Find a group $G$ with $ain G$ such that $|a|=6$ but $C_G(a)neq C_G(a^3)$.
This is part of Exercise 46 of Chapter 3 of Gallian's "Contemporary Abstract Algebra".
Notation 1: The centraliser of $g$ in a group $G$ is denoted $C_G(g)$.
Notation 2: The dihedral group $D_n$ has order $2n$.
I'm interested in how one would answer the following question using the tools available in the textbook prior to it.
Find a group $G$ with $ain G$ such that $|a|=6$ but $C_G(a)neq C_G(a^3)$.
NOTE: The textbook doesn't cover permutation groups prior to this.
Thoughts:
I understand that the group $H$ given by the presentation $$langle a, xmid a^6, a^3x=xa^3rangle$$ would suffice as $xin C_H(a^3)setminus C_H(a)$ by construction; however, presentations aren't covered in the preceding material. It's not clear to me how I would exploit my understanding of this perspective to construct/name a group in the spirit of the question.
Perhaps a dihedral group $D_n$ would do for some $n$ s.t. $6mid n$ but I'm not sure. These groups are defined in terms of reflections & rotations in the text, but the way I see them is, again, as in terms of a few different presentations for them. From my comment below: The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed. (Emphasis added.)
Please help :)
group-theory examples-counterexamples alternative-proof dihedral-groups
asked Jan 2 at 15:07
Shaun
8,810113680
|
show 1 more comment
This is part of Exercise 46 of Chapter 3 of Gallian's "Contemporary Abstract Algebra".
Notation 1: The centraliser of $g$ in a group $G$ is denoted $C_G(g)$.
Notation 2: The dihedral group $D_n$ has order $2n$.
I'm interested in how one would answer the following question using the tools available in the textbook prior to it.
Find a group $G$ with $ain G$ such that $|a|=6$ but $C_G(a)neq C_G(a^3)$.
NOTE: The textbook doesn't cover permutation groups prior to this.
Thoughts:
I understand that the group $H$ given by the presentation $$langle a, xmid a^6, a^3x=xa^3rangle$$ would suffice as $xin C_H(a^3)setminus C_H(a)$ by construction; however, presentations aren't covered in the preceding material. It's not clear to me how I would exploit my understanding of this perspective to construct/name a group in the spirit of the question.
Perhaps a dihedral group $D_n$ would do for some $n$ s.t. $6mid n$ but I'm not sure. These groups are defined in terms of reflections & rotations in the text, but the way I see them is, again, as in terms of a few different presentations for them. From my comment below: The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed. (Emphasis added.)
Please help :)
group-theory examples-counterexamples alternative-proof dihedral-groups
asked Jan 2 at 15:07
Shaun
8,810113680
Have you tried the dihedral group $D_6$ of order 12?
– Derek Holt
Jan 2 at 15:21
@DerekHolt Not yet. I'm on it now. The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed.
– Shaun
Jan 2 at 15:22
2
@Shaun I suspect that reasoning from the geometry will be easier than reasoning from the presentation.
– Ethan Bolker
Jan 2 at 15:45
4
Let $a$ be a rotation through $60^circ$. Then $a$ has order $6$ and does not commute with a reflection. But $a^3$ is a rotation through $180^circ$ and does commute with all reflections - it lies in the centrel of $D_6$.
– Derek Holt
Jan 2 at 16:11
2
Because thinking geometrically you can identify the orders of all the elements at a glance: flips have order $2$, rotations $2$ or $3$ or $6$. This information is buried when you use the presentation. See @DerekHolt 's comment.
– Ethan Bolker
Jan 2 at 16:13
|
show 1 more comment
This is part of Exercise 46 of Chapter 3 of Gallian's "Contemporary Abstract Algebra".
Notation 1: The centraliser of $g$ in a group $G$ is denoted $C_G(g)$.
Notation 2: The dihedral group $D_n$ has order $2n$.
I'm interested in how one would answer the following question using the tools available in the textbook prior to it.
