Proving Conjugate Subgroups K=H with Prime Number of Elements

  • Thread starter Thread starter *FaerieLight*
  • Start date Start date
  • Tags Tags
    Conjugate
Click For Summary
SUMMARY

In the discussion, the user seeks to prove that if K = gHg-1 for some g in G, where both K and H are subgroups of G with a prime number of elements, then K must equal H. The user initially attempts to demonstrate this by manipulating the subgroup relationships but realizes that the proof fails because H does not equal gHg-1 unless g is in the normalizer of H. The example provided using G = S3 illustrates that K and H can be conjugate subgroups of the same order without being equal.

PREREQUISITES
  • Understanding of group theory concepts, specifically subgroups and conjugation.
  • Familiarity with the symmetric group, particularly S3.
  • Knowledge of normalizers in group theory.
  • Basic proof techniques in abstract algebra.
NEXT STEPS
  • Study the properties of normalizers in group theory.
  • Learn about the classification of subgroups in symmetric groups.
  • Explore the concept of conjugacy classes in group theory.
  • Investigate the implications of Lagrange's theorem on subgroup orders.
USEFUL FOR

Students of abstract algebra, particularly those studying group theory, as well as educators and researchers looking to deepen their understanding of subgroup properties and conjugation in finite groups.

*FaerieLight*
Messages
43
Reaction score
0

Homework Statement


How would I go about proving that if K = gHg-1, for some g \inG, where K and H are both subgroups of G with a prime number of elements, then K = H?


Homework Equations


I've tried to prove it by saying that if K = gHg-1 then Kg = gH, and since H = gHg-1, then Hg = gH also, so Hg = Kg, and hence H = K. I don't think that this proof is valid, unfortunately. And I've just realized that H does not necessarily equal gHg-1 unless g is in the normaliser of H. :(

The Attempt at a Solution


This is actually something which I am attempting to prove in order to prove something else, so I'm not even sure if what I'm trying to prove holds at all. I just need K = H for my proof to work.
 
Physics news on Phys.org
Take G=S3, the symmetric group on three elements. Take K={e,(12)} and H={e,(13)}. The subgroups both have order 2 (a prime) and they are conjugate via g=(23), but they aren't equal.
 

Similar threads

Showing that subgroup of unique order implies normality
  • · Replies 3 ·
Replies
3
Views
2K
Conjugate Subgroups of a Finite Group
  • · Replies 23 ·
Replies
23
Views
3K
Conditions on H and K if H ∪ K is a subgroup
  • · Replies 1 ·
Replies
1
Views
1K
Prove: Limit Point of H ∪ K if p is Limit Point of H or K
  • · Replies 5 ·
Replies
5
Views
2K
Showing normality by showing it holds for generators
  • · Replies 3 ·
Replies
3
Views
1K
An exercise with the third isomorphism theorem in group theory
  • · Replies 5 ·
Replies
5
Views
3K
Showing that subgroups of G form a lattice
  • · Replies 4 ·
Replies
4
Views
1K
Conjugacy and Stabilizers in Group Actions
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad Differences between left vs right actions in some group theory questions
  • · Replies 1 ·
Replies
1
Views
1K
Subgroup of an arbitrary group
  • · Replies 5 ·
Replies
5
Views
2K