Can we decompose every group element to elements of order 2? (using Cayley's theorem to identificate the group with permutations)Finding the number of elements of order two in the symmetric group $S_4$Explicit isomorphism $S_4/V_4$ and $S_3$Group of order 24 with no element of order 6 is isomorphic to $S_4$Isomorphism problem involving the Symmetric GroupElementary Applications of Cayley's Theorem in Group TheoryAutomorphism on $S_n$ sends given k-transposition to a transpositionIf a subgroup of $G$ can be described as a set of permutations with certain cycle types, is it normal in $G$?Writing every element of $S_n$ as a product of permutations of order 2No homomorphism from $Z_16oplus Z_2$ onto $Z_4oplus Z_4$.Number of elements in the symmetric group with a particular transposition decomposition
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Can we decompose every group element to elements of order 2? (using Cayley's theorem to identificate the group with permutations)
Finding the number of elements of order two in the symmetric group $S_4$Explicit isomorphism $S_4/V_4$ and $S_3$Group of order 24 with no element of order 6 is isomorphic to $S_4$Isomorphism problem involving the Symmetric GroupElementary Applications of Cayley's Theorem in Group TheoryAutomorphism on $S_n$ sends given k-transposition to a transpositionIf a subgroup of $G$ can be described as a set of permutations with certain cycle types, is it normal in $G$?Writing every element of $S_n$ as a product of permutations of order 2No homomorphism from $Z_16oplus Z_2$ onto $Z_4oplus Z_4$.Number of elements in the symmetric group with a particular transposition decomposition
$begingroup$
I am struggling with this:
We know that every finite group $G$ is isomorphic to a subgroup of some symmetric group $S_n$ (Cayley's theorem).
But in the symmetric group $S_n$, every element - permutation can be written as a product of transpositions (cycles of length 2).
Order of a transposition is $2$. (For transposition $pi$ this means $pi ^2 = id$ = identity.)
So every element of the group $G$ can be written as a product of elements $a in G$ with order $2$ ($a^2 = 1$).
But this obviously can not hold as for example a cyclic group with an odd number of elements has no other element but $1$ for which it would be $a^2 = 1$.
Where is the mistake?
Isomorphism preserves the order of elements because it is injective.
EDIT:
I imagine something like this:
$G = mathbbZ_4 = 0, 1, 2, 3$
We hava a homomorphism $Phi: mathbbZ_4 rightarrow S_4$
(We get it using Cayly's theorem)
$Phi (0) = id$
$Phi (1) = (0123)$ that can be rewritten as $(03)(02)(12)$
$Phi (2) = (02)(13)$
$Phi (3) = (0321)$ that can be rewritten as $(13)(23)(01)$
$Phi$ is isomorphism on its image as it is already injective.
Now we have an isomorphism $Psi$ between $mathbbZ_4$ and $id, (03)(02)(12), (02)(13), (13)(23)(01)$, which is a subgroup of $S_4$.
$Psi ((02)(13)) = Psi ((02)) cdot Psi ((13))$
$Psi ((02)) $ and $ Psi ((13)) $ have the order $2$. So we have written $2 = Psi ((02)(13))$ as the product of two elements with order $2$.
F or $1 = Psi ((03)(02)(12)) = Psi ((03)) Psi ((02)) Psi ((12))$
Now it looks like we have written $1 in mathbbZ_4$ as the product of three alements of order $2$.
abstract-algebra group-theory finite-groups permutations
asked 8 hours ago
CoupeauCoupeau
1909
$endgroup$
|
show 1 more comment
$begingroup$
I am struggling with this:
We know that every finite group $G$ is isomorphic to a subgroup of some symmetric group $S_n$ (Cayley's theorem).
But in the symmetric group $S_n$, every element - permutation can be written as a product of transpositions (cycles of length 2).
Order of a transposition is $2$. (For transposition $pi$ this means $pi ^2 = id$ = identity.)