Find a group $G$ with $ain G$ such that $|a|=6$ but $C_G(a)neq C_G(a^3)$.
NOTE: The textbook doesn't cover permutation groups prior to this.
Thoughts:
I understand that the group $H$ given by the presentation $$langle a, xmid a^6, a^3x=xa^3rangle$$ would suffice as $xin C_H(a^3)setminus C_H(a)$ by construction; however, presentations aren't covered in the preceding material. It's not clear to me how I would exploit my understanding of this perspective to construct/name a group in the spirit of the question.
Perhaps a dihedral group $D_n$ would do for some $n$ s.t. $6mid n$ but I'm not sure. These groups are defined in terms of reflections & rotations in the text, but the way I see them is, again, as in terms of a few different presentations for them. From my comment below: The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed. (Emphasis added.)
Please help :)
group-theory examples-counterexamples alternative-proof dihedral-groups
asked Jan 2 at 15:07
Shaun
8,810113680
This is part of Exercise 46 of Chapter 3 of Gallian's "Contemporary Abstract Algebra".
Notation 1: The centraliser of $g$ in a group $G$ is denoted $C_G(g)$.
Notation 2: The dihedral group $D_n$ has order $2n$.
I'm interested in how one would answer the following question using the tools available in the textbook prior to it.
Find a group $G$ with $ain G$ such that $|a|=6$ but $C_G(a)neq C_G(a^3)$.
NOTE: The textbook doesn't cover permutation groups prior to this.
Thoughts:
I understand that the group $H$ given by the presentation $$langle a, xmid a^6, a^3x=xa^3rangle$$ would suffice as $xin C_H(a^3)setminus C_H(a)$ by construction; however, presentations aren't covered in the preceding material. It's not clear to me how I would exploit my understanding of this perspective to construct/name a group in the spirit of the question.
Perhaps a dihedral group $D_n$ would do for some $n$ s.t. $6mid n$ but I'm not sure. These groups are defined in terms of reflections & rotations in the text, but the way I see them is, again, as in terms of a few different presentations for them. From my comment below: The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed. (Emphasis added.)
Please help :)
group-theory examples-counterexamples alternative-proof dihedral-groups
group-theory examples-counterexamples alternative-proof dihedral-groups
asked Jan 2 at 15:07
Shaun
8,810113680
asked Jan 2 at 15:07
Shaun
8,810113680
edited Jan 3 at 17:08
asked Jan 2 at 15:07
Shaun
8,810113680
asked Jan 2 at 15:07
Shaun
8,810113680
asked Jan 2 at 15:07
Shaun
8,810113680
8,810113680
Have you tried the dihedral group $D_6$ of order 12?
– Derek Holt
Jan 2 at 15:21
@DerekHolt Not yet. I'm on it now. The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed.
– Shaun
Jan 2 at 15:22
2
@Shaun I suspect that reasoning from the geometry will be easier than reasoning from the presentation.
– Ethan Bolker
Jan 2 at 15:45
4
Let $a$ be a rotation through $60^circ$. Then $a$ has order $6$ and does not commute with a reflection. But $a^3$ is a rotation through $180^circ$ and does commute with all reflections - it lies in the centrel of $D_6$.
– Derek Holt
Jan 2 at 16:11
2
Because thinking geometrically you can identify the orders of all the elements at a glance: flips have order $2$, rotations $2$ or $3$ or $6$. This information is buried when you use the presentation. See @DerekHolt 's comment.
– Ethan Bolker
Jan 2 at 16:13
|
show 1 more comment
Have you tried the dihedral group $D_6$ of order 12?
– Derek Holt
Jan 2 at 15:21
@DerekHolt Not yet. I'm on it now. The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed.
– Shaun
Jan 2 at 15:22
2
@Shaun I suspect that reasoning from the geometry will be easier than reasoning from the presentation.
– Ethan Bolker
Jan 2 at 15:45
4
Let $a$ be a rotation through $60^circ$. Then $a$ has order $6$ and does not commute with a reflection. But $a^3$ is a rotation through $180^circ$ and does commute with all reflections - it lies in the centrel of $D_6$.
