Show that $mathbb{Z} = langle a, b mid a^{12} = b, ab = ba rangle$ has dead end elements
$begingroup$
This exercise is taken from the book "Office Hours with a Geometric Group Theorist" (Office Hour 15, exercise 8):
Exercise: Show that the group $mathbb{Z}$ has dead end elements with respect to the presentation $mathbb{Z} = langle a, b mid a^{12} = b, ab = ba rangle$. What are the possible depths of dead end elements with respect to this generating set?
Definition [from Bogopolski, Infinite commensurable hyperbolic groups
are bi-Lipschitz equivalent]: An element $g$ of a group $G$ is called dead in $G$ with respect to the finite generating set $X$ if $|gx| leq |g|$ $forall x in X^{pm 1}$. Here $|g|$ denotes the length of $g$ in the word metric with respect to $X$.
Intuitively, think of a dead end element as a dead end in a maze: no matter which direction you go, you'll be closer (or at least not farther away) from the exit (exit being the analogy for the neutral element here).
I have two questions:
Why does this presentation generate $mathbb{Z}$? If you take $a = 2$ and $b = 24$, I think it does not generate $mathbb{Z}$, since if you start from an odd number, you will only get odd numbers using $a$ and $b$.
Assuming $a = 1$ and $b = 12$, I think there is no dead end element: assume $g in mathbb{Z}$. Without loss of generality, we can assume $2 leq g leq 11$. Suppose that $|g| = n$. Then $|gb| = n + 1$ because the shortest path to go to $gb$ is to go from $0$ to $b$, and then use the same path from $0$ to $g$, but shifted by $b$. More formally, $g$ is of the form $g = a^i$, $2 leq i leq 6$ or $g = ba^{-j}$, $1 leq j leq 5$. Then $gb = ba^i$ or $g = b^2a^{-j}$, and both can't be reduced. So my question: is is possible that the exercise is actually impossible?
group-theory geometric-group-theory group-presentation finitely-generated
edited Dec 28 '18 at 5:45
Shaun
8,932113681
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
$endgroup$
|
show 4 more comments
$begingroup$
This exercise is taken from the book "Office Hours with a Geometric Group Theorist" (Office Hour 15, exercise 8):
Exercise: Show that the group $mathbb{Z}$ has dead end elements with respect to the presentation $mathbb{Z} = langle a, b mid a^{12} = b, ab = ba rangle$. What are the possible depths of dead end elements with respect to this generating set?
Definition [from Bogopolski, Infinite commensurable hyperbolic groups
are bi-Lipschitz equivalent]: An element $g$ of a group $G$ is called dead in $G$ with respect to the finite generating set $X$ if $|gx| leq |g|$ $forall x in X^{pm 1}$. Here $|g|$ denotes the length of $g$ in the word metric with respect to $X$.
Intuitively, think of a dead end element as a dead end in a maze: no matter which direction you go, you'll be closer (or at least not farther away) from the exit (exit being the analogy for the neutral element here).
I have two questions:
Why does this presentation generate $mathbb{Z}$? If you take $a = 2$ and $b = 24$, I think it does not generate $mathbb{Z}$, since if you start from an odd number, you will only get odd numbers using $a$ and $b$.
Assuming $a = 1$ and $b = 12$, I think there is no dead end element: assume $g in mathbb{Z}$. Without loss of generality, we can assume $2 leq g leq 11$. Suppose that $|g| = n$. Then $|gb| = n + 1$ because the shortest path to go to $gb$ is to go from $0$ to $b$, and then use the same path from $0$ to $g$, but shifted by $b$. More formally, $g$ is of the form $g = a^i$, $2 leq i leq 6$ or $g = ba^{-j}$, $1 leq j leq 5$. Then $gb = ba^i$ or $g = b^2a^{-j}$, and both can't be reduced. So my question: is is possible that the exercise is actually impossible?
group-theory geometric-group-theory group-presentation finitely-generated
edited Dec 28 '18 at 5:45
Shaun
8,932113681
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
$endgroup$
2
$begingroup$
Answering (1) the group, denote it by $G$, is $mathbb{Z}$ because it is generated by $a$ and it is not finite. As for (2) what is a dead end element?
