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2

$begingroup$

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:
$G = mathbb{Z_4} = {0, 1, 2, 3}$

We have a homomorphism $Phi: mathbb{Z_4} rightarrow S_4$

(We get it using Cayley'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 $mathbb{Z_4}$ 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 $1 in mathbb{Z_4}$ as the product of three elements of order $2$.

abstract-algebra group-theory finite-groups permutations

share|cite|improve this question

edited 51 mins ago

user21820

41k546166

asked 10 hours ago

CoupeauCoupeau

1859

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would recommend $A_3$ as an example to understand what is going on.
  $endgroup$
  – Thomas Shelby
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the order of a transposition may be two, but what is the order of (12)(13)?
  $endgroup$
  – graeme
  10 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
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
  $endgroup$
  – J. W. Tanner
  10 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
  10 hours ago
  
  
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

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:
$G = mathbb{Z_4} = {0, 1, 2, 3}$

We have a homomorphism $Phi: mathbb{Z_4} rightarrow S_4$

(We get it using Cayley'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 $mathbb{Z_4}$ 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 $1 in mathbb{Z_4}$ as the product of three elements of order $2$.

abstract-algebra group-theory finite-groups permutations

share|cite|improve this question

edited 51 mins ago

user21820

41k546166

asked 10 hours ago

CoupeauCoupeau

1859

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I would recommend $A_3$ as an example to understand what is going on.
  $endgroup$
  – Thomas Shelby
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the order of a transposition may be two, but what is the order of (12)(13)?
  $endgroup$
  – graeme
  10 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
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe $(12)$ is not in $H$ so $Phi((12))$ is not defined
  $endgroup$
  – J. W. Tanner
  10 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
  10 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:
$G = mathbb{Z_4} = {0, 1, 2, 3}$

We have a homomorphism $Phi: mathbb{Z_4} rightarrow S_4$

(We get it using Cayley'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 $mathbb{Z_4}$ 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 $1 in mathbb{Z_4}$ as the product of three elements of order $2$.

abstract-algebra group-theory finite-groups permutations

share|cite|improve this question

edited 51 mins ago

user21820

41k546166

asked 10 hours ago

CoupeauCoupeau

1859

$endgroup$

share|cite|improve this question

edited 51 mins ago

user21820

41k546166

asked 10 hours ago

CoupeauCoupeau

1859

share|cite|improve this question

edited 51 mins ago

user21820

41k546166

asked 10 hours ago

CoupeauCoupeau

1859

edited 51 mins ago

user21820

41k546166

edited 51 mins ago

user21820

41k546166

asked 10 hours ago

CoupeauCoupeau

1859

asked 10 hours ago

CoupeauCoupeau

1859

2 Answers
2

active

oldest

votes

7

$begingroup$

Every finite group is isomorphic to a subgroup of some $S_n$,

but the transpositions in $S_n$ are not necessarily in that subgroup.

share|cite|improve this answer

answered 10 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
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
  $endgroup$
  – Hagen von Eitzen
  10 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
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 $i not in Bbb R$?

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.8k13047

$endgroup$

add a comment | 

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7

$begingroup$

Every finite group is isomorphic to a subgroup of some $S_n$,

but the transpositions in $S_n$ are not necessarily in that subgroup.

share|cite|improve this answer

answered 10 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
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
  $endgroup$
  – Hagen von Eitzen
  10 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
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

Every finite group is isomorphic to a subgroup of some $S_n$,

but the transpositions in $S_n$ are not necessarily in that subgroup.

share|cite|improve this answer

answered 10 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
  10 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Coupeau Still, those transpositions are $in S_n$, but not necessairly $in G$
  $endgroup$
  – Hagen von Eitzen
  10 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
  10 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
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 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.

share|cite|improve this answer

answered 10 hours ago

J. W. TannerJ. W. Tanner

7,8441723

$endgroup$

share|cite|improve this answer

answered 10 hours ago

J. W. TannerJ. W. Tanner

7,8441723

answered 10 hours ago

J. W. TannerJ. W. Tanner

7,8441723

answered 10 hours ago

J. W. TannerJ. W. Tanner

7,8441723

2

$begingroup$

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 $i not in Bbb R$?

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.8k13047

$endgroup$

add a comment | 

2

$begingroup$

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 $i not in Bbb R$?

share|cite|improve this answer

answered 10 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 $i not in Bbb R$?

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.8k13047

$endgroup$

share|cite|improve this answer

answered 10 hours ago

ThéophileThéophile

20.8k13047

answered 10 hours ago

ThéophileThéophile

20.8k13047

answered 10 hours ago

ThéophileThéophile

20.8k13047