff. This separation condition is necessary: an indiscrete group with two elements has the one-element neighbourhood basis consisting of the whole group, but its topology cannot come from a metric.

Translations are homeomorphisms, since multiplication by a fixed element is continuous and its inverse is multiplication by the inverse element. Inversion is likewise a homeomorphism. If $W$ is an neighbourhood of the identity element, continuity of the triple product at $(e,e,e)$ supplies neighbourhoods of the identity element $A,B,C$ with $ABC\subset W$. Their intersection with their inverses contains a symmetric open neighbourhood of the identity element $V$, and $V^3\subset W$. This proves the neighbourhood shrinking property needed below rather than assuming it.

Let $(B_n)$ be a countable neighbourhood basis at the identity element, choosing open representatives inside its members. Starting with $U_0=G$, recursively choose symmetric open neighbourhoods of the identity element such thatThey are still a basis. Their intersection is $\{e\}$: for $x\ne e$, the Hausdorff condition gives an neighbourhood of the identity element not containing $x$, and some $U_n$ lies inside it.

Here is the dyadic product estimate for group neighbourhoods. If $g_j\in U_{n_j}$ and ${\sum }_j{2}^{-n_j}\le 2^{-n}$, then the ordered product lies in $U_{n-1}$, for $n\ge 1$. Prove this by induction on the number of factors. A single factor has $n_j\ge n$ and hence belongs to $U_n\subset U_{n-1}$. For a longer word, choose the factor crossing the half-total-cost point. The words preceding and following it each have cost at most half the total, hence at most $2^{-n-1}$. The induction hypothesis puts both in $U_n$, and the middle factor also lies in $U_n$. Their threefold product is in $U_{n-1}$. Empty side words cause no problem because $e\in U_n$.

Give an edge from $x_{j-1}$ to $x_j$ cost $2^{-n_j}$ whenever $x_{j-1}^{-1}{x}_j\in U_{n_j}$, and defineChains exist because $U_0=G$. Reversing a chain proves symmetry, concatenating chains proves the triangle inequality, and translating every vertex on the left leaves all edge differences unchanged. Thus $d$ is left invariant. The product estimate also givesThe first implication follows by choosing a chain with total cost below the bound and applying the estimate with index $n+1$. If $d(x,y)=0$, the first implication puts $x^{-1}y$ in every $U_n$, so $x=y$. Hence $d$ is a genuine left-invariant group metric. These two inclusions show that its balls and the original neighbourhoods of the identity element refine one another; translations then show equality of the two topologies everywhere. More explicitly, every metric ball contains a sufficiently small translated $U_n$, and every open neighbourhood contains a sufficiently small metric ball. This proves the sufficient direction of the Birkhoff-Kakutani theorem.

Conversely, if a compatible metric exists, the balls $B_d(e,1/n)$ form a countable neighbourhood basis at the identity element. Invariance is not even needed for this direction. Thus, with the necessary Hausdorff convention, a compatible left-invariant metric exists exactly when the identity element has a countable neighbourhood basis.

For the requested counterexample, take the orientation-preserving affine group of the real line, represented by $(a,b)$ with $a>0$ and product $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$. Its topology is metrizable, for instance by the ordinary Euclidean distance in coordinates $(\log a,b)$, and the product and inverse are continuous. Let $D_a=(a,0)$ and $T_b=(1,b)$. Direct multiplication gives $D_a{T}_b{D}_a^{-1}=T_{ab}$. A compatible bi-invariant group metric would make every conjugation an isometry fixing $e$. ThereforeBut $T_{1/n}\to e$ in the group's topology, a contradiction. This is the affine conjugation obstruction to a bi-invariant group metric: the group is metrizable, and even admits a compatible left-invariant metric, but admits no compatible metric invariant on both sides.

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.

Right translations are isometries for a right-invariant group metric. Given a compatible left-invariant group metric $d$, the formula $d_R(x,y)=d(x^{-1},y^{-1})$ defines a compatible right-invariant group metric, since inversion is a homeomorphism and $(xa{)}^{-1}=a^{-1}{x}^{-1}$. Compatible invariant metrics need not be Lipschitz equivalent; on the additive group $\mathbb{R}$, both $|x-y|$ and $\sqrt{|x-y|}$ are compatible invariant distances.