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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 $\mathbb{Z}/2mnr\mathbb{Z}$', \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=matrix(u,1;0,u^{-1}) a=\left( \begin{array}{cc} u & 1\\ 0 & u^{-1} \end{array} \right)$$and b=matrix{v,0;t,v^{-1}) b=\left( \begin{array}{cc}% v & 0\\ t & v^{-1}% \end{array} \right)$$ (elements of $SL_{2}(F_{q})$), where $t$ has been chosen so that $uv+t+u^{-1}v^{-1}=w+w^{-1}.$$uv+t+u^{-1}v^{-1}=w+w^{-1}.$$ 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 ab=\left( \begin{array}{cc} uv+t & v^{-1}\\ u^{-1}t & u^{-1}v^{-1}% \end{array} \right) , $$has characteristic polynomial %X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})$$ X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,} $$and so ab is similar to diag(w,w^{-1}). Therefore ab has order 2r. I don't know who found this beautiful proof. Apparently the original proof of G.A. Miller is very complicated; see mo24913MO24940. 6 deleted 2 characters in body; deleted 1 characters in body Let a and b be elements of a group G. If a has order m and b has order n, what can we say about the order of ab? The next theorem shows that we can say nothing at all. THEOREM: For any integers m,n,r>1, there exists a finite group G with elements a and b such that a has order m, b has order n, and ab has order r. PROOF: We shall show that, for a suitable prime power q, there exist elements a and b of SL_{2}(F_{q}) such that a, b, and ab have orders 2m, 2n, and 2r respectively. As -I is the unique element of order 2 in SL_{2}(F_{q}), the images of a, b, ab in SL_{2}(F_{q})/{\pm I} will then have orders m, n, and r as required. Let p be a prime number not dividing 2mnr. Then p is a unit in the finite ring \mathbb{Z}/2mnr\mathbb{Z}, \mathbb{Z}/2mnr\mathbb{Z}', and so some power of it, q say, is 1 in the ring. This means that 2mnr divides q-1. As the group F_{q}^{\times} has order q-1 and is cyclic, there exist elements u, v, and w of F_{q}^{\times} having orders 2m, 2n, and 2r respectively. Let$$a=\left( \begin{array}{cc} u & 1\\ 0 & u^{-1} \end{array} \right)$$a=matrix(u,1;0,u^{-1}) and$$b=\left( \begin{array}{cc}% v & 0\\ t & v^{-1}% \end{array} \right)$$(elements of SL_{2}(F_{q})), b=matrix{v,0;t,v^{-1}) where t has been chosen so that$$ uv+t+u^{-1}v^{-1}=w+w^{-1}. 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 $$ab=\left( \begin{array}{cc} uv+t & v^{-1}\\ u^{-1}t & u^{-1}v^{-1}% \end{array} \right) ,$$ ab$ has characteristic polynomial$$X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,} % X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1}) and so ab is similar to diag(w,w^{-1}). Therefore ab has order 2r. I don't know who found this beautiful proof. Apparently the original proof of G.A. Miller is very complicated; see MO24940mo24913. 5 deleted 1 characters in body Let a and b be elements of a group G. If a has order m and b has order n, what can we say about the order of ab? The next theorem shows that we can say nothing at all. THEOREM: For any integers m,n,r>1, there exists a finite group G with elements a and b such that a has order m, b has order n, and ab has order r. PROOF: We shall show that, for a suitable prime power q, there exist elements a and b of SL_{2}(F_{q}) such that a, b, and ab have orders 2m, 2n, and 2r respectively. As -I is the unique element of order 2 in SL_{2}(F_{q}), the images of a, b, ab in SL_{2}(F_{q})/{\pm I} will then have orders m, n, and r as required. Let p be a prime number not dividing 2mnr. Then p is a unit in the finite ring \mathbb{Z}/2mnr\mathbb{Z}', \mathbb{Z}/2mnr\mathbb{Z}, and so some power of it, q say, is 1 in the ring. This means that 2mnr divides q-1. As the group F_{q}^{\times} has order q-1 and is cyclic, there exist elements u, v, and w of F_{q}^{\times} having orders 2m, 2n, and 2r respectively. Let % a=matrix(u,1;0,u^{-1})  a=\left( \begin{array}{cc} u & 1\\ 0 & u^{-1} \end{array} \right)$$ and $b=matrix{v,0;t,v^{-1})$ $b=\left( \begin{array}{cc}% v & 0\\ t & v^{-1}% \end{array} \right)$$(elements of SL_{2}(F_{q})), where t has been chosen so that uv+t+u^{-1}v^{-1}=w+w^{-1}.$$ uv+t+u^{-1}v^{-1}=w+w^{-1}. $$The characteristic polynomial of a is (X-u)(X-u^{-1}), and so a is similar to \diag(u,u^{-1}). diag(u,u^{-1}). Therefore a has order 2m. Similarly b has order 2n. The matrix ab ab=\left( \begin{array}{cc} uv+t & v^{-1}\\ u^{-1}t & u^{-1}v^{-1}% \end{array} \right) ,$$ has characteristic polynomial %$X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})$$X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,}$$ and so $ab$ is similar to $\diag(w,w^{-1})$. 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 mo24913MO24940.

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