Congruence implying existence of a group of some order

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Homework Statement


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]).


Homework 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.
 

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