Finding a presentation of the quarternion group. (When do I know if I have given enough relations?)
I was working through presentation of the quaternion group (with element $8$), and I let $a = i$ and $b = j$. I immediately said $a^4 = b^4 = 1$, and $ab^2 a = 1$.
Since I have a relation for each generator and between the generator, I figured I have the whole presentation. However, when I looked up the presentation of the quaternion group, it was given as
$$Q=langle F{a,b}mid a^4=b^4=a^2b^2=1 , b^{-1} a d = a^{-1}rangle.tag{1}$$
It is hard for me to see whether my initial third relation is a mixture of 3rd or 4th relation given by $(1)$.
Also, when do I know if I have given enough relations? Do I have to just write it down and see?
Finding a presentation of a group seems quite tedious!
group-theory group-presentation
edited Dec 11 at 7:50
Shaun
8,339113578
asked Oct 2 '14 at 7:40
Quantization
20418
add a comment |
I was working through presentation of the quaternion group (with element $8$), and I let $a = i$ and $b = j$. I immediately said $a^4 = b^4 = 1$, and $ab^2 a = 1$.
Since I have a relation for each generator and between the generator, I figured I have the whole presentation. However, when I looked up the presentation of the quaternion group, it was given as
$$Q=langle F{a,b}mid a^4=b^4=a^2b^2=1 , b^{-1} a d = a^{-1}rangle.tag{1}$$
It is hard for me to see whether my initial third relation is a mixture of 3rd or 4th relation given by $(1)$.
Also, when do I know if I have given enough relations? Do I have to just write it down and see?
Finding a presentation of a group seems quite tedious!
group-theory group-presentation
edited Dec 11 at 7:50
Shaun
8,339113578
asked Oct 2 '14 at 7:40
Quantization
20418
4
You cannot talk about the presentation, because there are lots of presentations, but you do not have enough relations. In fact $b^4=1$ is a consequence of $a^4=1$ and $ab^2a=1$, so one of your relations is redundant. If you adjoined the extra relation $a^{-1}ba=b^{-1}$ then you would have a complete presentation. BTW, your displayed presentation for $Q$ doesn't make sense, because $c$ and $d$ are undefined.
– Derek Holt
Oct 2 '14 at 7:56
1
Sorry I meant $c =a$ &$d=b$
– Quantization
Oct 2 '14 at 8:25
add a comment |
I was working through presentation of the quaternion group (with element $8$), and I let $a = i$ and $b = j$. I immediately said $a^4 = b^4 = 1$, and $ab^2 a = 1$.
Since I have a relation for each generator and between the generator, I figured I have the whole presentation. However, when I looked up the presentation of the quaternion group, it was given as
$$Q=langle F{a,b}mid a^4=b^4=a^2b^2=1 , b^{-1} a d = a^{-1}rangle.tag{1}$$
It is hard for me to see whether my initial third relation is a mixture of 3rd or 4th relation given by $(1)$.
Also, when do I know if I have given enough relations? Do I have to just write it down and see?
Finding a presentation of a group seems quite tedious!
group-theory group-presentation
edited Dec 11 at 7:50
Shaun
8,339113578
asked Oct 2 '14 at 7:40
Quantization
20418
I was working through presentation of the quaternion group (with element $8$), and I let $a = i$ and $b = j$. I immediately said $a^4 = b^4 = 1$, and $ab^2 a = 1$.
Since I have a relation for each generator and between the generator, I figured I have the whole presentation. However, when I looked up the presentation of the quaternion group, it was given as
$$Q=langle F{a,b}mid a^4=b^4=a^2b^2=1 , b^{-1} a d = a^{-1}rangle.tag{1}$$
It is hard for me to see whether my initial third relation is a mixture of 3rd or 4th relation given by $(1)$.
Also, when do I know if I have given enough relations? Do I have to just write it down and see?
Finding a presentation of a group seems quite tedious!
group-theory group-presentation
group-theory group-presentation
edited Dec 11 at 7:50
Shaun
8,339113578
asked Oct 2 '14 at 7:40
Quantization
20418
edited Dec 11 at 7:50
Shaun
8,339113578
asked Oct 2 '14 at 7:40
Quantization
20418
edited Dec 11 at 7:50
Shaun
8,339113578
edited Dec 11 at 7:50
Shaun
8,339113578
edited Dec 11 at 7:50
Shaun
8,339113578
8,339113578
asked Oct 2 '14 at 7:40
Quantization
20418
asked Oct 2 '14 at 7:40
Quantization
20418
asked Oct 2 '14 at 7:40
Quantization
20418
20418
4
You cannot talk about the presentation, because there are lots of presentations, but you do not have enough relations. In fact $b^4=1$ is a consequence of $a^4=1$ and $ab^2a=1$, so one of your relations is redundant. If you adjoined the extra relation $a^{-1}ba=b^{-1}$ then you would have a complete presentation. BTW, your displayed presentation for $Q$ doesn't make sense, because $c$ and $d$ are undefined.
– Derek Holt
Oct 2 '14 at 7:56
1
Sorry I meant $c =a$ &$d=b$
– Quantization
Oct 2 '14 at 8:25
add a comment |
4
You cannot talk about the presentation, because there are lots of presentations, but you do not have enough relations. In fact $b^4=1$ is a consequence of $a^4=1$ and $ab^2a=1$, so one of your relations is redundant. If you adjoined the extra relation $a^{-1}ba=b^{-1}$ then you would have a complete presentation. BTW, your displayed presentation for $Q$ doesn't make sense, because $c$ and $d$ are undefined.
– Derek Holt
Oct 2 '14 at 7:56
1
Sorry I meant $c =a$ &$d=b$
– Quantization
Oct 2 '14 at 8:25
4
4
You cannot talk about the presentation, because there are lots of presentations, but you do not have enough relations. In fact $b^4=1$ is a consequence of $a^4=1$ and $ab^2a=1$, so one of your relations is redundant. If you adjoined the extra relation $a^{-1}ba=b^{-1}$ then you would have a complete presentation. BTW, your displayed presentation for $Q$ doesn't make sense, because $c$ and $d$ are undefined.
– Derek Holt
Oct 2 '14 at 7:56
You cannot talk about the presentation, because there are lots of presentations, but you do not have enough relations. In fact $b^4=1$ is a consequence of $a^4=1$ and $ab^2a=1$, so one of your relations is redundant. If you adjoined the extra relation $a^{-1}ba=b^{-1}$ then you would have a complete presentation. BTW, your displayed presentation for $Q$ doesn't make sense, because $c$ and $d$ are undefined.
– Derek Holt
Oct 2 '14 at 7:56
1
1
Sorry I meant $c =a$ &$d=b$
– Quantization
Oct 2 '14 at 8:25
Sorry I meant $c =a$ &$d=b$
– Quantization
Oct 2 '14 at 8:25
add a comment |
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You cannot talk about the presentation, because there are lots of presentations, but you do not have enough relations. In fact $b^4=1$ is a consequence of $a^4=1$ and $ab^2a=1$, so one of your relations is redundant. If you adjoined the extra relation $a^{-1}ba=b^{-1}$ then you would have a complete presentation. BTW, your displayed presentation for $Q$ doesn't make sense, because $c$ and $d$ are undefined.
– Derek Holt
Oct 2 '14 at 7:56
1
Sorry I meant $c =a$ &$d=b$
– Quantization
Oct 2 '14 at 8:25