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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 $g=1$.

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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 $(s_1,t_1),(s_2,t_2)\in S\times S$ 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 $g=1$.

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# Extending a bi-invariant metric from a set of generators to the whole group.

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 $G$?