Mixing
Checking the normality for finite versus infinite subgroups of a group
$begingroup$
My question has been touched upon already here and here.
Dummit and Foote give a flurry of exercises (3.1.27 to 3.1.29) starting with 'Let $N$ be a finite subgroup of a group $G$.'
The first of these asks to show that $gNg^{-1} subseteq N$ iff $gNg^{-1}=N.$ As I understand the previous posts, if N is finite, a counting argument can be used. Do I understand well that this is something along the following lines?
Assume $gNg^{-1} subseteq N$. One can devise a map $f$ from $N$ to $gNg^{-1}$ as $f(n)=gng^{-1}$. Since this map is injective, it also holds that $N subseteq gNg^{-1}$, proving the equality. Would this reasoning not hold in the infinite case?
Sincerely, I do not see why this argument would be simpler than the following, which also hold in the infinite case: since $gNg^{-1} subseteq N$ for all $g in G$, it holds as well that $g^{-1} Ngsubseteq N$, and hence $N subseteq gNg^{-1}$.
However, D&F continue the exercises for finite $N$. In brief, they ask to prove that normality of $N$ in $G$ can be checked from the generators of $N$ and of $G$.
So, my question is whether this would also hold for infinite $N$; I do not see why it would not. But, then, why do D&F insist of finiteness of $N$? This puzzles me all the more, because a preceding exercise (3.1.26.c) did not require finiteness of $N$.
abstract-algebra group-theory
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
$endgroup$
add a comment |
$begingroup$
My question has been touched upon already here and here.
Dummit and Foote give a flurry of exercises (3.1.27 to 3.1.29) starting with 'Let $N$ be a finite subgroup of a group $G$.'
The first of these asks to show that $gNg^{-1} subseteq N$ iff $gNg^{-1}=N.$ As I understand the previous posts, if N is finite, a counting argument can be used. Do I understand well that this is something along the following lines?
Assume $gNg^{-1} subseteq N$. One can devise a map $f$ from $N$ to $gNg^{-1}$ as $f(n)=gng^{-1}$. Since this map is injective, it also holds that $N subseteq gNg^{-1}$, proving the equality. Would this reasoning not hold in the infinite case?
Sincerely, I do not see why this argument would be simpler than the following, which also hold in the infinite case: since $gNg^{-1} subseteq N$ for all $g in G$, it holds as well that $g^{-1} Ngsubseteq N$, and hence $N subseteq gNg^{-1}$.
However, D&F continue the exercises for finite $N$. In brief, they ask to prove that normality of $N$ in $G$ can be checked from the generators of $N$ and of $G$.
So, my question is whether this would also hold for infinite $N$; I do not see why it would not. But, then, why do D&F insist of finiteness of $N$? This puzzles me all the more, because a preceding exercise (3.1.26.c) did not require finiteness of $N$.
abstract-algebra group-theory
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
$endgroup$
$begingroup$
As you noted (and found in the referenced questions), if $G$ is a group and $N$ a subset of it, then $forall gin Gcolon gNg^{-1}subseteq Niff forall gin Gcolon gNg^{-1}= N$. However, what youquoted from the prolem statement look smore like $forall gin Gcolon(gNg^{-1}subseteq Nleftrightarrow gNg^{-1}=N)$.
$endgroup$
– Hagen von Eitzen
Jan 14 at 22:34
$begingroup$
@HagenvonEitzen: If $N$ is a finite subset, then your second statement is also true.
$endgroup$
– Arturo Magidin
Jan 14 at 22:37
$begingroup$
The fact that the map $Nto gNg^{-1}$ given by $nmapsto gng^{-1}$ is injective tells you that $|N|leq |gNg^{-1}|$ (the fact that it is onto shows that you have equality of cardinality). Therefore, since $gNg^{-01}subseteq N$ shows $|gNg^{-1}|leq |N|$, you get equality of cardinalities. In the finite case, since $gNg^{-1}subseteq N$, equality of cardinality gives you equality of sets. The fact that the map is injective does not prove that $Nsubseteq gNg^{-1}$.
$endgroup$
– Arturo Magidin
Jan 14 at 22:41
add a comment |
$begingroup$
My question has been touched upon already here and here.
