Mixing
The Baumslag-Solitar Group $G=langle a, b, mid, b^{-1}ab=a^2rangle$.
Let $G=langle a, b, mid, b^{-1}ab=a^2rangle$.
I want to prove that the normal closure of $a$ is isomorphic with the additive group of all rational numbers whose denominators are powers of $2$. I set out $k/2^h mapsto b^{-h}a^kb^h$. To prove that this is an isomorphism I need to show that $o(a)=infty$, how can I do that?
(Of course, I know that $o(b)=infty$.)
abstract-algebra group-theory
edited Nov 21 '18 at 19:22
Shaun
8,820113681
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
add a comment |
Let $G=langle a, b, mid, b^{-1}ab=a^2rangle$.
I want to prove that the normal closure of $a$ is isomorphic with the additive group of all rational numbers whose denominators are powers of $2$. I set out $k/2^h mapsto b^{-h}a^kb^h$. To prove that this is an isomorphism I need to show that $o(a)=infty$, how can I do that?
(Of course, I know that $o(b)=infty$.)
abstract-algebra group-theory
edited Nov 21 '18 at 19:22
Shaun
8,820113681
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
add a comment |
Let $G=langle a, b, mid, b^{-1}ab=a^2rangle$.
I want to prove that the normal closure of $a$ is isomorphic with the additive group of all rational numbers whose denominators are powers of $2$. I set out $k/2^h mapsto b^{-h}a^kb^h$. To prove that this is an isomorphism I need to show that $o(a)=infty$, how can I do that?
(Of course, I know that $o(b)=infty$.)
abstract-algebra group-theory
edited Nov 21 '18 at 19:22
Shaun
8,820113681
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
Let $G=langle a, b, mid, b^{-1}ab=a^2rangle$.
I want to prove that the normal closure of $a$ is isomorphic with the additive group of all rational numbers whose denominators are powers of $2$. I set out $k/2^h mapsto b^{-h}a^kb^h$. To prove that this is an isomorphism I need to show that $o(a)=infty$, how can I do that?
(Of course, I know that $o(b)=infty$.)
abstract-algebra group-theory
abstract-algebra group-theory
edited Nov 21 '18 at 19:22
Shaun
8,820113681
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
edited Nov 21 '18 at 19:22
Shaun
8,820113681
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
edited Nov 21 '18 at 19:22
Shaun
8,820113681
edited Nov 21 '18 at 19:22
Shaun
8,820113681
edited Nov 21 '18 at 19:22
Shaun
8,820113681
8,820113681
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
asked Nov 26 '13 at 0:46
W4cc0W4cc0
1,89621227
1,89621227
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You can define a homomorphism $G to {rm GL}_2({mathbb Q})$ with
$b mapsto left(begin{array}{cc}2&0\0&1end{array}right)$,
$a mapsto left(begin{array}{cc}1&0\1&1end{array}right)$.
Just check that the images of $a$ and $b$ satisfy the group relation to verify that this defines a homomorphism. The image of $a$ has infinite order, and hence so does $a$. This map is actually injective, and the image of the normal closure of $a$ consists of all matrices of the form $left(begin{array}{cc}1&0\m/2^k&1end{array}right)$ with $m in {mathbb Z}$, $k ge 0$.
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can define a homomorphism $G to {rm GL}_2({mathbb Q})$ with
$b mapsto left(begin{array}{cc}2&0\0&1end{array}right)$,
$a mapsto left(begin{array}{cc}1&0\1&1end{array}right)$.
Just check that the images of $a$ and $b$ satisfy the group relation to verify that this defines a homomorphism. The image of $a$ has infinite order, and hence so does $a$. This map is actually injective, and the image of the normal closure of $a$ consists of all matrices of the form $left(begin{array}{cc}1&0\m/2^k&1end{array}right)$ with $m in {mathbb Z}$, $k ge 0$.
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
add a comment |
You can define a homomorphism $G to {rm GL}_2({mathbb Q})$ with
$b mapsto left(begin{array}{cc}2&0\0&1end{array}right)$,
$a mapsto left(begin{array}{cc}1&0\1&1end{array}right)$.
Just check that the images of $a$ and $b$ satisfy the group relation to verify that this defines a homomorphism. The image of $a$ has infinite order, and hence so does $a$. This map is actually injective, and the image of the normal closure of $a$ consists of all matrices of the form $left(begin{array}{cc}1&0\m/2^k&1end{array}right)$ with $m in {mathbb Z}$, $k ge 0$.
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
add a comment |
You can define a homomorphism $G to {rm GL}_2({mathbb Q})$ with
$b mapsto left(begin{array}{cc}2&0\0&1end{array}right)$,
$a mapsto left(begin{array}{cc}1&0\1&1end{array}right)$.
Just check that the images of $a$ and $b$ satisfy the group relation to verify that this defines a homomorphism. The image of $a$ has infinite order, and hence so does $a$. This map is actually injective, and the image of the normal closure of $a$ consists of all matrices of the form $left(begin{array}{cc}1&0\m/2^k&1end{array}right)$ with $m in {mathbb Z}$, $k ge 0$.
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
You can define a homomorphism $G to {rm GL}_2({mathbb Q})$ with
$b mapsto left(begin{array}{cc}2&0\0&1end{array}right)$,
$a mapsto left(begin{array}{cc}1&0\1&1end{array}right)$.
Just check that the images of $a$ and $b$ satisfy the group relation to verify that this defines a homomorphism. The image of $a$ has infinite order, and hence so does $a$. This map is actually injective, and the image of the normal closure of $a$ consists of all matrices of the form $left(begin{array}{cc}1&0\m/2^k&1end{array}right)$ with $m in {mathbb Z}$, $k ge 0$.
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
answered Nov 26 '13 at 8:34
Derek HoltDerek Holt
52.7k53570
52.7k53570
add a comment |
add a comment |
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