Two presentations of a group, one certainly finite. Need the other be?












2












$begingroup$


I know the answer to the question above is "no", quite flatly. The counter example is below:



$$mathbb{Z}conglangle a,bmid b^2a^{-1}ranglecong langle a,bmidlbrace b^{2^{n+1
}}a^{-2^n}:ninmathbb{N}rbracerangle.$$




So my question, more broadly, is that if $langle A|Rrangleconglangle A|R'rangle$ and we know that $R$ is finite, what can we say about $R'$?




I think we can say that $langle Rrangle^Ncong langle R'rangle^N$, viewed inside the free group on $A$, but I'm not sure.






group-theory finite-groups group-presentation combinatorial-group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can say that all but finitely many of the relators in $R'$ are redundant - i.e. they follow from the others. You haven't said what $N$ is and I don't know what you mean by "viewed inside the free group on $A$".
    $endgroup$
    – Derek Holt
    Sep 26 '18 at 20:51






  • 2




    $begingroup$
    I don't like that notation! It looks like the normal closure of $langle R rangle$ under some subgroup $N$. I prefer $langlelangle R ranglerangle$. Anyway, the answer is yes, $langlelangle R ranglerangle cong langlelangle R' ranglerangle$, becuase they are free groups whose rank is determined by $|langle A mid R rangle|$.
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 8:18






  • 1




    $begingroup$
    Although $langlelangle Rranglerangleconglanglelangle R'ranglerangle$, it is obvious that there are examples where $langlelangle Rranglerangleneq langlelangle R'ranglerangle$ (for example, $langle a, bmid arangleconglangle a, bmid brangle$). What is perhaps less obvious is that there are examples where the subgroups $langlelangle Rranglerangle$ and $langlelangle R'ranglerangle$ lie in different automorphic orbits of the ambient free group (and, surprisingly, in these examples you can take $R$ and $R'$ to consist of a single element).
    $endgroup$
    – user1729
    Sep 27 '18 at 8:33








  • 2




    $begingroup$
    @DerekHolt I've been thinking about the claim you made, that there is a reduction of $R'$. I believe it goes as follows: Since $langlelangle R ranglerangle cong langlelangle R' ranglerangle$ by some $phi$, we find that $langlelangle R' ranglerangle cong langlelangle phi(R) ranglerangle$. But, since $R$ is finite, $phi(R)$ is a collection of conjugates of some finite set $lbrace r_irbracesubset R'$. So, in fact, $langlelangle R' ranglerangle=langlelangle lbrace r_irbrace ranglerangle$ as desired.
    $endgroup$
    – Prototank
    Sep 27 '18 at 12:29








  • 2




    $begingroup$
    Yes that argument works!
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 12:54
















2












$begingroup$


I know the answer to the question above is "no", quite flatly. The counter example is below:



$$mathbb{Z}conglangle a,bmid b^2a^{-1}ranglecong langle a,bmidlbrace b^{2^{n+1
}}a^{-2^n}:ninmathbb{N}rbracerangle.$$




So my question, more broadly, is that if $langle A|Rrangleconglangle A|R'rangle$ and we know that $R$ is finite, what can we say about $R'$?




I think we can say that $langle Rrangle^Ncong langle R'rangle^N$, viewed inside the free group on $A$, but I'm not sure.






group-theory finite-groups group-presentation combinatorial-group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can say that all but finitely many of the relators in $R'$ are redundant - i.e. they follow from the others. You haven't said what $N$ is and I don't know what you mean by "viewed inside the free group on $A$".
    $endgroup$
    – Derek Holt
    Sep 26 '18 at 20:51






  • 2




    $begingroup$
    I don't like that notation! It looks like the normal closure of $langle R rangle$ under some subgroup $N$. I prefer $langlelangle R ranglerangle$. Anyway, the answer is yes, $langlelangle R ranglerangle cong langlelangle R' ranglerangle$, becuase they are free groups whose rank is determined by $|langle A mid R rangle|$.
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 8:18






  • 1




    $begingroup$
    Although $langlelangle Rranglerangleconglanglelangle R'ranglerangle$, it is obvious that there are examples where $langlelangle Rranglerangleneq langlelangle R'ranglerangle$ (for example, $langle a, bmid arangleconglangle a, bmid brangle$). What is perhaps less obvious is that there are examples where the subgroups $langlelangle Rranglerangle$ and $langlelangle R'ranglerangle$ lie in different automorphic orbits of the ambient free group (and, surprisingly, in these examples you can take $R$ and $R'$ to consist of a single element).
    $endgroup$
    – user1729
    Sep 27 '18 at 8:33








  • 2




    $begingroup$
    @DerekHolt I've been thinking about the claim you made, that there is a reduction of $R'$. I believe it goes as follows: Since $langlelangle R ranglerangle cong langlelangle R' ranglerangle$ by some $phi$, we find that $langlelangle R' ranglerangle cong langlelangle phi(R) ranglerangle$. But, since $R$ is finite, $phi(R)$ is a collection of conjugates of some finite set $lbrace r_irbracesubset R'$. So, in fact, $langlelangle R' ranglerangle=langlelangle lbrace r_irbrace ranglerangle$ as desired.
    $endgroup$
    – Prototank
    Sep 27 '18 at 12:29








  • 2




    $begingroup$
    Yes that argument works!
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 12:54














2












2








2


1



$begingroup$


I know the answer to the question above is "no", quite flatly. The counter example is below:



$$mathbb{Z}conglangle a,bmid b^2a^{-1}ranglecong langle a,bmidlbrace b^{2^{n+1
}}a^{-2^n}:ninmathbb{N}rbracerangle.$$




So my question, more broadly, is that if $langle A|Rrangleconglangle A|R'rangle$ and we know that $R$ is finite, what can we say about $R'$?