Dummit and Foote give a flurry of exercises (3.1.27 to 3.1.29) starting with 'Let $N$ be a finite subgroup of a group $G$.'
The first of these asks to show that $gNg^{-1} subseteq N$ iff $gNg^{-1}=N.$ As I understand the previous posts, if N is finite, a counting argument can be used. Do I understand well that this is something along the following lines?
Assume $gNg^{-1} subseteq N$. One can devise a map $f$ from $N$ to $gNg^{-1}$ as $f(n)=gng^{-1}$. Since this map is injective, it also holds that $N subseteq gNg^{-1}$, proving the equality. Would this reasoning not hold in the infinite case?
Sincerely, I do not see why this argument would be simpler than the following, which also hold in the infinite case: since $gNg^{-1} subseteq N$ for all $g in G$, it holds as well that $g^{-1} Ngsubseteq N$, and hence $N subseteq gNg^{-1}$.
However, D&F continue the exercises for finite $N$. In brief, they ask to prove that normality of $N$ in $G$ can be checked from the generators of $N$ and of $G$.
So, my question is whether this would also hold for infinite $N$; I do not see why it would not. But, then, why do D&F insist of finiteness of $N$? This puzzles me all the more, because a preceding exercise (3.1.26.c) did not require finiteness of $N$.
abstract-algebra group-theory
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
$endgroup$
My question has been touched upon already here and here.
Dummit and Foote give a flurry of exercises (3.1.27 to 3.1.29) starting with 'Let $N$ be a finite subgroup of a group $G$.'
The first of these asks to show that $gNg^{-1} subseteq N$ iff $gNg^{-1}=N.$ As I understand the previous posts, if N is finite, a counting argument can be used. Do I understand well that this is something along the following lines?
Assume $gNg^{-1} subseteq N$. One can devise a map $f$ from $N$ to $gNg^{-1}$ as $f(n)=gng^{-1}$. Since this map is injective, it also holds that $N subseteq gNg^{-1}$, proving the equality. Would this reasoning not hold in the infinite case?
Sincerely, I do not see why this argument would be simpler than the following, which also hold in the infinite case: since $gNg^{-1} subseteq N$ for all $g in G$, it holds as well that $g^{-1} Ngsubseteq N$, and hence $N subseteq gNg^{-1}$.
However, D&F continue the exercises for finite $N$. In brief, they ask to prove that normality of $N$ in $G$ can be checked from the generators of $N$ and of $G$.
So, my question is whether this would also hold for infinite $N$; I do not see why it would not. But, then, why do D&F insist of finiteness of $N$? This puzzles me all the more, because a preceding exercise (3.1.26.c) did not require finiteness of $N$.
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
asked Jan 14 at 22:22
Frank De GeeterFrank De Geeter
726
726
$begingroup$
As you noted (and found in the referenced questions), if $G$ is a group and $N$ a subset of it, then $forall gin Gcolon gNg^{-1}subseteq Niff forall gin Gcolon gNg^{-1}= N$. However, what youquoted from the prolem statement look smore like $forall gin Gcolon(gNg^{-1}subseteq Nleftrightarrow gNg^{-1}=N)$.
$endgroup$
– Hagen von Eitzen
Jan 14 at 22:34
$begingroup$
@HagenvonEitzen: If $N$ is a finite subset, then your second statement is also true.
$endgroup$
– Arturo Magidin
Jan 14 at 22:37
$begingroup$
The fact that the map $Nto gNg^{-1}$ given by $nmapsto gng^{-1}$ is injective tells you that $|N|leq |gNg^{-1}|$ (the fact that it is onto shows that you have equality of cardinality). Therefore, since $gNg^{-01}subseteq N$ shows $|gNg^{-1}|leq |N|$, you get equality of cardinalities. In the finite case, since $gNg^{-1}subseteq N$, equality of cardinality gives you equality of sets. The fact that the map is injective does not prove that $Nsubseteq gNg^{-1}$.
$endgroup$
– Arturo Magidin
Jan 14 at 22:41
add a comment |
$begingroup$
As you noted (and found in the referenced questions), if $G$ is a group and $N$ a subset of it, then $forall gin Gcolon gNg^{-1}subseteq Niff forall gin Gcolon gNg^{-1}= N$. However, what youquoted from the prolem statement look smore like $forall gin Gcolon(gNg^{-1}subseteq Nleftrightarrow gNg^{-1}=N)$.
