Normal subgroups

  • #1
DanielThrice
29
0
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?
 
Last edited:

Answers and Replies

  • #2
vr88
11
0
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.
 
  • #3
DanielThrice
29
0
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?
 
  • #4
kamil9876
6
0
Similair idea:

hkh-1k-1=(hkh-1)k-1
 

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