Reducing $ langle a,b vert ab^{-1}ab^{-1}=1 rangle $ into more useful form.
I am working on a question and have found the group presentation in the form of $$ langle a,b vert ab^{-1}ab^{-1}=1 rangle $$
I believe I need to reduce this further but am unsure how. I've tried to manipulate the relator to see where that leads me but I haven't managed anything useful so far.
Any help is appreciated
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 1:38
Shaun
8,507113580
asked May 24 '17 at 12:57
Andy
41829
add a comment |
I am working on a question and have found the group presentation in the form of $$ langle a,b vert ab^{-1}ab^{-1}=1 rangle $$
I believe I need to reduce this further but am unsure how. I've tried to manipulate the relator to see where that leads me but I haven't managed anything useful so far.
Any help is appreciated
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 1:38
Shaun
8,507113580
asked May 24 '17 at 12:57
Andy
41829
Rewrite using $x=ab^{-1}, y=a$.
– i. m. soloveichik
May 24 '17 at 13:01
3
If you change generators to $x=ab^{-1}$, $y=a$, then the presentation becomes $langle x,y mid x^2 rangle$, which is a free product $C_2 * {mathbb Z}$.
– Derek Holt
May 24 '17 at 13:02
add a comment |
I am working on a question and have found the group presentation in the form of $$ langle a,b vert ab^{-1}ab^{-1}=1 rangle $$
I believe I need to reduce this further but am unsure how. I've tried to manipulate the relator to see where that leads me but I haven't managed anything useful so far.
Any help is appreciated
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 1:38
Shaun
8,507113580
asked May 24 '17 at 12:57
Andy
41829
I am working on a question and have found the group presentation in the form of $$ langle a,b vert ab^{-1}ab^{-1}=1 rangle $$
I believe I need to reduce this further but am unsure how. I've tried to manipulate the relator to see where that leads me but I haven't managed anything useful so far.
Any help is appreciated
group-theory group-presentation combinatorial-group-theory
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 1:38
Shaun
8,507113580
asked May 24 '17 at 12:57
Andy
41829
edited Nov 30 at 1:38
Shaun
8,507113580
asked May 24 '17 at 12:57
Andy
41829
edited Nov 30 at 1:38
Shaun
8,507113580
edited Nov 30 at 1:38
Shaun
8,507113580
edited Nov 30 at 1:38
Shaun
8,507113580
8,507113580
asked May 24 '17 at 12:57
Andy
41829
asked May 24 '17 at 12:57
Andy
41829
asked May 24 '17 at 12:57
Andy
41829
41829
Rewrite using $x=ab^{-1}, y=a$.
– i. m. soloveichik
May 24 '17 at 13:01
3
If you change generators to $x=ab^{-1}$, $y=a$, then the presentation becomes $langle x,y mid x^2 rangle$, which is a free product $C_2 * {mathbb Z}$.
– Derek Holt
May 24 '17 at 13:02
add a comment |
Rewrite using $x=ab^{-1}, y=a$.
– i. m. soloveichik
May 24 '17 at 13:01
3
If you change generators to $x=ab^{-1}$, $y=a$, then the presentation becomes $langle x,y mid x^2 rangle$, which is a free product $C_2 * {mathbb Z}$.
– Derek Holt
May 24 '17 at 13:02
Rewrite using $x=ab^{-1}, y=a$.
– i. m. soloveichik
May 24 '17 at 13:01
Rewrite using $x=ab^{-1}, y=a$.
– i. m. soloveichik
May 24 '17 at 13:01
3
3
If you change generators to $x=ab^{-1}$, $y=a$, then the presentation becomes $langle x,y mid x^2 rangle$, which is a free product $C_2 * {mathbb Z}$.
– Derek Holt
May 24 '17 at 13:02
If you change generators to $x=ab^{-1}$, $y=a$, then the presentation becomes $langle x,y mid x^2 rangle$, which is a free product $C_2 * {mathbb Z}$.
– Derek Holt
May 24 '17 at 13:02
add a comment |
1 Answer
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oldest
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Let $x=ab^{-1}$. Then $bx=a$, so the presentation becomes
$$begin{align}
langle a, b, xmid x^2, bx=arangle &conglangle b,xmid x^2rangle \
&conglangle bmidrangleastlangle xmid x^2rangle \
&congBbb ZastBbb Z_2.
end{align}$$
answered Nov 30 at 1:51
Shaun
8,507113580
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
Let $x=ab^{-1}$. Then $bx=a$, so the presentation becomes
$$begin{align}
langle a, b, xmid x^2, bx=arangle &conglangle b,xmid x^2rangle \
&conglangle bmidrangleastlangle xmid x^2rangle \
&congBbb ZastBbb Z_2.
end{align}$$
answered Nov 30 at 1:51
Shaun
8,507113580
add a comment |
Let $x=ab^{-1}$. Then $bx=a$, so the presentation becomes
$$begin{align}
langle a, b, xmid x^2, bx=arangle &conglangle b,xmid x^2rangle \
&conglangle bmidrangleastlangle xmid x^2rangle \
&congBbb ZastBbb Z_2.
end{align}$$
answered Nov 30 at 1:51
Shaun
8,507113580
add a comment |
Let $x=ab^{-1}$. Then $bx=a$, so the presentation becomes
$$begin{align}
langle a, b, xmid x^2, bx=arangle &conglangle b,xmid x^2rangle \
&conglangle bmidrangleastlangle xmid x^2rangle \
&congBbb ZastBbb Z_2.
end{align}$$
answered Nov 30 at 1:51
Shaun
8,507113580
Let $x=ab^{-1}$. Then $bx=a$, so the presentation becomes
$$begin{align}
langle a, b, xmid x^2, bx=arangle &conglangle b,xmid x^2rangle \
&conglangle bmidrangleastlangle xmid x^2rangle \
&congBbb ZastBbb Z_2.
end{align}$$
answered Nov 30 at 1:51
Shaun
8,507113580
answered Nov 30 at 1:51
Shaun
8,507113580
answered Nov 30 at 1:51
Shaun
8,507113580
answered Nov 30 at 1:51
Shaun
8,507113580
8,507113580
add a comment |
add a comment |
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Rewrite using $x=ab^{-1}, y=a$.
– i. m. soloveichik
May 24 '17 at 13:01
3
If you change generators to $x=ab^{-1}$, $y=a$, then the presentation becomes $langle x,y mid x^2 rangle$, which is a free product $C_2 * {mathbb Z}$.
– Derek Holt
May 24 '17 at 13:02