Finite presentation of a cyclic group
$begingroup$
I'm struggling to understand group presentation.
There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.
Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.
Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?
abstract-algebra group-theory group-presentation
edited Dec 10 '18 at 23:39
Shaun
9,268113684
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
$endgroup$
add a comment |
$begingroup$
I'm struggling to understand group presentation.
There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.
Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.
Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?
abstract-algebra group-theory group-presentation
edited Dec 10 '18 at 23:39
Shaun
9,268113684
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
$endgroup$
4
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 18 '17 at 6:37
$begingroup$
Well this is the notation used in my class.
$endgroup$
– W.Scott
Apr 18 '17 at 7:02
$begingroup$
Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
$endgroup$
– Patrick Stevens
Apr 18 '17 at 7:09
$begingroup$
Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
$endgroup$
– W.Scott
Apr 18 '17 at 7:12
add a comment |
$begingroup$
I'm struggling to understand group presentation.
There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.
Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.
Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?
abstract-algebra group-theory group-presentation
edited Dec 10 '18 at 23:39
Shaun
9,268113684
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
$endgroup$
I'm struggling to understand group presentation.
There is a theorem that says, every group $G$ is the image of a suitable free group F (free upon a set $X$), and there must exist a homomorphism $pi$ from F to G, so by first isomorphism theorem, $G$ is isomorphic to the quotient group $F/ker(pi)$.
Question 1, I know we can denote $G$ as $<X mid R>$ where $X$ is the generating set of $F$. Is $R$ the generating set of $ker(pi)$? What about the normal closure of $R$? I am really confused here. Basically I want a concrete example demonstrating the difference between $R$, normal closure of $R$, and $ker(pi)$.
Question 2, $Z_7={0,1,2,3,4,5,6}$ is a cyclic group generated by ${1}$, which can also be written as $<x mid x^7>$. It seems to be the case that the generating set of the free group $F$ is $X={x}$, and $R={x^7}$. But then $G=F/ker(pi)$, what exactly is $F$ and $pi$ in this example?
abstract-algebra group-theory group-presentation
abstract-algebra group-theory group-presentation
edited Dec 10 '18 at 23:39
Shaun
9,268113684
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
edited Dec 10 '18 at 23:39
Shaun
9,268113684
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
edited Dec 10 '18 at 23:39
Shaun
9,268113684
edited Dec 10 '18 at 23:39
Shaun
9,268113684
edited Dec 10 '18 at 23:39
Shaun
9,268113684
9,268113684
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
asked Apr 18 '17 at 6:28
W.ScottW.Scott
1037
1037
4
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 18 '17 at 6:37
$begingroup$
Well this is the notation used in my class.
$endgroup$
– W.Scott
Apr 18 '17 at 7:02
$begingroup$
Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
$endgroup$
– Patrick Stevens
Apr 18 '17 at 7:09
$begingroup$
Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
$endgroup$
– W.Scott
Apr 18 '17 at 7:12
add a comment |
4
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 18 '17 at 6:37
$begingroup$
Well this is the notation used in my class.
$endgroup$
– W.Scott
Apr 18 '17 at 7:02
$begingroup$
Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
$endgroup$
– Patrick Stevens
Apr 18 '17 at 7:09
$begingroup$
Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
$endgroup$
– W.Scott
Apr 18 '17 at 7:12
4
4
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 18 '17 at 6:37
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 18 '17 at 6:37
$begingroup$
Well this is the notation used in my class.
$endgroup$
– W.Scott
Apr 18 '17 at 7:02
$begingroup$
Well this is the notation used in my class.
$endgroup$
– W.Scott
Apr 18 '17 at 7:02
$begingroup$
Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
$endgroup$
– Patrick Stevens
Apr 18 '17 at 7:09
$begingroup$
Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
$endgroup$
– Patrick Stevens
Apr 18 '17 at 7:09
$begingroup$
Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
$endgroup$
– W.Scott
Apr 18 '17 at 7:12
$begingroup$
Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
$endgroup$
– W.Scott
Apr 18 '17 at 7:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.
For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.
So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14
$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48
$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45
$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49
$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13
add a comment |
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1 Answer
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$begingroup$
In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.
For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.
So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14
$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48
$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45
$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49
$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13
add a comment |
$begingroup$
In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.
For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.
So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14
$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48
$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45
$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49
$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13
add a comment |
$begingroup$
In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.
For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.
So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
In the cyclic group $C_7$ with presentation $langle x mid x^7 rangle$, $F$ is the free group of rank $1$ with a single generator $x$ (so $F$ is an infinite cyclic group), $R = { x^7 }$, and $ker pi = langle R rangle =langle R^F rangle$ (the normal closure of $R$ in $F$). So in this example, $langle R rangle$ is equal to its normal closure. But that is not usually true.
For another example, consider $G = langle x,y mid r rangle$ with $r = [x,y] = xyx^{-1}y^{-1}$, which is a presentation of the free abelian group of rank $2$. Then $G = F/langle R^F rangle$ with $F$ a free group of rank $2$ and generators $x,y$, and $R = { r }$.
So $langle R rangle$ is a cyclic subgroup of $F$, but it is not normal. It's normal closure $langle R^F rangle$ is not even finitely generated. It contains all of the conjuagtes of $r$,like $[x^i,y^j]$ for all $i,j in {mathbb Z}$.
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
answered Apr 18 '17 at 8:04
Derek HoltDerek Holt
53.7k53571
53.7k53571
$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14
$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48
$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45
$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49
$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13
add a comment |
$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14
$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48
$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45
$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49
$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13
$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14
$begingroup$
How exactly do you compute normal closure of $R$?
$endgroup$
– W.Scott
Apr 18 '17 at 8:14
$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48
$begingroup$
$langle R^F rangle$ is the by definition the subgroup $langle frf^{-1} : r in R, f in F rangle$ of $F$. It is not usually finitely generated. What exactly do you mean by "compute"?
$endgroup$
– Derek Holt
Apr 18 '17 at 9:48
$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45
$begingroup$
So is the normal closure of $R$ always equal to the kernel of $pi$? Because on one hand, $G$ is isomorphic to $F/ker(pi)$, while on the other hand $G$ is isomorphic to $F/N$ where $N$ is the normal closure of $R$.
$endgroup$
– W.Scott
Apr 18 '17 at 11:45
$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49
$begingroup$
And for the $Z_7$ example, how can you show that $<R>=<R^F>$?
$endgroup$
– W.Scott
Apr 18 '17 at 11:49
$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13
$begingroup$
Yes, with your notation $N = {rm ker} pi$. In the $Z_7$ example $F$ is cyclic and hence abelian.
$endgroup$
– Derek Holt
Apr 18 '17 at 12:13
add a comment |
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I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
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– Patrick Stevens
Apr 18 '17 at 6:37
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Well this is the notation used in my class.
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– W.Scott
Apr 18 '17 at 7:02
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Really? They use $<X mid R>$ rather than $langle X mid R rangle$? That's unusual. $langle, rangle$ are angle brackets; $<, >$ are less than/greater than signs.
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– Patrick Stevens
Apr 18 '17 at 7:09
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Sorry my bad. Yes you are right. Can you please give me some insight to my two questions? Especially the first one.
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– W.Scott
Apr 18 '17 at 7:12