Does there exist a finite group with the following presentation?

Question

Let $G$ be a finite group (with only two generators and $m=n$) presented as

$$ G = \langle a, b : a^m = b^n = (W(a,b))^p= \ldots\text{other-such-relations}\ldots= 1 \rangle $$

where $m,n,p>1$ , and taking the smallest $p$ for each $W(a,b)$ which is made out of products of $a$ and $b$, e.g. $(ab)^2$, $(ab^2ab^{-1})^3$ etc.

I know three examples

1) Dihedral groups of order $n$: $ G = \langle a, b : a^2 = b^2 = (ab)^n= 1 \rangle $

2) Another two from the following paper (page 2) and presented as :

J. Howie, V. Metaftsis, and R. M. Thomas. Finite generalized triangle groups. Trans. Amer. Math. Soc., 347(9):3613–3623, 1995

$$ G = \langle a, b : a^3 = b^3 = (abab^2)^2= 1 \rangle $$ of order 180 and

$$ G = \langle a, b : a^3 = b^3 = (aba^2b^2)^2= 1 \rangle $$ of order 288.

Now, after going through the list of finite group presentations, I could not find any other finite group with such a presentation (i.e only two generators and $m=n$).

So, are there any other examples? Or is it possible to give arguments why there might not exist any other example?

References will also be useful.

Thank you.

Answer 1

You will find a lot of examples in

H.S.M.Coxeter, W.O.J.Moser, Generators and relations for discrete groups, 1972.

Answer 2

Is this cheating? $$G=\langle\,a,b:a^n=b^n=ab=1\,\rangle$$

Answer 3

If $p$ is not fixed (thanks @Herband for the clarification), then you can get any extra relation $W(a, b)$ in, just do $$ \left\langle a, b : a^m = b^m = 1, W(a,b)^p = W(a, b)^{p+1} = 1, \dots \right\rangle $$