Let $(G,d)$ be a compact metric space which as well is a group. Assume that $(x,y) \mapsto xy$ is continuous as a map $G \times G \to G$ and that group inversion $x \mapsto x^{-1}$ is continuous as a map $G \to G$. Define $\rho(x,y)=\inf_{g \in G} d(gx,gy)$.
First, show that $\rho(gx,gy)=\rho(x,y)$ $\forall x,y,g \in G$. Prove that the identity mapping $(G,\rho) \to (G,d)$ is continuous, and that the identity map $(G,d) \to (G,\rho)$ is Lipschitz (hence continuous). Then, deduce that $(G,d)$ and $(G,\rho)$ have the same topology.
To get things started, I first showed that $\rho$ properly defines a metric on $G$. However, I don't know how to get started to show that the metric is left invariant or that the identity mappings are continuous. What would be a way to get started with these?
If we know that the identity mappings are continuous, then it seems intuitive that both metric spaces have the same topology, since if $U$ is open in $(G,d)$, then $U$ will be open in $(G,\rho)$ (as it is the inverse image of the identity map, and we had that this was continuous). The other direction follows similarly, so both metric spaces give rise to the same topology.
Thanks for the help.
It suffices to show that given $g \in G$, the following equality holds: $$A \stackrel{\cdot}{=}\{ d(hgx,hgy) \mid h \in G \} = \{ d(hx,hy) \mid h \in G \} \stackrel{\cdot}{=} B.$$Taking $\inf_{h \in G}$ on both sides yields $\rho(gx,gy) = \rho(x,y)$.
Let $ z \in A$, so there is $h' \in G$ with $z = d(h'gx,h'gy)$. But $h \stackrel{\cdot}{=} h'g \in G$, and $z = d(hx,hy)$, so that $z \in B$. On the other hand, let $z \in B$, so that we have $h' \in G$ with $z = d(h'x,h'y)$. Putting $h \stackrel{\cdot}{=}h'g^{-1} \in G$, we have that $z = d(hgx,hgy)$, and finally $z \in A$. Done.
Now, notice that: $$\rho(x,y) = \inf_{g \in G} d(gx,gy) \leq d(1x,1y) = d(x,y),$$so that ${\rm id}\colon (G,d)\to (G, \rho)$ is Lipschitz-continuous. We only have to prove that ${\rm id}\colon (G, \rho) \to (G, d)$ is continuous now. Let $\epsilon > 0$, and choose $\delta = \epsilon > 0$. If $\rho(x,y) < \delta$, that is, $\inf_{g \in G}d(gx,gy) < \delta$, by compactness, there is $g \in G$ such that $d(gx,gy)<\delta$ but this actually reads $d(x,y) < \epsilon$.
You conclude that the topologies on $(G,d)$ and $(G,\rho)$ are the same because there is a homeomorphism between the spaces (which is essentialy what you've argued there).