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Just fundamental group of a graph of groups vs. fundamental group of a graph of groups with respect to a maximal subtree
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On page 87 of the book " Introductory to group theory by Bogopolski", they gave an example of the fundamental group of a graph of groups with respect to a maximal subtree, which I understood well. However, the problem below asks to find the fundamental group of a graph of groups without any mention to a maximal subtree, but I tried to solve it anyway. What changes about finding it without a maximal subtree? Is my way the right one? Thanks for your help!
Question: Find a presentation for the fundamental group of the graph of groups $(mathbbG, Y)$ defined as follows. The graph $Y$ is a triangle with 3 vertices $v_1, v_2, v_3$ and 6 edges $e_j$, $1le jle 6$ with $e_4=overlinee_1, e_5=overlinee_2 $ and $e_6=overlinee_3$; the initial vertices are $alpha(e_1)=v_1, alpha(e_2)=v_2, alpha(e_3)=v_3$ and the terminal vertices are $tau(e_1)=v_2, tau(e_2)=v_3, tau(e_3)=v_1$. Let $l,m,nge 2$ be integers. The vertex groups and edge groups are $$G_e_1=G_overlinee_1=mathbbZ_l=< a| a^l >,$$
$$G_e_2=G_overlinee_2=mathbbZ_m=<b|b^m>,$$
$$G_e_3=G_overlinee_3=mathbbZ_n=<c|c^n>,$$
$$G_v_1=mathbbZ_noplusmathbbZ_l=<c,a|cac^-1a^-1, c^n, a^l>,$$
$$G_v_2=mathbbZ_loplus mathbbZ_m=<a,b|bab^-1a^-1, a^l, b^m>,$$
$$G_v_3=mathbbz_moplusmathbbZ_n=<b,c|bcb^-1c^-1, b^m, c^n>.$$
The monomorphisms $alpha_e: G_eto G_alpha(e)$ are the obvious inclusions. For instance, $$alpha_e_1:mathbbZ_l=G_e_1to
G_alpha (e_1)=G_v_1=mathbbZ_noplusmathbbZ_l$$ is given by $alpha_e_1(x)=(1,x)$, where $1$ denotes the identity element of $mathbbZ_n$.
What I have tried:
$$pi_1(mathbbG,Y)=langle a,b,c,t_e_1, t_e_2, t_e_3|ca^-1ca^-1, c^n, a^l, bab^-1a^-1, b^m, bcb^-1c^-1, t^-1_e_1(1,a)t_e_1=(a,1), t^-1_e_2(1,b)t_e_2=(b,1), t^-1_e_3(1,c)t_e_3=(c,1) rangle$$
geometric-group-theory
asked Aug 1 at 15:17
Djinola
647
add a comment |Â
up vote
0
down vote
favorite
On page 87 of the book " Introductory to group theory by Bogopolski", they gave an example of the fundamental group of a graph of groups with respect to a maximal subtree, which I understood well. However, the problem below asks to find the fundamental group of a graph of groups without any mention to a maximal subtree, but I tried to solve it anyway. What changes about finding it without a maximal subtree? Is my way the right one? Thanks for your help!
Question: Find a presentation for the fundamental group of the graph of groups $(mathbbG, Y)$ defined as follows. The graph $Y$ is a triangle with 3 vertices $v_1, v_2, v_3$ and 6 edges $e_j$, $1le jle 6$ with $e_4=overlinee_1, e_5=overlinee_2 $ and $e_6=overlinee_3$; the initial vertices are $alpha(e_1)=v_1, alpha(e_2)=v_2, alpha(e_3)=v_3$ and the terminal vertices are $tau(e_1)=v_2, tau(e_2)=v_3, tau(e_3)=v_1$. Let $l,m,nge 2$ be integers. The vertex groups and edge groups are $$G_e_1=G_overlinee_1=mathbbZ_l=< a| a^l >,$$
$$G_e_2=G_overlinee_2=mathbbZ_m=<b|b^m>,$$
$$G_e_3=G_overlinee_3=mathbbZ_n=<c|c^n>,$$
$$G_v_1=mathbbZ_noplusmathbbZ_l=<c,a|cac^-1a^-1, c^n, a^l>,$$
$$G_v_2=mathbbZ_loplus mathbbZ_m=<a,b|bab^-1a^-1, a^l, b^m>,$$
$$G_v_3=mathbbz_moplusmathbbZ_n=<b,c|bcb^-1c^-1, b^m, c^n>.$$
The monomorphisms $alpha_e: G_eto G_alpha(e)$ are the obvious inclusions. For instance, $$alpha_e_1:mathbbZ_l=G_e_1to
G_alpha (e_1)=G_v_1=mathbbZ_noplusmathbbZ_l$$ is given by $alpha_e_1(x)=(1,x)$, where $1$ denotes the identity element of $mathbbZ_n$.
