Isomorphism in Groups with No Proper Subgroups and Absolute Value Greater than 1

  • Thread starter Thread starter kathrynag
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the properties of groups, specifically focusing on groups that have no proper, nontrivial subgroups and an absolute value greater than 1. The original poster attempts to prove that such a group must be isomorphic to Z_p for some prime p.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of G being cyclic and question the existence of nontrivial subgroups based on the number of elements in G. There is also discussion on the properties of cyclic groups and their subgroups, particularly in relation to their order.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of cyclic groups and their subgroups. Some guidance has been offered regarding the relationship between the order of a cyclic group and its subgroups, but no consensus has been reached on the proof itself.

Contextual Notes

Participants are considering the implications of the group being cyclic or not, and the constraints of having only trivial subgroups are under examination. The discussion also touches on the requirement for the group to be isomorphic to Z_p.

kathrynag
Messages
595
Reaction score
0

Homework Statement



Let G be any group with no proper , nontrivial subgroups and assume abs value(G)>1. Prove that G must be isomorphic to Z_p for some prime p.

Homework Equations





The Attempt at a Solution


I know we have an isomorphism if a group is 1-1, onto, and the homomorphism property holds.
 
Physics news on Phys.org


What happens if G is cyclic? In this case, if G does not contain p elements, can you find a nontrivial subgroup? What happens if G is not cyclic? In this case, can you find a nontrivial subgroup?
 


G is cyclic means G=<a>, where a is a generator.
 


Yes, but reason further. If G is cyclic and its number of elements is composite... is there a nontrivial subgroup?
 


I know a cyclic group of order n has exactly one subgroup of order m for each positive divisor m of n.
 


So can you prove that if G is cyclic with only trivial subgroups, then G is isomorphic to some Zp with p prime?

The next step after that is to prove that every group has a cyclic subgroup
 

Similar threads

Having all subgroups normal is isomorphism invariant
  • · Replies 1 ·
Replies
1
Views
2K
Showing that Q_8 can't be written as a direct product
  • · Replies 11 ·
Replies
11
Views
4K
Show that a group with no proper nontrivial subgroups is cyc
  • · Replies 1 ·
Replies
1
Views
3K
Isomorphism to certain Galois group and cyclic groups
  • · Replies 1 ·
Replies
1
Views
2K
Prove (Q+, *) is isomorphic to a proper subgroup of itself
  • · Replies 2 ·
Replies
2
Views
5K
Showing that GL(F_p) is isomorphic to an automorphism group
  • · Replies 1 ·
Replies
1
Views
2K
Normal group of order 60 isomorphic to A_5
  • · Replies 6 ·
Replies
6
Views
2K
Subgroup of Index ##n## for every ##n \in \Bbb{N}##.
  • · Replies 1 ·
Replies
1
Views
2K
Characterizing subgroups of a cyclic group
  • · Replies 9 ·
Replies
9
Views
2K
Showing that every finite group has a composition series
  • · Replies 3 ·
Replies
3
Views
3K