Every infinite cyclic group has non-trivial proper subgroups

Click For Summary

Homework Help Overview

The discussion revolves around the properties of infinite cyclic groups, specifically focusing on the existence of non-trivial proper subgroups. Participants are exploring the differences between finite and infinite cyclic groups and the implications of these differences on subgroup structure.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the definition of infinite cyclic groups and how they differ from finite cyclic groups. There is a focus on the implications of these definitions for subgroup existence.

Discussion Status

Some participants have provided insights into the nature of infinite cyclic groups and have begun to explore what constitutes a non-trivial subgroup. There is an ongoing inquiry into the characteristics of these groups and the conditions that define their subgroups.

Contextual Notes

Participants are questioning the assumptions related to subgroup definitions and the implications of group order, particularly in the context of infinite groups. There is an emphasis on understanding the structure of infinite cyclic groups without reaching a definitive conclusion.

Mr Davis 97
Messages
1,461
Reaction score
44

Homework Statement


Every infinite cyclic group has non-trivial proper subgroups

Homework Equations

The Attempt at a Solution


I know that if we have a finite cyclic group, it only has non-trivial proper subgroups if the order of the group is not prime. But I'm not sure how to make this argument with infinite groups
 
Physics news on Phys.org
What is a infinite cyclic group and how does it differ from a finite one?
 
fresh_42 said:
What is a infinite cyclic group and how does it differ from a finite one?
Doesn't it differ by the fact that there exists no integer ##n## for which ##a^n=e##, where ##a## is an element of the group besides e, the neutral element?
 
Mr Davis 97 said:
Doesn't it differ by the fact that there exists no integer ##n## for which ##a^n=e##, where ##a## is an element of the group besides e, the neutral element?
Yes. So we have ##C = \langle a^n\,\vert \,n \in \mathbb{Z} \rangle = \{\ldots , a^{-2},a^{-1},e,a,a^2,\ldots\}##.
Non-trivial subgroup ##U## means ##\{e\}\subsetneq U \subsetneq C##.
So what can we conclude from ##\{e\}\subsetneq U##?
What could be a subgroup?
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K