Connection between Freiheitssatz and Magnus property
$begingroup$
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?
group-theory free-groups
$endgroup$
add a comment |
$begingroup$
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?
group-theory free-groups
$endgroup$
$begingroup$
Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
$endgroup$
– Shaun
Jan 21 at 13:43
add a comment |
$begingroup$
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?
group-theory free-groups
$endgroup$
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
group-theory free-groups
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081873%2fconnection-between-freiheitssatz-and-magnus-property%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081873%2fconnection-between-freiheitssatz-and-magnus-property%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Please use a type of $LaTeX$ called MathJax here in future. Tutorials are freely available online.
$endgroup$
– Shaun
Jan 21 at 13:43