– Derek Holt
Jan 2 at 16:11
2
Because thinking geometrically you can identify the orders of all the elements at a glance: flips have order $2$, rotations $2$ or $3$ or $6$. This information is buried when you use the presentation. See @DerekHolt 's comment.
– Ethan Bolker
Jan 2 at 16:13
Have you tried the dihedral group $D_6$ of order 12?
– Derek Holt
Jan 2 at 15:21
Have you tried the dihedral group $D_6$ of order 12?
– Derek Holt
Jan 2 at 15:21
@DerekHolt Not yet. I'm on it now. The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed.
– Shaun
Jan 2 at 15:22
@DerekHolt Not yet. I'm on it now. The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed.
– Shaun
Jan 2 at 15:22
2
2
@Shaun I suspect that reasoning from the geometry will be easier than reasoning from the presentation.
– Ethan Bolker
Jan 2 at 15:45
@Shaun I suspect that reasoning from the geometry will be easier than reasoning from the presentation.
– Ethan Bolker
Jan 2 at 15:45
4
4
Let $a$ be a rotation through $60^circ$. Then $a$ has order $6$ and does not commute with a reflection. But $a^3$ is a rotation through $180^circ$ and does commute with all reflections - it lies in the centrel of $D_6$.
– Derek Holt
Jan 2 at 16:11
Let $a$ be a rotation through $60^circ$. Then $a$ has order $6$ and does not commute with a reflection. But $a^3$ is a rotation through $180^circ$ and does commute with all reflections - it lies in the centrel of $D_6$.
– Derek Holt
Jan 2 at 16:11
2
2
Because thinking geometrically you can identify the orders of all the elements at a glance: flips have order $2$, rotations $2$ or $3$ or $6$. This information is buried when you use the presentation. See @DerekHolt 's comment.
– Ethan Bolker
Jan 2 at 16:13
Because thinking geometrically you can identify the orders of all the elements at a glance: flips have order $2$, rotations $2$ or $3$ or $6$. This information is buried when you use the presentation. See @DerekHolt 's comment.
– Ethan Bolker
Jan 2 at 16:13
|
show 1 more comment
2 Answers
2
active
oldest
votes
Let $D_{12}$ be the dihedral group, presented as follows
$$D_{12}=leftlangle x,y;:;x^2=y^6=1,quad xyx=y^{-1}rightrangle.$$
Consider $a=xy$, it is easy to see that its order is $6$. Now, we have that $xyx=y^{-1}neq y$ but $xy^3x=y^{-3}=y^3$ (note that $x=x^{-1}$). Therefore, $x$ lies on the centrlizer of $y^3$ but not in the centralizer of $y$.
Not covered in the preceding material but:
Consider $G=S_8$ the symmetric group. Let $sigma=(1 2 3 4 5 6)$ be a permutation, it is clear that $|sigma|=6$, note that $sigma^3=(1 4)$. We have that $(5 6)$ lies in the centralizer of $sigma^3$ but does not lie on the centralizer of $sigma$.
edited Jan 2 at 16:02
Shaun
8,810113680
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
Permutation groups aren't covered in the textbook prior to the question.
– Shaun
Jan 2 at 15:36
Why are you studying a introductory book that does not cover that :S
– José Alejandro Aburto Araneda
Jan 2 at 15:37
1
It does, just not yet.
– Shaun
Jan 2 at 15:40
1
I've reviewed your book, and it studies the dihedral group, and if you look closer you can deduce this representation (he is using representations in a hidden way)
– José Alejandro Aburto Araneda
Jan 2 at 15:58
1
My bad. But it's not hard to develop that. I recommend you to "play" more with dihedral groups, anyways, it will be important as examples in your studies
– José Alejandro Aburto Araneda
Jan 2 at 16:05
|
show 6 more comments
Have you already covered the symmetric groups $S_n$ in your course? If you can use them, then an idea could be to employ the fact that non-overlapping cycles commute with each other. For example, if $a$ is a product of two independent cycles of lengths $2$ and $3$, then the latter one will disappear in $a^3$, and thus its centralizer will be different.
answered Jan 2 at 15:29
zipirovich
11.1k11631
No, they're not covered in the book so far. I have plenty of experience with them though. Thank you nonetheless.
