Find a group G that has subgroups isomorphic to $mathbbZ_n$ for all natural numbers n $geq$ 2

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I was thinking of using the theorem: " If G is finite of order n, then $G cong mathbbZ_n$ " But cannot find an appropriate group.

Any help is appreciated, thanks.

EDIT: The correct version of the theorem is "If G is cyclic AND finite of order n, then $G cong mathbbZ_n$ "

abstract-algebra group-theory finite-groups

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edited Jul 22 at 18:23

asked Jul 22 at 18:04

Wallace

625

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up vote
1
down vote

favorite

I was thinking of using the theorem: " If G is finite of order n, then $G cong mathbbZ_n$ " But cannot find an appropriate group.

Any help is appreciated, thanks.

EDIT: The correct version of the theorem is "If G is cyclic AND finite of order n, then $G cong mathbbZ_n$ "

abstract-algebra group-theory finite-groups

share|cite|improve this question

edited Jul 22 at 18:23

asked Jul 22 at 18:04

Wallace

625

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I was thinking of using the theorem: " If G is finite of order n, then $G cong mathbbZ_n$ " But cannot find an appropriate group.

Any help is appreciated, thanks.

EDIT: The correct version of the theorem is "If G is cyclic AND finite of order n, then $G cong mathbbZ_n$ "

abstract-algebra group-theory finite-groups

share|cite|improve this question

edited Jul 22 at 18:23

asked Jul 22 at 18:04

Wallace

625

I was thinking of using the theorem: " If G is finite of order n, then $G cong mathbbZ_n$ " But cannot find an appropriate group.

Any help is appreciated, thanks.

EDIT: The correct version of the theorem is "If G is cyclic AND finite of order n, then $G cong mathbbZ_n$ "

abstract-algebra group-theory finite-groups

share|cite|improve this question

edited Jul 22 at 18:23

asked Jul 22 at 18:04

Wallace

625

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 18:23

edited Jul 22 at 18:23

edited Jul 22 at 18:23

asked Jul 22 at 18:04

Wallace

625

asked Jul 22 at 18:04

Wallace

625

asked Jul 22 at 18:04

Wallace

625

625

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
6
down vote



accepted

An easy example is the multiplicative group of the complex numbers. For each $n$ it has exactly one cyclic subgroup of order $n$, consisting of the solutions of $z^n-1=0$.

share|cite|improve this answer

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate on why for each n it has exactly one cyclic subgroup of order n?
  – Wallace
  Jul 22 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think I'm starting to see why this is true, my knowledge on complex analysis is a bit groggy as of now.
  – Wallace
  Jul 22 at 18:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  All the solutions of $z^n=1$ are powers of $exp(2pi i/n)$. @Wallace
  – Lord Shark the Unknown
  Jul 22 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect! Thank you!
  – Wallace
  Jul 22 at 18:28
  

  
  
  
  

add a comment | 

up vote
3
down vote

It is not true that every finite group of order $n$ is isomorphic to $Bbb Z_n$.

However, you can take $G = displaystyle prod_n ge 2 Bbb Z_n$, where $prod$ means direct product.

share|cite|improve this answer

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm actually interested about your first statement. Could you provide me a counter example of a group with order n that is not isomorphic to $mathbbZ_n$?
  – Wallace
  Jul 22 at 18:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/List_of_small_groups
  – Kenny Lau
  Jul 22 at 18:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wallace You know that there are non-abelian groups. They give enough counterexamples...
  – Dietrich Burde
  Jul 22 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau, thank you for providing me with this. It appears I have misread the theorem. It requires G to be cyclic! Thank you for this clarification. I will edit my original post
  – Wallace
  Jul 22 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order of the group $mathbb Z_2 times mathbb Z_2$ is not isomorphic to $mathbb Z_4$
  – steven gregory
  Jul 22 at 23:38
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

Let $mathbb S = mathbb R/1$, the set of all real numbers modulo $1$, i.e. the set of all sets of the form $[x] = x + n $. Addition on $mathbb S$ is defined by $[x]+[y] =[x+y]$. Then $mathbb S$ is an abelian group with respect to addition. For any positive integer, $n$, the order of the group generated by $[frac 1n]$ is $n$.