$endgroup$
– freakish
Dec 5 '18 at 13:41
$begingroup$
@freakish I see, but why is it not required to have an isomorphism? I'll edit the question to add the definition of a dead end element.
$endgroup$
– Laurent Hayez
Dec 5 '18 at 13:45
1
$begingroup$
I've changed the argument to make it simplier. Anyway an epimorphism $f:mathbb{Z}to G$ is always an isomorphism if $f(1)$ is of infinite order.
$endgroup$
– freakish
Dec 5 '18 at 13:47
4
$begingroup$
I agree that this question is wrong. The group with this presentation is the cyclic group generated by $a$, and it has no dead end elements. Clearly the identity is not dead end, and positive powers of $a$ can be made longer by multiplying by $b$, whereas negative powers can be made longer by multiplying by $b^{-1}$.
$endgroup$
– Derek Holt
Dec 5 '18 at 16:17
1
$begingroup$
See math.stackexchange.com/a/2820476/35400 for an example of a finite generating subset of $mathbf{Z}$ for which the Cayley graph has a dead end.
$endgroup$
– YCor
Dec 5 '18 at 22:36
|
show 4 more comments
$begingroup$
This exercise is taken from the book "Office Hours with a Geometric Group Theorist" (Office Hour 15, exercise 8):
Exercise: Show that the group $mathbb{Z}$ has dead end elements with respect to the presentation $mathbb{Z} = langle a, b mid a^{12} = b, ab = ba rangle$. What are the possible depths of dead end elements with respect to this generating set?
Definition [from Bogopolski, Infinite commensurable hyperbolic groups
are bi-Lipschitz equivalent]: An element $g$ of a group $G$ is called dead in $G$ with respect to the finite generating set $X$ if $|gx| leq |g|$ $forall x in X^{pm 1}$. Here $|g|$ denotes the length of $g$ in the word metric with respect to $X$.
Intuitively, think of a dead end element as a dead end in a maze: no matter which direction you go, you'll be closer (or at least not farther away) from the exit (exit being the analogy for the neutral element here).
I have two questions:
Why does this presentation generate $mathbb{Z}$? If you take $a = 2$ and $b = 24$, I think it does not generate $mathbb{Z}$, since if you start from an odd number, you will only get odd numbers using $a$ and $b$.
Assuming $a = 1$ and $b = 12$, I think there is no dead end element: assume $g in mathbb{Z}$. Without loss of generality, we can assume $2 leq g leq 11$. Suppose that $|g| = n$. Then $|gb| = n + 1$ because the shortest path to go to $gb$ is to go from $0$ to $b$, and then use the same path from $0$ to $g$, but shifted by $b$. More formally, $g$ is of the form $g = a^i$, $2 leq i leq 6$ or $g = ba^{-j}$, $1 leq j leq 5$. Then $gb = ba^i$ or $g = b^2a^{-j}$, and both can't be reduced. So my question: is is possible that the exercise is actually impossible?
group-theory geometric-group-theory group-presentation finitely-generated
edited Dec 28 '18 at 5:45
Shaun
8,932113681
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
$endgroup$
This exercise is taken from the book "Office Hours with a Geometric Group Theorist" (Office Hour 15, exercise 8):
Exercise: Show that the group $mathbb{Z}$ has dead end elements with respect to the presentation $mathbb{Z} = langle a, b mid a^{12} = b, ab = ba rangle$. What are the possible depths of dead end elements with respect to this generating set?
Definition [from Bogopolski, Infinite commensurable hyperbolic groups
are bi-Lipschitz equivalent]: An element $g$ of a group $G$ is called dead in $G$ with respect to the finite generating set $X$ if $|gx| leq |g|$ $forall x in X^{pm 1}$. Here $|g|$ denotes the length of $g$ in the word metric with respect to $X$.
Intuitively, think of a dead end element as a dead end in a maze: no matter which direction you go, you'll be closer (or at least not farther away) from the exit (exit being the analogy for the neutral element here).
