Consider the order of the Union of hK where h is an element of H (All the left cosets) This will be equal to the |H|*|K|/|H intersecct K|
I think this will help especially since you know the intersection is of size one.
Let H and K be normal subgroups of G such that H intersect K=<e>. Show that hk=kh for all h in H and k in K.
H and K are normal so ghg^-1 is in H and gkg^-1 is in K.
want to show hk=kh. So basically I'm showing this is abelian.
Can I do ghg^-1=gkg^-1?
ghg^-1g=gkg^-1g
gh=gk
so that works if g=h