So every element of the group $G$ can be written as a product of elements $a in G$ with order $2$ ($a^2 = 1$).
But this obviously can not hold as for example a cyclic group with an odd number of elements has no other element but $1$ for which it would be $a^2 = 1$.
Where is the mistake?
Isomorphism preserves the order of elements because it is injective.
EDIT:
I imagine something like this:
$G = mathbbZ_4 = 0, 1, 2, 3$
We hava a homomorphism $Phi: mathbbZ_4 rightarrow S_4$
(We get it using Cayly's theorem)
$Phi (0) = id$
$Phi (1) = (0123)$ that can be rewritten as $(03)(02)(12)$
$Phi (2) = (02)(13)$
$Phi (3) = (0321)$ that can be rewritten as $(13)(23)(01)$
$Phi$ is isomorphism on its image as it is already injective.
Now we have an isomorphism $Psi$ between $mathbbZ_4$ and $id, (03)(02)(12), (02)(13), (13)(23)(01)$, which is a subgroup of $S_4$.
$Psi ((02)(13)) = Psi ((02)) cdot Psi ((13))$
$Psi ((02)) $ and $ Psi ((13)) $ have the order $2$. So we have written $2 = Psi ((02)(13))$ as the product of two elements with order $2$.
F or $1 = Psi ((03)(02)(12)) = Psi ((03)) Psi ((02)) Psi ((12))$
Now it looks like we have written $1 in mathbbZ_4$ as the product of three alements of order $2$.
abstract-algebra group-theory finite-groups permutations
asked 8 hours ago
CoupeauCoupeau
1909
$endgroup$
1
$begingroup$
I would recommend $A_3$ as an example to understand what is going on.
$endgroup$
– Thomas Shelby
8 hours ago
$begingroup$
the order of a transposition may be two, but what is the order of (12)(13)?
$endgroup$
– graeme
8 hours ago
$begingroup$
@graeme Yes, it is not $2$, but... Let $Phi$ be the isomorphism from $G$ to some subgroup $H leq S_n$. Isn't it then that $Phi ((12)(13)) = Phi ((12)) cdot Phi ((13))$ And order of $Phi ((12))$ is the same as of $(12)$, which is $2$? And the same for $Phi ((13))$
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
$endgroup$
– J. W. Tanner
7 hours ago
$begingroup$
Even if it was defined, it is not true. Elements of order $k$ are sent to elements of order $k$ by isomorphisms. (12)(13) does not have order 2, it has order 3, since it is equal to (1 3 2). I suggest you do some calculations on your own to convince yourself of this fact.
$endgroup$
– graeme
7 hours ago
|
show 1 more comment
$begingroup$
I am struggling with this:
We know that every finite group $G$ is isomorphic to a subgroup of some symmetric group $S_n$ (Cayley's theorem).
But in the symmetric group $S_n$, every element - permutation can be written as a product of transpositions (cycles of length 2).
Order of a transposition is $2$. (For transposition $pi$ this means $pi ^2 = id$ = identity.)
So every element of the group $G$ can be written as a product of elements $a in G$ with order $2$ ($a^2 = 1$).
But this obviously can not hold as for example a cyclic group with an odd number of elements has no other element but $1$ for which it would be $a^2 = 1$.
Where is the mistake?
Isomorphism preserves the order of elements because it is injective.
EDIT:
I imagine something like this:
$G = mathbbZ_4 = 0, 1, 2, 3$
We hava a homomorphism $Phi: mathbbZ_4 rightarrow S_4$
(We get it using Cayly's theorem)
$Phi (0) = id$
$Phi (1) = (0123)$ that can be rewritten as $(03)(02)(12)$
$Phi (2) = (02)(13)$
$Phi (3) = (0321)$ that can be rewritten as $(13)(23)(01)$
$Phi$ is isomorphism on its image as it is already injective.
Now we have an isomorphism $Psi$ between $mathbbZ_4$ and $id, (03)(02)(12), (02)(13), (13)(23)(01)$, which is a subgroup of $S_4$.
