Connection between Freiheitssatz and Magnus property












3












$begingroup$


I am currently studying the Magnus property:




Let $G$ be a group and $u, v in G$. If the normal closures
of $u$ and $v$ coincide, then $u$ is conjugate to $v$ or $v^{−1}$.




I was told that this property was named after W. Magnus who proved the Freiheitsatz and that free groups have this property in this text in pages 141-165. However, this text is in German, which I am not fluent in, and is very technical, using multiple lemma's and notation that I am not familiar with.



I do have a decent understanding of group theory and was hoping that some body could give some intuitive way of seeing how the Freiheitsatz implies free groups having the Magnus property?










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  • $begingroup$
    Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
    $endgroup$
    – Shaun
    Jan 21 at 13:43


















3












$begingroup$


I am currently studying the Magnus property:




Let $G$ be a group and $u, v in G$. If the normal closures
of $u$ and $v$ coincide, then $u$ is conjugate to $v$ or $v^{−1}$.




I was told that this property was named after W. Magnus who proved the Freiheitsatz and that free groups have this property in this text in pages 141-165. However, this text is in German, which I am not fluent in, and is very technical, using multiple lemma's and notation that I am not familiar with.



I do have a decent understanding of group theory and was hoping that some body could give some intuitive way of seeing how the Freiheitsatz implies free groups having the Magnus property?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
    $endgroup$
    – Shaun
    Jan 21 at 13:43
















3












3








3





$begingroup$


I am currently studying the Magnus property:




Let $G$ be a group and $u, v in G$. If the normal closures
of $u$ and $v$ coincide, then $u$ is conjugate to $v$ or $v^{−1}$.




I was told that this property was named after W. Magnus who proved the Freiheitsatz and that free groups have this property in this text in pages 141-165. However, this text is in German, which I am not fluent in, and is very technical, using multiple lemma's and notation that I am not familiar with.



I do have a decent understanding of group theory and was hoping that some body could give some intuitive way of seeing how the Freiheitsatz implies free groups having the Magnus property?










share|cite|improve this question











$endgroup$




I am currently studying the Magnus property:




Let $G$ be a group and $u, v in G$. If the normal closures
of $u$ and $v$ coincide, then $u$ is conjugate to $v$ or $v^{−1}$.




I was told that this property was named after W. Magnus who proved the Freiheitsatz and that free groups have this property in this text in pages 141-165. However, this text is in German, which I am not fluent in, and is very technical, using multiple lemma's and notation that I am not familiar with.



I do have a decent understanding of group theory and was hoping that some body could give some intuitive way of seeing how the Freiheitsatz implies free groups having the Magnus property?







group-theory free-groups






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edited Jan 21 at 14:29









user1729

17.2k64193




17.2k64193










asked Jan 21 at 13:20









MorierisMorieris

192




192












  • $begingroup$
    Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
    $endgroup$
    – Shaun
    Jan 21 at 13:43




















  • $begingroup$
    Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
    $endgroup$
    – Shaun
    Jan 21 at 13:43


















$begingroup$
Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
$endgroup$
– Shaun
Jan 21 at 13:43






$begingroup$
Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
$endgroup$
– Shaun
Jan 21 at 13:43












1 Answer
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2












$begingroup$

The result you state about free groups is Theorem 4.11 of the standard text Combinatorial group theory by Magnus, Karrass and Solitar* (p261 of my copy). The book includes an English-language proof of this result, as well as a proof of the Freiheitssatz itself. The proof in the book is 4.5 pages long, and the idea is to apply "Magnus' method". This method was first used by Magnus to study one-relator groups, and indeed is the method used in the text you link too.



However, these proofs are painful. They use something called "staggered presentations", which was Magnus' original way of applying his method. A more up-to-date version of Magnus' method, due to Moldavanski, uses HNN-extensions**. You can find detailed descriptions of this method in the paper McCool, Schupp, On one relator groups and HNN extensions, J. Aust. Math. Soc. 16(02):249--256 (1973), and also in these notes of Andy Putman. Neither resource contains a proof of the result that you are after, but I guess you can try to turn the handle if you are enthusiastic enough. Sounds like a nice master's project actually! :-)





