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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 $g=1$.

edited Jun 12 2011 at 10:31

asked Jun 11 2011 at 13:52

Valerio Capraro
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The new condition implies that S consists of just one element: if there are two different elements a and b in S then take pairs (a,a), (b,b) and g=ba^{-1}. Together with the assumption that S is symmetric it actually implies that G is the cyclic group of order 2. Maybe assuming $s_1 \neq t_1$ will make it more interesting. – Łukasz Grabowski Jun 11 2011 at 18:33

yes, of course. Thanks! – Valerio Capraro Jun 12 2011 at 10:30

However, with the new condition the original condition is empty, i.e. there are no elements $s,t,g$, $g\neq1$, such that $s,t, gs, gt \in S$. Therefore any metric on $S$ is bi-invariant - you sure you want it this way? Then the question could be more clearly phrased as: given a finite group $G$ and a set of generators $S$ fulfilling ..., is it true that any metric on $S$ can be extended to a bi-invariant metric on $G$? – Łukasz Grabowski Jun 12 2011 at 11:03

I'm doing a mess! Here is more or less my situation. Let $U(r)$ be the unitary group of rank $r$. I have a symmetric family $g_1,...g_n\in U(r)$ and I want to embed it (multiplicatively) into a finite group equipped with a bi-invariant metric which agrees with the metric induced by the 2-norm in $U(r)$. The idea would be easy: $g_1,...g_n$ generates a residually finite group and then they are multiplicatively embeddable into a finite group $G$. Inside $G$ they anyway carry the (pushforward of the) metric induced by the 2-norm and then I was trying to extend this metric to the whole $G$... – Valerio Capraro Jun 12 2011 at 15:34

I can't see a way to control a priori that $S$ is fulfilling for the $G$ constructed in this way.. – Valerio Capraro Jun 12 2011 at 15:40

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Łukasz Grabowski · Accepted Answer · 2011-06-11 14:38:02Z UTC

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 $S$.

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

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