Describing the group defined by $langle a,b mid (ab)^2=(abaa)^2=(abbb)^2=erangle.$
4
$begingroup$
I'm trying to find an isomorphism of a group with the following presentation:$$langle a,b mid (ab)^2=(abaa)^2=(abbb)^2=erangle$$
Basically, I'm not that experienced with groups so I'm wondering if I'm missing something obvious. I figured out some identities, but I can't seem to simplify the generators any further.
group-theory group-presentation group-isomorphism
edited Dec 10 '18 at 23:56
Shaun
9,025113682
asked Dec 29 '14 at 2:17
theelk801theelk801
233
$endgroup$
|
show 7 more comments
4
$begingroup$
I'm trying to find an isomorphism of a group with the following presentation:$$langle a,b mid (ab)^2=(abaa)^2=(abbb)^2=erangle$$
Basically, I'm not that experienced with groups so I'm wondering if I'm missing something obvious. I figured out some identities, but I can't seem to simplify the generators any further.
group-theory group-presentation group-isomorphism
edited Dec 10 '18 at 23:56
Shaun
9,025113682
asked Dec 29 '14 at 2:17
theelk801theelk801
233
$endgroup$
$begingroup$
$(ab)^2$ means $(ab)^2=e$ ?
$endgroup$
– mesel
Dec 29 '14 at 2:35
$begingroup$
Yeah, sorry. Edited.
$endgroup$
– theelk801
Dec 29 '14 at 2:40
1
$begingroup$
I don't know if it helps, but here it goes: First, $e=(ab)^2iff b'a'=ab$ and $e=(abaa)^2=(b'a'aa)^2=(b'a)^2iff b'a=a'b$. Then $abbb=b'a'bb=b'b'ab=b'b'b'a'implies (abbb)^2=e$, so you are actually trying to find $langle a,bmid b'a'=ab, b'a=a'brangle$.
$endgroup$
– user2345215
Dec 29 '14 at 9:29
1
$begingroup$
. . . as well as @janmarqz (and any other user for that matter).
$endgroup$
– Shaun
Dec 11 '18 at 0:02
1
$begingroup$
Yes your presentation defines a group that is isomorphic to the original.
$endgroup$
– Derek Holt
Dec 11 '18 at 8:13
|
show 7 more comments
4
4
4
1
$begingroup$
I'm trying to find an isomorphism of a group with the following presentation:$$langle a,b mid (ab)^2=(abaa)^2=(abbb)^2=erangle$$
Basically, I'm not that experienced with groups so I'm wondering if I'm missing something obvious. I figured out some identities, but I can't seem to simplify the generators any further.
group-theory group-presentation group-isomorphism
edited Dec 10 '18 at 23:56
Shaun
9,025113682
asked Dec 29 '14 at 2:17
theelk801theelk801
233
$endgroup$
I'm trying to find an isomorphism of a group with the following presentation:$$langle a,b mid (ab)^2=(abaa)^2=(abbb)^2=erangle$$
Basically, I'm not that experienced with groups so I'm wondering if I'm missing something obvious. I figured out some identities, but I can't seem to simplify the generators any further.
group-theory group-presentation group-isomorphism
group-theory group-presentation group-isomorphism
edited Dec 10 '18 at 23:56
Shaun
9,025113682
asked Dec 29 '14 at 2:17
theelk801theelk801
233
edited Dec 10 '18 at 23:56
Shaun
9,025113682
asked Dec 29 '14 at 2:17
theelk801theelk801
233
edited Dec 10 '18 at 23:56
Shaun
9,025113682
edited Dec 10 '18 at 23:56
Shaun
9,025113682
edited Dec 10 '18 at 23:56
Shaun
9,025113682
9,025113682
asked Dec 29 '14 at 2:17
theelk801theelk801
233
asked Dec 29 '14 at 2:17
theelk801theelk801
233
asked Dec 29 '14 at 2:17
theelk801theelk801
233
233
$begingroup$
$(ab)^2$ means $(ab)^2=e$ ?