I should say a few words about Magnus' method. Any proof which uses Magnus' method must apply the Freiheitssatz somewhere (in particular, the proof of Theorem 4.11 in Magnus, Karrass and Solitar - the proof that you are interested in - uses the Freiheitssatz, as you were told). Magnus' method is an inductive argument on the length of the word $R$. An explicit example of a single step in the HNN-version of the method is as follows: Suppose $R=ab^2c^2a^{-1}b^2c^2$ and consider the group $G=langle a, b, cmid Rrangle$. We have the following:
$$
begin{align*}
langle a, b, cmid ab^2c^2a^{-1}b^2c^2rangle&conglangle a, b_0, c_0, b_1, c_1mid ab_0^2c_0^2a^{-1}b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle\
&conglangle a, b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle
end{align*}
$$

By the Freiheitssatz, the subgroup $langle b, crangle$ of $G$ is free of rank two. Hence, $langle b_0, c_0rangle$ and $langle b_1, c_1rangle$ are also free of rank two. Therefore, the above working shows that $G$ is an HNN-extension of the group $G_1=langle b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2rangle$. This new group $G_1$ has a single defining relator which is shorter than the relator of the old group, so we have our induction step.





*This book is usually referred to as "Magnus, Karrass and Solitar" to distinguish it from "Lyndon and Schupp"; both sets of authors wrote books called "combinatorial group theory", and both books are still standard texts in geometric group theory.



**Jim Howie gave an alternative version of the method using "staggered complexes", while a third method used "pictures" (I cannot find out at the moment who first used this third method, but I know Hamish Short used it in his early work). All three methods are essentially equivalent, but HNN-extensions are much more standard in group theory and so the proofs are more readable to the generalist.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the reply! I'm a bit confused/worried that i may be on the wrong path here. I was under the impression that the Freiheitsatz was used to prove that free groups have the Magnus property, since most my sources about Magnus property mention both in the same sentence. Since neither of the sources you mention contain a proof of this format, do you think i misinterpretted my sources?
    $endgroup$
    – Morieris
    Jan 21 at 17:03










  • $begingroup$
    Sorry, I wasn't clear. The Freiheitssatz is used to prove this result; in fact, any proof which uses Magnus' method applies the Freiheitssatz somewhere! So you are on the right path.
    $endgroup$
    – user1729
    Jan 21 at 21:38











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1 Answer
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1 Answer
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2












$begingroup$

The result you state about free groups is Theorem 4.11 of the standard text Combinatorial group theory by Magnus, Karrass and Solitar* (p261 of my copy). The book includes an English-language proof of this result, as well as a proof of the Freiheitssatz itself. The proof in the book is 4.5 pages long, and the idea is to apply "Magnus' method". This method was first used by Magnus to study one-relator groups, and indeed is the method used in the text you link too.



However, these proofs are painful. They use something called "staggered presentations", which was Magnus' original way of applying his method. A more up-to-date version of Magnus' method, due to Moldavanski, uses HNN-extensions**. You can find detailed descriptions of this method in the paper McCool, Schupp, On one relator groups and HNN extensions, J. Aust. Math. Soc. 16(02):249--256 (1973), and also in these notes of Andy Putman. Neither resource contains a proof of the result that you are after, but I guess you can try to turn the handle if you are enthusiastic enough. Sounds like a nice master's project actually! :-)





I should say a few words about Magnus' method. Any proof which uses Magnus' method must apply the Freiheitssatz somewhere (in particular, the proof of Theorem 4.11 in Magnus, Karrass and Solitar - the proof that you are interested in - uses the Freiheitssatz, as you were told). Magnus' method is an inductive argument on the length of the word $R$. An explicit example of a single step in the HNN-version of the method is as follows: Suppose $R=ab^2c^2a^{-1}b^2c^2$ and consider the group $G=langle a, b, cmid Rrangle$. We have the following:
$$
begin{align*}
langle a, b, cmid ab^2c^2a^{-1}b^2c^2rangle&conglangle a, b_0, c_0, b_1, c_1mid ab_0^2c_0^2a^{-1}b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle\
&conglangle a, b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle
end{align*}
$$

By the Freiheitssatz, the subgroup $langle b, crangle$ of $G$ is free of rank two. Hence, $langle b_0, c_0rangle$ and $langle b_1, c_1rangle$ are also free of rank two. Therefore, the above working shows that $G$ is an HNN-extension of the group $G_1=langle b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2rangle$. This new group $G_1$ has a single defining relator which is shorter than the relator of the old group, so we have our induction step.





