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.

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- Dec 11th 2010, 06:48 AMDanielThriceNormal 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 AMOpalg
- Dec 11th 2010, 12:03 PMDanielThrice
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 PMOpalg
$\displaystyle hkh^{-1}k^{-1} = h(kh^{-1}k^{-1})$; 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 AMDanielThrice
'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 AMOpalg