$Psi ((02)(13)) = Psi ((02)) cdot Psi ((13))$
$Psi ((02)) $ and $ Psi ((13)) $ have the order $2$. So we have written $2 = Psi ((02)(13))$ as the product of two elements with order $2$.
F or $1 = Psi ((03)(02)(12)) = Psi ((03)) Psi ((02)) Psi ((12))$
Now it looks like we have written $1 in mathbbZ_4$ as the product of three alements of order $2$.
abstract-algebra group-theory finite-groups permutations
asked 8 hours ago
CoupeauCoupeau
1909
$endgroup$
I am struggling with this:
We know that every finite group $G$ is isomorphic to a subgroup of some symmetric group $S_n$ (Cayley's theorem).
But in the symmetric group $S_n$, every element - permutation can be written as a product of transpositions (cycles of length 2).
Order of a transposition is $2$. (For transposition $pi$ this means $pi ^2 = id$ = identity.)
So every element of the group $G$ can be written as a product of elements $a in G$ with order $2$ ($a^2 = 1$).
But this obviously can not hold as for example a cyclic group with an odd number of elements has no other element but $1$ for which it would be $a^2 = 1$.
Where is the mistake?
Isomorphism preserves the order of elements because it is injective.
EDIT:
I imagine something like this:
$G = mathbbZ_4 = 0, 1, 2, 3$
We hava a homomorphism $Phi: mathbbZ_4 rightarrow S_4$
(We get it using Cayly's theorem)
$Phi (0) = id$
$Phi (1) = (0123)$ that can be rewritten as $(03)(02)(12)$
$Phi (2) = (02)(13)$
$Phi (3) = (0321)$ that can be rewritten as $(13)(23)(01)$
$Phi$ is isomorphism on its image as it is already injective.
Now we have an isomorphism $Psi$ between $mathbbZ_4$ and $id, (03)(02)(12), (02)(13), (13)(23)(01)$, which is a subgroup of $S_4$.
$Psi ((02)(13)) = Psi ((02)) cdot Psi ((13))$
$Psi ((02)) $ and $ Psi ((13)) $ have the order $2$. So we have written $2 = Psi ((02)(13))$ as the product of two elements with order $2$.
F or $1 = Psi ((03)(02)(12)) = Psi ((03)) Psi ((02)) Psi ((12))$
Now it looks like we have written $1 in mathbbZ_4$ as the product of three alements of order $2$.
abstract-algebra group-theory finite-groups permutations
abstract-algebra group-theory finite-groups permutations
asked 8 hours ago
CoupeauCoupeau
1909
asked 8 hours ago
CoupeauCoupeau
1909
edited 7 hours ago
Coupeau
asked 8 hours ago
CoupeauCoupeau
1909
asked 8 hours ago
CoupeauCoupeau
1909
asked 8 hours ago
CoupeauCoupeau
1909
1909
1
$begingroup$
I would recommend $A_3$ as an example to understand what is going on.
$endgroup$
– Thomas Shelby
8 hours ago
$begingroup$
the order of a transposition may be two, but what is the order of (12)(13)?
$endgroup$
– graeme
8 hours ago
$begingroup$
@graeme Yes, it is not $2$, but... Let $Phi$ be the isomorphism from $G$ to some subgroup $H leq S_n$. Isn't it then that $Phi ((12)(13)) = Phi ((12)) cdot Phi ((13))$ And order of $Phi ((12))$ is the same as of $(12)$, which is $2$? And the same for $Phi ((13))$
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
$endgroup$
– J. W. Tanner
7 hours ago
$begingroup$
Even if it was defined, it is not true. Elements of order $k$ are sent to elements of order $k$ by isomorphisms. (12)(13) does not have order 2, it has order 3, since it is equal to (1 3 2). I suggest you do some calculations on your own to convince yourself of this fact.
$endgroup$
– graeme
7 hours ago
|
show 1 more comment
1
$begingroup$
I would recommend $A_3$ as an example to understand what is going on.
