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R.P.
R.P.
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There is a theorem in finite group theory, that if $a$, $b$, and $c$ are integers all greater than $1$, there exists a finite group $G$ with elements $x$ and $y$ such that: $x$ has order $a$, $y$ has order $b$, and $xy$ has order $c$. I think the first person to prove this was G.A. Miller, whose proof looked at lots of separate cases, and had tons of long, tedious calculations in symmetric groups (I will try and find the paper and post the reference later). I don't know who discovered the more modern proofs, but Derek Holt posted a proof on the group-pub that is one of the most elegant things I've ever seen. Unfortunately, it doesn't seem to be available on the archive of the list, so I will just post it here verbatim:

Let q$q$ be a prime power such that q-1$q-1$ is divisible by 2a$2a$, 2b$2b$, and 2c$2c$. We will construct elements x,y$x, y$ of SL(2,q)$\operatorname{SL}(2,q)$ such that x$x$, y$y$, and xy$xy$ have orders 2a$2a$, 2b$2b$, and 2c$2c$, and then the images of x,y,xy$x,y,xy$ in PSL(2,q)$\operatorname{PSL}(2,q)$ will have orders a$a$, b$b$, and c$c$ as required.

An element of SL(2,q)$\operatorname{SL}(2,q)$ with distinct eigenvalues is diagonalizable in GL(2,q)$\operatorname{GL}(2,q)$, and so its order is determined by its characteristic polynomial which is determined by its trace. In particular, since 2a,2b,2c > 2$2a,2b,2c > 2$, this applies to elements with these orders.

Let u$u$ and v$v$ be elements of the field F_q$\mathbb{F}_q$ with multiplicative orders 2a$2a$ and 2b$2b$, and let x = [ [u, 1], [0, u^-1] ]$x = \left( \begin{array}{cc} u & 1 \\ 0 & u^{-1} \end{array} \right)$ and y = [ [v, 0], [t, v^-1] ]$y = \left( \begin{array}{cc} v & 0 \\ t & v^{-1} \end{array} \right)$ be in SL(2,q)$\operatorname{SL}(2,q)$, where t$t$ remains to be chosen. Then x$x$ and y$y$ have orders 2a$2a$ and 2b$2b$.

The trace of xy$xy$ is uv + t + u^-1v^-1$uv + t + u^{-1}v^{-1}$, and so by suitable choice of t$t$, we can make this equal to any value we like. So we can make it equal to the trace of an element of SL(2,q)$\operatorname{SL}(2,q)$ with order 2c$2c$, and then xy$xy$ will have order 2c.

There is a theorem in finite group theory, that if $a$, $b$, and $c$ are integers all greater than $1$, there exists a finite group $G$ with elements $x$ and $y$ such that: $x$ has order $a$, $y$ has order $b$, and $xy$ has order $c$. I think the first person to prove this was G.A. Miller, whose proof looked at lots of separate cases, and had tons of long, tedious calculations in symmetric groups (I will try and find the paper and post the reference later). I don't know who discovered the more modern proofs, but Derek Holt posted a proof on the group-pub that is one of the most elegant things I've ever seen. Unfortunately, it doesn't seem to be available on the archive of the list, so I will just post it here verbatim:

Let q be a prime power such that q-1 is divisible by 2a, 2b, and 2c. We will construct elements x,y of SL(2,q) such that x, y, and xy have orders 2a, 2b, and 2c, and then the images of x,y,xy in PSL(2,q) will have orders a, b, and c as required.

An element of SL(2,q) with distinct eigenvalues is diagonalizable in GL(2,q), and so its order is determined by its characteristic polynomial which is determined by its trace. In particular, since 2a,2b,2c > 2, this applies to elements with these orders.

Let u and v be elements of the field F_q with multiplicative orders 2a and 2b, and let x = [ [u, 1], [0, u^-1] ] and y = [ [v, 0], [t, v^-1] ] be in SL(2,q), where t remains to be chosen. Then x and y have orders 2a and 2b.