$endgroup$
– Hagen von Eitzen
Jan 14 at 22:34
$begingroup$
@HagenvonEitzen: If $N$ is a finite subset, then your second statement is also true.
$endgroup$
– Arturo Magidin
Jan 14 at 22:37
$begingroup$
The fact that the map $Nto gNg^{-1}$ given by $nmapsto gng^{-1}$ is injective tells you that $|N|leq |gNg^{-1}|$ (the fact that it is onto shows that you have equality of cardinality). Therefore, since $gNg^{-01}subseteq N$ shows $|gNg^{-1}|leq |N|$, you get equality of cardinalities. In the finite case, since $gNg^{-1}subseteq N$, equality of cardinality gives you equality of sets. The fact that the map is injective does not prove that $Nsubseteq gNg^{-1}$.
$endgroup$
– Arturo Magidin
Jan 14 at 22:41
$begingroup$
As you noted (and found in the referenced questions), if $G$ is a group and $N$ a subset of it, then $forall gin Gcolon gNg^{-1}subseteq Niff forall gin Gcolon gNg^{-1}= N$. However, what youquoted from the prolem statement look smore like $forall gin Gcolon(gNg^{-1}subseteq Nleftrightarrow gNg^{-1}=N)$.
$endgroup$
– Hagen von Eitzen
Jan 14 at 22:34
$begingroup$
As you noted (and found in the referenced questions), if $G$ is a group and $N$ a subset of it, then $forall gin Gcolon gNg^{-1}subseteq Niff forall gin Gcolon gNg^{-1}= N$. However, what youquoted from the prolem statement look smore like $forall gin Gcolon(gNg^{-1}subseteq Nleftrightarrow gNg^{-1}=N)$.
$endgroup$
– Hagen von Eitzen
Jan 14 at 22:34
$begingroup$
@HagenvonEitzen: If $N$ is a finite subset, then your second statement is also true.
$endgroup$
– Arturo Magidin
Jan 14 at 22:37
$begingroup$
@HagenvonEitzen: If $N$ is a finite subset, then your second statement is also true.
$endgroup$
– Arturo Magidin
Jan 14 at 22:37
$begingroup$
The fact that the map $Nto gNg^{-1}$ given by $nmapsto gng^{-1}$ is injective tells you that $|N|leq |gNg^{-1}|$ (the fact that it is onto shows that you have equality of cardinality). Therefore, since $gNg^{-01}subseteq N$ shows $|gNg^{-1}|leq |N|$, you get equality of cardinalities. In the finite case, since $gNg^{-1}subseteq N$, equality of cardinality gives you equality of sets. The fact that the map is injective does not prove that $Nsubseteq gNg^{-1}$.
$endgroup$
– Arturo Magidin
Jan 14 at 22:41
$begingroup$
The fact that the map $Nto gNg^{-1}$ given by $nmapsto gng^{-1}$ is injective tells you that $|N|leq |gNg^{-1}|$ (the fact that it is onto shows that you have equality of cardinality). Therefore, since $gNg^{-01}subseteq N$ shows $|gNg^{-1}|leq |N|$, you get equality of cardinalities. In the finite case, since $gNg^{-1}subseteq N$, equality of cardinality gives you equality of sets. The fact that the map is injective does not prove that $Nsubseteq gNg^{-1}$.
$endgroup$
– Arturo Magidin
Jan 14 at 22:41
add a comment |
1 Answer
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You are incorrect in concluding that $Nsubseteq gNg^{-1}$ follows from the fact that the map $nmapsto gng^{-1}$ is injective. I mean, the map $[0,1]to[2,3]$ given by $xmapsto x+2$ is injective, but that doesn't tell me that $[0,1]subseteq [2,3]$.
Rather, the fact that the map is injective tells you that $|N|leq |gNg^{-1}|$ (in fact, you always get equality, because the map is also onto). Since $gNg^{-1}subseteq N$, you also get $|gNg^{-1}|leq|N|$, so now you have $|gNg^{-1}|=|N|$. In the finite case, since $gNg^{-1}subseteq N$, the fact that they have the same cardinality allows you to conclude that $gNg^{-1}=N$.