$endgroup$
– Thomas Shelby
8 hours ago
$begingroup$
the order of a transposition may be two, but what is the order of (12)(13)?
$endgroup$
– graeme
8 hours ago
$begingroup$
@graeme Yes, it is not $2$, but... Let $Phi$ be the isomorphism from $G$ to some subgroup $H leq S_n$. Isn't it then that $Phi ((12)(13)) = Phi ((12)) cdot Phi ((13))$ And order of $Phi ((12))$ is the same as of $(12)$, which is $2$? And the same for $Phi ((13))$
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
$endgroup$
– J. W. Tanner
7 hours ago
$begingroup$
Even if it was defined, it is not true. Elements of order $k$ are sent to elements of order $k$ by isomorphisms. (12)(13) does not have order 2, it has order 3, since it is equal to (1 3 2). I suggest you do some calculations on your own to convince yourself of this fact.
$endgroup$
– graeme
7 hours ago
1
1
$begingroup$
I would recommend $A_3$ as an example to understand what is going on.
$endgroup$
– Thomas Shelby
8 hours ago
$begingroup$
I would recommend $A_3$ as an example to understand what is going on.
$endgroup$
– Thomas Shelby
8 hours ago
$begingroup$
the order of a transposition may be two, but what is the order of (12)(13)?
$endgroup$
– graeme
8 hours ago
$begingroup$
the order of a transposition may be two, but what is the order of (12)(13)?
$endgroup$
– graeme
8 hours ago
$begingroup$
@graeme Yes, it is not $2$, but... Let $Phi$ be the isomorphism from $G$ to some subgroup $H leq S_n$. Isn't it then that $Phi ((12)(13)) = Phi ((12)) cdot Phi ((13))$ And order of $Phi ((12))$ is the same as of $(12)$, which is $2$? And the same for $Phi ((13))$
$endgroup$
– Coupeau
7 hours ago
$begingroup$
@graeme Yes, it is not $2$, but... Let $Phi$ be the isomorphism from $G$ to some subgroup $H leq S_n$. Isn't it then that $Phi ((12)(13)) = Phi ((12)) cdot Phi ((13))$ And order of $Phi ((12))$ is the same as of $(12)$, which is $2$? And the same for $Phi ((13))$
$endgroup$
– Coupeau
7 hours ago
1
1
$begingroup$
Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
$endgroup$
– J. W. Tanner
7 hours ago
$begingroup$
Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
$endgroup$
– J. W. Tanner
7 hours ago
$begingroup$
Even if it was defined, it is not true. Elements of order $k$ are sent to elements of order $k$ by isomorphisms. (12)(13) does not have order 2, it has order 3, since it is equal to (1 3 2). I suggest you do some calculations on your own to convince yourself of this fact.
$endgroup$
– graeme
7 hours ago
$begingroup$
Even if it was defined, it is not true. Elements of order $k$ are sent to elements of order $k$ by isomorphisms. (12)(13) does not have order 2, it has order 3, since it is equal to (1 3 2). I suggest you do some calculations on your own to convince yourself of this fact.
$endgroup$
– graeme
7 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Every finite group is isomorphic to a subgroup of some $S_n$,
but the transpositions in $S_n$ are not necessarily in that subgroup.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
$endgroup$
$begingroup$
We can rewrite any permutation from $S_n$ as a product of transpositions. So also every permutation from that subgroup that group $G$ is isomorphic to, is a product of transpositions. So, of elements with order $2$.
$endgroup$
– Coupeau
8 hours ago
1
$begingroup$
@Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
$endgroup$
– Hagen von Eitzen
8 hours ago
$begingroup$
Yes, but there is an isomorphism, let's say $phi$ that maps them to elements from $G$. And isomorphism preserves the order of elements. I am missing something here but I do not really see what.
