Let be a group homomorphism. We proved in this question that its kernel is infinite. We are proving now that it is also uncountable.
Before we proceed with the proof we are stating that not all homomorphisms are of the form . We can find non trivial homomorphisms. But all continuous are of the above form.
Let be the group of roots of unity. Both groups and are uniquely divisible and thus are vector spaces over .
Since the positive reals are closed under multiplication, it’s easy to see that
Using the axiom of choice , we construct a group homomorphism
Hence the kernel is uncountable.
Note: is a direct summand of . That is because
( since is injective ) and so the exact sequence