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