*This book is usually referred to as "Magnus, Karrass and Solitar" to distinguish it from "Lyndon and Schupp"; both sets of authors wrote books called "combinatorial group theory", and both books are still standard texts in geometric group theory.



**Jim Howie gave an alternative version of the method using "staggered complexes", while a third method used "pictures" (I cannot find out at the moment who first used this third method, but I know Hamish Short used it in his early work). All three methods are essentially equivalent, but HNN-extensions are much more standard in group theory and so the proofs are more readable to the generalist.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the reply! I'm a bit confused/worried that i may be on the wrong path here. I was under the impression that the Freiheitsatz was used to prove that free groups have the Magnus property, since most my sources about Magnus property mention both in the same sentence. Since neither of the sources you mention contain a proof of this format, do you think i misinterpretted my sources?
    $endgroup$
    – Morieris
    Jan 21 at 17:03










  • $begingroup$
    Sorry, I wasn't clear. The Freiheitssatz is used to prove this result; in fact, any proof which uses Magnus' method applies the Freiheitssatz somewhere! So you are on the right path.
    $endgroup$
    – user1729
    Jan 21 at 21:38
















2












$begingroup$

The result you state about free groups is Theorem 4.11 of the standard text Combinatorial group theory by Magnus, Karrass and Solitar* (p261 of my copy). The book includes an English-language proof of this result, as well as a proof of the Freiheitssatz itself. The proof in the book is 4.5 pages long, and the idea is to apply "Magnus' method". This method was first used by Magnus to study one-relator groups, and indeed is the method used in the text you link too.



However, these proofs are painful. They use something called "staggered presentations", which was Magnus' original way of applying his method. A more up-to-date version of Magnus' method, due to Moldavanski, uses HNN-extensions**. You can find detailed descriptions of this method in the paper McCool, Schupp, On one relator groups and HNN extensions, J. Aust. Math. Soc. 16(02):249--256 (1973), and also in these notes of Andy Putman. Neither resource contains a proof of the result that you are after, but I guess you can try to turn the handle if you are enthusiastic enough. Sounds like a nice master's project actually! :-)





I should say a few words about Magnus' method. Any proof which uses Magnus' method must apply the Freiheitssatz somewhere (in particular, the proof of Theorem 4.11 in Magnus, Karrass and Solitar - the proof that you are interested in - uses the Freiheitssatz, as you were told). Magnus' method is an inductive argument on the length of the word $R$. An explicit example of a single step in the HNN-version of the method is as follows: Suppose $R=ab^2c^2a^{-1}b^2c^2$ and consider the group $G=langle a, b, cmid Rrangle$. We have the following:
$$
begin{align*}
langle a, b, cmid ab^2c^2a^{-1}b^2c^2rangle&conglangle a, b_0, c_0, b_1, c_1mid ab_0^2c_0^2a^{-1}b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle\
&conglangle a, b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle
end{align*}
$$

By the Freiheitssatz, the subgroup $langle b, crangle$ of $G$ is free of rank two. Hence, $langle b_0, c_0rangle$ and $langle b_1, c_1rangle$ are also free of rank two. Therefore, the above working shows that $G$ is an HNN-extension of the group $G_1=langle b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2rangle$. This new group $G_1$ has a single defining relator which is shorter than the relator of the old group, so we have our induction step.





*This book is usually referred to as "Magnus, Karrass and Solitar" to distinguish it from "Lyndon and Schupp"; both sets of authors wrote books called "combinatorial group theory", and both books are still standard texts in geometric group theory.



**Jim Howie gave an alternative version of the method using "staggered complexes", while a third method used "pictures" (I cannot find out at the moment who first used this third method, but I know Hamish Short used it in his early work). All three methods are essentially equivalent, but HNN-extensions are much more standard in group theory and so the proofs are more readable to the generalist.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the reply! I'm a bit confused/worried that i may be on the wrong path here. I was under the impression that the Freiheitsatz was used to prove that free groups have the Magnus property, since most my sources about Magnus property mention both in the same sentence. Since neither of the sources you mention contain a proof of this format, do you think i misinterpretted my sources?
    $endgroup$
    – Morieris
    Jan 21 at 17:03










