Proving Group Equality for Normal Subgroups

[DanielThrice]

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?

 

[vr88]

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.

 

[DanielThrice]

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?

 

[kamil9876]

Similair idea:

hkh^-1k^-1=(hkh^-1)k^-1