Subgroups of direct products of free groups












1














I am reading the following paper of Miller: http://researchers.ms.unimelb.edu.au/



He says that if $G= F_{1} times F_{2}$ is a direct product of two free groups and $H$ is a subgroup of $G$, then it can be assumed that the projection maps $$text{p}_{i} colon H rightarrow F_{i}$$ are surjective.



I do not understand why is this trivial. If ${f_{1},cdots,f_{n}}$ is a basis of the free group $F_{1}$, why should we have elements of the form $(f_{i},h_{i})$ in $H$ for all $iin {1,cdots,n}$? He simply says that this follows because subgroups of free groups are free.






group-theory free-groups direct-product







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  • 3




    Let $H_isubset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $Hsubset H_1times H_2$
    – Max
    2 days ago










  • Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :)
    – Karen
    2 days ago






  • 1




    Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica".
    – Shaun
    2 days ago
















1














I am reading the following paper of Miller: http://researchers.ms.unimelb.edu.au/



He says that if $G= F_{1} times F_{2}$ is a direct product of two free groups and $H$ is a subgroup of $G$, then it can be assumed that the projection maps $$text{p}_{i} colon H rightarrow F_{i}$$ are surjective.



I do not understand why is this trivial. If ${f_{1},cdots,f_{n}}$ is a basis of the free group $F_{1}$, why should we have elements of the form $(f_{i},h_{i})$ in $H$ for all $iin {1,cdots,n}$? He simply says that this follows because subgroups of free groups are free.






group-theory free-groups direct-product







share|cite|improve this question


















  • 3




    Let $H_isubset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $Hsubset H_1times H_2$
    – Max
    2 days ago










  • Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :)
    – Karen
    2 days ago






  • 1




    Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica".
    – Shaun
    2 days ago














1












1








1


1





I am reading the following paper of Miller: http://researchers.ms.unimelb.edu.au/



He says that if $G= F_{1} times F_{2}$ is a direct product of two free groups and $H$ is a subgroup of $G$, then it can be assumed that the projection maps $$text{p}_{i} colon H rightarrow F_{i}$$ are surjective.



I do not understand why is this trivial. If ${f_{1},cdots,f_{n}}$ is a basis of the free group $F_{1}$, why should we have elements of the form $(f_{i},h_{i})$ in $H$ for all $iin {1,cdots,n}$? He simply says that this follows because subgroups of free groups are free.






group-theory free-groups direct-product







share|cite|improve this question













I am reading the following paper of Miller: http://researchers.ms.unimelb.edu.au/



He says that if $G= F_{1} times F_{2}$ is a direct product of two free groups and $H$ is a subgroup of $G$, then it can be assumed that the projection maps $$text{p}_{i} colon H rightarrow F_{i}$$ are surjective.



I do not understand why is this trivial. If ${f_{1},cdots,f_{n}}$ is a basis of the free group $F_{1}$, why should we have elements of the form $(f_{i},h_{i})$ in $H$ for all $iin {1,cdots,n}$? He simply says that this follows because subgroups of free groups are free.






group-theory free-groups direct-product




group-theory free-groups direct-product






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









Karen

906




906








  • 3




    Let $H_isubset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $Hsubset H_1times H_2$
    – Max
    2 days ago










  • Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :)
    – Karen
    2 days ago






  • 1




    Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica".
    – Shaun
    2 days ago














  • 3




    Let $H_isubset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $Hsubset H_1times H_2$
    – Max
    2 days ago










  • Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :)
    – Karen
    2 days ago






  • 1




    Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica".
    – Shaun
    2 days ago








3




3




Let $H_isubset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $Hsubset H_1times H_2$
– Max
2 days ago




Let $H_isubset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $Hsubset H_1times H_2$
– Max
2 days ago












Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :)
– Karen
2 days ago




Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :)
– Karen
2 days ago




1




1




Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica".
– Shaun
2 days ago




Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica".
– Shaun
2 days ago










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