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Yes, there is an isomorphism between $G$ and a subgroup of $S_n$ but it could be that the transpositions in $S_n$ are not involved in the isomorphism; i.e., there are no elements in $G$ corresponding to transpositions in $S_n,$ so there are no elements of order $2$ in $G$
$endgroup$
– J. W. Tanner
7 hours ago
1
$begingroup$
Oh okay. I think I understand now where is the problem. So I can not 'expand' $phi (x)$ to $ phi (ab)$ and then to $ phi (a) phi (b)$ if $a$ and $b$ are not in the domain of the function $phi$ even though $ab = x$. Thank you.
$endgroup$
– Coupeau
6 hours ago
add a comment |
$begingroup$
Here's an analogy: consider the group of even numbers under addition. Now, every even number can be described as the sum of two odd numbers. How is that possible when there are no odd numbers in the group?
Or consider the field of real numbers. Every real $x$ can be written as $i^4x$: how is that possible when $i not in Bbb R$?
answered 8 hours ago
ThéophileThéophile
20.8k13047
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Every finite group is isomorphic to a subgroup of some $S_n$,
but the transpositions in $S_n$ are not necessarily in that subgroup.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
$endgroup$
$begingroup$
We can rewrite any permutation from $S_n$ as a product of transpositions. So also every permutation from that subgroup that group $G$ is isomorphic to, is a product of transpositions. So, of elements with order $2$.
$endgroup$
– Coupeau
8 hours ago
1
$begingroup$
@Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
$endgroup$
– Hagen von Eitzen
8 hours ago
$begingroup$
Yes, but there is an isomorphism, let's say $phi$ that maps them to elements from $G$. And isomorphism preserves the order of elements. I am missing something here but I do not really see what.
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Yes, there is an isomorphism between $G$ and a subgroup of $S_n$ but it could be that the transpositions in $S_n$ are not involved in the isomorphism; i.e., there are no elements in $G$ corresponding to transpositions in $S_n,$ so there are no elements of order $2$ in $G$
$endgroup$
– J. W. Tanner
7 hours ago
1
$begingroup$
Oh okay. I think I understand now where is the problem. So I can not 'expand' $phi (x)$ to $ phi (ab)$ and then to $ phi (a) phi (b)$ if $a$ and $b$ are not in the domain of the function $phi$ even though $ab = x$. Thank you.
$endgroup$
– Coupeau
6 hours ago
add a comment |
$begingroup$
Every finite group is isomorphic to a subgroup of some $S_n$,
but the transpositions in $S_n$ are not necessarily in that subgroup.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
$endgroup$
$begingroup$
We can rewrite any permutation from $S_n$ as a product of transpositions. So also every permutation from that subgroup that group $G$ is isomorphic to, is a product of transpositions. So, of elements with order $2$.
$endgroup$
– Coupeau
8 hours ago
1
$begingroup$
@Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
$endgroup$
– Hagen von Eitzen
8 hours ago
$begingroup$
Yes, but there is an isomorphism, let's say $phi$ that maps them to elements from $G$. And isomorphism preserves the order of elements. I am missing something here but I do not really see what.
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Yes, there is an isomorphism between $G$ and a subgroup of $S_n$ but it could be that the transpositions in $S_n$ are not involved in the isomorphism; i.e., there are no elements in $G$ corresponding to transpositions in $S_n,$ so there are no elements of order $2$ in $G$
$endgroup$
– J. W. Tanner
7 hours ago
1
$begingroup$
Oh okay. I think I understand now where is the problem. So I can not 'expand' $phi (x)$ to $ phi (ab)$ and then to $ phi (a) phi (b)$ if $a$ and $b$ are not in the domain of the function $phi$ even though $ab = x$. Thank you.
$endgroup$
– Coupeau
6 hours ago
add a comment |
$begingroup$
Every finite group is isomorphic to a subgroup of some $S_n$,
but the transpositions in $S_n$ are not necessarily in that subgroup.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
$endgroup$
Every finite group is isomorphic to a subgroup of some $S_n$,
but the transpositions in $S_n$ are not necessarily in that subgroup.
