Multiplication in a Dihedral Group

A dihedral group is a set $D$ with a multiplication operation $\ast$ defined on $D$. The set $D$ is generated by two elements $a$ and $b$. The multiplication operation $\ast$ will not be explicitly written, so that a*a will be indicated by $aa$ or by a^2. Similarly, a*a*b will be indicated by a^2b^1 or simply a^2b.

The dihedral group $D$ has two integer parameters $m$ and $n$, with m \geq 2 and n \geq 2, and so we write the dihedral group as D_{m,n}. The relations that define D_{m,n} are:

a^m = a^0 = 1, b^n = b^0 = 1, and ba = a^{m-1}b.

Thus D_{m,n} has $mn$ elements, namely

D_{m,n} = \{ a^j b^k | 0 \leq j < m, 0 \leq k < n \} = \{ a^0b^0, a^1b^0, a^2b^0, \ldots, a^{m-1}b^0, \ldots, a^{m-1}b^{n-1} \}.

For example, select m = 7 and n = 2. The elements of D_{7,2} are

\{ a^0b^0, a^1b^0, a^2b^0, a^3b^0, a^4b^0, a^5b^0, a^6b^0, a^0b^1, a^1b^1, a^2b^1, a^3b^1, a^4b^1, a^5b^1, a^6b^1 \}.

Multiplication is not commutative, and so ba \neq ab. In fact ba = a^6b, according to the defining relations of D_{7,2}. Also, a^j a^k = a^{(j+k) \% m}, and b^j b^k = b^{(j+k) \% n}. Thus

(a^jb^k)(a^pb^q) \neq (a^{(j+p) \% m}b^{(k+q) \% n}).

To properly multiply the two elements (a^3b^1) and (a^2b^1) of D_{7,2}, we proceed as follows:

(a^3b^1)(a^2b^1) = (a^3b^0)(ba)(a^1b^1) = (a^3b^0)(a^6b^1)(a^1b^1),

since ba = a^6b.

Also, since a^3b^0a^6 = a^3a^6 = a^{(3+6) \% 7} = a^2, we can multiply the first two factors to get (a^3b^0)(a^6b^1) = (a^2b^1).

Thus

(a^3b^0)(a^6b^1)(a^1b^1) = (a^2b^1)(a^1b^1).

Now we can rewrite this as follows:

(a^2b^1)(a^1b^1) = (a^2b^0)(ba)(a^0b^1) = (a^2b^0)(a^6b^1)(a^0b^1) = (a^8b^2) = (a^1b^0).

Thus, multiplying the two elements (a^3b^1) and (a^2b^1) gives (a^3b^1)(a^2b^1) = (a^1b^0).

Your task is to write a computer program that, given $m$, $n$, and any two elements (a^j b^k) and (a^p b^q) of the dihedral group D_{m,n}, computes the correct product of the two.