If $operatorname{Sym}_X simeq operatorname{Sym}_Y$ then there is bijection from $X to Y$
up vote
4
down vote
favorite
If $operatorname{Sym}_X simeq operatorname{Sym}_Y$ then there is bijection from $X to Y$ ? ,
I proved the other way around, i think i need to build $f$ from $psi : operatorname{Sym}_X tooperatorname{Sym}_Y$.
abstract-algebra group-theory symmetric-groups
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
asked Nov 24 at 16:17
Ahmad
2,4921625
add a comment |
up vote
4
down vote
favorite
If $operatorname{Sym}_X simeq operatorname{Sym}_Y$ then there is bijection from $X to Y$ ? ,
I proved the other way around, i think i need to build $f$ from $psi : operatorname{Sym}_X tooperatorname{Sym}_Y$.
abstract-algebra group-theory symmetric-groups
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
asked Nov 24 at 16:17
Ahmad
2,4921625
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
– Alex Vong
Nov 24 at 18:26
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
If $operatorname{Sym}_X simeq operatorname{Sym}_Y$ then there is bijection from $X to Y$ ? ,
I proved the other way around, i think i need to build $f$ from $psi : operatorname{Sym}_X tooperatorname{Sym}_Y$.
abstract-algebra group-theory symmetric-groups
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
asked Nov 24 at 16:17
Ahmad
2,4921625
If $operatorname{Sym}_X simeq operatorname{Sym}_Y$ then there is bijection from $X to Y$ ? ,
I proved the other way around, i think i need to build $f$ from $psi : operatorname{Sym}_X tooperatorname{Sym}_Y$.
abstract-algebra group-theory symmetric-groups
abstract-algebra group-theory symmetric-groups
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
asked Nov 24 at 16:17
Ahmad
2,4921625
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
asked Nov 24 at 16:17
Ahmad
2,4921625
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
edited Dec 7 at 19:00
Davide Giraudo
125k16150259
125k16150259
asked Nov 24 at 16:17
Ahmad
2,4921625
asked Nov 24 at 16:17
Ahmad
2,4921625
asked Nov 24 at 16:17
Ahmad
2,4921625
2,4921625
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
– Alex Vong
Nov 24 at 18:26
add a comment |
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
– Alex Vong
Nov 24 at 18:26
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
– Alex Vong
Nov 24 at 18:26
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
– Alex Vong
Nov 24 at 18:26
add a comment |
1 Answer
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up vote
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We have to recover $|X|$ from $operatorname{Sym}(X)$.
Note that $operatorname{Sym}(X)$ is finite iff $X$ is finite, and in that case, the cardinality of $operatorname{Sym}(X)$ uniquely determines that of $X$.
So we may assume that $operatorname{Sym}(X)$ is infinite (equivalently $X$ is infinite).
Let $Hsubseteq 2^{operatorname{Sym}(X)}$ consist of all elementary Abelian 2-subgroups in $operatorname{Sym}(X)$. These are isomorphic to additive groups of vector spaces over the 2-element field, and all of them contain the identity element and pairwise commuting involutions, i.e., possibly infinite products of transpositions with pairwise disjoint support. Let $kappa$ be the maximum dimension occurring over all these vector spaces (more precisely, we should define it as the supremum of all such dimensions, but it will turn out that there is an actual maximum, as well).
We claim that $kappa= |X|$.
First of all, it is possible to partition $|X|$ into $|X|$ pairs: the transpositions corresponding to these pairs form the basis of a vector space in $H$. So $kappageq |X|$.
For the reverse direction, note that in a vector space $Vin H$, for any two $f,gin V$ and $xin X$ it is not possible that $x, f(x)$ and $g(x)$ are three different elements, otherwise $f$ and $g$ do not commute. So for all $xin X$ either $x$ is a fixed point of the whole $V$, or there exists a unique $yin X$ such that "half" of the elements of $V$ fix $x$ and the other "half" transposes $x$ and $y$.
After dropping the fixed points, we have a unique $y$ for all $x$. But then the set of these transpositions $(x~y)$ certainly generates $V$. The cardinality of such a set of transpositions is clearly at most $|X|$, so $dim(V)leq |X|$ for all $Vin H$, i.e., $kappaleq |X|$.
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
answered Nov 24 at 18:14
A. Pongrácz
5,228725
You forgot $0!=1!$.
– user10354138
Nov 24 at 21:08
Absolutely, thank you. I concentrated on the interesting case, when $X$ is infinite. So in fact, the statement itself is false: the fully symmetric group acting on the emptyset is isomorphic to the one acting on a 1-element set. But this is not a very impoortant issue, and easy to fix.
– A. Pongrácz
Nov 26 at 13:57
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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up vote
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We have to recover $|X|$ from $operatorname{Sym}(X)$.
Note that $operatorname{Sym}(X)$ is finite iff $X$ is finite, and in that case, the cardinality of $operatorname{Sym}(X)$ uniquely determines that of $X$.