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
answered 8 hours ago
J. W. TannerJ. W. Tanner
7,8441723
7,8441723
$begingroup$
We can rewrite any permutation from $S_n$ as a product of transpositions. So also every permutation from that subgroup that group $G$ is isomorphic to, is a product of transpositions. So, of elements with order $2$.
$endgroup$
– Coupeau
8 hours ago
1
$begingroup$
@Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
$endgroup$
– Hagen von Eitzen
8 hours ago
$begingroup$
Yes, but there is an isomorphism, let's say $phi$ that maps them to elements from $G$. And isomorphism preserves the order of elements. I am missing something here but I do not really see what.
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Yes, there is an isomorphism between $G$ and a subgroup of $S_n$ but it could be that the transpositions in $S_n$ are not involved in the isomorphism; i.e., there are no elements in $G$ corresponding to transpositions in $S_n,$ so there are no elements of order $2$ in $G$
$endgroup$
– J. W. Tanner
7 hours ago
1
$begingroup$
Oh okay. I think I understand now where is the problem. So I can not 'expand' $phi (x)$ to $ phi (ab)$ and then to $ phi (a) phi (b)$ if $a$ and $b$ are not in the domain of the function $phi$ even though $ab = x$. Thank you.
$endgroup$
– Coupeau
6 hours ago
add a comment |
$begingroup$
We can rewrite any permutation from $S_n$ as a product of transpositions. So also every permutation from that subgroup that group $G$ is isomorphic to, is a product of transpositions. So, of elements with order $2$.
$endgroup$
– Coupeau
8 hours ago
1
$begingroup$
@Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
$endgroup$
– Hagen von Eitzen
8 hours ago
$begingroup$
Yes, but there is an isomorphism, let's say $phi$ that maps them to elements from $G$. And isomorphism preserves the order of elements. I am missing something here but I do not really see what.
$endgroup$
– Coupeau
7 hours ago
1
$begingroup$
Yes, there is an isomorphism between $G$ and a subgroup of $S_n$ but it could be that the transpositions in $S_n$ are not involved in the isomorphism; i.e., there are no elements in $G$ corresponding to transpositions in $S_n,$ so there are no elements of order $2$ in $G$
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– J. W. Tanner
7 hours ago
1
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Oh okay. I think I understand now where is the problem. So I can not 'expand' $phi (x)$ to $ phi (ab)$ and then to $ phi (a) phi (b)$ if $a$ and $b$ are not in the domain of the function $phi$ even though $ab = x$. Thank you.
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– Coupeau
6 hours ago
$begingroup$
We can rewrite any permutation from $S_n$ as a product of transpositions. So also every permutation from that subgroup that group $G$ is isomorphic to, is a product of transpositions. So, of elements with order $2$.
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– Coupeau
8 hours ago
$begingroup$
We can rewrite any permutation from $S_n$ as a product of transpositions. So also every permutation from that subgroup that group $G$ is isomorphic to, is a product of transpositions. So, of elements with order $2$.
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– Coupeau
8 hours ago
1
1
$begingroup$
@Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
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– Hagen von Eitzen
8 hours ago
$begingroup$
@Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
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– Hagen von Eitzen
8 hours ago
$begingroup$
Yes, but there is an isomorphism, let's say $phi$ that maps them to elements from $G$. And isomorphism preserves the order of elements. I am missing something here but I do not really see what.
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– Coupeau
7 hours ago
$begingroup$
Yes, but there is an isomorphism, let's say $phi$ that maps them to elements from $G$. And isomorphism preserves the order of elements. I am missing something here but I do not really see what.