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answered Jul 22 at 18:23

steven gregory

16.4k22055

add a comment | 

up vote
1
down vote

It's a bit silly, but you can use Cayley's theorem that states that every finite group is embedded in some symmetric group $S_n$, and then
$$
G = prod_n in mathbb N S_n
$$
will be a group that contains any finite group as a subgroup. This also is true for the symmetric group of the natural numbers $S(mathbb N)$, since it contains all $S_n$ as subgroups.

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answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think you can take the subgroup of $G$ consisting of elements with finite support.
  – Kenny Lau
  Jul 23 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, then we'd even be countable!
  – AlgebraicsAnonymous
  Jul 28 at 7:06
  

  
  
  
  

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote



accepted

An easy example is the multiplicative group of the complex numbers. For each $n$ it has exactly one cyclic subgroup of order $n$, consisting of the solutions of $z^n-1=0$.

share|cite|improve this answer

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate on why for each n it has exactly one cyclic subgroup of order n?
  – Wallace
  Jul 22 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think I'm starting to see why this is true, my knowledge on complex analysis is a bit groggy as of now.
  – Wallace
  Jul 22 at 18:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  All the solutions of $z^n=1$ are powers of $exp(2pi i/n)$. @Wallace
  – Lord Shark the Unknown
  Jul 22 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect! Thank you!
  – Wallace
  Jul 22 at 18:28
  

  
  
  
  

add a comment | 

up vote
6
down vote



accepted

An easy example is the multiplicative group of the complex numbers. For each $n$ it has exactly one cyclic subgroup of order $n$, consisting of the solutions of $z^n-1=0$.

share|cite|improve this answer

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate on why for each n it has exactly one cyclic subgroup of order n?
  – Wallace
  Jul 22 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think I'm starting to see why this is true, my knowledge on complex analysis is a bit groggy as of now.
  – Wallace
  Jul 22 at 18:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  All the solutions of $z^n=1$ are powers of $exp(2pi i/n)$. @Wallace
  – Lord Shark the Unknown
  Jul 22 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect! Thank you!
  – Wallace
  Jul 22 at 18:28
  

  
  
  
  

add a comment | 

up vote
6
down vote



accepted

up vote
6
down vote



accepted

An easy example is the multiplicative group of the complex numbers. For each $n$ it has exactly one cyclic subgroup of order $n$, consisting of the solutions of $z^n-1=0$.

share|cite|improve this answer

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

An easy example is the multiplicative group of the complex numbers. For each $n$ it has exactly one cyclic subgroup of order $n$, consisting of the solutions of $z^n-1=0$.

share|cite|improve this answer

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

answered Jul 22 at 18:08

Lord Shark the Unknown

85.2k950111

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate on why for each n it has exactly one cyclic subgroup of order n?
  – Wallace
  Jul 22 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think I'm starting to see why this is true, my knowledge on complex analysis is a bit groggy as of now.
  – Wallace
  Jul 22 at 18:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  All the solutions of $z^n=1$ are powers of $exp(2pi i/n)$. @Wallace
  – Lord Shark the Unknown
  Jul 22 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect! Thank you!
  – Wallace
  Jul 22 at 18:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate on why for each n it has exactly one cyclic subgroup of order n?
  – Wallace
  Jul 22 at 18:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think I'm starting to see why this is true, my knowledge on complex analysis is a bit groggy as of now.
  – Wallace
  Jul 22 at 18:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  All the solutions of $z^n=1$ are powers of $exp(2pi i/n)$. @Wallace
  – Lord Shark the Unknown
  Jul 22 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perfect! Thank you!
  – Wallace
  Jul 22 at 18:28
  