What I have tried:
$$pi_1(mathbbG,Y)=langle a,b,c,t_e_1, t_e_2, t_e_3|ca^-1ca^-1, c^n, a^l, bab^-1a^-1, b^m, bcb^-1c^-1, t^-1_e_1(1,a)t_e_1=(a,1), t^-1_e_2(1,b)t_e_2=(b,1), t^-1_e_3(1,c)t_e_3=(c,1) rangle$$
geometric-group-theory
asked Aug 1 at 15:17
Djinola
647
The way to do it without being given a maximal tree is... first compute a maximal tree.
– Lee Mosher
Aug 2 at 3:28
@Lee Mosher, how to compute a maximal tree in this particular case? I have been learning the subject.
– Djinola
Aug 2 at 3:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
On page 87 of the book " Introductory to group theory by Bogopolski", they gave an example of the fundamental group of a graph of groups with respect to a maximal subtree, which I understood well. However, the problem below asks to find the fundamental group of a graph of groups without any mention to a maximal subtree, but I tried to solve it anyway. What changes about finding it without a maximal subtree? Is my way the right one? Thanks for your help!
Question: Find a presentation for the fundamental group of the graph of groups $(mathbbG, Y)$ defined as follows. The graph $Y$ is a triangle with 3 vertices $v_1, v_2, v_3$ and 6 edges $e_j$, $1le jle 6$ with $e_4=overlinee_1, e_5=overlinee_2 $ and $e_6=overlinee_3$; the initial vertices are $alpha(e_1)=v_1, alpha(e_2)=v_2, alpha(e_3)=v_3$ and the terminal vertices are $tau(e_1)=v_2, tau(e_2)=v_3, tau(e_3)=v_1$. Let $l,m,nge 2$ be integers. The vertex groups and edge groups are $$G_e_1=G_overlinee_1=mathbbZ_l=< a| a^l >,$$
$$G_e_2=G_overlinee_2=mathbbZ_m=<b|b^m>,$$
$$G_e_3=G_overlinee_3=mathbbZ_n=<c|c^n>,$$
$$G_v_1=mathbbZ_noplusmathbbZ_l=<c,a|cac^-1a^-1, c^n, a^l>,$$
$$G_v_2=mathbbZ_loplus mathbbZ_m=<a,b|bab^-1a^-1, a^l, b^m>,$$
$$G_v_3=mathbbz_moplusmathbbZ_n=<b,c|bcb^-1c^-1, b^m, c^n>.$$
The monomorphisms $alpha_e: G_eto G_alpha(e)$ are the obvious inclusions. For instance, $$alpha_e_1:mathbbZ_l=G_e_1to
G_alpha (e_1)=G_v_1=mathbbZ_noplusmathbbZ_l$$ is given by $alpha_e_1(x)=(1,x)$, where $1$ denotes the identity element of $mathbbZ_n$.
What I have tried:
$$pi_1(mathbbG,Y)=langle a,b,c,t_e_1, t_e_2, t_e_3|ca^-1ca^-1, c^n, a^l, bab^-1a^-1, b^m, bcb^-1c^-1, t^-1_e_1(1,a)t_e_1=(a,1), t^-1_e_2(1,b)t_e_2=(b,1), t^-1_e_3(1,c)t_e_3=(c,1) rangle$$
geometric-group-theory
asked Aug 1 at 15:17
Djinola
647
On page 87 of the book " Introductory to group theory by Bogopolski", they gave an example of the fundamental group of a graph of groups with respect to a maximal subtree, which I understood well. However, the problem below asks to find the fundamental group of a graph of groups without any mention to a maximal subtree, but I tried to solve it anyway. What changes about finding it without a maximal subtree? Is my way the right one? Thanks for your help!
