# Normal subgroups

• Dec 11th 2010, 06:48 AM
DanielThrice
Normal subgroups
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.

This was in our study guide for the final and a couple friends of mine and I couldn't figure it out.
• Dec 11th 2010, 07:06 AM
Opalg
Quote:

Originally Posted by DanielThrice
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.

This was in our study guide for the final and a couple friends of mine and I couldn't figure it out.

Hint: $\displaystyle hkh^{-1}k^{-1} = h(kh^{-1}k^{-1}) = (hkh^{-1})k^{-1} \in H\cap K$.
• Dec 11th 2010, 12:03 PM
DanielThrice
So Essentially what we want to show is that hkh-1k-1=e.
But H∩K={e}, so we just have to show that hkh-1k-1∈ H and hkh-1k-1∈ K, but how do we do this?
• Dec 11th 2010, 01:19 PM
Opalg
Quote:

Originally Posted by DanielThrice
So we just have to show that hkh-1k-1∈ H and hkh-1k-1∈ K, but how do we do this?

$\displaystyle hkh^{-1}k^{-1} = h(kh^{-1}k^{-1})$; and H is normal, which tells you that $\displaystyle kh^{-1}k^{-1}\in H$. So $\displaystyle hkh^{-1}k^{-1}$ is the product of two elements of H and is therefore in H. Bracketing it the other way shows that the same element is also in K.
• Dec 12th 2010, 06:48 AM
DanielThrice
'm sorry I've tried a couple ways to solve it for K also but it's not working for this same set up
• Dec 12th 2010, 07:16 AM
Opalg
Quote:

Originally Posted by DanielThrice
'm sorry I've tried a couple ways to solve it for K also but it's not working for this same set up

$\displaystyle hkh^{-1}k^{-1} = (hkh^{-1})k^{-1}$.