  
  
  
  

Could you elaborate on why for each n it has exactly one cyclic subgroup of order n?
– Wallace
Jul 22 at 18:17

Could you elaborate on why for each n it has exactly one cyclic subgroup of order n?
– Wallace
Jul 22 at 18:17

I think I'm starting to see why this is true, my knowledge on complex analysis is a bit groggy as of now.
– Wallace
Jul 22 at 18:26

I think I'm starting to see why this is true, my knowledge on complex analysis is a bit groggy as of now.
– Wallace
Jul 22 at 18:26

All the solutions of $z^n=1$ are powers of $exp(2pi i/n)$. @Wallace
– Lord Shark the Unknown
Jul 22 at 18:27

All the solutions of $z^n=1$ are powers of $exp(2pi i/n)$. @Wallace
– Lord Shark the Unknown
Jul 22 at 18:27

Perfect! Thank you!
– Wallace
Jul 22 at 18:28

Perfect! Thank you!
– Wallace
Jul 22 at 18:28

add a comment | 

up vote
3
down vote

It is not true that every finite group of order $n$ is isomorphic to $Bbb Z_n$.

However, you can take $G = displaystyle prod_n ge 2 Bbb Z_n$, where $prod$ means direct product.

share|cite|improve this answer

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm actually interested about your first statement. Could you provide me a counter example of a group with order n that is not isomorphic to $mathbbZ_n$?
  – Wallace
  Jul 22 at 18:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/List_of_small_groups
  – Kenny Lau
  Jul 22 at 18:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wallace You know that there are non-abelian groups. They give enough counterexamples...
  – Dietrich Burde
  Jul 22 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau, thank you for providing me with this. It appears I have misread the theorem. It requires G to be cyclic! Thank you for this clarification. I will edit my original post
  – Wallace
  Jul 22 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order of the group $mathbb Z_2 times mathbb Z_2$ is not isomorphic to $mathbb Z_4$
  – steven gregory
  Jul 22 at 23:38
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

It is not true that every finite group of order $n$ is isomorphic to $Bbb Z_n$.

However, you can take $G = displaystyle prod_n ge 2 Bbb Z_n$, where $prod$ means direct product.

share|cite|improve this answer

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm actually interested about your first statement. Could you provide me a counter example of a group with order n that is not isomorphic to $mathbbZ_n$?
  – Wallace
  Jul 22 at 18:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/List_of_small_groups
  – Kenny Lau
  Jul 22 at 18:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wallace You know that there are non-abelian groups. They give enough counterexamples...
  – Dietrich Burde
  Jul 22 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau, thank you for providing me with this. It appears I have misread the theorem. It requires G to be cyclic! Thank you for this clarification. I will edit my original post
  – Wallace
  Jul 22 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order of the group $mathbb Z_2 times mathbb Z_2$ is not isomorphic to $mathbb Z_4$
  – steven gregory
  Jul 22 at 23:38
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

It is not true that every finite group of order $n$ is isomorphic to $Bbb Z_n$.

However, you can take $G = displaystyle prod_n ge 2 Bbb Z_n$, where $prod$ means direct product.

share|cite|improve this answer

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

It is not true that every finite group of order $n$ is isomorphic to $Bbb Z_n$.

However, you can take $G = displaystyle prod_n ge 2 Bbb Z_n$, where $prod$ means direct product.