Question: Find a presentation for the fundamental group of the graph of groups $(mathbbG, Y)$ defined as follows. The graph $Y$ is a triangle with 3 vertices $v_1, v_2, v_3$ and 6 edges $e_j$, $1le jle 6$ with $e_4=overlinee_1, e_5=overlinee_2 $ and $e_6=overlinee_3$; the initial vertices are $alpha(e_1)=v_1, alpha(e_2)=v_2, alpha(e_3)=v_3$ and the terminal vertices are $tau(e_1)=v_2, tau(e_2)=v_3, tau(e_3)=v_1$. Let $l,m,nge 2$ be integers. The vertex groups and edge groups are $$G_e_1=G_overlinee_1=mathbbZ_l=< a| a^l >,$$
$$G_e_2=G_overlinee_2=mathbbZ_m=<b|b^m>,$$
$$G_e_3=G_overlinee_3=mathbbZ_n=<c|c^n>,$$
$$G_v_1=mathbbZ_noplusmathbbZ_l=<c,a|cac^-1a^-1, c^n, a^l>,$$
$$G_v_2=mathbbZ_loplus mathbbZ_m=<a,b|bab^-1a^-1, a^l, b^m>,$$
$$G_v_3=mathbbz_moplusmathbbZ_n=<b,c|bcb^-1c^-1, b^m, c^n>.$$
The monomorphisms $alpha_e: G_eto G_alpha(e)$ are the obvious inclusions. For instance, $$alpha_e_1:mathbbZ_l=G_e_1to
G_alpha (e_1)=G_v_1=mathbbZ_noplusmathbbZ_l$$ is given by $alpha_e_1(x)=(1,x)$, where $1$ denotes the identity element of $mathbbZ_n$.
What I have tried:
$$pi_1(mathbbG,Y)=langle a,b,c,t_e_1, t_e_2, t_e_3|ca^-1ca^-1, c^n, a^l, bab^-1a^-1, b^m, bcb^-1c^-1, t^-1_e_1(1,a)t_e_1=(a,1), t^-1_e_2(1,b)t_e_2=(b,1), t^-1_e_3(1,c)t_e_3=(c,1) rangle$$
geometric-group-theory
asked Aug 1 at 15:17
Djinola
647
asked Aug 1 at 15:17
Djinola
647
asked Aug 1 at 15:17
Djinola
647
asked Aug 1 at 15:17
Djinola
647
647
The way to do it without being given a maximal tree is... first compute a maximal tree.
– Lee Mosher
Aug 2 at 3:28
@Lee Mosher, how to compute a maximal tree in this particular case? I have been learning the subject.
– Djinola
Aug 2 at 3:52
add a comment |Â
The way to do it without being given a maximal tree is... first compute a maximal tree.
– Lee Mosher
Aug 2 at 3:28
@Lee Mosher, how to compute a maximal tree in this particular case? I have been learning the subject.
– Djinola
Aug 2 at 3:52
The way to do it without being given a maximal tree is... first compute a maximal tree.
– Lee Mosher
Aug 2 at 3:28
The way to do it without being given a maximal tree is... first compute a maximal tree.
– Lee Mosher
Aug 2 at 3:28
@Lee Mosher, how to compute a maximal tree in this particular case? I have been learning the subject.
– Djinola
Aug 2 at 3:52
@Lee Mosher, how to compute a maximal tree in this particular case? I have been learning the subject.
– Djinola
Aug 2 at 3:52
add a comment |Â
1 Answer
1
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oldest
votes
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0
down vote
accepted
If you are not given a maximal tree, then you have to compute one for yourself.
You can compute maximal trees, also called "spanning trees", using the depth first search algorithm. I'll carry out the algorithm on your example, but, I strongly advise you to draw a picture of the graph in order to learn to see maximal trees.
A maximal tree $T$ in a connected graph $G$ will contain each vertex. Roughly speaking, the depth first search algorithm builds $T$ inductively, starting with one vertex, and then adding one edge at a time as long as the new edge connects one vertex already in the graph with another vertex not yet in the graph. Since $G$ is connected, the algorithm is guaranteed to stop after a number of steps equal to the number of vertices.
We'll build $T$ inductively, starting with
- Step 1: $T_1 = v_1$
Next, we'll ask: whats the first edge that connects $T_1$ to one of the other vertices $v_2,v_3$? The answer is $e_1$, because $alpha(e_1)=v_1$ and $tau(e_1)=v_2$. So, we not throw in $v_2$ and $e_1$, and so
- Step 2: $T_2 = v_1 cup e_1 cup v_2$
Next, we'll ask: what's the first edge that connects $T_2$ to the remaining vertex $v_3$? The answer is $e_2$, because $alpha(e_2)=v_2$ and $tau(e_3)=v_3$, and so
- Step 3: $T = T_3 = v_1 cup e_1 cup v_2 cup e_2 cup v_3$
and the algorithm is done.
