Proving Group Equality for Normal Subgroups

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Discussion Overview

The discussion revolves around proving a property related to normal subgroups in group theory. Specifically, participants explore the implications of the intersection of two normal subgroups H and K being trivial (i.e., containing only the identity element) and how this affects the commutativity of elements from these subgroups within a larger group G.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a claim that if H and K are normal subgroups of G with H∩K = {e}, then it follows that hk = kh for all h ∈ H and k ∈ K.
  • Another participant challenges the first equality presented, suggesting that the goal should be to show hkh⁻¹k⁻¹ = e, and emphasizes that since H∩K = {e}, it suffices to demonstrate that hkh⁻¹k⁻¹ belongs to both H and K.
  • A subsequent reply indicates that since H is normal, the expression kh⁻¹k⁻¹ is in H, leading to the conclusion that hkh⁻¹k⁻¹ is in H as well.
  • Another participant echoes a similar idea, restating the expression hkh⁻¹k⁻¹ in a different form.

Areas of Agreement / Disagreement

The discussion contains disagreement regarding the correctness of the initial equality and the steps needed to prove the claim. Participants are refining their understanding of the implications of normality and the properties of the intersection of subgroups, but no consensus is reached on the validity of the initial approach.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the proof, and there are assumptions about the properties of normal subgroups that remain unexamined.

DanielThrice
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No. This isn't homework. And I think I am right there with this one.

I'm interested in the intersection of groups and what they equal, my professor proposed starting with something like this:

Show that if H and K are normal subgroups of a group G such that H∩K = {e}, then hk = kh for all h ∈ H and for all k ∈ K.

I've gotten this far:

kh-1k-1 = h(kh-1k-1) = (hkh-1)k-1 ∈ H∩K

Am I allowed to assume what I just did?
 
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The first equality is wrong.
Essentially what you want to show is that hkh-1k-1=e.
But H∩K={e}, so just show that hkh-1k-1∈ H and hkh-1k-1∈ K.
 
hkh-1k-1 = h(kh-1k-1; and H is normal, which tells you that kh-1k-1is in H. So hkh-1k-1 is the product of two elements of H and is therefore in H.
Is this correct? What about for K?
 
Similair idea:

hkh-1k-1=(hkh-1)k-1
 

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