  • $begingroup$
    Sorry, I wasn't clear. The Freiheitssatz is used to prove this result; in fact, any proof which uses Magnus' method applies the Freiheitssatz somewhere! So you are on the right path.
    $endgroup$
    – user1729
    Jan 21 at 21:38














2












2








2





$begingroup$

The result you state about free groups is Theorem 4.11 of the standard text Combinatorial group theory by Magnus, Karrass and Solitar* (p261 of my copy). The book includes an English-language proof of this result, as well as a proof of the Freiheitssatz itself. The proof in the book is 4.5 pages long, and the idea is to apply "Magnus' method". This method was first used by Magnus to study one-relator groups, and indeed is the method used in the text you link too.



However, these proofs are painful. They use something called "staggered presentations", which was Magnus' original way of applying his method. A more up-to-date version of Magnus' method, due to Moldavanski, uses HNN-extensions**. You can find detailed descriptions of this method in the paper McCool, Schupp, On one relator groups and HNN extensions, J. Aust. Math. Soc. 16(02):249--256 (1973), and also in these notes of Andy Putman. Neither resource contains a proof of the result that you are after, but I guess you can try to turn the handle if you are enthusiastic enough. Sounds like a nice master's project actually! :-)





I should say a few words about Magnus' method. Any proof which uses Magnus' method must apply the Freiheitssatz somewhere (in particular, the proof of Theorem 4.11 in Magnus, Karrass and Solitar - the proof that you are interested in - uses the Freiheitssatz, as you were told). Magnus' method is an inductive argument on the length of the word $R$. An explicit example of a single step in the HNN-version of the method is as follows: Suppose $R=ab^2c^2a^{-1}b^2c^2$ and consider the group $G=langle a, b, cmid Rrangle$. We have the following:
$$
begin{align*}
langle a, b, cmid ab^2c^2a^{-1}b^2c^2rangle&conglangle a, b_0, c_0, b_1, c_1mid ab_0^2c_0^2a^{-1}b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle\
&conglangle a, b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle
end{align*}
$$

By the Freiheitssatz, the subgroup $langle b, crangle$ of $G$ is free of rank two. Hence, $langle b_0, c_0rangle$ and $langle b_1, c_1rangle$ are also free of rank two. Therefore, the above working shows that $G$ is an HNN-extension of the group $G_1=langle b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2rangle$. This new group $G_1$ has a single defining relator which is shorter than the relator of the old group, so we have our induction step.





*This book is usually referred to as "Magnus, Karrass and Solitar" to distinguish it from "Lyndon and Schupp"; both sets of authors wrote books called "combinatorial group theory", and both books are still standard texts in geometric group theory.



**Jim Howie gave an alternative version of the method using "staggered complexes", while a third method used "pictures" (I cannot find out at the moment who first used this third method, but I know Hamish Short used it in his early work). All three methods are essentially equivalent, but HNN-extensions are much more standard in group theory and so the proofs are more readable to the generalist.






share|cite|improve this answer











$endgroup$



The result you state about free groups is Theorem 4.11 of the standard text Combinatorial group theory by Magnus, Karrass and Solitar* (p261 of my copy). The book includes an English-language proof of this result, as well as a proof of the Freiheitssatz itself. The proof in the book is 4.5 pages long, and the idea is to apply "Magnus' method". This method was first used by Magnus to study one-relator groups, and indeed is the method used in the text you link too.



However, these proofs are painful. They use something called "staggered presentations", which was Magnus' original way of applying his method. A more up-to-date version of Magnus' method, due to Moldavanski, uses HNN-extensions**. You can find detailed descriptions of this method in the paper McCool, Schupp, On one relator groups and HNN extensions, J. Aust. Math. Soc. 16(02):249--256 (1973), and also in these notes of Andy Putman. Neither resource contains a proof of the result that you are after, but I guess you can try to turn the handle if you are enthusiastic enough. Sounds like a nice master's project actually! :-)





