However, it turns out that need not be normal in . It can have a weaker property but yet .
Theorem: If has the property that for any and then is a group.
(Do you see why this would prove you case? Meaning, if is normal then certainly . So of course it works).
Proof: All I will show is closure of which is the only part which takes a little work. Let and we want to show . But this means and for and . Thus, . But is this product in ? Not necessarily, so we need to use the additional condition in the theorem. Which says , i.e. for some . Thus, which certainly is in . Q.E.D.