Is $langle a,b vert aba=bab, abab=baba rangle$ a presentation of the free group on a single generator?
Is the following a presentation of the free group generated by a single element?
$langle a,b vert aba=bab, abab=baba rangle.$
My thinking is the following:
$abab = baba=b(bab)=b^2ab$ by substituting the first relation into the second. Simplifying, we get $a=b$. Since these steps give equivalent statements, the above presentation is in fact $langle a,b vert a=b rangle$, ie. the free group on one generator.
Is this correct?
group-theory proof-verification group-presentation combinatorial-group-theory
edited Nov 30 '18 at 2:38
Shaun
8,820113681
asked Feb 5 '17 at 18:50
JoeJoe
458211
add a comment |
Is the following a presentation of the free group generated by a single element?
$langle a,b vert aba=bab, abab=baba rangle.$
My thinking is the following:
$abab = baba=b(bab)=b^2ab$ by substituting the first relation into the second. Simplifying, we get $a=b$. Since these steps give equivalent statements, the above presentation is in fact $langle a,b vert a=b rangle$, ie. the free group on one generator.
Is this correct?
group-theory proof-verification group-presentation combinatorial-group-theory
edited Nov 30 '18 at 2:38
Shaun
8,820113681
asked Feb 5 '17 at 18:50
JoeJoe
458211
8
Yes , What you have said its correct
– Chirantan Chowdhury
Feb 5 '17 at 18:52
add a comment |
Is the following a presentation of the free group generated by a single element?
$langle a,b vert aba=bab, abab=baba rangle.$
My thinking is the following:
$abab = baba=b(bab)=b^2ab$ by substituting the first relation into the second. Simplifying, we get $a=b$. Since these steps give equivalent statements, the above presentation is in fact $langle a,b vert a=b rangle$, ie. the free group on one generator.
Is this correct?
group-theory proof-verification group-presentation combinatorial-group-theory
edited Nov 30 '18 at 2:38
Shaun
8,820113681
asked Feb 5 '17 at 18:50
JoeJoe
458211
Is the following a presentation of the free group generated by a single element?
$langle a,b vert aba=bab, abab=baba rangle.$
My thinking is the following:
$abab = baba=b(bab)=b^2ab$ by substituting the first relation into the second. Simplifying, we get $a=b$. Since these steps give equivalent statements, the above presentation is in fact $langle a,b vert a=b rangle$, ie. the free group on one generator.
Is this correct?
group-theory proof-verification group-presentation combinatorial-group-theory
group-theory proof-verification group-presentation combinatorial-group-theory
edited Nov 30 '18 at 2:38
Shaun
8,820113681
asked Feb 5 '17 at 18:50
JoeJoe
458211
edited Nov 30 '18 at 2:38
Shaun
8,820113681
asked Feb 5 '17 at 18:50
JoeJoe
458211
edited Nov 30 '18 at 2:38
Shaun
8,820113681
edited Nov 30 '18 at 2:38
Shaun
8,820113681
edited Nov 30 '18 at 2:38
Shaun
8,820113681
8,820113681
asked Feb 5 '17 at 18:50
JoeJoe
458211
asked Feb 5 '17 at 18:50
JoeJoe
458211
asked Feb 5 '17 at 18:50
JoeJoe
458211
458211
8
Yes , What you have said its correct
– Chirantan Chowdhury
Feb 5 '17 at 18:52
add a comment |
8
Yes , What you have said its correct
– Chirantan Chowdhury
Feb 5 '17 at 18:52
8
8
Yes , What you have said its correct
– Chirantan Chowdhury
Feb 5 '17 at 18:52
Yes , What you have said its correct
– Chirantan Chowdhury
Feb 5 '17 at 18:52
add a comment |
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Yes , What you have said its correct
– Chirantan Chowdhury
Feb 5 '17 at 18:52