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7
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edited Sep 23 2011 at 22:12 |
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{,}
$$
and so $ab$ is similar to $diag(w,w^{-1})$. Therefore $ab$ has order
$2r$.
I don't know who found this beautiful proof. Apparently the
original proof of G.A. Miller is very complicated; see mo24913MO24940.
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6
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edited Sep 23 2011 at 22:08 |
Let $a$ and $b$ be elements of a group $G$. If $a$ has order $m$ and $b$ has
order $n$, what can we say about the order of $ab$? The next theorem shows
that we can say nothing at all.
THEOREM: For any integers $m,n,r>1$, there exists a finite group $G$ with
elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$
has order $r$.
PROOF: We shall show that, for a suitable prime power $q$, there exist elements $a$
and $b$ of $SL_{2}(F_{q})$ such that $a$, $b$, and $ab$ have
orders $2m$, $2n$, and $2r$ respectively. As $-I$ is the unique element of
order $2$ in $SL_{2}(F_{q})$, the images of $a$, $b$, $ab$ in
$SL_{2}(F_{q})/{\pm I}$ will then have orders $m$, $n$, and $r$
as required.
Let $p$ be a prime number not dividing $2mnr$. Then $p$ is a unit in the
finite ring $\mathbb{Z}/2mnr\mathbb{Z}$, \mathbb{Z}/2mnr\mathbb{Z}$', and so some power of it, $q$ say,
is $1$ in the ring. This means that $2mnr$ divides $q-1$. As the group
$F_{q}^{\times}$ has order $q-1$ and is cyclic,
there exist elements $u$, $v$, and $w$ of $F_{q}^{\times}$ having
orders $2m$, $2n$, and $2r$ respectively.
Let $$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$$
X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,}
$%
$X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})$
and so $ab$ is similar to $diag(w,w^{-1})$. Therefore $ab$ has order
$2r$.
I don't know who found this beautiful proof. Apparently the
original proof of G.A. Miller is very complicated; see MO24940mo24913.
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5
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edited Sep 23 2011 at 22:08 |
Let $a$ and $b$ be elements of a group $G$. If $a$ has order $m$ and $b$ has
order $n$, what can we say about the order of $ab$? The next theorem shows
that we can say nothing at all.
THEOREM: For any integers $m,n,r>1$, there exists a finite group $G$ with
elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$
has order $r$.
PROOF: We shall show that, for a suitable prime power $q$, there exist elements $a$
and $b$ of $SL_{2}(F_{q})$ such that $a$, $b$, and $ab$ have
orders $2m$, $2n$, and $2r$ respectively. As $-I$ is the unique element of
order $2$ in $SL_{2}(F_{q})$, the images of $a$, $b$, $ab$ in
$SL_{2}(F_{q})/{\pm I}$ will then have orders $m$, $n$, and $r$
as required.
Let $p$ be a prime number not dividing $2mnr$. Then $p$ is a unit in the
finite ring $\mathbb{Z}/2mnr\mathbb{Z}$', \mathbb{Z}/2mnr\mathbb{Z}$, and so some power of it, $q$ say,
is $1$ in the ring. This means that $2mnr$ divides $q-1$. As the group
$F_{q}^{\times}$ has order $q-1$ and is cyclic,
there exist elements $u$, $v$, and $w$ of $F_{q}^{\times}$ having
orders $2m$, $2n$, and $2r$ respectively.
Let
%
$a=matrix(u,1;0,u^{-1})$ $
a=\left(
\begin{array}{cc}
u & 1\\
0 & u^{-1}
\end{array}
\right)$$
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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4
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edited Sep 23 2011 at 22:07 |
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)$$
a=matrix(u,1;0,u^{-1})$ and $$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}.
$$
uv+t+u^{-1}v^{-1}=w+w^{-1}.$
The characteristic polynomial of $a$ is $(X-u)(X-u^{-1})$, and so $a$ is
similar to $diag(u,u^{-1})$. \diag(u,u^{-1})$. Therefore $a$ has order $2m$. Similarly $b$ has
order $2n$. The matrix $$
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 $diag(w,w^{-1})$. \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.
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3
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edited Sep 23 2011 at 22:07 |
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
$$
a=\left(
\begin{array}
[c]{cc}%
begin{array}{cc}
u & 1& 1\\
0 & u^{-1}%
& u^{-1}
\end{array}
\right)$right)$$
and $b=\left(
$b=\left(
\begin{array}
[c]{cc}%
begin{array}{cc}%
v & 0& 0\\
t & & v^{-1}%
\end{array}
\right)$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}
[c]{cc}%
begin{array}{cc}
uv+t & v^{-1}& v^{-1}\\
u^{-1}t & & u^{-1}v^{-1}%
\end{array}
\right) ,
]
$$
has characteristic polynomial
%
[
X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,}%
]
$$
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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2
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edited Sep 23 2011 at 22:00 |
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 $SL_{2}(\mathbb{F}{}{q})$ 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 $SL{2}(\mathbb{F}{}{q})$, SL_{2}(F_{q})$, the images of $a$, $b$, $ab$ in
$\SL{2}(\mathbb{F}{}_{q})/{\pm SL_{2}(F_{q})/{\pm I}$ 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
$\mathbb{F}{}{q}^{\times}$ F_{q}^{\times}$ has order $q-1$ and is cyclic,
there exist elements $u$, $v$, and $w$ of $\mathbb{F}{}{q}^{\times}$ F_{q}^{\times}$ having
orders $2m$, $2n$, and $2r$ respectively.
Let
%
[
$
a=\left(
\begin{array}
[c]{cc}%
u & 1\
0 & u^{-1}%
\end{array}
\right) \text{ right)$
and }$b=\left(
\begin{array}
[c]{cc}%
v & 0\
t & v^{-1}%
\end{array}
\right) \quad\text{(elements right)$
(elements of }SL_{2}(\mathbb{F}{}_{q})\text{)},
$SL_{2}(F_{q})$),
]
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}
[c]{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 mo24913.
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1
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answered Sep 23 2011 at 21:55 |
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 $SL_{2}(\mathbb{F}{}{q})$ such that $a$, $b$, and $ab$ have
orders $2m$, $2n$, and $2r$ respectively. As $-I$ is the unique element of
order $2$ in $SL{2}(\mathbb{F}{}{q})$, the images of $a$, $b$, $ab$ in
$\SL{2}(\mathbb{F}{}_{q})/{\pm I}$ 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
$\mathbb{F}{}{q}^{\times}$ has order $q-1$ and is cyclic,
there exist elements $u$, $v$, and $w$ of $\mathbb{F}{}{q}^{\times}$ having
orders $2m$, $2n$, and $2r$ respectively. Let%
[
a=\left(
\begin{array}
[c]{cc}%
u & 1\
0 & u^{-1}%
\end{array}
\right) \text{ and }b=\left(
\begin{array}
[c]{cc}%
v & 0\
t & v^{-1}%
\end{array}
\right) \quad\text{(elements of }SL_{2}(\mathbb{F}{}_{q})\text{)},
]
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}
[c]{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 mo24913.
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