– Shaun
Jan 2 at 15:30
2
It was a helpful answer nonetheless, I don't think the downvote is fair. You only specified that you would like to use the tools available to you, but not really which set of tools you have. You instead can choose to not vote.
– Wesley Strik
Jan 2 at 15:35
2
Answers should address the question as asked. If somebody finds a question unsuitable they are free not to answer it, and even to vote negatively on the question. What is not alright is to answer the question that one would like to answer (even if it is a reasonable one). The restriction to answer an exercise from a text in the way(s) intended is a coherent restriction. It'd been better to be right away more explicit, but "I'm interested in how one would answer the following question using the tools available in the text prior to it." is in principle clear (though implicit).
– quid♦
Jan 2 at 15:52
1
I've seen the book, and in that chapter he has only seen abelian groups, but the dihedral, but it seems he does not understand the representation given in the book
– José Alejandro Aburto Araneda
Jan 2 at 16:02
1
@zipirovich it was there in the very first revision It is possible that the absolutely first version did not contain it as there is a grace period for edits, but it definitely was there well before an answer was given.
– quid♦
Jan 2 at 16:10
|
show 3 more comments
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2 Answers
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2 Answers
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Let $D_{12}$ be the dihedral group, presented as follows
$$D_{12}=leftlangle x,y;:;x^2=y^6=1,quad xyx=y^{-1}rightrangle.$$
Consider $a=xy$, it is easy to see that its order is $6$. Now, we have that $xyx=y^{-1}neq y$ but $xy^3x=y^{-3}=y^3$ (note that $x=x^{-1}$). Therefore, $x$ lies on the centrlizer of $y^3$ but not in the centralizer of $y$.
Not covered in the preceding material but:
Consider $G=S_8$ the symmetric group. Let $sigma=(1 2 3 4 5 6)$ be a permutation, it is clear that $|sigma|=6$, note that $sigma^3=(1 4)$. We have that $(5 6)$ lies in the centralizer of $sigma^3$ but does not lie on the centralizer of $sigma$.
edited Jan 2 at 16:02
Shaun
8,810113680
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
Permutation groups aren't covered in the textbook prior to the question.
– Shaun
Jan 2 at 15:36
Why are you studying a introductory book that does not cover that :S
– José Alejandro Aburto Araneda
Jan 2 at 15:37
1
It does, just not yet.
– Shaun
Jan 2 at 15:40
1
I've reviewed your book, and it studies the dihedral group, and if you look closer you can deduce this representation (he is using representations in a hidden way)
– José Alejandro Aburto Araneda
Jan 2 at 15:58
1
My bad. But it's not hard to develop that. I recommend you to "play" more with dihedral groups, anyways, it will be important as examples in your studies
– José Alejandro Aburto Araneda
Jan 2 at 16:05
|
show 6 more comments
Let $D_{12}$ be the dihedral group, presented as follows
$$D_{12}=leftlangle x,y;:;x^2=y^6=1,quad xyx=y^{-1}rightrangle.$$
Consider $a=xy$, it is easy to see that its order is $6$. Now, we have that $xyx=y^{-1}neq y$ but $xy^3x=y^{-3}=y^3$ (note that $x=x^{-1}$). Therefore, $x$ lies on the centrlizer of $y^3$ but not in the centralizer of $y$.
Not covered in the preceding material but:
Consider $G=S_8$ the symmetric group. Let $sigma=(1 2 3 4 5 6)$ be a permutation, it is clear that $|sigma|=6$, note that $sigma^3=(1 4)$. We have that $(5 6)$ lies in the centralizer of $sigma^3$ but does not lie on the centralizer of $sigma$.
edited Jan 2 at 16:02
Shaun
8,810113680
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
Permutation groups aren't covered in the textbook prior to the question.
