It is correct except for a phrase you use "H has no proper normal subgroup", why not?

It does not matter. Just choose . We really do not care if it has proper normal subgroup (by the way it does, it has ) and then apply the theorem and get your result. But then you say something that is again not necessary "so the quotient group of H is not defined", yes it is! Certainly with . It is defined, it is just not a very interesting group.