Congruence implying existence of a group of some order

[Barre]

Homework Statement

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 $m$).

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 $a^ib^j, i = 0, \ldots, m-1, j = 0, \ldots, n-1$ 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.

 

[lippyka]

The Attempt at a SolutionSuppose d^n \equiv 1 (mod m). Let a^ib^j, i = 0, \ldots, m-1, j = 0, \ldots, n-1 be arbitrary elements of G. If i_1 < i_2 then a^{i_1}b^j a^{i_2 - i_1}b^j = a^{i_2}b^j and so distinct elements of the form a^ib^j are not affected by addition of k \in \mathbb{Z} to i. Similarly, if j_1 < j_2 then b^{j_1}a^i b^{j_2 - j_1}a^i = b^{j_2}a^i and so distinct elements of the form a^ib^j are not affected by addition of k \in \mathbb{Z} to j. So in order to prove that elements a^ib^j, i = 0, \ldots, m-1, j = 0, \ldots, n-1 are distinct it is sufficient to show that for all i and j, a^ib^j \neq a^kb^l for any k \neq i and any l \neq j. Let us assume that a^ib^j = a^kb^l for some i,j,k,l with k \neq i and l \neq j. Then we can apply the relation ba = a^db over and over again, starting from the left to obtain b^la^i = a^kb^l = a^db^la^i = a^{d^2}b^{2l}a^i = \ldots = a^{d^j}b^{jl}a^i. But since d^n \equiv 1 (mod m) and j \leq n-1, d^j \equiv 1 (mod m) and so a^{d^j} \equiv 1 (mod m). Therefore b^{jl}a^i = a^i, hence b^l = a^{-i}. However, since a^m = 1 and i \leq m-1, a^i \neq b^l, which is