In the infinite case, this is no longer true. And in fact, it is possible for $gNg^{-1}subseteq N$ to hold for every $g$ in a generating set, and yet for $N$ to not be normal because the inclusion is proper.
For example, if $G$ is the free group in two letters $x$ and $y$, and $H$ is the group generated by all elements of the form $uxu^{-1}$, where $u$ is a positive word (a group elements written as a product of $x$s and $y$s where all the exponents are nonnegative), then $xHx^{-1}subseteq H$ and $yHy^{-1}subseteq H$, ${x,y}$ is a generating set for $G$, but $y^{-1}xynotin H$, so $H$ is not normal. The problem here is that even though $|yHy^{-1}|=|H|$ and $yHy^{-1}subseteq H$, you still have $yHy^{-1}neq H$.
Consider the following four statements:
- For all $gin G$, $gNg^{-1}=N$.
- For all $gin G$, $gNg^{-1}subseteq N$.
$langle Srangle = G$, and for all $sin S$, $sNs^{-1}=N$.
$langle Srangle = G$ and for all $sin S$, $sNs^{-1}subseteq N$.
Turns out that statements 1, 2, and 3 are equivalent, so they all imply normality (statement 1). However, statement 4 is weaker in the general case (though equivalent when $N$ is finite).
Clearly 1 implies the remaining three, 2 implies 4, and 3 implies 4.
You've noticed already that 2 implies 1: using the given statement with $g^{-1}$ instead of $g$, we get $g^{-1}N(g^{-1})^{-1} subseteq N$, which gives $g^{-1}Ngsubseteq N$ for any $gin G$; then multiplying on the left by $g$ and on the right by $g^{-1}$, we conclude that for all $gin G$, $Nsubseteq gNg^{-1}$; together with 2, this yields condition 1.
To see that 3 implies 1, first note that for every $sin S$ we get $N=s^{-1}Ns$ by taking $sNs^{-1}=N$ and multiplying on the left by $s^{-1}$ and on the right by $s$; fromt his, given an arbitrary $gin G$, write it as a product of elements of $S$ and their inverses and use 3 to conclude that $gNg^{-1}=N$.
The problem for 4, however, is that you can't get that $s^{-1}Nssubseteq N$ for $sin S$ from the property given. So you can never make the jump to either 2 or 3 in the general case. It is only when $N$ is finite (or $S$ is closed under inverses) that you can get the equality you need to push the argument through.
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
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add a comment |
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$begingroup$
You are incorrect in concluding that $Nsubseteq gNg^{-1}$ follows from the fact that the map $nmapsto gng^{-1}$ is injective. I mean, the map $[0,1]to[2,3]$ given by $xmapsto x+2$ is injective, but that doesn't tell me that $[0,1]subseteq [2,3]$.
Rather, the fact that the map is injective tells you that $|N|leq |gNg^{-1}|$ (in fact, you always get equality, because the map is also onto). Since $gNg^{-1}subseteq N$, you also get $|gNg^{-1}|leq|N|$, so now you have $|gNg^{-1}|=|N|$. In the finite case, since $gNg^{-1}subseteq N$, the fact that they have the same cardinality allows you to conclude that $gNg^{-1}=N$.
In the infinite case, this is no longer true. And in fact, it is possible for $gNg^{-1}subseteq N$ to hold for every $g$ in a generating set, and yet for $N$ to not be normal because the inclusion is proper.
For example, if $G$ is the free group in two letters $x$ and $y$, and $H$ is the group generated by all elements of the form $uxu^{-1}$, where $u$ is a positive word (a group elements written as a product of $x$s and $y$s where all the exponents are nonnegative), then $xHx^{-1}subseteq H$ and $yHy^{-1}subseteq H$, ${x,y}$ is a generating set for $G$, but $y^{-1}xynotin H$, so $H$ is not normal. The problem here is that even though $|yHy^{-1}|=|H|$ and $yHy^{-1}subseteq H$, you still have $yHy^{-1}neq H$.
