# Congruence implying existence of a group of some order

1. Oct 9, 2012

### Barre

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

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 $a^ib^j, i = 0, \ldots, m-1, j = 0, \ldots, n-1$ 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 $a^ib^j, i = 0, \ldots, m-1, j = 0, \ldots, n-1$ are all distinct. Could this be possible?

For example, it is clear that the condition is necessary. In $G$ we have $a = 1\cdot a= b^na$ and by repeated application of the relation $ba = a^db$ we obtain $a = b^na = a^{d^n}b^n = a^{d^n}$, hence if $d^n \not\equiv 1$ (mod $m$) then clearly $a$ has order less than $m$, and the number of distinct elements in $G$ is bounded from above by $ord(a)\cdot ord(b) \leq mn$. 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.