$endgroup$
– mesel
Dec 29 '14 at 2:35
$begingroup$
Yeah, sorry. Edited.
$endgroup$
– theelk801
Dec 29 '14 at 2:40
1
$begingroup$
I don't know if it helps, but here it goes: First, $e=(ab)^2iff b'a'=ab$ and $e=(abaa)^2=(b'a'aa)^2=(b'a)^2iff b'a=a'b$. Then $abbb=b'a'bb=b'b'ab=b'b'b'a'implies (abbb)^2=e$, so you are actually trying to find $langle a,bmid b'a'=ab, b'a=a'brangle$.
$endgroup$
– user2345215
Dec 29 '14 at 9:29
1
$begingroup$
. . . as well as @janmarqz (and any other user for that matter).
$endgroup$
– Shaun
Dec 11 '18 at 0:02
1
$begingroup$
Yes your presentation defines a group that is isomorphic to the original.
$endgroup$
– Derek Holt
Dec 11 '18 at 8:13
|
show 7 more comments
$begingroup$
$(ab)^2$ means $(ab)^2=e$ ?
$endgroup$
– mesel
Dec 29 '14 at 2:35
$begingroup$
Yeah, sorry. Edited.
$endgroup$
– theelk801
Dec 29 '14 at 2:40
1
$begingroup$
I don't know if it helps, but here it goes: First, $e=(ab)^2iff b'a'=ab$ and $e=(abaa)^2=(b'a'aa)^2=(b'a)^2iff b'a=a'b$. Then $abbb=b'a'bb=b'b'ab=b'b'b'a'implies (abbb)^2=e$, so you are actually trying to find $langle a,bmid b'a'=ab, b'a=a'brangle$.
$endgroup$
– user2345215
Dec 29 '14 at 9:29
1
$begingroup$
. . . as well as @janmarqz (and any other user for that matter).
$endgroup$
– Shaun
Dec 11 '18 at 0:02
1
$begingroup$
Yes your presentation defines a group that is isomorphic to the original.
$endgroup$
– Derek Holt
Dec 11 '18 at 8:13
$begingroup$
$(ab)^2$ means $(ab)^2=e$ ?
$endgroup$
– mesel
Dec 29 '14 at 2:35
$begingroup$
$(ab)^2$ means $(ab)^2=e$ ?
$endgroup$
– mesel
Dec 29 '14 at 2:35
$begingroup$
Yeah, sorry. Edited.
$endgroup$
– theelk801
Dec 29 '14 at 2:40
$begingroup$
Yeah, sorry. Edited.
$endgroup$
– theelk801
Dec 29 '14 at 2:40
1
1
$begingroup$
I don't know if it helps, but here it goes: First, $e=(ab)^2iff b'a'=ab$ and $e=(abaa)^2=(b'a'aa)^2=(b'a)^2iff b'a=a'b$. Then $abbb=b'a'bb=b'b'ab=b'b'b'a'implies (abbb)^2=e$, so you are actually trying to find $langle a,bmid b'a'=ab, b'a=a'brangle$.
$endgroup$
– user2345215
Dec 29 '14 at 9:29
$begingroup$
I don't know if it helps, but here it goes: First, $e=(ab)^2iff b'a'=ab$ and $e=(abaa)^2=(b'a'aa)^2=(b'a)^2iff b'a=a'b$. Then $abbb=b'a'bb=b'b'ab=b'b'b'a'implies (abbb)^2=e$, so you are actually trying to find $langle a,bmid b'a'=ab, b'a=a'brangle$.
$endgroup$
– user2345215
Dec 29 '14 at 9:29
1
1
$begingroup$
. . . as well as @janmarqz (and any other user for that matter).
$endgroup$
– Shaun
Dec 11 '18 at 0:02
$begingroup$
. . . as well as @janmarqz (and any other user for that matter).