I have two questions:
Why does this presentation generate $mathbb{Z}$? If you take $a = 2$ and $b = 24$, I think it does not generate $mathbb{Z}$, since if you start from an odd number, you will only get odd numbers using $a$ and $b$.
Assuming $a = 1$ and $b = 12$, I think there is no dead end element: assume $g in mathbb{Z}$. Without loss of generality, we can assume $2 leq g leq 11$. Suppose that $|g| = n$. Then $|gb| = n + 1$ because the shortest path to go to $gb$ is to go from $0$ to $b$, and then use the same path from $0$ to $g$, but shifted by $b$. More formally, $g$ is of the form $g = a^i$, $2 leq i leq 6$ or $g = ba^{-j}$, $1 leq j leq 5$. Then $gb = ba^i$ or $g = b^2a^{-j}$, and both can't be reduced. So my question: is is possible that the exercise is actually impossible?
group-theory geometric-group-theory group-presentation finitely-generated
group-theory geometric-group-theory group-presentation finitely-generated
edited Dec 28 '18 at 5:45
Shaun
8,932113681
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
edited Dec 28 '18 at 5:45
Shaun
8,932113681
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
edited Dec 28 '18 at 5:45
Shaun
8,932113681
edited Dec 28 '18 at 5:45
Shaun
8,932113681
edited Dec 28 '18 at 5:45
Shaun
8,932113681
8,932113681
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
asked Dec 5 '18 at 13:32
Laurent HayezLaurent Hayez
635415
635415
2
$begingroup$
Answering (1) the group, denote it by $G$, is $mathbb{Z}$ because it is generated by $a$ and it is not finite. As for (2) what is a dead end element?
$endgroup$
– freakish
Dec 5 '18 at 13:41
$begingroup$
@freakish I see, but why is it not required to have an isomorphism? I'll edit the question to add the definition of a dead end element.
$endgroup$
– Laurent Hayez
Dec 5 '18 at 13:45
1
$begingroup$
I've changed the argument to make it simplier. Anyway an epimorphism $f:mathbb{Z}to G$ is always an isomorphism if $f(1)$ is of infinite order.
$endgroup$
– freakish
Dec 5 '18 at 13:47
4
$begingroup$
I agree that this question is wrong. The group with this presentation is the cyclic group generated by $a$, and it has no dead end elements. Clearly the identity is not dead end, and positive powers of $a$ can be made longer by multiplying by $b$, whereas negative powers can be made longer by multiplying by $b^{-1}$.
$endgroup$
– Derek Holt
Dec 5 '18 at 16:17
1
$begingroup$
See math.stackexchange.com/a/2820476/35400 for an example of a finite generating subset of $mathbf{Z}$ for which the Cayley graph has a dead end.
$endgroup$
– YCor
Dec 5 '18 at 22:36
|
show 4 more comments
2
$begingroup$
Answering (1) the group, denote it by $G$, is $mathbb{Z}$ because it is generated by $a$ and it is not finite. As for (2) what is a dead end element?
$endgroup$
– freakish
Dec 5 '18 at 13:41
$begingroup$
@freakish I see, but why is it not required to have an isomorphism? I'll edit the question to add the definition of a dead end element.
$endgroup$
– Laurent Hayez
Dec 5 '18 at 13:45
1
$begingroup$
I've changed the argument to make it simplier. Anyway an epimorphism $f:mathbb{Z}to G$ is always an isomorphism if $f(1)$ is of infinite order.
$endgroup$
– freakish
Dec 5 '18 at 13:47
4
$begingroup$
I agree that this question is wrong. The group with this presentation is the cyclic group generated by $a$, and it has no dead end elements. Clearly the identity is not dead end, and positive powers of $a$ can be made longer by multiplying by $b$, whereas negative powers can be made longer by multiplying by $b^{-1}$.
$endgroup$
– Derek Holt
Dec 5 '18 at 16:17
1
$begingroup$
See math.stackexchange.com/a/2820476/35400 for an example of a finite generating subset of $mathbf{Z}$ for which the Cayley graph has a dead end.
