I do not believe this is a proof. You assume
, but this is not necessarily so. Essentially, you have produced a generic proof which says that a proper subgroup of a group cannot be isomorphic to the group itself. This is not so. An obvious counter-example is the infinite cartesian product of the integers.
.
However, the result you used is true, but needs to be proven.
Notice that for
,
. So the result now just needs to be shown for the denominators.
as the kernel is trivial.
Thus,
as required.
Or something along those lines...