In the additive group $\Bbb Z$ we can fairly unambiguously say $1$ is the smallest difference between two elements.

I guess a more rigorous statement might be to give $\Bbb Z$ its topology as a subspace of $\Bbb R$ and then to say that the pair $\{1,-1\}$ are the set that share the smallest absolute value: $\{1,-1\}=\{x\in \Bbb Z:\lvert x\rvert=\min\{\lvert x\rvert:x\in \Bbb Z\}\}$.

Is there a general form for the smallest interval in a group? Does it equal the smallest nonzero element in general?

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    $\begingroup$ No, because there is no canonical definition of "small" in an arbitrary group (when is $g < h$?) $\endgroup$ – AnalysisStudent0414 Dec 12 '19 at 13:39
  • $\begingroup$ In the group of the rationals (or the group of the reals) there is not even a smallest difference. $\endgroup$ – Peter Dec 12 '19 at 13:45
  • $\begingroup$ @AnalysisStudent0414 the answer need not be canonical, only general. $\endgroup$ – samerivertwice Dec 14 '19 at 5:52

In general, groups that can be endowed with an order that is compatible with the binary operation - that is, left- and/or right-invariant, in the sense that $$\forall a,b,c,\qquad a<b\implies ca<cb\quad(\text{respectively, }ac<bc)$$ are very sparse in the landscape of groups. Apart from $\Bbb Z$, all subgroups of $\Bbb R$ can be - and you can already see that in $\Bbb Q$ and $\Bbb R$ there is no smallest difference (or smallest element).

The braid group is one quite non-trivial example where this can be done, although the Dehornoy order is only left-invariant (no order on the braid group $B_n,n\geq 3$ can be both left- and right-invariant).

In the rare cases where there is such an order, then you can define positive to mean larger than the identity element, and it is immediate that the "smallest difference", if any, is going to be equivalent to the smallest element of the group in the sense that any difference between two elements can be translated into the difference between some element and the identity.

In the group $\Bbb S^1$ for instance, you can see that there is no smallest difference, and that it is impossible to define a compatible order. In the group $\Bbb Z^2$, you could define the smallest difference by using the distance $\sqrt{x^2+y^2}$; this would be compatible with the binary operation (translations don't distort distances), but not with any order.

  • $\begingroup$ Thanks Arnaud. That bijection from $B_n\to\Bbb Z$ looks very interesting to me. $\endgroup$ – samerivertwice Dec 12 '19 at 16:12
  • $\begingroup$ ...and I guess the Cantor n-tuple function combined with any bijection from $\Bbb N\to\Bbb Z$ can order $\Bbb Z^n$, right? $\endgroup$ – samerivertwice Dec 12 '19 at 21:44
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    $\begingroup$ @samerivertwice Wait, a bijection between a group and $\Bbb Z$ will order the underlying set of the group, but it most likely won't be an order compatible with the group operation. $\endgroup$ – Arnaud Mortier Dec 12 '19 at 21:50
  • $\begingroup$ Ah yes, I can see how that falls apart. $\endgroup$ – samerivertwice Dec 12 '19 at 21:59

For a general group, no, there's no notion of a "smallest difference" between two elements. In order to speak of such a thing, we need some additional structure, namely a way to compare sizes of group elements. This is the notion of a length function on the group, also known as a normed group. If we have a length function on our group, then we can define distances between group elements, making the group into a metric space.

However, even if we have a normed group, there isn't necessarily a smallest distance between two distinct elements; as was pointed out in a comment, in the group of rationals or reals (where the usual absolute value is a length function), there are arbitrarily close pairs of distinct elements.


This answer mainly comes from Dale Rolfsen’s Ordered Groups and Topology.

Amazingly, sometimes groups do have smallest elements!

Before we ask questions about the smallest element, we should define what we mean.

A left-ordering on a group $G$ is a total order $<$ on the group so that if $g<h$ then $xg < xh$ for every $x\in G$.

Not all groups are left-orderable! Notice that the invariance under multiplication shows that any left-orderable group has no finite order elements. (Prove this yourself!). Given the identity element $e\in G$, let’s define the smallest element of $G$ to be the smallest element of the set

$$\mathcal{P}^+ = \{g\in G ~|~ e<g\}.$$

This is called the positive cone of $G$. Some examples of left-orderable groups are:

  • Free groups
  • Surface groups
  • Mapping class groups of surfaces with boundary
  • In particular, braid groups

and all their subgroups. Now, a sufficient condition to have a smallest element is for the order $<$ to be a well order.

An order $<$ on $G$ is a well order if any subset $S$ of $G$ has an element $s\in S$ such that $s<x$ for all other $x\in S$.

This property, applied to the positive cone $\mathcal{P}^+$, allows us to prove that $G$ would have a smallest element.

Fact: The positive pure braid group is well ordered under the Magnus ordering. Thus, it has a smallest element. This is also true for the positive braid group under the Dehornoy ordering.

However, the full (pure) braid group itself under this order is order dense; that is, for any $g<h$ there is some $x$ so that $g<x<h$. Thus, it can never have a smallest element.

It isn’t clear to me if there are algebraic obstructions to having a well ordering, what the smallest element is in the two above examples, or whether you can have a smallest element without having a well order.

  • $\begingroup$ Finite order elements would make the order antisymmetric i guess - a contradiction. $\endgroup$ – samerivertwice Dec 18 '19 at 5:16
  • $\begingroup$ @samerivertwice Can you say a bit more about the “antisymmetry”? $\endgroup$ – Santana Afton Dec 18 '19 at 5:17
  • $\begingroup$ Sorry, it's a standard property of an order which says if a is less than b then b can't be less than a. Else I think it's relegated to being a preorder, which can have loops. $\endgroup$ – samerivertwice Dec 18 '19 at 8:19

If $G$ is a finitely generated group then there is a natural metric on $G$, called the word metric: If $G=\langle S\rangle$ then $d_S(g, h):=\min\{|w|, w\in F(S)\mid w=_Gg^{-1}h\}$. That is, the distance from $g$ to $h$ is the minimum length of a word in the generators which represents the group element $g^{-1}h$.

If we take the generating set $\{1\}$ of $\mathbb{Z}$ then $d_{\{1\}}(m, n)=|m-n|$, and so the word metric corresponds naturally to subtraction. So then to answer your question in the setting of word metrics: the smallest distance between two non-equal elements is $1$, so $d_S(g, h)\geq1$ unless $g=h$.

Word metrics are an important tool in geometric group theory. For example, hyperbolic groups and automatic groups are both defined using word metrics (and if these definitions hold for some finite generating set then they hold for any finite generating set).

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    $\begingroup$ It seems as though using the word metric allows you to “choose” your smallest elements, and that the smallest distance will always be $1$. $\endgroup$ – Santana Afton Dec 12 '19 at 14:52
  • $\begingroup$ @SantanaAfton Yes, the smallest distance is always $1$ (see my second paragraph). $\endgroup$ – user1729 Dec 12 '19 at 15:43
  • $\begingroup$ Also, yes you can "choose" your smallest elements. However, strangely, when we study the asymptotics of the number of elements of length $n$, so consider the function $f_S(n):=|\{g\in G\mid d_S(g, 1)=n\}|$ for large $n$, then changing the shortest elements (that is, changing the generating set) does not alter the asymptotics of this function (the key phrase is "growth of groups"). That is, the definition is robust enough to study the number of "large" elements in your group, even when you mess around with the small ones. $\endgroup$ – user1729 Dec 12 '19 at 15:46
  • $\begingroup$ Thanks @user1729 that's really helpful. $\endgroup$ – samerivertwice Dec 12 '19 at 16:12

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