So we may assume that $operatorname{Sym}(X)$ is infinite (equivalently $X$ is infinite).
Let $Hsubseteq 2^{operatorname{Sym}(X)}$ consist of all elementary Abelian 2-subgroups in $operatorname{Sym}(X)$. These are isomorphic to additive groups of vector spaces over the 2-element field, and all of them contain the identity element and pairwise commuting involutions, i.e., possibly infinite products of transpositions with pairwise disjoint support. Let $kappa$ be the maximum dimension occurring over all these vector spaces (more precisely, we should define it as the supremum of all such dimensions, but it will turn out that there is an actual maximum, as well).
We claim that $kappa= |X|$.
First of all, it is possible to partition $|X|$ into $|X|$ pairs: the transpositions corresponding to these pairs form the basis of a vector space in $H$. So $kappageq |X|$.
For the reverse direction, note that in a vector space $Vin H$, for any two $f,gin V$ and $xin X$ it is not possible that $x, f(x)$ and $g(x)$ are three different elements, otherwise $f$ and $g$ do not commute. So for all $xin X$ either $x$ is a fixed point of the whole $V$, or there exists a unique $yin X$ such that "half" of the elements of $V$ fix $x$ and the other "half" transposes $x$ and $y$.
After dropping the fixed points, we have a unique $y$ for all $x$. But then the set of these transpositions $(x~y)$ certainly generates $V$. The cardinality of such a set of transpositions is clearly at most $|X|$, so $dim(V)leq |X|$ for all $Vin H$, i.e., $kappaleq |X|$.
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
answered Nov 24 at 18:14
A. Pongrácz
5,228725
You forgot $0!=1!$.
– user10354138
Nov 24 at 21:08
Absolutely, thank you. I concentrated on the interesting case, when $X$ is infinite. So in fact, the statement itself is false: the fully symmetric group acting on the emptyset is isomorphic to the one acting on a 1-element set. But this is not a very impoortant issue, and easy to fix.
– A. Pongrácz
Nov 26 at 13:57
add a comment |
up vote
3
down vote
We have to recover $|X|$ from $operatorname{Sym}(X)$.
Note that $operatorname{Sym}(X)$ is finite iff $X$ is finite, and in that case, the cardinality of $operatorname{Sym}(X)$ uniquely determines that of $X$.
So we may assume that $operatorname{Sym}(X)$ is infinite (equivalently $X$ is infinite).
Let $Hsubseteq 2^{operatorname{Sym}(X)}$ consist of all elementary Abelian 2-subgroups in $operatorname{Sym}(X)$. These are isomorphic to additive groups of vector spaces over the 2-element field, and all of them contain the identity element and pairwise commuting involutions, i.e., possibly infinite products of transpositions with pairwise disjoint support. Let $kappa$ be the maximum dimension occurring over all these vector spaces (more precisely, we should define it as the supremum of all such dimensions, but it will turn out that there is an actual maximum, as well).
We claim that $kappa= |X|$.
First of all, it is possible to partition $|X|$ into $|X|$ pairs: the transpositions corresponding to these pairs form the basis of a vector space in $H$. So $kappageq |X|$.
For the reverse direction, note that in a vector space $Vin H$, for any two $f,gin V$ and $xin X$ it is not possible that $x, f(x)$ and $g(x)$ are three different elements, otherwise $f$ and $g$ do not commute. So for all $xin X$ either $x$ is a fixed point of the whole $V$, or there exists a unique $yin X$ such that "half" of the elements of $V$ fix $x$ and the other "half" transposes $x$ and $y$.
After dropping the fixed points, we have a unique $y$ for all $x$. But then the set of these transpositions $(x~y)$ certainly generates $V$. The cardinality of such a set of transpositions is clearly at most $|X|$, so $dim(V)leq |X|$ for all $Vin H$, i.e., $kappaleq |X|$.
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
answered Nov 24 at 18:14
A. Pongrácz
5,228725
You forgot $0!=1!$.
– user10354138
Nov 24 at 21:08
Absolutely, thank you. I concentrated on the interesting case, when $X$ is infinite. So in fact, the statement itself is false: the fully symmetric group acting on the emptyset is isomorphic to the one acting on a 1-element set. But this is not a very impoortant issue, and easy to fix.
– A. Pongrácz
Nov 26 at 13:57
add a comment |
up vote
3
down vote
up vote
3
down vote
We have to recover $|X|$ from $operatorname{Sym}(X)$.
Note that $operatorname{Sym}(X)$ is finite iff $X$ is finite, and in that case, the cardinality of $operatorname{Sym}(X)$ uniquely determines that of $X$.
So we may assume that $operatorname{Sym}(X)$ is infinite (equivalently $X$ is infinite).
Let $Hsubseteq 2^{operatorname{Sym}(X)}$ consist of all elementary Abelian 2-subgroups in $operatorname{Sym}(X)$. These are isomorphic to additive groups of vector spaces over the 2-element field, and all of them contain the identity element and pairwise commuting involutions, i.e., possibly infinite products of transpositions with pairwise disjoint support. Let $kappa$ be the maximum dimension occurring over all these vector spaces (more precisely, we should define it as the supremum of all such dimensions, but it will turn out that there is an actual maximum, as well).
