1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Congruence implying existence of a group of some order

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A group presentation [itex]G = (a,b : a^m = b^n = 1, ba = a^db)[/itex] defines a group of order [itex]mn[/itex] if and only if [itex]d^n \equiv 1[/itex] (mod [itex]m[/itex]).


    2. Relevant equations
    One book that I read presents a solution in a way of constructing a group of said order by defining associative binary operation on set of formal products [itex]a^ib^j, i = 0, \ldots, m-1, j = 0, \ldots, n-1[/itex] in a complicated way, and I am sort of unsatisfied with that proof. I would much rather work with the presentation itself and for example show that given the valid congruence, the elements of set [itex]a^ib^j, i = 0, \ldots, m-1, j = 0, \ldots, n-1[/itex] are all distinct. Could this be possible?

    For example, it is clear that the condition is necessary. In [itex]G[/itex] we have [itex]a = 1\cdot a= b^na[/itex] and by repeated application of the relation [itex]ba = a^db[/itex] we obtain [itex]a = b^na = a^{d^n}b^n = a^{d^n}[/itex], hence if [itex]d^n \not\equiv 1[/itex] (mod [itex]m[/itex]) then clearly [itex]a[/itex] has order less than [itex]m[/itex], and the number of distinct elements in [itex]G[/itex] is bounded from above by [itex]ord(a)\cdot ord(b) \leq mn[/itex]. It is not as clear to me if one can go the other way around as easily.

    Perhaps anyone know of anywhere on the internet where I can read more about this result? It is apparently quite famous and due to Cayley, but afaik neither Hungerford or Dummit and Foote mention it in their books.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Congruence implying existence of a group of some order
Loading...