Understanding Subgroups of Integers: Explained Simply

  • Context: Undergrad 
  • Thread starter Thread starter matheinste
  • Start date Start date
  • Tags Tags
    Integers
Click For Summary

Discussion Overview

The discussion revolves around the concept of subgroups within the set of integers, specifically focusing on the conditions that define a subset S of integers as a subgroup of Z. Participants explore the implications of the subgroup definition and the properties of elements within such subsets.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Matheinste expresses confusion about the subgroup definition, particularly regarding the implications of having a non-empty subset S and how this relates to the inclusion of 0 and the additive inverses of elements.
  • Radou suggests that since S is non-empty, if an element a is in S, then its inverse must also be in S, leading to the conclusion that 0 is included.
  • Matheinste questions the assumption that the non-empty nature of S allows for conclusions about its structure, seeking further clarification.
  • Matheinste later acknowledges understanding that if x is in S, then x-x=0 must also be in S, which supports the subgroup conditions.
  • HallsofIvy raises a question about the necessity of having x+x in S, prompting further reflection on the properties of elements in S.

Areas of Agreement / Disagreement

Participants do not reach a consensus on all points, as there are ongoing questions about the implications of the subgroup definition and the structure of S. Some participants provide clarification while others express uncertainty.

Contextual Notes

The discussion highlights the need for a clear understanding of subgroup properties and the assumptions made about subsets, which may not be universally agreed upon by all participants.

matheinste
Messages
1,068
Reaction score
0
Hello all.

I reluctantly ask this question because it is probably,as the text states easy, but my desire to clear this point up overides my fear of looking a fool.

I quote word for word but will use words instead of the belongs to symbol.

A subset S of the set Z of integers is a subgroup of Z if 0 is in S, -x is in S,and x+y is in S for all x and y in S. It is easy to see that a non empty subset S of Z is a subgroup of Z if and only if x-y is in S for all x and y in S.

I understand the definition and I can see that 0 is in S. I can only assume that somehow because -x ( the additive inverse of x ) is in S that this guarantees that x is in S. If S were a subgroup of Z of course -x being in the subgroup means that x was in it. But we are not assuming that S is a subgroup but testing for it.

Help!

Matheinste.
 
Physics news on Phys.org
Well, first of all, how "can" you see that 0 is in S? S is non empty. So a is in S. And you know that aa^-1 = e ( = 0) is in S.
 
Thaks radou. I am not being deliberately awkward, I just need to completely grasp the basics which I used to find uninteresting but now regard as most interesting. As S is defined only as a subset ( non empty } and has therefore no structure defined I cannot see how we can assume anything about the set from its being non empty.

More help please.

Matheinste.
 
matheinste said:
As S is defined only as a subset ( non empty } and has therefore no structure defined I cannot see how we can assume anything about the set from its being non empty.

More help please.

Matheinste.

Ok, what exactly is your question? Does this bother you?

matheinste said:
It is easy to see that a non empty subset S of Z is a subgroup of Z if and only if x-y is in S for all x and y in S.

If so, you already know that 0 is in S, from the previous post. Now, (since we're proving direction "<=", and we assumed that x-y is in S for all x and y in S), what can we "do" with 0 and a?
 
Thanks radou. I think I may be a little less stupid today.

I see now that if we have at least one element in S and call it x then we have x-x=0 is in S by what we are given. Then we can have 0-x is in S and so
-x is in S and this is our inverse element and so x-(-x)=x+x is in S. This fulfils the requirements stated subgroup.

Is that OK.

Matheinste.
 
matheinste said:
Thanks radou. I think I may be a little less stupid today.

I see now that if we have at least one element in S and call it x then we have x-x=0 is in S by what we are given. Then we can have 0-x is in S and so
-x is in S and this is our inverse element and so x-(-x)=x+x is in S. This fulfils the requirements stated subgroup.

Is that OK.

Matheinste.
I'm puzzled by that last statement. Why is it important that x+ x be in S?
More to the point is that if x and y are in S, then so is -y and x-(-y)= x+ y is in S.
 
Thankyou HallsofIvy. I take note of what you have said. I am now happy with the answer.

Matheinste.
 

Similar threads

Undergrad Proof for subgroup -- How prove it is a subgroup of Z^m?
  • · Replies 11 ·
Replies
11
Views
3K
Undergrad Differences between left vs right actions in some group theory questions
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad Trying to get the point of some Group Theory Lemmas
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Trouble with a passage in Zariski & Samuel's Commutative algebra text
  • · Replies 5 ·
Replies
5
Views
2K
High School Solving a Linear Programming Problem which Requires Integer Solutions
  • · Replies 17 ·
Replies
17
Views
3K
Undergrad Did I figure-out z-translation of 2D-coordinates correctly?
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad How to show ##p(x)=g(x)x\pm 1\in\Bbb{Q}[x]## is irreducible in ##\Bbb{Q}_{\Bbb{Z}}[x]##?
  • · Replies 48 ·
2
Replies
48
Views
6K
Graduate What are the conjugacy classes of subgroups of Z X Z?
  • · Replies 6 ·
Replies
6
Views
7K
Undergrad Question about "A Group Epimorphism is Surjective"
  • · Replies 13 ·
Replies
13
Views
2K
Undergrad On a bound on the norm of a matrix with a simple pole
  • · Replies 8 ·
Replies
8
Views
3K