We claim that $kappa= |X|$.
First of all, it is possible to partition $|X|$ into $|X|$ pairs: the transpositions corresponding to these pairs form the basis of a vector space in $H$. So $kappageq |X|$.
For the reverse direction, note that in a vector space $Vin H$, for any two $f,gin V$ and $xin X$ it is not possible that $x, f(x)$ and $g(x)$ are three different elements, otherwise $f$ and $g$ do not commute. So for all $xin X$ either $x$ is a fixed point of the whole $V$, or there exists a unique $yin X$ such that "half" of the elements of $V$ fix $x$ and the other "half" transposes $x$ and $y$.
After dropping the fixed points, we have a unique $y$ for all $x$. But then the set of these transpositions $(x~y)$ certainly generates $V$. The cardinality of such a set of transpositions is clearly at most $|X|$, so $dim(V)leq |X|$ for all $Vin H$, i.e., $kappaleq |X|$.
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
answered Nov 24 at 18:14
A. Pongrácz
5,228725
We have to recover $|X|$ from $operatorname{Sym}(X)$.
Note that $operatorname{Sym}(X)$ is finite iff $X$ is finite, and in that case, the cardinality of $operatorname{Sym}(X)$ uniquely determines that of $X$.
So we may assume that $operatorname{Sym}(X)$ is infinite (equivalently $X$ is infinite).
Let $Hsubseteq 2^{operatorname{Sym}(X)}$ consist of all elementary Abelian 2-subgroups in $operatorname{Sym}(X)$. These are isomorphic to additive groups of vector spaces over the 2-element field, and all of them contain the identity element and pairwise commuting involutions, i.e., possibly infinite products of transpositions with pairwise disjoint support. Let $kappa$ be the maximum dimension occurring over all these vector spaces (more precisely, we should define it as the supremum of all such dimensions, but it will turn out that there is an actual maximum, as well).
We claim that $kappa= |X|$.
First of all, it is possible to partition $|X|$ into $|X|$ pairs: the transpositions corresponding to these pairs form the basis of a vector space in $H$. So $kappageq |X|$.
For the reverse direction, note that in a vector space $Vin H$, for any two $f,gin V$ and $xin X$ it is not possible that $x, f(x)$ and $g(x)$ are three different elements, otherwise $f$ and $g$ do not commute. So for all $xin X$ either $x$ is a fixed point of the whole $V$, or there exists a unique $yin X$ such that "half" of the elements of $V$ fix $x$ and the other "half" transposes $x$ and $y$.
After dropping the fixed points, we have a unique $y$ for all $x$. But then the set of these transpositions $(x~y)$ certainly generates $V$. The cardinality of such a set of transpositions is clearly at most $|X|$, so $dim(V)leq |X|$ for all $Vin H$, i.e., $kappaleq |X|$.
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
answered Nov 24 at 18:14
A. Pongrácz
5,228725
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
edited Dec 7 at 18:59
Davide Giraudo
125k16150259
125k16150259
answered Nov 24 at 18:14
A. Pongrácz
5,228725
answered Nov 24 at 18:14
A. Pongrácz
5,228725
answered Nov 24 at 18:14
A. Pongrácz
5,228725
5,228725
You forgot $0!=1!$.
– user10354138
Nov 24 at 21:08
Absolutely, thank you. I concentrated on the interesting case, when $X$ is infinite. So in fact, the statement itself is false: the fully symmetric group acting on the emptyset is isomorphic to the one acting on a 1-element set. But this is not a very impoortant issue, and easy to fix.
– A. Pongrácz
Nov 26 at 13:57
add a comment |
You forgot $0!=1!$.
– user10354138
Nov 24 at 21:08
Absolutely, thank you. I concentrated on the interesting case, when $X$ is infinite. So in fact, the statement itself is false: the fully symmetric group acting on the emptyset is isomorphic to the one acting on a 1-element set. But this is not a very impoortant issue, and easy to fix.
– A. Pongrácz
Nov 26 at 13:57
You forgot $0!=1!$.
– user10354138
Nov 24 at 21:08
You forgot $0!=1!$.
– user10354138
Nov 24 at 21:08
Absolutely, thank you. I concentrated on the interesting case, when $X$ is infinite. So in fact, the statement itself is false: the fully symmetric group acting on the emptyset is isomorphic to the one acting on a 1-element set. But this is not a very impoortant issue, and easy to fix.
– A. Pongrácz
Nov 26 at 13:57
Absolutely, thank you. I concentrated on the interesting case, when $X$ is infinite. So in fact, the statement itself is false: the fully symmetric group acting on the emptyset is isomorphic to the one acting on a 1-element set. But this is not a very impoortant issue, and easy to fix.
– A. Pongrácz
Nov 26 at 13:57
add a comment |
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Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
– Alex Vong
Nov 24 at 18:26