Prove that finite and infinite presentations of Thompson group $F$ are isomorphic.
Let $$
G=langle x_0,x_1,dotsmid x_jx_i=x_ix_{j+1}text{ for }i<jrangle,
$$
$$
H=langle a,bmid [ab^{-1},a^{-1}ba],[ab^{-1},a^{-2}ba^2]rangle,
$$
where $[x,y]$ is commutator.
These are both presentations of Thompson group $F$ and I want to show that they are indeed isomorphic. I can prove that $phi:Hto G$ when $phi(a)=x_0$ and $phi(b)=x_1$ is homomorphism. I define inverse as
$$psi(x_0)=a,psi(x_1)=b,psi(x_n)=a^{1-n}ba^{n-1}.$$
And here I have problem.
It is easy to show that it works for $i=0$. Having that we can always assume that $i=1$, so all we need is to check that homomorphism works for all $j$.
It is easy to check that it works for $j=2,3$ because $$psi(x_1^{-1})psi(x_j)psi(x_1)(psi(x_{j+1}))^{-1}$$ is one of the relators of $H$.
But I do not know how to work out induction for any larger $j$.
I would be thankful for help or recommending some source where it is proved.
group-theory group-isomorphism group-presentation combinatorial-group-theory
add a comment |
Let $$
G=langle x_0,x_1,dotsmid x_jx_i=x_ix_{j+1}text{ for }i<jrangle,
$$
$$
H=langle a,bmid [ab^{-1},a^{-1}ba],[ab^{-1},a^{-2}ba^2]rangle,
$$
where $[x,y]$ is commutator.
These are both presentations of Thompson group $F$ and I want to show that they are indeed isomorphic. I can prove that $phi:Hto G$ when $phi(a)=x_0$ and $phi(b)=x_1$ is homomorphism. I define inverse as
$$psi(x_0)=a,psi(x_1)=b,psi(x_n)=a^{1-n}ba^{n-1}.$$
And here I have problem.
It is easy to show that it works for $i=0$. Having that we can always assume that $i=1$, so all we need is to check that homomorphism works for all $j$.
It is easy to check that it works for $j=2,3$ because $$psi(x_1^{-1})psi(x_j)psi(x_1)(psi(x_{j+1}))^{-1}$$ is one of the relators of $H$.
But I do not know how to work out induction for any larger $j$.
I would be thankful for help or recommending some source where it is proved.
group-theory group-isomorphism group-presentation combinatorial-group-theory
1
The presentations are not really isomorphic, rather they define isomorphic groups.
– YCor
May 1 '18 at 17:38
1
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
– Derek Holt
May 2 '18 at 9:22
add a comment |
Let $$
G=langle x_0,x_1,dotsmid x_jx_i=x_ix_{j+1}text{ for }i<jrangle,
$$
$$
H=langle a,bmid [ab^{-1},a^{-1}ba],[ab^{-1},a^{-2}ba^2]rangle,
$$
where $[x,y]$ is commutator.
These are both presentations of Thompson group $F$ and I want to show that they are indeed isomorphic. I can prove that $phi:Hto G$ when $phi(a)=x_0$ and $phi(b)=x_1$ is homomorphism. I define inverse as
$$psi(x_0)=a,psi(x_1)=b,psi(x_n)=a^{1-n}ba^{n-1}.$$
And here I have problem.
It is easy to show that it works for $i=0$. Having that we can always assume that $i=1$, so all we need is to check that homomorphism works for all $j$.
It is easy to check that it works for $j=2,3$ because $$psi(x_1^{-1})psi(x_j)psi(x_1)(psi(x_{j+1}))^{-1}$$ is one of the relators of $H$.
But I do not know how to work out induction for any larger $j$.
I would be thankful for help or recommending some source where it is proved.
group-theory group-isomorphism group-presentation combinatorial-group-theory
Let $$
G=langle x_0,x_1,dotsmid x_jx_i=x_ix_{j+1}text{ for }i<jrangle,
$$
$$
H=langle a,bmid [ab^{-1},a^{-1}ba],[ab^{-1},a^{-2}ba^2]rangle,
$$
where $[x,y]$ is commutator.
These are both presentations of Thompson group $F$ and I want to show that they are indeed isomorphic. I can prove that $phi:Hto G$ when $phi(a)=x_0$ and $phi(b)=x_1$ is homomorphism. I define inverse as
$$psi(x_0)=a,psi(x_1)=b,psi(x_n)=a^{1-n}ba^{n-1}.$$
And here I have problem.
It is easy to show that it works for $i=0$. Having that we can always assume that $i=1$, so all we need is to check that homomorphism works for all $j$.
It is easy to check that it works for $j=2,3$ because $$psi(x_1^{-1})psi(x_j)psi(x_1)(psi(x_{j+1}))^{-1}$$ is one of the relators of $H$.
But I do not know how to work out induction for any larger $j$.
I would be thankful for help or recommending some source where it is proved.
group-theory group-isomorphism group-presentation combinatorial-group-theory
group-theory group-isomorphism group-presentation combinatorial-group-theory
edited Dec 3 '18 at 0:48
Shaun
8,810113680
8,810113680
asked May 1 '18 at 16:40
SekstusEmpiryk
1,171416
1,171416
1
The presentations are not really isomorphic, rather they define isomorphic groups.
– YCor
May 1 '18 at 17:38
1
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
– Derek Holt
May 2 '18 at 9:22
add a comment |
1
The presentations are not really isomorphic, rather they define isomorphic groups.
– YCor
May 1 '18 at 17:38
1
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
– Derek Holt
May 2 '18 at 9:22
1
1
The presentations are not really isomorphic, rather they define isomorphic groups.
– YCor
May 1 '18 at 17:38
The presentations are not really isomorphic, rather they define isomorphic groups.
– YCor
May 1 '18 at 17:38
1
1
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
– Derek Holt
May 2 '18 at 9:22
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
– Derek Holt
May 2 '18 at 9:22
add a comment |
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1
The presentations are not really isomorphic, rather they define isomorphic groups.
– YCor
May 1 '18 at 17:38
1
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
– Derek Holt
May 2 '18 at 9:22