Let $r,s,t>1$ be positive integers. Must there exist a finite group $G$ with elements $x$ and $y$ such that $ord(x)=r$, $ord(y)=s$, and $ord(xy)=t$?

The answer is probably "yes." Is there a nice description of such a $G$?

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    $\begingroup$ tea.mathoverflow.net/discussion/1145/… $\endgroup$ – Anton Geraschenko Sep 23 '11 at 19:21
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    $\begingroup$ At any rate, the way to build $G$ is as follows. Take the $r,s,t$--triangle group $T = \langle x, y \ | \ x^r = y^s = (xy)^t \rangle$, in which $x$, $y$, and $xy$ have the correct orders, but which, as Anton mentions, is usually infinite. $T$ is, however, residually finite (since it is linear). Now just pick some finite quotient $G$ of $T$ in which all of the nontrivial powers of $x$, $y$, and $xy$ survive. $\endgroup$ – Autumn Kent Sep 23 '11 at 19:35
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    $\begingroup$ See here: mathoverflow.net/questions/24913/quick-proofs-of-hard-theorems/… $\endgroup$ – Steve D Sep 23 '11 at 19:50
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    $\begingroup$ @Anton: Finitely generated linear groups are always residually finite: this is a result of Malcev; I sketched (one of many possible) proof here: mathoverflow.net/questions/9628/… $\endgroup$ – Steve D Sep 23 '11 at 19:51
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    $\begingroup$ @Anton: $T$ is either finite or a discrete group of orientation preserving isometries of the Euclidean or hyperbolic plane. These isometry groups are $\mathbb{R}^2 \ltimes \mathrm{SL}_2(\mathbb{R})$ and $\mathrm{PSL}_2(\mathbb{R}) \cong \mathrm{SO}_0(2,1)$, where $\mathrm{SO}_0(2,1)$ is the identity component of $\mathrm{SO}_0(2,1)$. $\endgroup$ – Autumn Kent Sep 23 '11 at 19:53

Let $a$ and $b$ be elements of a group $G$. If $a$ has order $m$ and $b$ has order $n$, what can we say about the order of $ab$? The next theorem shows that we can say nothing at all.

THEOREM: For any integers $m,n,r>1$, there exists a finite group $G$ with elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$ has order $r$.

PROOF: We shall show that, for a suitable prime power $q$, there exist elements $a$ and $b$ of $SL_{2}(F_{q})$ such that $a$, $b$, and $ab$ have orders $2m$, $2n$, and $2r$ respectively. As $-I$ is the unique element of order $2$ in $SL_{2}(F_{q})$, the images of $a$, $b$, $ab$ in $SL_{2}(F_{q})/\{\pm I\}$ will then have orders $m$, $n$, and $r$ as required.

Let $p$ be a prime number not dividing $2mnr$. Then $p$ is a unit in the finite ring $\mathbb{Z}/2mnr\mathbb{Z}$, and so some power of it, $q$ say, is $1$ in the ring. This means that $2mnr$ divides $q-1$. As the group $F_{q}^{\times}$ has order $q-1$ and is cyclic, there exist elements $u$, $v$, and $w$ of $F_{q}^{\times}$ having orders $2m$, $2n$, and $2r$ respectively.

Let $$ a=\left( \begin{array}{cc} u & 1\\ 0 & u^{-1} \end{array} \right)$$ and $$b=\left( \begin{array}{cc}% v & 0\\ t & v^{-1}% \end{array} \right)$$ (elements of $SL_{2}(F_{q})$), where $t$ has been chosen so that $$ uv+t+u^{-1}v^{-1}=w+w^{-1}. $$

The characteristic polynomial of $a$ is $(X-u)(X-u^{-1})$, and so $a$ is similar to $diag(u,u^{-1})$. Therefore $a$ has order $2m$. Similarly $b$ has order $2n$. The matrix $$ ab=\left( \begin{array}{cc} uv+t & v^{-1}\\ u^{-1}t & u^{-1}v^{-1}% \end{array} \right) , $$ has characteristic polynomial $$ X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,} $$ and so $ab$ is similar to $diag(w,w^{-1})$. Therefore $ab$ has order $2r$.

I don't know who found this beautiful proof. Apparently the original proof of G.A. Miller is very complicated; see MO24940.

  • $\begingroup$ I give up --- I'll someone else fix the TeX. $\endgroup$ – anon Sep 23 '11 at 22:12
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    $\begingroup$ ok, done. I hope I didn't mess anything else up. $\endgroup$ – Anton Geraschenko Sep 23 '11 at 22:13
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    $\begingroup$ Thanks Anton -- beautiful! (Maybe I should read the instructions.) $\endgroup$ – anon Sep 24 '11 at 14:35
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    $\begingroup$ Nice! I don't know who first gave this proof either, but I first read it from a post by Derek Holt: mathoverflow.net/questions/24913/quick-proofs-of-hard-theorems/… $\endgroup$ – Steve D Sep 24 '11 at 19:43

Here's my comment as an answer:

Take the $r,s,t$--(ordinary) triangle group $T(r,s,t)=\langle x,y \ | \ x^r=y^s=(xy)^t = 1 \rangle$, in which $x$, $y$, and $xy$ have the correct orders. See the section on ``von Dyck" groups here.

As Anton mentions in his comment, $T(r,s,t)$ is infinite when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} \leq 1$. However, $T(r,s,t)$ is residually finite. The easiest way to see this is to use the facts that finitely generated linear groups are residually finite (due to Malcev, as Steve D mentions), and the fact that $T(r,s,t)$ is linear.

To see that $T(r,s,t)$ is linear, note that when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} = 1$, it is a discrete subgroup of the affine group $\mathbb{R}^2 \rtimes \mathrm{SL}_2(\mathbb{R})$, and when $\frac{1}{r} + \frac{1}{s} + \frac{1}{t} < 1$, it is a discrete subgroup of $\mathrm{Isom}^+(\mathbb{H}^2) \cong \mathrm{PSL}_2(\mathbb{R}) \cong \mathrm{SO}_0(2,1)$, where $\mathrm{SO}_0(2,1)$ is the identity component of $\mathrm{SO}(2,1)$. See this again.

Now, since $T(r,s,t)$ is residually finite, there is a quotient $G$ in which $$x, x^2, \ldots, x^{r-1}, y, y^2, \ldots, y^{s-1}, (xy), (xy)^2, \ldots, (xy)^{t-1}$$ are all nontrivial. This is the $G$ you seek.

Also see Steve D's answer here.


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