share|cite|improve this answer

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

answered Jul 22 at 18:08

Kenny Lau

18.7k2157

18.7k2157

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm actually interested about your first statement. Could you provide me a counter example of a group with order n that is not isomorphic to $mathbbZ_n$?
  – Wallace
  Jul 22 at 18:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/List_of_small_groups
  – Kenny Lau
  Jul 22 at 18:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wallace You know that there are non-abelian groups. They give enough counterexamples...
  – Dietrich Burde
  Jul 22 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau, thank you for providing me with this. It appears I have misread the theorem. It requires G to be cyclic! Thank you for this clarification. I will edit my original post
  – Wallace
  Jul 22 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order of the group $mathbb Z_2 times mathbb Z_2$ is not isomorphic to $mathbb Z_4$
  – steven gregory
  Jul 22 at 23:38
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm actually interested about your first statement. Could you provide me a counter example of a group with order n that is not isomorphic to $mathbbZ_n$?
  – Wallace
  Jul 22 at 18:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  en.wikipedia.org/wiki/List_of_small_groups
  – Kenny Lau
  Jul 22 at 18:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Wallace You know that there are non-abelian groups. They give enough counterexamples...
  – Dietrich Burde
  Jul 22 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KennyLau, thank you for providing me with this. It appears I have misread the theorem. It requires G to be cyclic! Thank you for this clarification. I will edit my original post
  – Wallace
  Jul 22 at 18:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The order of the group $mathbb Z_2 times mathbb Z_2$ is not isomorphic to $mathbb Z_4$
  – steven gregory
  Jul 22 at 23:38
  
  

  
  
  
  

I'm actually interested about your first statement. Could you provide me a counter example of a group with order n that is not isomorphic to $mathbbZ_n$?
– Wallace
Jul 22 at 18:18

I'm actually interested about your first statement. Could you provide me a counter example of a group with order n that is not isomorphic to $mathbbZ_n$?
– Wallace
Jul 22 at 18:18

en.wikipedia.org/wiki/List_of_small_groups
– Kenny Lau
Jul 22 at 18:19

en.wikipedia.org/wiki/List_of_small_groups
– Kenny Lau
Jul 22 at 18:19

@Wallace You know that there are non-abelian groups. They give enough counterexamples...
– Dietrich Burde
Jul 22 at 18:20

@Wallace You know that there are non-abelian groups. They give enough counterexamples...
– Dietrich Burde
Jul 22 at 18:20

@KennyLau, thank you for providing me with this. It appears I have misread the theorem. It requires G to be cyclic! Thank you for this clarification. I will edit my original post
– Wallace
Jul 22 at 18:22

@KennyLau, thank you for providing me with this. It appears I have misread the theorem. It requires G to be cyclic! Thank you for this clarification. I will edit my original post
– Wallace
Jul 22 at 18:22

The order of the group $mathbb Z_2 times mathbb Z_2$ is not isomorphic to $mathbb Z_4$
– steven gregory
Jul 22 at 23:38

The order of the group $mathbb Z_2 times mathbb Z_2$ is not isomorphic to $mathbb Z_4$
– steven gregory
Jul 22 at 23:38

add a comment | 

up vote
2
down vote

Let $mathbb S = mathbb R/1$, the set of all real numbers modulo $1$, i.e. the set of all sets of the form $[x] = x + n $. Addition on $mathbb S$ is defined by $[x]+[y] =[x+y]$. Then $mathbb S$ is an abelian group with respect to addition. For any positive integer, $n$, the order of the group generated by $[frac 1n]$ is $n$.

share|cite|improve this answer

answered Jul 22 at 18:23

steven gregory

16.4k22055

add a comment | 

up vote
2
down vote

Let $mathbb S = mathbb R/1$, the set of all real numbers modulo $1$, i.e. the set of all sets of the form $[x] = x + n $. Addition on $mathbb S$ is defined by $[x]+[y] =[x+y]$. Then $mathbb S$ is an abelian group with respect to addition. For any positive integer, $n$, the order of the group generated by $[frac 1n]$ is $n$.

share|cite|improve this answer

answered Jul 22 at 18:23

steven gregory

16.4k22055

add a comment | 

up vote
2
down vote

up vote
2
down vote

Let $mathbb S = mathbb R/1$, the set of all real numbers modulo $1$, i.e. the set of all sets of the form $[x] = x + n $. Addition on $mathbb S$ is defined by $[x]+[y] =[x+y]$. Then $mathbb S$ is an abelian group with respect to addition. For any positive integer, $n$, the order of the group generated by $[frac 1n]$ is $n$.