– Shaun
Jan 2 at 15:36
Why are you studying a introductory book that does not cover that :S
– José Alejandro Aburto Araneda
Jan 2 at 15:37
1
It does, just not yet.
– Shaun
Jan 2 at 15:40
1
I've reviewed your book, and it studies the dihedral group, and if you look closer you can deduce this representation (he is using representations in a hidden way)
– José Alejandro Aburto Araneda
Jan 2 at 15:58
1
My bad. But it's not hard to develop that. I recommend you to "play" more with dihedral groups, anyways, it will be important as examples in your studies
– José Alejandro Aburto Araneda
Jan 2 at 16:05
|
show 6 more comments
Let $D_{12}$ be the dihedral group, presented as follows
$$D_{12}=leftlangle x,y;:;x^2=y^6=1,quad xyx=y^{-1}rightrangle.$$
Consider $a=xy$, it is easy to see that its order is $6$. Now, we have that $xyx=y^{-1}neq y$ but $xy^3x=y^{-3}=y^3$ (note that $x=x^{-1}$). Therefore, $x$ lies on the centrlizer of $y^3$ but not in the centralizer of $y$.
Not covered in the preceding material but:
Consider $G=S_8$ the symmetric group. Let $sigma=(1 2 3 4 5 6)$ be a permutation, it is clear that $|sigma|=6$, note that $sigma^3=(1 4)$. We have that $(5 6)$ lies in the centralizer of $sigma^3$ but does not lie on the centralizer of $sigma$.
edited Jan 2 at 16:02
Shaun
8,810113680
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
Let $D_{12}$ be the dihedral group, presented as follows
$$D_{12}=leftlangle x,y;:;x^2=y^6=1,quad xyx=y^{-1}rightrangle.$$
Consider $a=xy$, it is easy to see that its order is $6$. Now, we have that $xyx=y^{-1}neq y$ but $xy^3x=y^{-3}=y^3$ (note that $x=x^{-1}$). Therefore, $x$ lies on the centrlizer of $y^3$ but not in the centralizer of $y$.
Not covered in the preceding material but:
Consider $G=S_8$ the symmetric group. Let $sigma=(1 2 3 4 5 6)$ be a permutation, it is clear that $|sigma|=6$, note that $sigma^3=(1 4)$. We have that $(5 6)$ lies in the centralizer of $sigma^3$ but does not lie on the centralizer of $sigma$.
edited Jan 2 at 16:02
Shaun
8,810113680
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
edited Jan 2 at 16:02
Shaun
8,810113680
edited Jan 2 at 16:02
Shaun
8,810113680
edited Jan 2 at 16:02
Shaun
8,810113680
8,810113680
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
answered Jan 2 at 15:36
José Alejandro Aburto Araneda
808110
808110
Permutation groups aren't covered in the textbook prior to the question.
– Shaun
Jan 2 at 15:36
Why are you studying a introductory book that does not cover that :S
– José Alejandro Aburto Araneda
Jan 2 at 15:37
1
It does, just not yet.
– Shaun
Jan 2 at 15:40
1
I've reviewed your book, and it studies the dihedral group, and if you look closer you can deduce this representation (he is using representations in a hidden way)
– José Alejandro Aburto Araneda
Jan 2 at 15:58
1
My bad. But it's not hard to develop that. I recommend you to "play" more with dihedral groups, anyways, it will be important as examples in your studies
– José Alejandro Aburto Araneda
Jan 2 at 16:05
|
show 6 more comments
Permutation groups aren't covered in the textbook prior to the question.
– Shaun
Jan 2 at 15:36
Why are you studying a introductory book that does not cover that :S
– José Alejandro Aburto Araneda
Jan 2 at 15:37
1
It does, just not yet.