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
I have understood the algorithm! So, the maximal subtree contains all vertices and edges of the graph Y except the edge $e_3$.
– Djinola
Aug 2 at 14:37
1
The maximal subtree I constructed is like that, yet. However, note that this is not the only maximal subtree. If you changed the indexing of the vertices and edges, the algorithm might well produce a different maximal subtree.
– Lee Mosher
Aug 2 at 18:13
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If you are not given a maximal tree, then you have to compute one for yourself.
You can compute maximal trees, also called "spanning trees", using the depth first search algorithm. I'll carry out the algorithm on your example, but, I strongly advise you to draw a picture of the graph in order to learn to see maximal trees.
A maximal tree $T$ in a connected graph $G$ will contain each vertex. Roughly speaking, the depth first search algorithm builds $T$ inductively, starting with one vertex, and then adding one edge at a time as long as the new edge connects one vertex already in the graph with another vertex not yet in the graph. Since $G$ is connected, the algorithm is guaranteed to stop after a number of steps equal to the number of vertices.
We'll build $T$ inductively, starting with
- Step 1: $T_1 = v_1$
Next, we'll ask: whats the first edge that connects $T_1$ to one of the other vertices $v_2,v_3$? The answer is $e_1$, because $alpha(e_1)=v_1$ and $tau(e_1)=v_2$. So, we not throw in $v_2$ and $e_1$, and so
- Step 2: $T_2 = v_1 cup e_1 cup v_2$
Next, we'll ask: what's the first edge that connects $T_2$ to the remaining vertex $v_3$? The answer is $e_2$, because $alpha(e_2)=v_2$ and $tau(e_3)=v_3$, and so
- Step 3: $T = T_3 = v_1 cup e_1 cup v_2 cup e_2 cup v_3$
and the algorithm is done.
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
I have understood the algorithm! So, the maximal subtree contains all vertices and edges of the graph Y except the edge $e_3$.
– Djinola
Aug 2 at 14:37
1
The maximal subtree I constructed is like that, yet. However, note that this is not the only maximal subtree. If you changed the indexing of the vertices and edges, the algorithm might well produce a different maximal subtree.
– Lee Mosher
Aug 2 at 18:13
add a comment |Â
up vote
0
down vote
accepted
If you are not given a maximal tree, then you have to compute one for yourself.
You can compute maximal trees, also called "spanning trees", using the depth first search algorithm. I'll carry out the algorithm on your example, but, I strongly advise you to draw a picture of the graph in order to learn to see maximal trees.
A maximal tree $T$ in a connected graph $G$ will contain each vertex. Roughly speaking, the depth first search algorithm builds $T$ inductively, starting with one vertex, and then adding one edge at a time as long as the new edge connects one vertex already in the graph with another vertex not yet in the graph. Since $G$ is connected, the algorithm is guaranteed to stop after a number of steps equal to the number of vertices.
We'll build $T$ inductively, starting with
- Step 1: $T_1 = v_1$
Next, we'll ask: whats the first edge that connects $T_1$ to one of the other vertices $v_2,v_3$? The answer is $e_1$, because $alpha(e_1)=v_1$ and $tau(e_1)=v_2$. So, we not throw in $v_2$ and $e_1$, and so
- Step 2: $T_2 = v_1 cup e_1 cup v_2$
Next, we'll ask: what's the first edge that connects $T_2$ to the remaining vertex $v_3$? The answer is $e_2$, because $alpha(e_2)=v_2$ and $tau(e_3)=v_3$, and so
- Step 3: $T = T_3 = v_1 cup e_1 cup v_2 cup e_2 cup v_3$
and the algorithm is done.
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
I have understood the algorithm! So, the maximal subtree contains all vertices and edges of the graph Y except the edge $e_3$.
– Djinola
Aug 2 at 14:37
1
The maximal subtree I constructed is like that, yet. However, note that this is not the only maximal subtree. If you changed the indexing of the vertices and edges, the algorithm might well produce a different maximal subtree.
– Lee Mosher
Aug 2 at 18:13
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If you are not given a maximal tree, then you have to compute one for yourself.
You can compute maximal trees, also called "spanning trees", using the depth first search algorithm. I'll carry out the algorithm on your example, but, I strongly advise you to draw a picture of the graph in order to learn to see maximal trees.