I should say a few words about Magnus' method. Any proof which uses Magnus' method must apply the Freiheitssatz somewhere (in particular, the proof of Theorem 4.11 in Magnus, Karrass and Solitar - the proof that you are interested in - uses the Freiheitssatz, as you were told). Magnus' method is an inductive argument on the length of the word $R$. An explicit example of a single step in the HNN-version of the method is as follows: Suppose $R=ab^2c^2a^{-1}b^2c^2$ and consider the group $G=langle a, b, cmid Rrangle$. We have the following:
$$
begin{align*}
langle a, b, cmid ab^2c^2a^{-1}b^2c^2rangle&conglangle a, b_0, c_0, b_1, c_1mid ab_0^2c_0^2a^{-1}b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle\
&conglangle a, b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2, ab_0a^{-1}=b_1, ac_0a^{-1}=c_1rangle
end{align*}
$$

By the Freiheitssatz, the subgroup $langle b, crangle$ of $G$ is free of rank two. Hence, $langle b_0, c_0rangle$ and $langle b_1, c_1rangle$ are also free of rank two. Therefore, the above working shows that $G$ is an HNN-extension of the group $G_1=langle b_0, c_0, b_1, c_1mid b_1^2c_1^2b_0^2c_0^2rangle$. This new group $G_1$ has a single defining relator which is shorter than the relator of the old group, so we have our induction step.





*This book is usually referred to as "Magnus, Karrass and Solitar" to distinguish it from "Lyndon and Schupp"; both sets of authors wrote books called "combinatorial group theory", and both books are still standard texts in geometric group theory.



**Jim Howie gave an alternative version of the method using "staggered complexes", while a third method used "pictures" (I cannot find out at the moment who first used this third method, but I know Hamish Short used it in his early work). All three methods are essentially equivalent, but HNN-extensions are much more standard in group theory and so the proofs are more readable to the generalist.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 22 at 12:48

























answered Jan 21 at 14:28









user1729user1729

17.2k64193




17.2k64193












  • $begingroup$
    Thanks for the reply! I'm a bit confused/worried that i may be on the wrong path here. I was under the impression that the Freiheitsatz was used to prove that free groups have the Magnus property, since most my sources about Magnus property mention both in the same sentence. Since neither of the sources you mention contain a proof of this format, do you think i misinterpretted my sources?
    $endgroup$
    – Morieris
    Jan 21 at 17:03










  • $begingroup$
    Sorry, I wasn't clear. The Freiheitssatz is used to prove this result; in fact, any proof which uses Magnus' method applies the Freiheitssatz somewhere! So you are on the right path.
    $endgroup$
    – user1729
    Jan 21 at 21:38


















  • $begingroup$
    Thanks for the reply! I'm a bit confused/worried that i may be on the wrong path here. I was under the impression that the Freiheitsatz was used to prove that free groups have the Magnus property, since most my sources about Magnus property mention both in the same sentence. Since neither of the sources you mention contain a proof of this format, do you think i misinterpretted my sources?
    $endgroup$
    – Morieris
    Jan 21 at 17:03










  • $begingroup$
    Sorry, I wasn't clear. The Freiheitssatz is used to prove this result; in fact, any proof which uses Magnus' method applies the Freiheitssatz somewhere! So you are on the right path.
    $endgroup$
    – user1729
    Jan 21 at 21:38
















$begingroup$
Thanks for the reply! I'm a bit confused/worried that i may be on the wrong path here. I was under the impression that the Freiheitsatz was used to prove that free groups have the Magnus property, since most my sources about Magnus property mention both in the same sentence. Since neither of the sources you mention contain a proof of this format, do you think i misinterpretted my sources?
$endgroup$
– Morieris
Jan 21 at 17:03




$begingroup$
Thanks for the reply! I'm a bit confused/worried that i may be on the wrong path here. I was under the impression that the Freiheitsatz was used to prove that free groups have the Magnus property, since most my sources about Magnus property mention both in the same sentence. Since neither of the sources you mention contain a proof of this format, do you think i misinterpretted my sources?
$endgroup$
– Morieris
Jan 21 at 17:03












$begingroup$
Sorry, I wasn't clear. The Freiheitssatz is used to prove this result; in fact, any proof which uses Magnus' method applies the Freiheitssatz somewhere! So you are on the right path.
$endgroup$
– user1729
Jan 21 at 21:38




$begingroup$
Sorry, I wasn't clear. The Freiheitssatz is used to prove this result; in fact, any proof which uses Magnus' method applies the Freiheitssatz somewhere! So you are on the right path.
$endgroup$
– user1729
Jan 21 at 21:38


















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