Consider the following four statements:
- For all $gin G$, $gNg^{-1}=N$.
- For all $gin G$, $gNg^{-1}subseteq N$.
$langle Srangle = G$, and for all $sin S$, $sNs^{-1}=N$.
$langle Srangle = G$ and for all $sin S$, $sNs^{-1}subseteq N$.
Turns out that statements 1, 2, and 3 are equivalent, so they all imply normality (statement 1). However, statement 4 is weaker in the general case (though equivalent when $N$ is finite).
Clearly 1 implies the remaining three, 2 implies 4, and 3 implies 4.
You've noticed already that 2 implies 1: using the given statement with $g^{-1}$ instead of $g$, we get $g^{-1}N(g^{-1})^{-1} subseteq N$, which gives $g^{-1}Ngsubseteq N$ for any $gin G$; then multiplying on the left by $g$ and on the right by $g^{-1}$, we conclude that for all $gin G$, $Nsubseteq gNg^{-1}$; together with 2, this yields condition 1.
To see that 3 implies 1, first note that for every $sin S$ we get $N=s^{-1}Ns$ by taking $sNs^{-1}=N$ and multiplying on the left by $s^{-1}$ and on the right by $s$; fromt his, given an arbitrary $gin G$, write it as a product of elements of $S$ and their inverses and use 3 to conclude that $gNg^{-1}=N$.
The problem for 4, however, is that you can't get that $s^{-1}Nssubseteq N$ for $sin S$ from the property given. So you can never make the jump to either 2 or 3 in the general case. It is only when $N$ is finite (or $S$ is closed under inverses) that you can get the equality you need to push the argument through.
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
add a comment |
$begingroup$
You are incorrect in concluding that $Nsubseteq gNg^{-1}$ follows from the fact that the map $nmapsto gng^{-1}$ is injective. I mean, the map $[0,1]to[2,3]$ given by $xmapsto x+2$ is injective, but that doesn't tell me that $[0,1]subseteq [2,3]$.
Rather, the fact that the map is injective tells you that $|N|leq |gNg^{-1}|$ (in fact, you always get equality, because the map is also onto). Since $gNg^{-1}subseteq N$, you also get $|gNg^{-1}|leq|N|$, so now you have $|gNg^{-1}|=|N|$. In the finite case, since $gNg^{-1}subseteq N$, the fact that they have the same cardinality allows you to conclude that $gNg^{-1}=N$.
In the infinite case, this is no longer true. And in fact, it is possible for $gNg^{-1}subseteq N$ to hold for every $g$ in a generating set, and yet for $N$ to not be normal because the inclusion is proper.
For example, if $G$ is the free group in two letters $x$ and $y$, and $H$ is the group generated by all elements of the form $uxu^{-1}$, where $u$ is a positive word (a group elements written as a product of $x$s and $y$s where all the exponents are nonnegative), then $xHx^{-1}subseteq H$ and $yHy^{-1}subseteq H$, ${x,y}$ is a generating set for $G$, but $y^{-1}xynotin H$, so $H$ is not normal. The problem here is that even though $|yHy^{-1}|=|H|$ and $yHy^{-1}subseteq H$, you still have $yHy^{-1}neq H$.
Consider the following four statements:
- For all $gin G$, $gNg^{-1}=N$.
- For all $gin G$, $gNg^{-1}subseteq N$.
$langle Srangle = G$, and for all $sin S$, $sNs^{-1}=N$.
$langle Srangle = G$ and for all $sin S$, $sNs^{-1}subseteq N$.
Turns out that statements 1, 2, and 3 are equivalent, so they all imply normality (statement 1). However, statement 4 is weaker in the general case (though equivalent when $N$ is finite).
Clearly 1 implies the remaining three, 2 implies 4, and 3 implies 4.
You've noticed already that 2 implies 1: using the given statement with $g^{-1}$ instead of $g$, we get $g^{-1}N(g^{-1})^{-1} subseteq N$, which gives $g^{-1}Ngsubseteq N$ for any $gin G$; then multiplying on the left by $g$ and on the right by $g^{-1}$, we conclude that for all $gin G$, $Nsubseteq gNg^{-1}$; together with 2, this yields condition 1.