I think we can say that $langle Rrangle^Ncong langle R'rangle^N$, viewed inside the free group on $A$, but I'm not sure.






group-theory finite-groups group-presentation combinatorial-group-theory







share|cite|improve this question











$endgroup$




I know the answer to the question above is "no", quite flatly. The counter example is below:



$$mathbb{Z}conglangle a,bmid b^2a^{-1}ranglecong langle a,bmidlbrace b^{2^{n+1
}}a^{-2^n}:ninmathbb{N}rbracerangle.$$




So my question, more broadly, is that if $langle A|Rrangleconglangle A|R'rangle$ and we know that $R$ is finite, what can we say about $R'$?




I think we can say that $langle Rrangle^Ncong langle R'rangle^N$, viewed inside the free group on $A$, but I'm not sure.






group-theory finite-groups group-presentation combinatorial-group-theory




group-theory finite-groups group-presentation combinatorial-group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 1:04









Shaun

8,883113681




8,883113681










asked Sep 26 '18 at 20:06









PrototankPrototank

1,030820




1,030820








  • 1




    $begingroup$
    You can say that all but finitely many of the relators in $R'$ are redundant - i.e. they follow from the others. You haven't said what $N$ is and I don't know what you mean by "viewed inside the free group on $A$".
    $endgroup$
    – Derek Holt
    Sep 26 '18 at 20:51






  • 2




    $begingroup$
    I don't like that notation! It looks like the normal closure of $langle R rangle$ under some subgroup $N$. I prefer $langlelangle R ranglerangle$. Anyway, the answer is yes, $langlelangle R ranglerangle cong langlelangle R' ranglerangle$, becuase they are free groups whose rank is determined by $|langle A mid R rangle|$.
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 8:18






  • 1




    $begingroup$
    Although $langlelangle Rranglerangleconglanglelangle R'ranglerangle$, it is obvious that there are examples where $langlelangle Rranglerangleneq langlelangle R'ranglerangle$ (for example, $langle a, bmid arangleconglangle a, bmid brangle$). What is perhaps less obvious is that there are examples where the subgroups $langlelangle Rranglerangle$ and $langlelangle R'ranglerangle$ lie in different automorphic orbits of the ambient free group (and, surprisingly, in these examples you can take $R$ and $R'$ to consist of a single element).
    $endgroup$
    – user1729
    Sep 27 '18 at 8:33








  • 2




    $begingroup$
    @DerekHolt I've been thinking about the claim you made, that there is a reduction of $R'$. I believe it goes as follows: Since $langlelangle R ranglerangle cong langlelangle R' ranglerangle$ by some $phi$, we find that $langlelangle R' ranglerangle cong langlelangle phi(R) ranglerangle$. But, since $R$ is finite, $phi(R)$ is a collection of conjugates of some finite set $lbrace r_irbracesubset R'$. So, in fact, $langlelangle R' ranglerangle=langlelangle lbrace r_irbrace ranglerangle$ as desired.
    $endgroup$
    – Prototank
    Sep 27 '18 at 12:29








  • 2




    $begingroup$
    Yes that argument works!
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 12:54














  • 1




    $begingroup$
    You can say that all but finitely many of the relators in $R'$ are redundant - i.e. they follow from the others. You haven't said what $N$ is and I don't know what you mean by "viewed inside the free group on $A$".
    $endgroup$
    – Derek Holt
    Sep 26 '18 at 20:51






  • 2




    $begingroup$
    I don't like that notation! It looks like the normal closure of $langle R rangle$ under some subgroup $N$. I prefer $langlelangle R ranglerangle$. Anyway, the answer is yes, $langlelangle R ranglerangle cong langlelangle R' ranglerangle$, becuase they are free groups whose rank is determined by $|langle A mid R rangle|$.
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 8:18






  • 1




    $begingroup$
    Although $langlelangle Rranglerangleconglanglelangle R'ranglerangle$, it is obvious that there are examples where $langlelangle Rranglerangleneq langlelangle R'ranglerangle$ (for example, $langle a, bmid arangleconglangle a, bmid brangle$). What is perhaps less obvious is that there are examples where the subgroups $langlelangle Rranglerangle$ and $langlelangle R'ranglerangle$ lie in different automorphic orbits of the ambient free group (and, surprisingly, in these examples you can take $R$ and $R'$ to consist of a single element).
    $endgroup$
    – user1729
    Sep 27 '18 at 8:33








