# Does there exist a finite group with the following presentation?

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.

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Just one question, is $p$ fixed, that is, do you require it the same exponent $p$ for all other-such-relations? –  Andreas Caranti Mar 10 '13 at 9:34
I've removed the representation-theory tag. Presentations and representations aren't the same thing @Herband. –  Alexander Gruber Mar 10 '13 at 9:38
@AndreasCaranti P is not fixed(I will avoid infinity though) and please also note that other-such-relations may not be necessary. –  Herband Mar 10 '13 at 9:40
@Alexander Thanks though I added it to have a wider audience. –  Herband Mar 10 '13 at 9:43
@Herband Shouldn't be a problem - I added abstract-algebra representation-theory and reference-request for you, each of which have many more followers anyhow. –  Alexander Gruber Mar 10 '13 at 9:57

You will find a lot of examples in

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

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Thanks, yes many examples. –  Herband Mar 10 '13 at 17:48

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

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thanks but I shall edit the question with p>1 –  Herband Mar 10 '13 at 9:45
a comment: these groups are not finite. In general for one W(a,b), such presentations give the generalized triangle groups which are finite only if 1/m + 1/n + 1/p > 1. –  Herband Mar 10 '13 at 9:56

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

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Thanks, now I see the reason why but how would I know that the resulting groups are finite. Searching on classification of finite groups, I have not found any with such presentation. So, I will presume that like this group is not finite. –  Herband Mar 10 '13 at 9:59
My point is very simple. If $p$ is allowed to vary, then you can get arbitrary presentations on two generators, with the constraints $a^m = b^m = 1$. Not all such presentations will give you a finite group, but on the other hand many finite groups admit such presentations. –  Andreas Caranti Mar 10 '13 at 10:05
Thank you for the clarification. I will now fix p for each W(a,b). –  Herband Mar 10 '13 at 10:14
@Herband: Rather than fixing $p$, you could say require every word to not be a proper power and that if $W(a, b)^p$ and $W(a, b)^q$ are two relators then $p=q$. This also gets around Andreas Caranti's example, but gives you more groups. –  user1729 Mar 12 '13 at 10:48