Proving normality of a subset from its generators
$begingroup$
Exercise 3.26 in Dummit and Foote reads:
Let $N = langle Srangle$ for some subset $S$ of $G$. Prove that $N trianglelefteq G$ if $ gSg^{-1} subseteq N$ for all $g in G$.
If $S$ is a finite set, I presume the proof is straightforward. Suppose $ gSg^{-1} subseteq N$ for all $g in G$. Any element of $N$ can be written as a product of elements of $S$. The properties of conjugation therefore make that the conjugate of an element of $N$ is a product of conjugates of elements of $S$. Since the latter by the assumption are all in $N$, conjugation will be closed in $N$.
I have two questions, however. First, how to go about an infinite $S$? An element of $N$ could be a product with infinitely many factors, for which the properties of conjugation have not been proven.
Second, I would think that the question might have read:
"Prove that $N trianglelefteq G$ if and only if $ gSg^{-1} subseteq N$ for all $g in G$."
Indeed, if $N trianglelefteq G$, since $S subseteq N$, $gSg^{-1} subseteq N$ for all $g in G$, proving the reverse implication.
Did I make a mistake here? (I am suspicious because it is my impression that it is not very like. D&F not to make the most comprehensive statements.)
abstract-algebra group-theory
edited Jan 13 at 12:41
Shaun
9,442113684
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
$endgroup$
add a comment |
$begingroup$
Exercise 3.26 in Dummit and Foote reads:
Let $N = langle Srangle$ for some subset $S$ of $G$. Prove that $N trianglelefteq G$ if $ gSg^{-1} subseteq N$ for all $g in G$.
If $S$ is a finite set, I presume the proof is straightforward. Suppose $ gSg^{-1} subseteq N$ for all $g in G$. Any element of $N$ can be written as a product of elements of $S$. The properties of conjugation therefore make that the conjugate of an element of $N$ is a product of conjugates of elements of $S$. Since the latter by the assumption are all in $N$, conjugation will be closed in $N$.
I have two questions, however. First, how to go about an infinite $S$? An element of $N$ could be a product with infinitely many factors, for which the properties of conjugation have not been proven.
Second, I would think that the question might have read:
"Prove that $N trianglelefteq G$ if and only if $ gSg^{-1} subseteq N$ for all $g in G$."
Indeed, if $N trianglelefteq G$, since $S subseteq N$, $gSg^{-1} subseteq N$ for all $g in G$, proving the reverse implication.
Did I make a mistake here? (I am suspicious because it is my impression that it is not very like. D&F not to make the most comprehensive statements.)
abstract-algebra group-theory
edited Jan 13 at 12:41
Shaun
9,442113684
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
$endgroup$
1
$begingroup$
How can an element of $N$ be the product of an infinite number of factors? How is that even defined? Cheers!
$endgroup$
– Robert Lewis
Jan 12 at 19:54
add a comment |
$begingroup$
Exercise 3.26 in Dummit and Foote reads:
Let $N = langle Srangle$ for some subset $S$ of $G$. Prove that $N trianglelefteq G$ if $ gSg^{-1} subseteq N$ for all $g in G$.
If $S$ is a finite set, I presume the proof is straightforward. Suppose $ gSg^{-1} subseteq N$ for all $g in G$. Any element of $N$ can be written as a product of elements of $S$. The properties of conjugation therefore make that the conjugate of an element of $N$ is a product of conjugates of elements of $S$. Since the latter by the assumption are all in $N$, conjugation will be closed in $N$.
I have two questions, however. First, how to go about an infinite $S$? An element of $N$ could be a product with infinitely many factors, for which the properties of conjugation have not been proven.
Second, I would think that the question might have read:
"Prove that $N trianglelefteq G$ if and only if $ gSg^{-1} subseteq N$ for all $g in G$."
Indeed, if $N trianglelefteq G$, since $S subseteq N$, $gSg^{-1} subseteq N$ for all $g in G$, proving the reverse implication.
Did I make a mistake here? (I am suspicious because it is my impression that it is not very like. D&F not to make the most comprehensive statements.)
abstract-algebra group-theory
edited Jan 13 at 12:41
Shaun
9,442113684
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
$endgroup$
Exercise 3.26 in Dummit and Foote reads:
Let $N = langle Srangle$ for some subset $S$ of $G$. Prove that $N trianglelefteq G$ if $ gSg^{-1} subseteq N$ for all $g in G$.
If $S$ is a finite set, I presume the proof is straightforward. Suppose $ gSg^{-1} subseteq N$ for all $g in G$. Any element of $N$ can be written as a product of elements of $S$. The properties of conjugation therefore make that the conjugate of an element of $N$ is a product of conjugates of elements of $S$. Since the latter by the assumption are all in $N$, conjugation will be closed in $N$.
