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Definition of Subgroup

Group proof/disproofProve a group is an abelian groupInfinite groups admitting field structureAdditive non-abelian group?Restricting Binary Operator $*$ To A SubsetSubgroup of finitely generated abelian group is finitely generatedeach finite group with abelian chain induces cyclic chainGroup and subgroupWhy is the commutator subgroup of the group associated to a finite quandle finitely generated?Ring theory (plz guide me what the meaning of statement 1,2 and 3)nontrivial solvable group contains a nontrivial characteristic abelian subgroup

1

$begingroup$

I know that an abelian group is a set G, with a binary operation $$ + : G times G rightarrow G \ (a,b) mapsto+(a,b)=: a+b$$ such that for every $a,b,c in G$

i) $exists 0 in G $ such that $ a+0 = a$

ii) $exists a'$ such that $a + a' = 0$

iii) $a + (b + c) = (a +b) + c$

iv) $a+b = b+a$

Now, I need a little help with the concept of subgroup. Intuitively I think of that as a subset of a group that is also a group. So because we select elements from a group G, iii) and iv) are automatically satisfied. With that idea I would define a subgroup as:

A subgroup of a group $G$ is a subset $G_0$ such that for every $a in G_0$ :

i) $exists 0 in G_0 $ such that $ a+0 = a$

ii) $exists a'$ such that $a + a' = 0$

But instead I have the following definition

An abelian group, $ A_0 subseteq A$ it said to be an abelian subgroup if

i) $0 in A_0$

ii) $forall a,b in A_0 a-b in A_0$

Are those equivalent? or there is something that I am missing?

abstract-algebra

share|cite|improve this question

asked 22 hours ago

José MarínJosé Marín

183210

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Regarding subgroup criterion, see my answer there, though there the group is written multiplicatively (not necessarily Abelian)
  $endgroup$
  – J. W. Tanner
  22 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose $A$ must be an abelian group? (The grammar seems confusing to me.)
  $endgroup$
  – Diglett
  15 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I know that an abelian group is a set G, with a binary operation $$ + : G times G rightarrow G \ (a,b) mapsto+(a,b)=: a+b$$ such that for every $a,b,c in G$

i) $exists 0 in G $ such that $ a+0 = a$

ii) $exists a'$ such that $a + a' = 0$

iii) $a + (b + c) = (a +b) + c$

iv) $a+b = b+a$

Now, I need a little help with the concept of subgroup. Intuitively I think of that as a subset of a group that is also a group. So because we select elements from a group G, iii) and iv) are automatically satisfied. With that idea I would define a subgroup as:

A subgroup of a group $G$ is a subset $G_0$ such that for every $a in G_0$ :

i) $exists 0 in G_0 $ such that $ a+0 = a$

ii) $exists a'$ such that $a + a' = 0$

But instead I have the following definition

An abelian group, $ A_0 subseteq A$ it said to be an abelian subgroup if

i) $0 in A_0$

ii) $forall a,b in A_0 a-b in A_0$

Are those equivalent? or there is something that I am missing?

abstract-algebra

share|cite|improve this question

asked 22 hours ago

José MarínJosé Marín

183210

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Regarding subgroup criterion, see my answer there, though there the group is written multiplicatively (not necessarily Abelian)
  $endgroup$
  – J. W. Tanner
  22 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose $A$ must be an abelian group? (The grammar seems confusing to me.)
  $endgroup$
  – Diglett
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I know that an abelian group is a set G, with a binary operation $$ + : G times G rightarrow G \ (a,b) mapsto+(a,b)=: a+b$$ such that for every $a,b,c in G$

i) $exists 0 in G $ such that $ a+0 = a$

ii) $exists a'$ such that $a + a' = 0$

iii) $a + (b + c) = (a +b) + c$

iv) $a+b = b+a$

Now, I need a little help with the concept of subgroup. Intuitively I think of that as a subset of a group that is also a group. So because we select elements from a group G, iii) and iv) are automatically satisfied. With that idea I would define a subgroup as:

A subgroup of a group $G$ is a subset $G_0$ such that for every $a in G_0$ :

i) $exists 0 in G_0 $ such that $ a+0 = a$

ii) $exists a'$ such that $a + a' = 0$

But instead I have the following definition

An abelian group, $ A_0 subseteq A$ it said to be an abelian subgroup if

i) $0 in A_0$

ii) $forall a,b in A_0 a-b in A_0$

Are those equivalent? or there is something that I am missing?

abstract-algebra

share|cite|improve this question

asked 22 hours ago

José MarínJosé Marín

183210

$endgroup$

share|cite|improve this question

asked 22 hours ago

José MarínJosé Marín

183210

share|cite|improve this question

asked 22 hours ago

José MarínJosé Marín

183210

asked 22 hours ago

José MarínJosé Marín

183210

asked 22 hours ago

José MarínJosé Marín

183210

2 Answers
2

active

oldest

votes

2

$begingroup$

Your definition would allow $-1, 0, 1$ to be a subgroup of the integers under addition. It's not since it doesn't contain $1+1$.

share|cite|improve this answer

answered 22 hours ago

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

add a comment | 

0

$begingroup$

An abelian subgroup satisfies $a+b=b+a$ rather than $a-b in A_0$. Any non-empty subset of a group $A_0$ satisfying $a-b in A_0$ is in fact a subgroup and this property is called the subgroup criterion. It doesn't not force the subgroup to be abelian.

share|cite|improve this answer

answered 22 hours ago

CyclotomicFieldCyclotomicField

2,4431314

$endgroup$

add a comment | 

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2

$begingroup$

Your definition would allow $-1, 0, 1$ to be a subgroup of the integers under addition. It's not since it doesn't contain $1+1$.

share|cite|improve this answer

answered 22 hours ago

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

add a comment | 

2

$begingroup$

Your definition would allow $-1, 0, 1$ to be a subgroup of the integers under addition. It's not since it doesn't contain $1+1$.

share|cite|improve this answer

answered 22 hours ago

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

add a comment | 

$begingroup$

Your definition would allow $-1, 0, 1$ to be a subgroup of the integers under addition. It's not since it doesn't contain $1+1$.

share|cite|improve this answer

answered 22 hours ago

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

share|cite|improve this answer

answered 22 hours ago

Ethan BolkerEthan Bolker

45.4k553120

answered 22 hours ago

Ethan BolkerEthan Bolker

45.4k553120

answered 22 hours ago

Ethan BolkerEthan Bolker

45.4k553120

0

$begingroup$

An abelian subgroup satisfies $a+b=b+a$ rather than $a-b in A_0$. Any non-empty subset of a group $A_0$ satisfying $a-b in A_0$ is in fact a subgroup and this property is called the subgroup criterion. It doesn't not force the subgroup to be abelian.

share|cite|improve this answer

answered 22 hours ago

CyclotomicFieldCyclotomicField

2,4431314

$endgroup$

add a comment | 

0

$begingroup$

An abelian subgroup satisfies $a+b=b+a$ rather than $a-b in A_0$. Any non-empty subset of a group $A_0$ satisfying $a-b in A_0$ is in fact a subgroup and this property is called the subgroup criterion. It doesn't not force the subgroup to be abelian.

share|cite|improve this answer

answered 22 hours ago

CyclotomicFieldCyclotomicField

2,4431314

$endgroup$

add a comment | 

$begingroup$

An abelian subgroup satisfies $a+b=b+a$ rather than $a-b in A_0$. Any non-empty subset of a group $A_0$ satisfying $a-b in A_0$ is in fact a subgroup and this property is called the subgroup criterion. It doesn't not force the subgroup to be abelian.

share|cite|improve this answer

answered 22 hours ago

CyclotomicFieldCyclotomicField

2,4431314

$endgroup$

share|cite|improve this answer

answered 22 hours ago

CyclotomicFieldCyclotomicField

2,4431314

answered 22 hours ago

CyclotomicFieldCyclotomicField

2,4431314

answered 22 hours ago

CyclotomicFieldCyclotomicField

2,4431314