share|cite|improve this answer

answered Jul 22 at 18:23

steven gregory

16.4k22055

Let $mathbb S = mathbb R/1$, the set of all real numbers modulo $1$, i.e. the set of all sets of the form $[x] = x + n $. Addition on $mathbb S$ is defined by $[x]+[y] =[x+y]$. Then $mathbb S$ is an abelian group with respect to addition. For any positive integer, $n$, the order of the group generated by $[frac 1n]$ is $n$.

share|cite|improve this answer

answered Jul 22 at 18:23

steven gregory

16.4k22055

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 18:23

steven gregory

16.4k22055

answered Jul 22 at 18:23

steven gregory

16.4k22055

answered Jul 22 at 18:23

steven gregory

16.4k22055

16.4k22055

add a comment | 

add a comment | 

up vote
1
down vote

It's a bit silly, but you can use Cayley's theorem that states that every finite group is embedded in some symmetric group $S_n$, and then
$$
G = prod_n in mathbb N S_n
$$
will be a group that contains any finite group as a subgroup. This also is true for the symmetric group of the natural numbers $S(mathbb N)$, since it contains all $S_n$ as subgroups.

share|cite|improve this answer

answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think you can take the subgroup of $G$ consisting of elements with finite support.
  – Kenny Lau
  Jul 23 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, then we'd even be countable!
  – AlgebraicsAnonymous
  Jul 28 at 7:06
  

  
  
  
  

add a comment | 

up vote
1
down vote

It's a bit silly, but you can use Cayley's theorem that states that every finite group is embedded in some symmetric group $S_n$, and then
$$
G = prod_n in mathbb N S_n
$$
will be a group that contains any finite group as a subgroup. This also is true for the symmetric group of the natural numbers $S(mathbb N)$, since it contains all $S_n$ as subgroups.

share|cite|improve this answer

answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think you can take the subgroup of $G$ consisting of elements with finite support.
  – Kenny Lau
  Jul 23 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, then we'd even be countable!
  – AlgebraicsAnonymous
  Jul 28 at 7:06
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

It's a bit silly, but you can use Cayley's theorem that states that every finite group is embedded in some symmetric group $S_n$, and then
$$
G = prod_n in mathbb N S_n
$$
will be a group that contains any finite group as a subgroup. This also is true for the symmetric group of the natural numbers $S(mathbb N)$, since it contains all $S_n$ as subgroups.

share|cite|improve this answer

answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

It's a bit silly, but you can use Cayley's theorem that states that every finite group is embedded in some symmetric group $S_n$, and then
$$
G = prod_n in mathbb N S_n
$$
will be a group that contains any finite group as a subgroup. This also is true for the symmetric group of the natural numbers $S(mathbb N)$, since it contains all $S_n$ as subgroups.

share|cite|improve this answer

answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

answered Jul 22 at 18:43

AlgebraicsAnonymous

69111

69111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think you can take the subgroup of $G$ consisting of elements with finite support.
  – Kenny Lau
  Jul 23 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, then we'd even be countable!
  – AlgebraicsAnonymous
  Jul 28 at 7:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think you can take the subgroup of $G$ consisting of elements with finite support.
  – Kenny Lau
  Jul 23 at 1:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, then we'd even be countable!
  – AlgebraicsAnonymous
  Jul 28 at 7:06
  

  
  
  
  

1

1

I think you can take the subgroup of $G$ consisting of elements with finite support.
– Kenny Lau
Jul 23 at 1:53

I think you can take the subgroup of $G$ consisting of elements with finite support.
– Kenny Lau
Jul 23 at 1:53

Yes, then we'd even be countable!
– AlgebraicsAnonymous
Jul 28 at 7:06

Yes, then we'd even be countable!
– AlgebraicsAnonymous
Jul 28 at 7:06

add a comment | 

 

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