I have two questions, however. First, how to go about an infinite $S$? An element of $N$ could be a product with infinitely many factors, for which the properties of conjugation have not been proven.
Second, I would think that the question might have read:
"Prove that $N trianglelefteq G$ if and only if $ gSg^{-1} subseteq N$ for all $g in G$."
Indeed, if $N trianglelefteq G$, since $S subseteq N$, $gSg^{-1} subseteq N$ for all $g in G$, proving the reverse implication.
Did I make a mistake here? (I am suspicious because it is my impression that it is not very like. D&F not to make the most comprehensive statements.)
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 13 at 12:41
Shaun
9,442113684
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
edited Jan 13 at 12:41
Shaun
9,442113684
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
edited Jan 13 at 12:41
Shaun
9,442113684
edited Jan 13 at 12:41
Shaun
9,442113684
edited Jan 13 at 12:41
Shaun
9,442113684
9,442113684
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
asked Jan 12 at 19:49
Frank De GeeterFrank De Geeter
776
776
1
$begingroup$
How can an element of $N$ be the product of an infinite number of factors? How is that even defined? Cheers!
$endgroup$
– Robert Lewis
Jan 12 at 19:54
add a comment |
1
$begingroup$
How can an element of $N$ be the product of an infinite number of factors? How is that even defined? Cheers!
$endgroup$
– Robert Lewis
Jan 12 at 19:54
1
1
$begingroup$
How can an element of $N$ be the product of an infinite number of factors? How is that even defined? Cheers!
$endgroup$
– Robert Lewis
Jan 12 at 19:54
$begingroup$
How can an element of $N$ be the product of an infinite number of factors? How is that even defined? Cheers!
$endgroup$
– Robert Lewis
Jan 12 at 19:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$N=langle Srangle$ means by definition that every element in $N$ can be written as a finite product of elements in $S$. Even if $S$ is infinite, only finite products are involved.
The converse direction also holds, you did not make a mistake. But the authors did not mention it because it is not so interesting, since it is a rather trivial statement.
edited Jan 13 at 12:43
Shaun
9,442113684
answered Jan 12 at 21:04
PedroPedro
2,9291720
$endgroup$
add a comment |
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$begingroup$
$N=langle Srangle$ means by definition that every element in $N$ can be written as a finite product of elements in $S$. Even if $S$ is infinite, only finite products are involved.
The converse direction also holds, you did not make a mistake. But the authors did not mention it because it is not so interesting, since it is a rather trivial statement.
edited Jan 13 at 12:43
Shaun
9,442113684
answered Jan 12 at 21:04
PedroPedro
2,9291720
$endgroup$
add a comment |
$begingroup$
$N=langle Srangle$ means by definition that every element in $N$ can be written as a finite product of elements in $S$. Even if $S$ is infinite, only finite products are involved.
The converse direction also holds, you did not make a mistake. But the authors did not mention it because it is not so interesting, since it is a rather trivial statement.
edited Jan 13 at 12:43
Shaun
9,442113684
answered Jan 12 at 21:04
PedroPedro
2,9291720
$endgroup$
add a comment |
$begingroup$
$N=langle Srangle$ means by definition that every element in $N$ can be written as a finite product of elements in $S$. Even if $S$ is infinite, only finite products are involved.
The converse direction also holds, you did not make a mistake. But the authors did not mention it because it is not so interesting, since it is a rather trivial statement.
edited Jan 13 at 12:43
Shaun
9,442113684
answered Jan 12 at 21:04
PedroPedro
2,9291720
$endgroup$
$N=langle Srangle$ means by definition that every element in $N$ can be written as a finite product of elements in $S$. Even if $S$ is infinite, only finite products are involved.
The converse direction also holds, you did not make a mistake. But the authors did not mention it because it is not so interesting, since it is a rather trivial statement.
edited Jan 13 at 12:43
Shaun
9,442113684
answered Jan 12 at 21:04
PedroPedro
2,9291720
edited Jan 13 at 12:43
Shaun
9,442113684
edited Jan 13 at 12:43
Shaun
9,442113684
edited Jan 13 at 12:43
Shaun
9,442113684
9,442113684
answered Jan 12 at 21:04
PedroPedro
2,9291720
answered Jan 12 at 21:04
PedroPedro
2,9291720
answered Jan 12 at 21:04
PedroPedro
2,9291720
2,9291720
add a comment |
add a comment |
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$begingroup$
How can an element of $N$ be the product of an infinite number of factors? How is that even defined? Cheers!
$endgroup$
– Robert Lewis
Jan 12 at 19:54