Difference between the product group, the direct sum and the free group
The following is from Hatcher's Algebraic Topology:
Suppose one is given a collection of groups $G_{alpha}$ and one wishes to construct a single group containing all these groups as subgroups. One way to do this would be to take the product group $prod_{alpha}G_{alpha}$, whose elements can be regarded as the functions $alphamapsto g_{alpha}in G_{alpha}$. Or one could restrict to functions taking on nonidentity values at most finitely often, forming the direct sum $bigoplus_{alpha}G_{alpha}$. Both these constructions produce groups containing all the $G_{alpha}$'s as subgroups, but with the property that elements of different subgroups $G_{alpha}$ commute with each other. In the realm of nonabelian groups this commutativity is unnatural, and so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$. Since the sum $bigoplus_{alpha}G_{alpha}$ is smaller and presumably simpler than $prod_{alpha}G_{alpha}$, it sould be easier to construct a nonabelian version of $bigoplus_{alpha}G_{alpha}$, and this is what the free product achieves.
Can someone explain what he means by "but with the property that elements of different subgroups $G_{alpha}$ commute with each other."?
algebraic-topology
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
add a comment |
The following is from Hatcher's Algebraic Topology:
Suppose one is given a collection of groups $G_{alpha}$ and one wishes to construct a single group containing all these groups as subgroups. One way to do this would be to take the product group $prod_{alpha}G_{alpha}$, whose elements can be regarded as the functions $alphamapsto g_{alpha}in G_{alpha}$. Or one could restrict to functions taking on nonidentity values at most finitely often, forming the direct sum $bigoplus_{alpha}G_{alpha}$. Both these constructions produce groups containing all the $G_{alpha}$'s as subgroups, but with the property that elements of different subgroups $G_{alpha}$ commute with each other. In the realm of nonabelian groups this commutativity is unnatural, and so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$. Since the sum $bigoplus_{alpha}G_{alpha}$ is smaller and presumably simpler than $prod_{alpha}G_{alpha}$, it sould be easier to construct a nonabelian version of $bigoplus_{alpha}G_{alpha}$, and this is what the free product achieves.
Can someone explain what he means by "but with the property that elements of different subgroups $G_{alpha}$ commute with each other."?
algebraic-topology
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
add a comment |
The following is from Hatcher's Algebraic Topology:
Suppose one is given a collection of groups $G_{alpha}$ and one wishes to construct a single group containing all these groups as subgroups. One way to do this would be to take the product group $prod_{alpha}G_{alpha}$, whose elements can be regarded as the functions $alphamapsto g_{alpha}in G_{alpha}$. Or one could restrict to functions taking on nonidentity values at most finitely often, forming the direct sum $bigoplus_{alpha}G_{alpha}$. Both these constructions produce groups containing all the $G_{alpha}$'s as subgroups, but with the property that elements of different subgroups $G_{alpha}$ commute with each other. In the realm of nonabelian groups this commutativity is unnatural, and so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$. Since the sum $bigoplus_{alpha}G_{alpha}$ is smaller and presumably simpler than $prod_{alpha}G_{alpha}$, it sould be easier to construct a nonabelian version of $bigoplus_{alpha}G_{alpha}$, and this is what the free product achieves.
Can someone explain what he means by "but with the property that elements of different subgroups $G_{alpha}$ commute with each other."?
algebraic-topology
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
The following is from Hatcher's Algebraic Topology:
Suppose one is given a collection of groups $G_{alpha}$ and one wishes to construct a single group containing all these groups as subgroups. One way to do this would be to take the product group $prod_{alpha}G_{alpha}$, whose elements can be regarded as the functions $alphamapsto g_{alpha}in G_{alpha}$. Or one could restrict to functions taking on nonidentity values at most finitely often, forming the direct sum $bigoplus_{alpha}G_{alpha}$. Both these constructions produce groups containing all the $G_{alpha}$'s as subgroups, but with the property that elements of different subgroups $G_{alpha}$ commute with each other. In the realm of nonabelian groups this commutativity is unnatural, and so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$. Since the sum $bigoplus_{alpha}G_{alpha}$ is smaller and presumably simpler than $prod_{alpha}G_{alpha}$, it sould be easier to construct a nonabelian version of $bigoplus_{alpha}G_{alpha}$, and this is what the free product achieves.
Can someone explain what he means by "but with the property that elements of different subgroups $G_{alpha}$ commute with each other."?
algebraic-topology
algebraic-topology
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
asked 2 days ago
gladimetcampbellsgladimetcampbells
38211
38211
add a comment |
add a comment |
1 Answer
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If $gin G$ and $hin H$, then these correspond to $(g,1_H)$ and $(1_G,h)$ in the direct product or sum $Goplus H$. We have
$$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
Thank you. I see what he means now. By "so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$", he does not mean these two groups are abelian for nonabelian $G_{alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
– gladimetcampbells
2 days ago
add a comment |
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1 Answer
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1 Answer
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If $gin G$ and $hin H$, then these correspond to $(g,1_H)$ and $(1_G,h)$ in the direct product or sum $Goplus H$. We have
$$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
Thank you. I see what he means now. By "so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$", he does not mean these two groups are abelian for nonabelian $G_{alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
– gladimetcampbells
2 days ago
add a comment |
If $gin G$ and $hin H$, then these correspond to $(g,1_H)$ and $(1_G,h)$ in the direct product or sum $Goplus H$. We have
$$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
Thank you. I see what he means now. By "so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$", he does not mean these two groups are abelian for nonabelian $G_{alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
– gladimetcampbells
2 days ago
add a comment |
If $gin G$ and $hin H$, then these correspond to $(g,1_H)$ and $(1_G,h)$ in the direct product or sum $Goplus H$. We have
$$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
If $gin G$ and $hin H$, then these correspond to $(g,1_H)$ and $(1_G,h)$ in the direct product or sum $Goplus H$. We have
$$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
276k21269496
Thank you. I see what he means now. By "so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$", he does not mean these two groups are abelian for nonabelian $G_{alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
– gladimetcampbells
2 days ago
add a comment |
Thank you. I see what he means now. By "so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$", he does not mean these two groups are abelian for nonabelian $G_{alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
– gladimetcampbells
2 days ago
Thank you. I see what he means now. By "so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$", he does not mean these two groups are abelian for nonabelian $G_{alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
– gladimetcampbells
2 days ago
Thank you. I see what he means now. By "so one would like a nonabelian version of $prod_{alpha}G_{alpha}$ or $bigoplus_{alpha}G_{alpha}$", he does not mean these two groups are abelian for nonabelian $G_{alpha}$'s, correct? Does this mean when $G$ and $H$ are nonabelian, the commutativity of elements $(g,1_G)$ and $(1_H, h)$ are unnatural, so we would like something else in which $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
– gladimetcampbells
2 days ago
add a comment |
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