$endgroup$
– YCor
Dec 5 '18 at 22:36
2
2
$begingroup$
Answering (1) the group, denote it by $G$, is $mathbb{Z}$ because it is generated by $a$ and it is not finite. As for (2) what is a dead end element?
$endgroup$
– freakish
Dec 5 '18 at 13:41
$begingroup$
Answering (1) the group, denote it by $G$, is $mathbb{Z}$ because it is generated by $a$ and it is not finite. As for (2) what is a dead end element?
$endgroup$
– freakish
Dec 5 '18 at 13:41
$begingroup$
@freakish I see, but why is it not required to have an isomorphism? I'll edit the question to add the definition of a dead end element.
$endgroup$
– Laurent Hayez
Dec 5 '18 at 13:45
$begingroup$
@freakish I see, but why is it not required to have an isomorphism? I'll edit the question to add the definition of a dead end element.
$endgroup$
– Laurent Hayez
Dec 5 '18 at 13:45
1
1
$begingroup$
I've changed the argument to make it simplier. Anyway an epimorphism $f:mathbb{Z}to G$ is always an isomorphism if $f(1)$ is of infinite order.
$endgroup$
– freakish
Dec 5 '18 at 13:47
$begingroup$
I've changed the argument to make it simplier. Anyway an epimorphism $f:mathbb{Z}to G$ is always an isomorphism if $f(1)$ is of infinite order.
$endgroup$
– freakish
Dec 5 '18 at 13:47
4
4
$begingroup$
I agree that this question is wrong. The group with this presentation is the cyclic group generated by $a$, and it has no dead end elements. Clearly the identity is not dead end, and positive powers of $a$ can be made longer by multiplying by $b$, whereas negative powers can be made longer by multiplying by $b^{-1}$.
$endgroup$
– Derek Holt
Dec 5 '18 at 16:17
$begingroup$
I agree that this question is wrong. The group with this presentation is the cyclic group generated by $a$, and it has no dead end elements. Clearly the identity is not dead end, and positive powers of $a$ can be made longer by multiplying by $b$, whereas negative powers can be made longer by multiplying by $b^{-1}$.
$endgroup$
– Derek Holt
Dec 5 '18 at 16:17
1
1
$begingroup$
See math.stackexchange.com/a/2820476/35400 for an example of a finite generating subset of $mathbf{Z}$ for which the Cayley graph has a dead end.
$endgroup$
– YCor
Dec 5 '18 at 22:36
$begingroup$
See math.stackexchange.com/a/2820476/35400 for an example of a finite generating subset of $mathbf{Z}$ for which the Cayley graph has a dead end.
$endgroup$
– YCor
Dec 5 '18 at 22:36
|
show 4 more comments
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$begingroup$
Answering (1) the group, denote it by $G$, is $mathbb{Z}$ because it is generated by $a$ and it is not finite. As for (2) what is a dead end element?
$endgroup$
– freakish
Dec 5 '18 at 13:41
$begingroup$
@freakish I see, but why is it not required to have an isomorphism? I'll edit the question to add the definition of a dead end element.
$endgroup$
– Laurent Hayez
Dec 5 '18 at 13:45
1
$begingroup$
I've changed the argument to make it simplier. Anyway an epimorphism $f:mathbb{Z}to G$ is always an isomorphism if $f(1)$ is of infinite order.
$endgroup$
– freakish
Dec 5 '18 at 13:47
4
$begingroup$
I agree that this question is wrong. The group with this presentation is the cyclic group generated by $a$, and it has no dead end elements. Clearly the identity is not dead end, and positive powers of $a$ can be made longer by multiplying by $b$, whereas negative powers can be made longer by multiplying by $b^{-1}$.
$endgroup$
– Derek Holt
Dec 5 '18 at 16:17
1
$begingroup$
See math.stackexchange.com/a/2820476/35400 for an example of a finite generating subset of $mathbf{Z}$ for which the Cayley graph has a dead end.
$endgroup$
– YCor
Dec 5 '18 at 22:36