$endgroup$
– Coupeau
7 hours ago
1
1
$begingroup$
Yes, there is an isomorphism between $G$ and a subgroup of $S_n$ but it could be that the transpositions in $S_n$ are not involved in the isomorphism; i.e., there are no elements in $G$ corresponding to transpositions in $S_n,$ so there are no elements of order $2$ in $G$
$endgroup$
– J. W. Tanner
7 hours ago
$begingroup$
Yes, there is an isomorphism between $G$ and a subgroup of $S_n$ but it could be that the transpositions in $S_n$ are not involved in the isomorphism; i.e., there are no elements in $G$ corresponding to transpositions in $S_n,$ so there are no elements of order $2$ in $G$
$endgroup$
– J. W. Tanner
7 hours ago
1
1
$begingroup$
Oh okay. I think I understand now where is the problem. So I can not 'expand' $phi (x)$ to $ phi (ab)$ and then to $ phi (a) phi (b)$ if $a$ and $b$ are not in the domain of the function $phi$ even though $ab = x$. Thank you.
$endgroup$
– Coupeau
6 hours ago
$begingroup$
Oh okay. I think I understand now where is the problem. So I can not 'expand' $phi (x)$ to $ phi (ab)$ and then to $ phi (a) phi (b)$ if $a$ and $b$ are not in the domain of the function $phi$ even though $ab = x$. Thank you.
$endgroup$
– Coupeau
6 hours ago
add a comment |
$begingroup$
Here's an analogy: consider the group of even numbers under addition. Now, every even number can be described as the sum of two odd numbers. How is that possible when there are no odd numbers in the group?
Or consider the field of real numbers. Every real $x$ can be written as $i^4x$: how is that possible when $i not in Bbb R$?
answered 8 hours ago
ThéophileThéophile
20.8k13047
$endgroup$
add a comment |
$begingroup$
Here's an analogy: consider the group of even numbers under addition. Now, every even number can be described as the sum of two odd numbers. How is that possible when there are no odd numbers in the group?
Or consider the field of real numbers. Every real $x$ can be written as $i^4x$: how is that possible when $i not in Bbb R$?
answered 8 hours ago
ThéophileThéophile
20.8k13047
$endgroup$
add a comment |
$begingroup$
Here's an analogy: consider the group of even numbers under addition. Now, every even number can be described as the sum of two odd numbers. How is that possible when there are no odd numbers in the group?
Or consider the field of real numbers. Every real $x$ can be written as $i^4x$: how is that possible when $i not in Bbb R$?
answered 8 hours ago
ThéophileThéophile
20.8k13047
$endgroup$
Here's an analogy: consider the group of even numbers under addition. Now, every even number can be described as the sum of two odd numbers. How is that possible when there are no odd numbers in the group?
Or consider the field of real numbers. Every real $x$ can be written as $i^4x$: how is that possible when $i not in Bbb R$?
answered 8 hours ago
ThéophileThéophile
20.8k13047
answered 8 hours ago
ThéophileThéophile
20.8k13047
answered 8 hours ago
ThéophileThéophile
20.8k13047
answered 8 hours ago
ThéophileThéophile
20.8k13047
20.8k13047
add a comment |
add a comment |
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1
$begingroup$
I would recommend $A_3$ as an example to understand what is going on.
$endgroup$
– Thomas Shelby
8 hours ago
$begingroup$
the order of a transposition may be two, but what is the order of (12)(13)?
$endgroup$
– graeme
8 hours ago
$begingroup$
@graeme Yes, it is not $2$, but... Let $Phi$ be the isomorphism from $G$ to some subgroup $H leq S_n$. Isn't it then that $Phi ((12)(13)) = Phi ((12)) cdot Phi ((13))$ And order of $Phi ((12))$ is the same as of $(12)$, which is $2$? And the same for $Phi ((13))$
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– Coupeau
7 hours ago
1
$begingroup$
Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
$endgroup$
– J. W. Tanner
7 hours ago
$begingroup$
Even if it was defined, it is not true. Elements of order $k$ are sent to elements of order $k$ by isomorphisms. (12)(13) does not have order 2, it has order 3, since it is equal to (1 3 2). I suggest you do some calculations on your own to convince yourself of this fact.
$endgroup$
– graeme
7 hours ago