To see that 3 implies 1, first note that for every $sin S$ we get $N=s^{-1}Ns$ by taking $sNs^{-1}=N$ and multiplying on the left by $s^{-1}$ and on the right by $s$; fromt his, given an arbitrary $gin G$, write it as a product of elements of $S$ and their inverses and use 3 to conclude that $gNg^{-1}=N$.
The problem for 4, however, is that you can't get that $s^{-1}Nssubseteq N$ for $sin S$ from the property given. So you can never make the jump to either 2 or 3 in the general case. It is only when $N$ is finite (or $S$ is closed under inverses) that you can get the equality you need to push the argument through.
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
add a comment |
$begingroup$
You are incorrect in concluding that $Nsubseteq gNg^{-1}$ follows from the fact that the map $nmapsto gng^{-1}$ is injective. I mean, the map $[0,1]to[2,3]$ given by $xmapsto x+2$ is injective, but that doesn't tell me that $[0,1]subseteq [2,3]$.
Rather, the fact that the map is injective tells you that $|N|leq |gNg^{-1}|$ (in fact, you always get equality, because the map is also onto). Since $gNg^{-1}subseteq N$, you also get $|gNg^{-1}|leq|N|$, so now you have $|gNg^{-1}|=|N|$. In the finite case, since $gNg^{-1}subseteq N$, the fact that they have the same cardinality allows you to conclude that $gNg^{-1}=N$.
In the infinite case, this is no longer true. And in fact, it is possible for $gNg^{-1}subseteq N$ to hold for every $g$ in a generating set, and yet for $N$ to not be normal because the inclusion is proper.
For example, if $G$ is the free group in two letters $x$ and $y$, and $H$ is the group generated by all elements of the form $uxu^{-1}$, where $u$ is a positive word (a group elements written as a product of $x$s and $y$s where all the exponents are nonnegative), then $xHx^{-1}subseteq H$ and $yHy^{-1}subseteq H$, ${x,y}$ is a generating set for $G$, but $y^{-1}xynotin H$, so $H$ is not normal. The problem here is that even though $|yHy^{-1}|=|H|$ and $yHy^{-1}subseteq H$, you still have $yHy^{-1}neq H$.
Consider the following four statements:
- For all $gin G$, $gNg^{-1}=N$.
- For all $gin G$, $gNg^{-1}subseteq N$.
$langle Srangle = G$, and for all $sin S$, $sNs^{-1}=N$.
$langle Srangle = G$ and for all $sin S$, $sNs^{-1}subseteq N$.
Turns out that statements 1, 2, and 3 are equivalent, so they all imply normality (statement 1). However, statement 4 is weaker in the general case (though equivalent when $N$ is finite).
Clearly 1 implies the remaining three, 2 implies 4, and 3 implies 4.
You've noticed already that 2 implies 1: using the given statement with $g^{-1}$ instead of $g$, we get $g^{-1}N(g^{-1})^{-1} subseteq N$, which gives $g^{-1}Ngsubseteq N$ for any $gin G$; then multiplying on the left by $g$ and on the right by $g^{-1}$, we conclude that for all $gin G$, $Nsubseteq gNg^{-1}$; together with 2, this yields condition 1.
To see that 3 implies 1, first note that for every $sin S$ we get $N=s^{-1}Ns$ by taking $sNs^{-1}=N$ and multiplying on the left by $s^{-1}$ and on the right by $s$; fromt his, given an arbitrary $gin G$, write it as a product of elements of $S$ and their inverses and use 3 to conclude that $gNg^{-1}=N$.
The problem for 4, however, is that you can't get that $s^{-1}Nssubseteq N$ for $sin S$ from the property given. So you can never make the jump to either 2 or 3 in the general case. It is only when $N$ is finite (or $S$ is closed under inverses) that you can get the equality you need to push the argument through.
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
$endgroup$
You are incorrect in concluding that $Nsubseteq gNg^{-1}$ follows from the fact that the map $nmapsto gng^{-1}$ is injective. I mean, the map $[0,1]to[2,3]$ given by $xmapsto x+2$ is injective, but that doesn't tell me that $[0,1]subseteq [2,3]$.
