The two subgroups are H1={1,x} and H2={1,y}, essentially (with x2=y2=1). They're not really disjoint, as all subgroups contain 1, so I assume we mean that they're disjoint except for that.
So we get in H the elements 1,x,y,xy,yx,xyx,yxy,…There is no natural upper limit (because xy and yx need not have order 2). I think we can go as high as we like using x and y reflections in the plane, and xy becomes a rotation, and we can get lots of finite orders for that...)
If H1 or H2 is normal, then statement is true, in fact in that case H=H1H2 and |H1H2|=4. In general, the set H1H2 has 4 elements, since |H1H2|=|H1| |H2| / | H1∩H2|=2.2/ 1= 4, and of course H1H2⊆H. So H has at least 4 elements.
Reference : Stackoverflow