A maximal tree $T$ in a connected graph $G$ will contain each vertex. Roughly speaking, the depth first search algorithm builds $T$ inductively, starting with one vertex, and then adding one edge at a time as long as the new edge connects one vertex already in the graph with another vertex not yet in the graph. Since $G$ is connected, the algorithm is guaranteed to stop after a number of steps equal to the number of vertices.
We'll build $T$ inductively, starting with
- Step 1: $T_1 = v_1$
Next, we'll ask: whats the first edge that connects $T_1$ to one of the other vertices $v_2,v_3$? The answer is $e_1$, because $alpha(e_1)=v_1$ and $tau(e_1)=v_2$. So, we not throw in $v_2$ and $e_1$, and so
- Step 2: $T_2 = v_1 cup e_1 cup v_2$
Next, we'll ask: what's the first edge that connects $T_2$ to the remaining vertex $v_3$? The answer is $e_2$, because $alpha(e_2)=v_2$ and $tau(e_3)=v_3$, and so
- Step 3: $T = T_3 = v_1 cup e_1 cup v_2 cup e_2 cup v_3$
and the algorithm is done.
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
If you are not given a maximal tree, then you have to compute one for yourself.
You can compute maximal trees, also called "spanning trees", using the depth first search algorithm. I'll carry out the algorithm on your example, but, I strongly advise you to draw a picture of the graph in order to learn to see maximal trees.
A maximal tree $T$ in a connected graph $G$ will contain each vertex. Roughly speaking, the depth first search algorithm builds $T$ inductively, starting with one vertex, and then adding one edge at a time as long as the new edge connects one vertex already in the graph with another vertex not yet in the graph. Since $G$ is connected, the algorithm is guaranteed to stop after a number of steps equal to the number of vertices.
We'll build $T$ inductively, starting with
- Step 1: $T_1 = v_1$
Next, we'll ask: whats the first edge that connects $T_1$ to one of the other vertices $v_2,v_3$? The answer is $e_1$, because $alpha(e_1)=v_1$ and $tau(e_1)=v_2$. So, we not throw in $v_2$ and $e_1$, and so
- Step 2: $T_2 = v_1 cup e_1 cup v_2$
Next, we'll ask: what's the first edge that connects $T_2$ to the remaining vertex $v_3$? The answer is $e_2$, because $alpha(e_2)=v_2$ and $tau(e_3)=v_3$, and so
- Step 3: $T = T_3 = v_1 cup e_1 cup v_2 cup e_2 cup v_3$
and the algorithm is done.
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
answered Aug 2 at 13:03
Lee Mosher
45.3k33478
45.3k33478
I have understood the algorithm! So, the maximal subtree contains all vertices and edges of the graph Y except the edge $e_3$.
– Djinola
Aug 2 at 14:37
1
The maximal subtree I constructed is like that, yet. However, note that this is not the only maximal subtree. If you changed the indexing of the vertices and edges, the algorithm might well produce a different maximal subtree.
– Lee Mosher
Aug 2 at 18:13
add a comment |Â
I have understood the algorithm! So, the maximal subtree contains all vertices and edges of the graph Y except the edge $e_3$.
– Djinola
Aug 2 at 14:37
1
The maximal subtree I constructed is like that, yet. However, note that this is not the only maximal subtree. If you changed the indexing of the vertices and edges, the algorithm might well produce a different maximal subtree.
– Lee Mosher
Aug 2 at 18:13
I have understood the algorithm! So, the maximal subtree contains all vertices and edges of the graph Y except the edge $e_3$.
– Djinola
Aug 2 at 14:37
I have understood the algorithm! So, the maximal subtree contains all vertices and edges of the graph Y except the edge $e_3$.
– Djinola
Aug 2 at 14:37
1
1
The maximal subtree I constructed is like that, yet. However, note that this is not the only maximal subtree. If you changed the indexing of the vertices and edges, the algorithm might well produce a different maximal subtree.
– Lee Mosher
Aug 2 at 18:13
The maximal subtree I constructed is like that, yet. However, note that this is not the only maximal subtree. If you changed the indexing of the vertices and edges, the algorithm might well produce a different maximal subtree.
– Lee Mosher
Aug 2 at 18:13
add a comment |Â
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The way to do it without being given a maximal tree is... first compute a maximal tree.
– Lee Mosher
Aug 2 at 3:28
@Lee Mosher, how to compute a maximal tree in this particular case? I have been learning the subject.
– Djinola
Aug 2 at 3:52