Covers of the wedge of three circles
I'm trying to come up with covering spaces of $X= S^1 lor S^1 lor S^1$ corresponding to two subgroups of $pi_1(X)= F_3=langle a,b,c rangle$. First, the subgroup $H=langle a^2, b^2, c^2 rangle$ and second, the normal subgroup $N= langle langle a^2, b^2, c^2 rangle rangle$ (the smallest normal subgroup containing $H$).
I've got some candidate covers in mind, but I'm not sure if I'm correct. Below is my idea for the cover corresponding to $N$.
This cover shown is normal, since the map swapping the vertices is a covering space transformation, and so the subgroups corresponding to this cover is normal in $F_3$, and it contains $H$.
If this is the correct cover, how can I show this corresponds to the smallest normal subgroup containing $H$? What about the cover corresponding to $H$ itself? No finite degree cover has worked, and I can't seem to cook up an infinite cover that works.
Any hints would be appreciated!
algebraic-topology
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
add a comment |
I'm trying to come up with covering spaces of $X= S^1 lor S^1 lor S^1$ corresponding to two subgroups of $pi_1(X)= F_3=langle a,b,c rangle$. First, the subgroup $H=langle a^2, b^2, c^2 rangle$ and second, the normal subgroup $N= langle langle a^2, b^2, c^2 rangle rangle$ (the smallest normal subgroup containing $H$).
I've got some candidate covers in mind, but I'm not sure if I'm correct. Below is my idea for the cover corresponding to $N$.
This cover shown is normal, since the map swapping the vertices is a covering space transformation, and so the subgroups corresponding to this cover is normal in $F_3$, and it contains $H$.
If this is the correct cover, how can I show this corresponds to the smallest normal subgroup containing $H$? What about the cover corresponding to $H$ itself? No finite degree cover has worked, and I can't seem to cook up an infinite cover that works.
Any hints would be appreciated!
algebraic-topology
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
1
The number of vertices is equal to the number of cosets of the subgroup. In both cases, the index of your subgroup is infinite, so the corresponding cover has infinitely many vertices.
– Hempelicious
2 days ago
add a comment |
I'm trying to come up with covering spaces of $X= S^1 lor S^1 lor S^1$ corresponding to two subgroups of $pi_1(X)= F_3=langle a,b,c rangle$. First, the subgroup $H=langle a^2, b^2, c^2 rangle$ and second, the normal subgroup $N= langle langle a^2, b^2, c^2 rangle rangle$ (the smallest normal subgroup containing $H$).
I've got some candidate covers in mind, but I'm not sure if I'm correct. Below is my idea for the cover corresponding to $N$.
This cover shown is normal, since the map swapping the vertices is a covering space transformation, and so the subgroups corresponding to this cover is normal in $F_3$, and it contains $H$.
If this is the correct cover, how can I show this corresponds to the smallest normal subgroup containing $H$? What about the cover corresponding to $H$ itself? No finite degree cover has worked, and I can't seem to cook up an infinite cover that works.
Any hints would be appreciated!
algebraic-topology
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
I'm trying to come up with covering spaces of $X= S^1 lor S^1 lor S^1$ corresponding to two subgroups of $pi_1(X)= F_3=langle a,b,c rangle$. First, the subgroup $H=langle a^2, b^2, c^2 rangle$ and second, the normal subgroup $N= langle langle a^2, b^2, c^2 rangle rangle$ (the smallest normal subgroup containing $H$).
I've got some candidate covers in mind, but I'm not sure if I'm correct. Below is my idea for the cover corresponding to $N$.
This cover shown is normal, since the map swapping the vertices is a covering space transformation, and so the subgroups corresponding to this cover is normal in $F_3$, and it contains $H$.
If this is the correct cover, how can I show this corresponds to the smallest normal subgroup containing $H$? What about the cover corresponding to $H$ itself? No finite degree cover has worked, and I can't seem to cook up an infinite cover that works.
Any hints would be appreciated!
algebraic-topology
algebraic-topology
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
asked Jan 5 at 21:50
languidlobsterlanguidlobster
1218
1218
1
The number of vertices is equal to the number of cosets of the subgroup. In both cases, the index of your subgroup is infinite, so the corresponding cover has infinitely many vertices.
– Hempelicious
2 days ago
add a comment |
1
The number of vertices is equal to the number of cosets of the subgroup. In both cases, the index of your subgroup is infinite, so the corresponding cover has infinitely many vertices.
– Hempelicious
2 days ago
1
1
The number of vertices is equal to the number of cosets of the subgroup. In both cases, the index of your subgroup is infinite, so the corresponding cover has infinitely many vertices.
– Hempelicious
2 days ago
The number of vertices is equal to the number of cosets of the subgroup. In both cases, the index of your subgroup is infinite, so the corresponding cover has infinitely many vertices.
– Hempelicious
2 days ago
add a comment |
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The number of vertices is equal to the number of cosets of the subgroup. In both cases, the index of your subgroup is infinite, so the corresponding cover has infinitely many vertices.
– Hempelicious
2 days ago