$endgroup$
– Shaun
Dec 11 '18 at 0:02
1
1
$begingroup$
Yes your presentation defines a group that is isomorphic to the original.
$endgroup$
– Derek Holt
Dec 11 '18 at 8:13
$begingroup$
Yes your presentation defines a group that is isomorphic to the original.
$endgroup$
– Derek Holt
Dec 11 '18 at 8:13
|
show 7 more comments
1 Answer
1
active
oldest
votes
2
$begingroup$
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$langle a, ymid y^2, y^{-1}a^2=a^{-2}yrangle,$$
where $y=ab$. This presentation is equivalent (via Tietze transformations) to
$$langle a, ymid y^2, y^{-1}a^2y=a^{-2}rangle,tag{P}$$
which defines the group
$$boxed{operatorname{BS}(2, -2)/Y,}tag{1}$$
where $operatorname{BS}(2, -2)$ is a Baumslag-Solitar group and $Y$ is given by the normal subgroup $$langle y^2rangle^{operatorname{BS}(2, -2)}$$ of $operatorname{BS}(2, -2)$ generated by $y^2$.
Since $G$ and $(1)$ are defined by equivalent presentations, they are isomorphic.
NB: I could be wrong.
NB 2: I think $(text{P})$ could be rewritten as an amalgamated free product. That might define a better (or even more accurate) group that $G$ is isomorphic to.
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
$endgroup$
2
$begingroup$
This group splits as $1rightarrowmathbb{Z}rightarrow Grightarrow D_{infty}rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=langle a^2rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group).
$endgroup$
– user1729
Dec 12 '18 at 12:28
1
$begingroup$
@janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup...").
$endgroup$
– user1729
Dec 13 '18 at 15:05
1
$begingroup$
@Shaun I believe the normal closure of $y^2$, $Y_p:=langlelangle y^2ranglerangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $langle arangle$, and if $Y_pcaplangle arangle^gneq1$ for some $gin BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_pcaplangle arangleneq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment.
$endgroup$
– user1729
Dec 13 '18 at 15:07
1
$begingroup$
@janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $langle a, y; y^{-1}a^2y=a^2rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $langle y^2rangle^{BS(2, -2)}$ (I much prefer the notation $langlelangle y^2ranglerangle$, but anyway). The question in Shaun's comment is: describe this kernel.
$endgroup$
– user1729
Dec 13 '18 at 15:21
1
$begingroup$
(If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).)
$endgroup$
– user1729
Dec 13 '18 at 15:24
|
show 10 more comments
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1 Answer
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1 Answer
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2
$begingroup$
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$langle a, ymid y^2, y^{-1}a^2=a^{-2}yrangle,$$
where $y=ab$. This presentation is equivalent (via Tietze transformations) to
$$langle a, ymid y^2, y^{-1}a^2y=a^{-2}rangle,tag{P}$$
which defines the group
$$boxed{operatorname{BS}(2, -2)/Y,}tag{1}$$
where $operatorname{BS}(2, -2)$ is a Baumslag-Solitar group and $Y$ is given by the normal subgroup $$langle y^2rangle^{operatorname{BS}(2, -2)}$$ of $operatorname{BS}(2, -2)$ generated by $y^2$.
Since $G$ and $(1)$ are defined by equivalent presentations, they are isomorphic.
NB: I could be wrong.
NB 2: I think $(text{P})$ could be rewritten as an amalgamated free product. That might define a better (or even more accurate) group that $G$ is isomorphic to.
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
$endgroup$
2
$begingroup$
This group splits as $1rightarrowmathbb{Z}rightarrow Grightarrow D_{infty}rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=langle a^2rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group).
$endgroup$
– user1729
Dec 12 '18 at 12:28
1
$begingroup$
@janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup...").
