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Let $G$ be a finite group. Define $d:G\times G\longrightarrow\mathbb{N}$ by $d(x,y)=o(xy^{-1})-1, \forall\, x,y\in G$. Then $d$ is a metric on $G$ if and only if $$(*)\hspace{5mm}o(ab)<o(a)+o(b), \forall\, a,b\in G.$$It is easy to see that an abelian group satisfies $(*)$ if and only if it is a $p$-group. Moreover, if an arbitrary group $G$ satisfies $(*)$, then all its elements must be of prime power order, i.e. $G$ must be a $CP$-group (a classification of these groups can be found here: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.45.3738&rep=rep1&type=pdf). I have some difficulties to check whether a $CP$-group satisfies $(*)$. Is there another way to proceed? Also, is it interesting to study such a metric on finite groups?

Additional question: Is it true that all finite groups satisfying $(*)$ are solvable? By the answer below it follows that this class contains no nonabelian simple CP-group (in fact, no nonabelian simple group). Is it possible that a group in the other class (ii) of nonsolvable CP-groups to satisfy $(*)$?

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    $\begingroup$ What is $o(g)$? $\endgroup$
    – Alex Degtyarev
    Commented Mar 22, 2014 at 8:11
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    $\begingroup$ $o(g)$ is the order of $g\in G$. $\endgroup$
    – Marius Tarnauceanu
    Commented Mar 22, 2014 at 11:37

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The group ${\rm PSL}(2,7)$ seems to be a $CP$ group according to your definition, but it contains an element $a$ of order $2$ and an element $b$ of order $3$ whose product $ab$ has order $7,$ so you do not have a metric in this group. In fact, as I write, I realise that in a group $G$ where $d$ defines a metric, the product of any two elements of order $2$ can have order at most $3.$ This seems to exclude $G = {\rm PSL}(2,q)$ as a possibility whenever $q >3.$ Also, the Suzuki groups (the only non-Abelian simple groups of order prime to $3$) are excluded, as they always contain a dihedral subgroup of order $10.$

Later edit: note that the structure of a Sylow $2$-subgroup $S$ of a group $G$ such that $d$ defines a metric is extremely restricted. The elements of order at most $4$ in $S$ form a subgroup, and all involutions of $S$ commute with each other. Thus $S$ has no dihedral subgroup of order $8$ and no generalized quaternion subgroup of order $16$ or more. Even later edit: note also that by an earlier remark, if $t$ and $u$ are involutions of $G,$ then either $t$ and $u$ are conjugate (via an element of order $3$), or else $t$ commutes with every $G$-conjugate of $u.$

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  • $\begingroup$ Thank you very much. I continue to check whether ${\rm PSL}(3,4)$, $M_{10}$, or the groups in (ii)-(v) satisfy $(*)$. Remark: $(*)$ is weaker than the condition studied in one of my previous papers (see math.uaic.ro/~martar/pdf/articole/articol%2049.pdf). $\endgroup$
    – Marius Tarnauceanu
    Commented Mar 22, 2014 at 16:49
  • $\begingroup$ Well, $M_{10}$ contains $A_{6}$ which contains $A_{5},$ which does not satisfy $*$. Also ${\rm PSL}(3,4)$ contains ${\rm PSL}(2,4) \cong A_{5},$ which does not satisfy $*$. $\endgroup$
    – Geoff Robinson
    Commented Mar 22, 2014 at 17:05
  • $\begingroup$ Thanks again. In this way, the class of groups satisfying $(*)$ is getting smaller! $\endgroup$
    – Marius Tarnauceanu
    Commented Mar 22, 2014 at 18:00

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