– Shaun
Jan 2 at 15:40
1
I've reviewed your book, and it studies the dihedral group, and if you look closer you can deduce this representation (he is using representations in a hidden way)
– José Alejandro Aburto Araneda
Jan 2 at 15:58
1
My bad. But it's not hard to develop that. I recommend you to "play" more with dihedral groups, anyways, it will be important as examples in your studies
– José Alejandro Aburto Araneda
Jan 2 at 16:05
Permutation groups aren't covered in the textbook prior to the question.
– Shaun
Jan 2 at 15:36
Permutation groups aren't covered in the textbook prior to the question.
– Shaun
Jan 2 at 15:36
Why are you studying a introductory book that does not cover that :S
– José Alejandro Aburto Araneda
Jan 2 at 15:37
Why are you studying a introductory book that does not cover that :S
– José Alejandro Aburto Araneda
Jan 2 at 15:37
1
1
It does, just not yet.
– Shaun
Jan 2 at 15:40
It does, just not yet.
– Shaun
Jan 2 at 15:40
1
1
I've reviewed your book, and it studies the dihedral group, and if you look closer you can deduce this representation (he is using representations in a hidden way)
– José Alejandro Aburto Araneda
Jan 2 at 15:58
I've reviewed your book, and it studies the dihedral group, and if you look closer you can deduce this representation (he is using representations in a hidden way)
– José Alejandro Aburto Araneda
Jan 2 at 15:58
1
1
My bad. But it's not hard to develop that. I recommend you to "play" more with dihedral groups, anyways, it will be important as examples in your studies
– José Alejandro Aburto Araneda
Jan 2 at 16:05
My bad. But it's not hard to develop that. I recommend you to "play" more with dihedral groups, anyways, it will be important as examples in your studies
– José Alejandro Aburto Araneda
Jan 2 at 16:05
|
show 6 more comments
Have you already covered the symmetric groups $S_n$ in your course? If you can use them, then an idea could be to employ the fact that non-overlapping cycles commute with each other. For example, if $a$ is a product of two independent cycles of lengths $2$ and $3$, then the latter one will disappear in $a^3$, and thus its centralizer will be different.
answered Jan 2 at 15:29
zipirovich
11.1k11631
No, they're not covered in the book so far. I have plenty of experience with them though. Thank you nonetheless.
– Shaun
Jan 2 at 15:30
2
It was a helpful answer nonetheless, I don't think the downvote is fair. You only specified that you would like to use the tools available to you, but not really which set of tools you have. You instead can choose to not vote.
– Wesley Strik
Jan 2 at 15:35
2
Answers should address the question as asked. If somebody finds a question unsuitable they are free not to answer it, and even to vote negatively on the question. What is not alright is to answer the question that one would like to answer (even if it is a reasonable one). The restriction to answer an exercise from a text in the way(s) intended is a coherent restriction. It'd been better to be right away more explicit, but "I'm interested in how one would answer the following question using the tools available in the text prior to it." is in principle clear (though implicit).
– quid♦
Jan 2 at 15:52
1
I've seen the book, and in that chapter he has only seen abelian groups, but the dihedral, but it seems he does not understand the representation given in the book
– José Alejandro Aburto Araneda
Jan 2 at 16:02
1
@zipirovich it was there in the very first revision It is possible that the absolutely first version did not contain it as there is a grace period for edits, but it definitely was there well before an answer was given.
– quid♦
Jan 2 at 16:10
|
show 3 more comments
Have you already covered the symmetric groups $S_n$ in your course? If you can use them, then an idea could be to employ the fact that non-overlapping cycles commute with each other. For example, if $a$ is a product of two independent cycles of lengths $2$ and $3$, then the latter one will disappear in $a^3$, and thus its centralizer will be different.
answered Jan 2 at 15:29
zipirovich
11.1k11631
No, they're not covered in the book so far. I have plenty of experience with them though. Thank you nonetheless.
– Shaun
Jan 2 at 15:30
2
It was a helpful answer nonetheless, I don't think the downvote is fair. You only specified that you would like to use the tools available to you, but not really which set of tools you have. You instead can choose to not vote.