The trace of xy is uv + t + u^-1v^-1, and so by suitable choice of t, we can make this equal to any value we like. So we can make it equal to the trace of an element of SL(2,q) with order 2c, and then xy will have order 2c.

There is a theorem in finite group theory, that if $a$, $b$, and $c$ are integers all greater than $1$, there exists a finite group $G$ with elements $x$ and $y$ such that: $x$ has order $a$, $y$ has order $b$, and $xy$ has order $c$. I think the first person to prove this was G.A. Miller, whose proof looked at lots of separate cases, and had tons of long, tedious calculations in symmetric groups (I will try and find the paper and post the reference later). I don't know who discovered the more modern proofs, but Derek Holt posted a proof on the group-pub that is one of the most elegant things I've ever seen. Unfortunately, it doesn't seem to be available on the archive of the list, so I will just post it here verbatim:

Let $q$ be a prime power such that $q-1$ is divisible by $2a$, $2b$, and $2c$. We will construct elements $x, y$ of $\operatorname{SL}(2,q)$ such that $x$, $y$, and $xy$ have orders $2a$, $2b$, and $2c$, and then the images of $x,y,xy$ in $\operatorname{PSL}(2,q)$ will have orders $a$, $b$, and $c$ as required.

An element of $\operatorname{SL}(2,q)$ with distinct eigenvalues is diagonalizable in $\operatorname{GL}(2,q)$, and so its order is determined by its characteristic polynomial which is determined by its trace. In particular, since $2a,2b,2c > 2$, this applies to elements with these orders.

Let $u$ and $v$ be elements of the field $\mathbb{F}_q$ with multiplicative orders $2a$ and $2b$, and let $x = \left( \begin{array}{cc} u & 1 \\ 0 & u^{-1} \end{array} \right)$ and $y = \left( \begin{array}{cc} v & 0 \\ t & v^{-1} \end{array} \right)$ be in $\operatorname{SL}(2,q)$, where $t$ remains to be chosen. Then $x$ and $y$ have orders $2a$ and $2b$.

The trace of $xy$ is $uv + t + u^{-1}v^{-1}$, and so by suitable choice of $t$, we can make this equal to any value we like. So we can make it equal to the trace of an element of $\operatorname{SL}(2,q)$ with order $2c$, and then $xy$ will have order 2c.

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Steve D
Steve D
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There is a theorem in finite group theory, that if $a$, $b$, and $c$ are integers all greater than $1$, there exists a finite group $G$ with elements $x$ and $y$ such that: $x$ has order $a$, $y$ has order $b$, and $xy$ has order $c$. I think the first person to prove this was G.A. Miller, whose proof looked at lots of separate cases, and had tons of long, tedious calculations in symmetric groups (I will try and find the paper and post the reference later). I don't know who discovered the more modern proofs, but Derek Holt posted a proof on the group-pub that is one of the most elegant things I've ever seen. Unfortunately, it doesn't seem to be available on the archive of the list, so I will just post it here verbatim:

Let q be a prime power such that q-1 is divisible by 2a, 2b, and 2c. We will construct elements x,y of SL(2,q) such that x, y, and xy have orders 2a, 2b, and 2c, and then the images of x,y,xy in PSL(2,q) will have orders a, b, and c as required.

An element of SL(2,q) with distinct eigenvalues is diagonalizable in GL(2,q), and so its order is determined by its characteristic polynomial which is determined by its trace. In particular, since 2a,2b,2c > 2, this applies to elements with these orders.

Let u and v be elements of the field F_q with multiplicative orders 2a and 2b, and let x = [ [u, 1], [0, u^-1] ] and y = [ [v, 0], [t, v^-1] ] be in SL(2,q), where t remains to be chosen. Then x and y have orders 2a and 2b.

The trace of xy is uv + t + u^-1v^-1, and so by suitable choice of t, we can make this equal to any value we like. So we can make it equal to the trace of an element of SL(2,q) with order 2c, and then xy will have order 2c.