If $operatorname{Sym}_X simeq operatorname{Sym}_Y$ then there is bijection from $X to Y$
$begingroup$
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 '18 at 19:00
Davide Giraudo
126k16150261
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
$endgroup$
add a comment |
$begingroup$
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 '18 at 19:00
Davide Giraudo
126k16150261
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
$endgroup$
$begingroup$
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
$endgroup$
– Alex Vong
Nov 24 '18 at 18:26
add a comment |
$begingroup$
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 '18 at 19:00
Davide Giraudo
126k16150261
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
$endgroup$
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 '18 at 19:00
Davide Giraudo
126k16150261
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
edited Dec 7 '18 at 19:00
Davide Giraudo
126k16150261
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
edited Dec 7 '18 at 19:00
Davide Giraudo
126k16150261
edited Dec 7 '18 at 19:00
Davide Giraudo
126k16150261
edited Dec 7 '18 at 19:00
Davide Giraudo
126k16150261
126k16150261
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
asked Nov 24 '18 at 16:17
AhmadAhmad
2,5401625
2,5401625
$begingroup$
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
$endgroup$
– Alex Vong
Nov 24 '18 at 18:26
add a comment |
$begingroup$
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
$endgroup$
– Alex Vong
Nov 24 '18 at 18:26
$begingroup$
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
$endgroup$
– Alex Vong
Nov 24 '18 at 18:26
$begingroup$
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
$endgroup$
– Alex Vong
Nov 24 '18 at 18:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 18:59
Davide Giraudo
126k16150261
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
$endgroup$
$begingroup$
You forgot $0!=1!$.
$endgroup$
– user10354138
Nov 24 '18 at 21:08
$begingroup$
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.
$endgroup$
– A. Pongrácz
Nov 26 '18 at 13:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 '18 at 18:59
Davide Giraudo
126k16150261
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
$endgroup$
$begingroup$
You forgot $0!=1!$.
$endgroup$
– user10354138
Nov 24 '18 at 21:08
$begingroup$
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.
$endgroup$
– A. Pongrácz
Nov 26 '18 at 13:57
add a comment |
$begingroup$
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 '18 at 18:59
Davide Giraudo
126k16150261
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
$endgroup$
$begingroup$
You forgot $0!=1!$.
$endgroup$
– user10354138
Nov 24 '18 at 21:08
$begingroup$
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.
$endgroup$
– A. Pongrácz
Nov 26 '18 at 13:57
add a comment |
$begingroup$
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 '18 at 18:59
Davide Giraudo
126k16150261
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
$endgroup$
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 '18 at 18:59
Davide Giraudo
126k16150261
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
edited Dec 7 '18 at 18:59
Davide Giraudo
126k16150261
edited Dec 7 '18 at 18:59
Davide Giraudo
126k16150261
edited Dec 7 '18 at 18:59
Davide Giraudo
126k16150261
126k16150261
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
answered Nov 24 '18 at 18:14
A. PongráczA. Pongrácz
5,9631929
5,9631929
$begingroup$
You forgot $0!=1!$.
$endgroup$
– user10354138
Nov 24 '18 at 21:08
$begingroup$
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.
$endgroup$
– A. Pongrácz
Nov 26 '18 at 13:57
add a comment |
$begingroup$
You forgot $0!=1!$.
$endgroup$
– user10354138
Nov 24 '18 at 21:08
$begingroup$
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.
$endgroup$
– A. Pongrácz
Nov 26 '18 at 13:57
$begingroup$
You forgot $0!=1!$.
$endgroup$
– user10354138
Nov 24 '18 at 21:08
$begingroup$
You forgot $0!=1!$.
$endgroup$
– user10354138
Nov 24 '18 at 21:08
$begingroup$
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.
$endgroup$
– A. Pongrácz
Nov 26 '18 at 13:57
$begingroup$
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.
$endgroup$
– A. Pongrácz
Nov 26 '18 at 13:57
add a comment |
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$begingroup$
Related: Can the symmetric groups on sets of different cardinalities be isomorphic?
$endgroup$
– Alex Vong
Nov 24 '18 at 18:26