Existing of groups for different cases.

Question

Here my question, does there always exist such groups in which product of two elements (namely , $a,b$ with $|a|=m,|b|=n$) with order $k$? Then what will be the structure of that group?

Here $m,n,k$ may distinct or not but runs over all natural numbers.

I can think for just of some particular cases, can't think arbitrarily.

If $m=2,n=2$ then for every $k$ we will have dihedral group $D_k$(with order $2k$) in our hand.

If $m=2,k=2$ then for every $n$ we will have dihedral group $D_n$ in our hand.

.........

Is there any suitable way to think?

Do those question have an affirmative answer?

Answer 1

Here is sketch of a completely different proof. Note that we need to assume that $m,n,k>1$.

Let $K$ be any field containing primitive $i$th roots of $1$ for $i=2m,2n,2k$. Then the group ${\rm PSL}(2,K)$ contains elements $a,b$ of order $m$ and $n$ such that $ab$ has order $k$. (So by choosing $K$ to be a finite field, we get finite examples.)

The construction makes use of the fact that the order of an element of ${\rm SL}(2,K)$ with distinct eigenvalues is determined by its trace.

So we can take $A= \left(\begin{array}{cc}\lambda&1\\0&\lambda^{-1}\end{array}\right)$ of order $2m$, $B = \left(\begin{array}{cc}\mu&0\\t&\mu^{-1}\end{array}\right)$ of order $2n$, where we choose $t \in K$ such that the trace of $AB$ is equal to $\nu + \nu^{-1}$, where $\nu$ has order $2k$ in $K$. Then $AB$ has order $2k$, and we define $a$ and $b$ to be the images of $A$ and $B$ in ${\rm PSL}(2,K)$.

(I have a feeling that this construction has been given here before.)

Example: $m=3$, $n=4$, $k=3$. We choose $\lambda$ and $\mu$ to be primitive $2m$th and $2n$th roots of unity, so 6th and 8th in this case.

Let $\nu$ be a primitive $2k$th root of unity (so 6th in this example).

The trace of $AB$ is $t+\lambda\mu + \lambda^{-1}\mu^{-1}$, so choose $t = \nu + \nu^{-1} - \lambda\mu - \lambda^{-1}\mu^{-1}$.

You can always choose $K = {\mathbb C}$. If you want a finite example, then choose $K = {\mathbb F}_q$, where $q-1$ is divisible by $2m$, $2n$ and $2k$. So in this case you could choose $q=25$, or $49$, or $73$, etc.

Answer 2

The way to approach this problem is to construct groups with with the properties you want. The standard examples are triangle groups*, which are groups of orientation-preserving actions of tilings of planes (spherical, Euclidean, or hyperbolic as appropriate) by triangles.

For example, if you want elements $a$ of order $2$ and $b$ of order $3$ such that $ab$ has order $6$ then the group of orientation-preserving symmetries of the Euclidean plane by triangle with angles $\pi/2$, $\pi/3$ and $\pi/6$ is what you want:

Here, $\pi/2+\pi/3+\pi/6=\pi$, which is what you're after for tiling the Euclidean plane (angles adding up to $\pi$ radians, so $180$ degrees). Therefore, dividing the above by $\pi$, we see that if your numbers $m,n,k$ are such that $1/n+1/m+1/k=1$ then there is a tiling of the Euclidean plan with triangles which has the properties you want. We can generalise: if $1/n+1/m+1/k<1$ then the tiling is of the hyperbolic plane, while if $1/n+1/m+1/k>1$ then the tiling is of the sphere. See the Wikipedia link above for more pictures/examples of these tilings.

*Wikipedia calls these "von Dyck groups", using triangle groups for the non-orientation-preserving groups, but the papers and books I read called these groups triangle groups so I'm sticking with that...!