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Does Cayley's Theorem imply all groups are countable?

  1. Jan 4, 2014 #1
    My question is exactly what is stated in the title: Does Cayley's theorem imply that all groups are countable?

    I don't see how a well defined transitive action of an uncountable group on itself. How could you possibly find a set of permutations sending a single element to every other element in it, much less each element of a group to every other element. Such a set of permutations would have to be uncountably uncountable, if that makes any sense. I don't see how such a set could have any well defined "order" (I don't know if this is the correct term), not in the sense of magnitude but in the sense of structure imposed.

    In other words, what I mean by "order" is that in mathematics, we deal spaces, or sets on which we have added some structure by way of axioms which I see as increasing the "order". We add "order" to arbitrary sets to distinguish them up to isomorphism and also to in some sense approximate what may be true in sets with no order. However, in these "unordered" sets, no structure is imposed so that the tools of conventional mathematics are not necessarily well defined becuse it is not clear as to whether or not we can distinguish between elements, whether or not individual elements even exist since such a set would transcend completeness, or whether or not either of these actually even makes any sense. So, we can't even begin to partition the set and define equivalence classes, moreover, defining relations such as inclusion may be totally ambiguous and may not even be useful or make any sense whatsoever.

    In this sense of "order", I don't see how a transitive action of an uncountable group on itself can be well defined. Take for instance the real line acting on itself by left addition. We can define very simple permutations of all the elements of this group:

    Fix any [itex]\epsilon > 0[/itex], and define a group by

    [itex]G = \left\{\pm g|g\geq\epsilon\right\}[/itex]

    and define its action on the real line [itex]\forall g\in G[/itex] by

    [itex]g\cdot x=g + x [/itex]

    This should define every permutation of the real line under the action right?

    However, the epsilon method of analysis requires that you have fixed such an epsilon, so that in effect, what we are considering here, since epsilon is arbitrary, is the union of an arbitrary collection of actions, where each is defined to be one of an arbitrary collection uncountable groups, uniquely determined by its arbitrary but fixed minimal positive non-zero element, acting on the real line. The collection of all of such elements and zero is most definitely not the real line since the real line has no minimal element, while fixing elements element requires that a minimal group element exists. In effect, what I'm saying is that in fixing epsilon we set a precision and any number smaller in magnitude than this precision is zero, this is the the epsilon principle. Thus, by completeness,

    [itex](\forall\epsilon > 0) (\exists c \in R \ni: 0< c < \epsilon)[/itex]

    so that there is some c is in the kernel of the action of the real line on itself no matter how small you fix [itex]\epsilon[/itex], and so the relative complement of the intersection of all such groups determined by their respective [itex]\epsilon[/itex] in the real line is not empty and it follows that in combining all of this the epsilon delta principle implies continuity: for each x and g,

    [itex](\forall\epsilon > 0) ((\exists\delta>0 \ni: 0 < |x-c| < \delta) \Rightarrow
    (|g\cdot (x-c) - g\cdot x|=|(g-c)\cdot x - g\cdot x|=|((g-c)+x) - (g + x)|=c < \epsilon))[/itex]

    It also follows that since on any closed and bounded subset of R the function defined by the action is uniformly continuous. So, for all x and y in some [itex]\delta[/itex]-ball dependent on the fixed [itex]\epsilon[/itex], permutations of x and y are not distinct. The inverse of the action, which should also be an action since the real line under addition is a group, is therefore ambiguous, and not well-defined.

    I don't see any way around this. Can anyone help me out? Because the real line under addition should definitely be a group...
     
    Last edited: Jan 4, 2014
  2. jcsd
  3. Jan 4, 2014 #2

    WWGD

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    I think you're misunderstanding ( or misunderestimting ;) ) Cayley's theorem, which says that every group can be expressed/represented as a permutation group, i.e., a sa group of permutations. A permutation group may be infinite.
     
  4. Jan 4, 2014 #3
    Where's your identity element?
     
  5. Jan 5, 2014 #4
    I meant union with 0.
     
  6. Jan 5, 2014 #5
    I understand that it may be infinite. But i just don't see how it is possible to have a group acting on itself transitively if it is uncountable. If it's uncountable it can't be indexed by any subset of the natural numbers, so a set of permutations which permutes every element to every other element really makes no sense.
     
  7. Jan 5, 2014 #6
    What I'm saying for the real numbers, is that every time you try to fix some labeling of them, there exist some elements in the real line not in this labeling. How then can a transitive action, which just means that the action has a single orbit containing all permutations of the group, be defined?
     
  8. Jan 5, 2014 #7
    Note that under union with 0 this is in fact a group, if x is in G and y is in R and |x-y|<epsilon then x is equivalent to y if you consider y to be an element of G. In other words, once you set the precision, to consider such a y as an element of G also, the part of y that is smaller than epsilon is dropped because it is 0 with respect to your precision. So, the group consists of all integer multiples of epsilon, and is therefore countable.
     
    Last edited: Jan 5, 2014
  9. Jan 5, 2014 #8
    With what operation? Certainly not addition; it isn't closed. Let [itex]x > \epsilon[/itex], then [itex]x - (x + \frac{\epsilon}{2})[/itex] is not in G.

    Only if there are countably many labels.

    Let R act on itself by left addition. Every real number a induces a permutation of R (since the function f(x) = a+x is a bijection). Further, given any real numbers x and y, there exists some a such that y = a+x. It follows that the action is transitive. Done. The reals are isomorphic to an (uncountable) group of permutations.
     
    Last edited: Jan 5, 2014
  10. Jan 5, 2014 #9
    Maybe the definition of permutation is what's causing your misunderstanding. A permutation on a set X is simply a bijection ##f : X \to X##. Often we deal with permutations of finite sets of positive integers, but this need not be the case. For instance, the function ##f: \mathbb{R} \to \mathbb{R}## given by ##f(x) = x^3## is one-to-one and onto, hence a permutation of ##\mathbb{R}##.
     
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