– Wesley Strik
Jan 2 at 15:35
2
Answers should address the question as asked. If somebody finds a question unsuitable they are free not to answer it, and even to vote negatively on the question. What is not alright is to answer the question that one would like to answer (even if it is a reasonable one). The restriction to answer an exercise from a text in the way(s) intended is a coherent restriction. It'd been better to be right away more explicit, but "I'm interested in how one would answer the following question using the tools available in the text prior to it." is in principle clear (though implicit).
– quid♦
Jan 2 at 15:52
1
I've seen the book, and in that chapter he has only seen abelian groups, but the dihedral, but it seems he does not understand the representation given in the book
– José Alejandro Aburto Araneda
Jan 2 at 16:02
1
@zipirovich it was there in the very first revision It is possible that the absolutely first version did not contain it as there is a grace period for edits, but it definitely was there well before an answer was given.
– quid♦
Jan 2 at 16:10
|
show 3 more comments
Have you already covered the symmetric groups $S_n$ in your course? If you can use them, then an idea could be to employ the fact that non-overlapping cycles commute with each other. For example, if $a$ is a product of two independent cycles of lengths $2$ and $3$, then the latter one will disappear in $a^3$, and thus its centralizer will be different.
answered Jan 2 at 15:29
zipirovich
11.1k11631
Have you already covered the symmetric groups $S_n$ in your course? If you can use them, then an idea could be to employ the fact that non-overlapping cycles commute with each other. For example, if $a$ is a product of two independent cycles of lengths $2$ and $3$, then the latter one will disappear in $a^3$, and thus its centralizer will be different.
answered Jan 2 at 15:29
zipirovich
11.1k11631
answered Jan 2 at 15:29
zipirovich
11.1k11631
answered Jan 2 at 15:29
zipirovich
11.1k11631
answered Jan 2 at 15:29
zipirovich
11.1k11631
11.1k11631
No, they're not covered in the book so far. I have plenty of experience with them though. Thank you nonetheless.
– Shaun
Jan 2 at 15:30
2
It was a helpful answer nonetheless, I don't think the downvote is fair. You only specified that you would like to use the tools available to you, but not really which set of tools you have. You instead can choose to not vote.
– Wesley Strik
Jan 2 at 15:35
2
Answers should address the question as asked. If somebody finds a question unsuitable they are free not to answer it, and even to vote negatively on the question. What is not alright is to answer the question that one would like to answer (even if it is a reasonable one). The restriction to answer an exercise from a text in the way(s) intended is a coherent restriction. It'd been better to be right away more explicit, but "I'm interested in how one would answer the following question using the tools available in the text prior to it." is in principle clear (though implicit).
– quid♦
Jan 2 at 15:52
1
I've seen the book, and in that chapter he has only seen abelian groups, but the dihedral, but it seems he does not understand the representation given in the book
– José Alejandro Aburto Araneda
Jan 2 at 16:02
1
@zipirovich it was there in the very first revision It is possible that the absolutely first version did not contain it as there is a grace period for edits, but it definitely was there well before an answer was given.
– quid♦
Jan 2 at 16:10
|
show 3 more comments
No, they're not covered in the book so far. I have plenty of experience with them though. Thank you nonetheless.
– Shaun
Jan 2 at 15:30
2
It was a helpful answer nonetheless, I don't think the downvote is fair. You only specified that you would like to use the tools available to you, but not really which set of tools you have. You instead can choose to not vote.
– Wesley Strik
Jan 2 at 15:35
2
Answers should address the question as asked. If somebody finds a question unsuitable they are free not to answer it, and even to vote negatively on the question. What is not alright is to answer the question that one would like to answer (even if it is a reasonable one). The restriction to answer an exercise from a text in the way(s) intended is a coherent restriction. It'd been better to be right away more explicit, but "I'm interested in how one would answer the following question using the tools available in the text prior to it." is in principle clear (though implicit).