$endgroup$
– user1729
Dec 13 '18 at 15:05
1
$begingroup$
@Shaun I believe the normal closure of $y^2$, $Y_p:=langlelangle y^2ranglerangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $langle arangle$, and if $Y_pcaplangle arangle^gneq1$ for some $gin BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_pcaplangle arangleneq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment.
$endgroup$
– user1729
Dec 13 '18 at 15:07
1
$begingroup$
@janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $langle a, y; y^{-1}a^2y=a^2rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $langle y^2rangle^{BS(2, -2)}$ (I much prefer the notation $langlelangle y^2ranglerangle$, but anyway). The question in Shaun's comment is: describe this kernel.
$endgroup$
– user1729
Dec 13 '18 at 15:21
1
$begingroup$
(If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).)
$endgroup$
– user1729
Dec 13 '18 at 15:24
|
show 10 more comments
2
$begingroup$
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$langle a, ymid y^2, y^{-1}a^2=a^{-2}yrangle,$$
where $y=ab$. This presentation is equivalent (via Tietze transformations) to
$$langle a, ymid y^2, y^{-1}a^2y=a^{-2}rangle,tag{P}$$
which defines the group
$$boxed{operatorname{BS}(2, -2)/Y,}tag{1}$$
where $operatorname{BS}(2, -2)$ is a Baumslag-Solitar group and $Y$ is given by the normal subgroup $$langle y^2rangle^{operatorname{BS}(2, -2)}$$ of $operatorname{BS}(2, -2)$ generated by $y^2$.
Since $G$ and $(1)$ are defined by equivalent presentations, they are isomorphic.
NB: I could be wrong.
NB 2: I think $(text{P})$ could be rewritten as an amalgamated free product. That might define a better (or even more accurate) group that $G$ is isomorphic to.
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
$endgroup$
2
$begingroup$
This group splits as $1rightarrowmathbb{Z}rightarrow Grightarrow D_{infty}rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=langle a^2rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group).
$endgroup$
– user1729
Dec 12 '18 at 12:28
1
$begingroup$
@janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup...").
$endgroup$
– user1729
Dec 13 '18 at 15:05
1
$begingroup$
@Shaun I believe the normal closure of $y^2$, $Y_p:=langlelangle y^2ranglerangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $langle arangle$, and if $Y_pcaplangle arangle^gneq1$ for some $gin BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_pcaplangle arangleneq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment.
$endgroup$
– user1729
Dec 13 '18 at 15:07
1
$begingroup$
@janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $langle a, y; y^{-1}a^2y=a^2rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $langle y^2rangle^{BS(2, -2)}$ (I much prefer the notation $langlelangle y^2ranglerangle$, but anyway). The question in Shaun's comment is: describe this kernel.
$endgroup$
– user1729
Dec 13 '18 at 15:21
1
$begingroup$
(If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).)
$endgroup$
– user1729
Dec 13 '18 at 15:24
|
show 10 more comments
2
2
2
$begingroup$
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$langle a, ymid y^2, y^{-1}a^2=a^{-2}yrangle,$$
where $y=ab$. This presentation is equivalent (via Tietze transformations) to
$$langle a, ymid y^2, y^{-1}a^2y=a^{-2}rangle,tag{P}$$
which defines the group
$$boxed{operatorname{BS}(2, -2)/Y,}tag{1}$$
where $operatorname{BS}(2, -2)$ is a Baumslag-Solitar group and $Y$ is given by the normal subgroup $$langle y^2rangle^{operatorname{BS}(2, -2)}$$ of $operatorname{BS}(2, -2)$ generated by $y^2$.
Since $G$ and $(1)$ are defined by equivalent presentations, they are isomorphic.
NB: I could be wrong.
NB 2: I think $(text{P})$ could be rewritten as an amalgamated free product. That might define a better (or even more accurate) group that $G$ is isomorphic to.
