Difference between the product group, the direct sum and the free group












2












$begingroup$


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







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$endgroup$

















    2












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Jan 5 at 16:17









      gladimetcampbellsgladimetcampbells

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          1 Answer
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          $begingroup$

          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
          $$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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 $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
            $endgroup$
            – gladimetcampbells
            Jan 5 at 16:35












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          active

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          2












          $begingroup$

          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
          $$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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 $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
            $endgroup$
            – gladimetcampbells
            Jan 5 at 16:35
















          2












          $begingroup$

          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
          $$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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 $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
            $endgroup$
            – gladimetcampbells
            Jan 5 at 16:35














          2












          2








          2





          $begingroup$

          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
          $$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$






          share|cite|improve this answer









          $endgroup$



          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
          $$ (g,1_H)cdot(1_G,h)=(gcdot_G1_G,1_Hcdot_H h)=(1_Gcdot_Gg,hcdot_H 1_G)=(1_G,h)cdot(g,1_H)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 16:23









          Hagen von EitzenHagen von Eitzen

          283k23273508




          283k23273508












          • $begingroup$
            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 $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
            $endgroup$
            – gladimetcampbells
            Jan 5 at 16:35


















          • $begingroup$
            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 $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
            $endgroup$
            – gladimetcampbells
            Jan 5 at 16:35
















          $begingroup$
          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 $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
          $endgroup$
          – gladimetcampbells
          Jan 5 at 16:35




          $begingroup$
          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 $(g,1_G)dot(1_H, h)≠(1_H, h)dot(g,1_G)$?
          $endgroup$
          – gladimetcampbells
          Jan 5 at 16:35


















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