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
edited Dec 3 '18 at 1:04
Shaun
8,883113681
asked Sep 26 '18 at 20:06
PrototankPrototank
1,030820
$endgroup$
|
show 1 more comment
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
edited Dec 3 '18 at 1:04
Shaun
8,883113681
asked Sep 26 '18 at 20:06
PrototankPrototank
1,030820
$endgroup$
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
|
show 1 more comment
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
edited Dec 3 '18 at 1:04
Shaun
8,883113681
asked Sep 26 '18 at 20:06
PrototankPrototank
1,030820
$endgroup$
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
edited Dec 3 '18 at 1:04
Shaun
8,883113681
asked Sep 26 '18 at 20:06
PrototankPrototank
1,030820
edited Dec 3 '18 at 1:04
Shaun
8,883113681
asked Sep 26 '18 at 20:06
PrototankPrototank
1,030820
edited Dec 3 '18 at 1:04
Shaun
8,883113681
edited Dec 3 '18 at 1:04
Shaun
8,883113681
edited Dec 3 '18 at 1:04
Shaun
8,883113681
8,883113681
asked Sep 26 '18 at 20:06
PrototankPrototank
1,030820
asked Sep 26 '18 at 20:06
PrototankPrototank
1,030820
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
|
show 1 more comment
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
|
show 1 more comment
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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