Let $S$ be a symmetric set of generators of the (finite) group $G$. Having a bi-invariant metric $d$ on $S$ (meaning that whenever $s,t,g,gs,gt,sg,tg\in S$, then $d(s,t)=d(sg,tg)=d(gs,gt)$), is it always possible to extend $d$ to a bi-invariant metric on $G$?


**Update:** It has been shown below by Lucasz that the answer is negative. Let me try to add a condition on $S$ which hopefully makes the question more interesting. Suppose that $S$ verifies the following condition: whenever there are $(s\_1,t\_1),(s\_2,t\_2)\in S\times S$ with $s\_1\neq t\_1$ and $g\in G$ such that either $gs\_1=s\_2, gt\_1=t\_2$ or $s\_1g=s\_2,t\_1g=t\_2$, then $g=1$.



Extending a bi-invariant metric from a set of generators to the whole group.
No, it's not always possible. Take $G$ to be the cyclic group of order $20$, let $g$ be its generator. Let $S =$ { $g,g^3,g^{17},g^{19}$ }. Define $d$ on $S$ by putting $d(g,g^3)=2$ and all the other distances equal to $1$. 


Now, the pair $(g,g^3)$ is not in the same "$(S\times S)$-orbit" as $(g^{17},g^{19})$, and therefore the above defines a bi-invariant metric on $S$, but it is in the same $G\times G$-orbit, so it cannot be extended to a bi-invariant metric on $S$.