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
$endgroup$
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$langle a, ymid y^2, y^{-1}a^2=a^{-2}yrangle,$$
where $y=ab$. This presentation is equivalent (via Tietze transformations) to
$$langle a, ymid y^2, y^{-1}a^2y=a^{-2}rangle,tag{P}$$
which defines the group
$$boxed{operatorname{BS}(2, -2)/Y,}tag{1}$$
where $operatorname{BS}(2, -2)$ is a Baumslag-Solitar group and $Y$ is given by the normal subgroup $$langle y^2rangle^{operatorname{BS}(2, -2)}$$ of $operatorname{BS}(2, -2)$ generated by $y^2$.
Since $G$ and $(1)$ are defined by equivalent presentations, they are isomorphic.
NB: I could be wrong.
NB 2: I think $(text{P})$ could be rewritten as an amalgamated free product. That might define a better (or even more accurate) group that $G$ is isomorphic to.
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
edited Dec 12 '18 at 11:01
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
answered Dec 10 '18 at 23:57
ShaunShaun
9,025113682
9,025113682
2
$begingroup$
This group splits as $1rightarrowmathbb{Z}rightarrow Grightarrow D_{infty}rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=langle a^2rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group).
$endgroup$
– user1729
Dec 12 '18 at 12:28
1
$begingroup$
@janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup...").
$endgroup$
– user1729
Dec 13 '18 at 15:05
1
$begingroup$
@Shaun I believe the normal closure of $y^2$, $Y_p:=langlelangle y^2ranglerangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $langle arangle$, and if $Y_pcaplangle arangle^gneq1$ for some $gin BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_pcaplangle arangleneq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment.
$endgroup$
– user1729
Dec 13 '18 at 15:07
1
$begingroup$
@janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $langle a, y; y^{-1}a^2y=a^2rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $langle y^2rangle^{BS(2, -2)}$ (I much prefer the notation $langlelangle y^2ranglerangle$, but anyway). The question in Shaun's comment is: describe this kernel.
$endgroup$
– user1729
Dec 13 '18 at 15:21
1
$begingroup$
(If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).)
$endgroup$
– user1729
Dec 13 '18 at 15:24
|
show 10 more comments
2
$begingroup$
This group splits as $1rightarrowmathbb{Z}rightarrow Grightarrow D_{infty}rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=langle a^2rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group).
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– user1729
Dec 12 '18 at 12:28
1
$begingroup$
@janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup...").
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– user1729
Dec 13 '18 at 15:05
1
$begingroup$
@Shaun I believe the normal closure of $y^2$, $Y_p:=langlelangle y^2ranglerangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $langle arangle$, and if $Y_pcaplangle arangle^gneq1$ for some $gin BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_pcaplangle arangleneq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment.
$endgroup$
– user1729
Dec 13 '18 at 15:07
1
$begingroup$
@janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $langle a, y; y^{-1}a^2y=a^2rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $langle y^2rangle^{BS(2, -2)}$ (I much prefer the notation $langlelangle y^2ranglerangle$, but anyway). The question in Shaun's comment is: describe this kernel.
$endgroup$
– user1729
Dec 13 '18 at 15:21
1
$begingroup$
(If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).)
$endgroup$
– user1729
Dec 13 '18 at 15:24
2
2
$begingroup$
This group splits as $1rightarrowmathbb{Z}rightarrow Grightarrow D_{infty}rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=langle a^2rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group).
$endgroup$
– user1729
Dec 12 '18 at 12:28
$begingroup$
This group splits as $1rightarrowmathbb{Z}rightarrow Grightarrow D_{infty}rightarrow1$. In particular, $G$ is not virtually cyclic. To see this splitting, begin with the presentation (P). It is relatively easy to show that $a^2$ generates a normal subgroup $N$ of $G$ ($y^2=1$ so $y$ normalises $N=langle a^2rangle$, and obviously $a$ normalises $N$). Killing $a^2$ in the presentation (P) gives you the free product of two cyclic groups of order two, which is just $D_{infty}$ under a different name. This sequence does not split (the element $a$ has order two in the image group).
