Prove that finite and infinite presentations of Thompson group $F$ are isomorphic.












4












$begingroup$


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.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The presentations are not really isomorphic, rather they define isomorphic groups.
    $endgroup$
    – YCor
    May 1 '18 at 17:38






  • 1




    $begingroup$
    This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
    $endgroup$
    – Derek Holt
    May 2 '18 at 9:22
















4












$begingroup$


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.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The presentations are not really isomorphic, rather they define isomorphic groups.
    $endgroup$
    – YCor
    May 1 '18 at 17:38






  • 1




    $begingroup$
    This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
    $endgroup$
    – Derek Holt
    May 2 '18 at 9:22














4












4








4


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 0:48









Shaun

8,883113681




8,883113681










asked May 1 '18 at 16:40









SekstusEmpirykSekstusEmpiryk

1,176416




1,176416








  • 1




    $begingroup$
    The presentations are not really isomorphic, rather they define isomorphic groups.
    $endgroup$
    – YCor
    May 1 '18 at 17:38






  • 1




    $begingroup$
    This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
    $endgroup$
    – Derek Holt
    May 2 '18 at 9:22














  • 1




    $begingroup$
    The presentations are not really isomorphic, rather they define isomorphic groups.
    $endgroup$
    – YCor
    May 1 '18 at 17:38






  • 1




    $begingroup$
    This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
    $endgroup$
    – Derek Holt
    May 2 '18 at 9:22








1




1




$begingroup$
The presentations are not really isomorphic, rather they define isomorphic groups.
$endgroup$
– YCor
May 1 '18 at 17:38




$begingroup$
The presentations are not really isomorphic, rather they define isomorphic groups.
$endgroup$
– YCor
May 1 '18 at 17:38




1




1




$begingroup$
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
$endgroup$
– Derek Holt
May 2 '18 at 9:22




$begingroup$
This honours thesis has a moderately detailed proof in Section 2.1, although it has a few typos.
$endgroup$
– Derek Holt
May 2 '18 at 9:22










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