  • 2




    $begingroup$
    @DerekHolt I've been thinking about the claim you made, that there is a reduction of $R'$. I believe it goes as follows: Since $langlelangle R ranglerangle cong langlelangle R' ranglerangle$ by some $phi$, we find that $langlelangle R' ranglerangle cong langlelangle phi(R) ranglerangle$. But, since $R$ is finite, $phi(R)$ is a collection of conjugates of some finite set $lbrace r_irbracesubset R'$. So, in fact, $langlelangle R' ranglerangle=langlelangle lbrace r_irbrace ranglerangle$ as desired.
    $endgroup$
    – Prototank
    Sep 27 '18 at 12:29








  • 2




    $begingroup$
    Yes that argument works!
    $endgroup$
    – Derek Holt
    Sep 27 '18 at 12:54








1




1




$begingroup$
You can say that all but finitely many of the relators in $R'$ are redundant - i.e. they follow from the others. You haven't said what $N$ is and I don't know what you mean by "viewed inside the free group on $A$".
$endgroup$
– Derek Holt
Sep 26 '18 at 20:51




$begingroup$
You can say that all but finitely many of the relators in $R'$ are redundant - i.e. they follow from the others. You haven't said what $N$ is and I don't know what you mean by "viewed inside the free group on $A$".
$endgroup$
– Derek Holt
Sep 26 '18 at 20:51




2




2




$begingroup$
I don't like that notation! It looks like the normal closure of $langle R rangle$ under some subgroup $N$. I prefer $langlelangle R ranglerangle$. Anyway, the answer is yes, $langlelangle R ranglerangle cong langlelangle R' ranglerangle$, becuase they are free groups whose rank is determined by $|langle A mid R rangle|$.
$endgroup$
– Derek Holt
Sep 27 '18 at 8:18




$begingroup$
I don't like that notation! It looks like the normal closure of $langle R rangle$ under some subgroup $N$. I prefer $langlelangle R ranglerangle$. Anyway, the answer is yes, $langlelangle R ranglerangle cong langlelangle R' ranglerangle$, becuase they are free groups whose rank is determined by $|langle A mid R rangle|$.
$endgroup$
– Derek Holt
Sep 27 '18 at 8:18




1




1




$begingroup$
Although $langlelangle Rranglerangleconglanglelangle R'ranglerangle$, it is obvious that there are examples where $langlelangle Rranglerangleneq langlelangle R'ranglerangle$ (for example, $langle a, bmid arangleconglangle a, bmid brangle$). What is perhaps less obvious is that there are examples where the subgroups $langlelangle Rranglerangle$ and $langlelangle R'ranglerangle$ lie in different automorphic orbits of the ambient free group (and, surprisingly, in these examples you can take $R$ and $R'$ to consist of a single element).
$endgroup$
– user1729
Sep 27 '18 at 8:33






$begingroup$
Although $langlelangle Rranglerangleconglanglelangle R'ranglerangle$, it is obvious that there are examples where $langlelangle Rranglerangleneq langlelangle R'ranglerangle$ (for example, $langle a, bmid arangleconglangle a, bmid brangle$). What is perhaps less obvious is that there are examples where the subgroups $langlelangle Rranglerangle$ and $langlelangle R'ranglerangle$ lie in different automorphic orbits of the ambient free group (and, surprisingly, in these examples you can take $R$ and $R'$ to consist of a single element).
$endgroup$
– user1729
Sep 27 '18 at 8:33






2




2




$begingroup$
@DerekHolt I've been thinking about the claim you made, that there is a reduction of $R'$. I believe it goes as follows: Since $langlelangle R ranglerangle cong langlelangle R' ranglerangle$ by some $phi$, we find that $langlelangle R' ranglerangle cong langlelangle phi(R) ranglerangle$. But, since $R$ is finite, $phi(R)$ is a collection of conjugates of some finite set $lbrace r_irbracesubset R'$. So, in fact, $langlelangle R' ranglerangle=langlelangle lbrace r_irbrace ranglerangle$ as desired.
$endgroup$
– Prototank
Sep 27 '18 at 12:29






$begingroup$
@DerekHolt I've been thinking about the claim you made, that there is a reduction of $R'$. I believe it goes as follows: Since $langlelangle R ranglerangle cong langlelangle R' ranglerangle$ by some $phi$, we find that $langlelangle R' ranglerangle cong langlelangle phi(R) ranglerangle$. But, since $R$ is finite, $phi(R)$ is a collection of conjugates of some finite set $lbrace r_irbracesubset R'$. So, in fact, $langlelangle R' ranglerangle=langlelangle lbrace r_irbrace ranglerangle$ as desired.
$endgroup$
– Prototank
Sep 27 '18 at 12:29






2




2




$begingroup$
Yes that argument works!
$endgroup$
– Derek Holt
Sep 27 '18 at 12:54




$begingroup$
Yes that argument works!
$endgroup$
– Derek Holt
Sep 27 '18 at 12:54










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