$endgroup$
– user1729
Dec 12 '18 at 12:28
1
1
$begingroup$
@janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup...").
$endgroup$
– user1729
Dec 13 '18 at 15:05
$begingroup$
@janmarqz $H^G$ is sometimes used to denote the normal closure of the subgroup $H$. This, I believe, is the meaning here (as he says "the normal subgroup...").
$endgroup$
– user1729
Dec 13 '18 at 15:05
1
1
$begingroup$
@Shaun I believe the normal closure of $y^2$, $Y_p:=langlelangle y^2ranglerangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $langle arangle$, and if $Y_pcaplangle arangle^gneq1$ for some $gin BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_pcaplangle arangleneq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment.
$endgroup$
– user1729
Dec 13 '18 at 15:07
$begingroup$
@Shaun I believe the normal closure of $y^2$, $Y_p:=langlelangle y^2ranglerangle$, is free of infinite rank. I am pretty sure it is free: as $BS(2, -2)$ is an HNN-extension any non-free subgroup must intersect a conjugate of the base group. Here the base group is $langle arangle$, and if $Y_pcaplangle arangle^gneq1$ for some $gin BS(2, -2)$ then (by normality of $Y_p$) we have that $Y_pcaplangle arangleneq1$. So $a$ has finite order in $G=BS(2, -2)/Y_p$. Now, I am not positive that $a$ has infinite order in $G$, but I was yesterday when I wrote the above comment.
$endgroup$
– user1729
Dec 13 '18 at 15:07
1
1
$begingroup$
@janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $langle a, y; y^{-1}a^2y=a^2rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $langle y^2rangle^{BS(2, -2)}$ (I much prefer the notation $langlelangle y^2ranglerangle$, but anyway). The question in Shaun's comment is: describe this kernel.
$endgroup$
– user1729
Dec 13 '18 at 15:21
$begingroup$
@janmarqz We are not working in the group $G$ here. Let $BS(2, -2)$ denote the group with presentation $langle a, y; y^{-1}a^2y=a^2rangle$. Then $BS(2, -2)$ surjects onto $G$, and this map has kernel $langle y^2rangle^{BS(2, -2)}$ (I much prefer the notation $langlelangle y^2ranglerangle$, but anyway). The question in Shaun's comment is: describe this kernel.
$endgroup$
– user1729
Dec 13 '18 at 15:21
1
1
$begingroup$
(If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).)
$endgroup$
– user1729
Dec 13 '18 at 15:24
$begingroup$
(If $Y_p$ is free then one should be able to prove that it is free of infinite rank by similar methods to the proof that normal subgroups of infinite index in free groups are free of infinite rank: see here or here (includes pictures).)
$endgroup$
– user1729
Dec 13 '18 at 15:24
|
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$begingroup$
$(ab)^2$ means $(ab)^2=e$ ?
$endgroup$
– mesel
Dec 29 '14 at 2:35
$begingroup$
Yeah, sorry. Edited.
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– theelk801
Dec 29 '14 at 2:40
1
$begingroup$
I don't know if it helps, but here it goes: First, $e=(ab)^2iff b'a'=ab$ and $e=(abaa)^2=(b'a'aa)^2=(b'a)^2iff b'a=a'b$. Then $abbb=b'a'bb=b'b'ab=b'b'b'a'implies (abbb)^2=e$, so you are actually trying to find $langle a,bmid b'a'=ab, b'a=a'brangle$.
$endgroup$
– user2345215
Dec 29 '14 at 9:29
1
$begingroup$
. . . as well as @janmarqz (and any other user for that matter).
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– Shaun
Dec 11 '18 at 0:02
1
$begingroup$
Yes your presentation defines a group that is isomorphic to the original.
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– Derek Holt
Dec 11 '18 at 8:13