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

 

[blue_raver22]

I would say that you have made a valid assumption and have taken a logical approach to the problem. However, in order to fully prove the statement, you would need to provide a more formal and rigorous proof. This could involve using properties of normal subgroups, the definition of group equality, and other relevant theorems and definitions. It may also be helpful to provide a clear explanation of your thought process and reasoning behind each step. Overall, your approach seems to be on the right track and with some more work, you could potentially provide a solid proof for the given statement.