– quid♦
Jan 2 at 15:52
1
I've seen the book, and in that chapter he has only seen abelian groups, but the dihedral, but it seems he does not understand the representation given in the book
– José Alejandro Aburto Araneda
Jan 2 at 16:02
1
@zipirovich it was there in the very first revision It is possible that the absolutely first version did not contain it as there is a grace period for edits, but it definitely was there well before an answer was given.
– quid♦
Jan 2 at 16:10
No, they're not covered in the book so far. I have plenty of experience with them though. Thank you nonetheless.
– Shaun
Jan 2 at 15:30
No, they're not covered in the book so far. I have plenty of experience with them though. Thank you nonetheless.
– Shaun
Jan 2 at 15:30
2
2
It was a helpful answer nonetheless, I don't think the downvote is fair. You only specified that you would like to use the tools available to you, but not really which set of tools you have. You instead can choose to not vote.
– Wesley Strik
Jan 2 at 15:35
It was a helpful answer nonetheless, I don't think the downvote is fair. You only specified that you would like to use the tools available to you, but not really which set of tools you have. You instead can choose to not vote.
– Wesley Strik
Jan 2 at 15:35
2
2
Answers should address the question as asked. If somebody finds a question unsuitable they are free not to answer it, and even to vote negatively on the question. What is not alright is to answer the question that one would like to answer (even if it is a reasonable one). The restriction to answer an exercise from a text in the way(s) intended is a coherent restriction. It'd been better to be right away more explicit, but "I'm interested in how one would answer the following question using the tools available in the text prior to it." is in principle clear (though implicit).
– quid♦
Jan 2 at 15:52
Answers should address the question as asked. If somebody finds a question unsuitable they are free not to answer it, and even to vote negatively on the question. What is not alright is to answer the question that one would like to answer (even if it is a reasonable one). The restriction to answer an exercise from a text in the way(s) intended is a coherent restriction. It'd been better to be right away more explicit, but "I'm interested in how one would answer the following question using the tools available in the text prior to it." is in principle clear (though implicit).
– quid♦
Jan 2 at 15:52
1
1
I've seen the book, and in that chapter he has only seen abelian groups, but the dihedral, but it seems he does not understand the representation given in the book
– José Alejandro Aburto Araneda
Jan 2 at 16:02
I've seen the book, and in that chapter he has only seen abelian groups, but the dihedral, but it seems he does not understand the representation given in the book
– José Alejandro Aburto Araneda
Jan 2 at 16:02
1
1
@zipirovich it was there in the very first revision It is possible that the absolutely first version did not contain it as there is a grace period for edits, but it definitely was there well before an answer was given.
– quid♦
Jan 2 at 16:10
@zipirovich it was there in the very first revision It is possible that the absolutely first version did not contain it as there is a grace period for edits, but it definitely was there well before an answer was given.
– quid♦
Jan 2 at 16:10
|
show 3 more comments
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Have you tried the dihedral group $D_6$ of order 12?
– Derek Holt
Jan 2 at 15:21
@DerekHolt Not yet. I'm on it now. The problem is that it's defined in terms of a regular $n$-gon in this textbook. I'm used to just using a presentation. If I'm to use $D_6$ here, then I'm to argue from the perspective of the book; I'm not so sure how to proceed.
– Shaun
Jan 2 at 15:22
2
@Shaun I suspect that reasoning from the geometry will be easier than reasoning from the presentation.
– Ethan Bolker
Jan 2 at 15:45
4
Let $a$ be a rotation through $60^circ$. Then $a$ has order $6$ and does not commute with a reflection. But $a^3$ is a rotation through $180^circ$ and does commute with all reflections - it lies in the centrel of $D_6$.
– Derek Holt
Jan 2 at 16:11
2
Because thinking geometrically you can identify the orders of all the elements at a glance: flips have order $2$, rotations $2$ or $3$ or $6$. This information is buried when you use the presentation. See @DerekHolt 's comment.
– Ethan Bolker
Jan 2 at 16:13