Rather, the fact that the map is injective tells you that $|N|leq |gNg^{-1}|$ (in fact, you always get equality, because the map is also onto). Since $gNg^{-1}subseteq N$, you also get $|gNg^{-1}|leq|N|$, so now you have $|gNg^{-1}|=|N|$. In the finite case, since $gNg^{-1}subseteq N$, the fact that they have the same cardinality allows you to conclude that $gNg^{-1}=N$.
In the infinite case, this is no longer true. And in fact, it is possible for $gNg^{-1}subseteq N$ to hold for every $g$ in a generating set, and yet for $N$ to not be normal because the inclusion is proper.
For example, if $G$ is the free group in two letters $x$ and $y$, and $H$ is the group generated by all elements of the form $uxu^{-1}$, where $u$ is a positive word (a group elements written as a product of $x$s and $y$s where all the exponents are nonnegative), then $xHx^{-1}subseteq H$ and $yHy^{-1}subseteq H$, ${x,y}$ is a generating set for $G$, but $y^{-1}xynotin H$, so $H$ is not normal. The problem here is that even though $|yHy^{-1}|=|H|$ and $yHy^{-1}subseteq H$, you still have $yHy^{-1}neq H$.
Consider the following four statements:
- For all $gin G$, $gNg^{-1}=N$.
- For all $gin G$, $gNg^{-1}subseteq N$.
$langle Srangle = G$, and for all $sin S$, $sNs^{-1}=N$.
$langle Srangle = G$ and for all $sin S$, $sNs^{-1}subseteq N$.
Turns out that statements 1, 2, and 3 are equivalent, so they all imply normality (statement 1). However, statement 4 is weaker in the general case (though equivalent when $N$ is finite).
Clearly 1 implies the remaining three, 2 implies 4, and 3 implies 4.
You've noticed already that 2 implies 1: using the given statement with $g^{-1}$ instead of $g$, we get $g^{-1}N(g^{-1})^{-1} subseteq N$, which gives $g^{-1}Ngsubseteq N$ for any $gin G$; then multiplying on the left by $g$ and on the right by $g^{-1}$, we conclude that for all $gin G$, $Nsubseteq gNg^{-1}$; together with 2, this yields condition 1.
To see that 3 implies 1, first note that for every $sin S$ we get $N=s^{-1}Ns$ by taking $sNs^{-1}=N$ and multiplying on the left by $s^{-1}$ and on the right by $s$; fromt his, given an arbitrary $gin G$, write it as a product of elements of $S$ and their inverses and use 3 to conclude that $gNg^{-1}=N$.
The problem for 4, however, is that you can't get that $s^{-1}Nssubseteq N$ for $sin S$ from the property given. So you can never make the jump to either 2 or 3 in the general case. It is only when $N$ is finite (or $S$ is closed under inverses) that you can get the equality you need to push the argument through.
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
edited Jan 14 at 23:07
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
answered Jan 14 at 22:58
Arturo MagidinArturo Magidin
263k34587911
263k34587911
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$begingroup$
As you noted (and found in the referenced questions), if $G$ is a group and $N$ a subset of it, then $forall gin Gcolon gNg^{-1}subseteq Niff forall gin Gcolon gNg^{-1}= N$. However, what youquoted from the prolem statement look smore like $forall gin Gcolon(gNg^{-1}subseteq Nleftrightarrow gNg^{-1}=N)$.
$endgroup$
– Hagen von Eitzen
Jan 14 at 22:34
$begingroup$
@HagenvonEitzen: If $N$ is a finite subset, then your second statement is also true.
$endgroup$
– Arturo Magidin
Jan 14 at 22:37
$begingroup$
The fact that the map $Nto gNg^{-1}$ given by $nmapsto gng^{-1}$ is injective tells you that $|N|leq |gNg^{-1}|$ (the fact that it is onto shows that you have equality of cardinality). Therefore, since $gNg^{-01}subseteq N$ shows $|gNg^{-1}|leq |N|$, you get equality of cardinalities. In the finite case, since $gNg^{-1}subseteq N$, equality of cardinality gives you equality of sets. The fact that the map is injective does not prove that $Nsubseteq gNg^{-1}$.
$endgroup$
– Arturo Magidin
Jan 14 at 22:41