Let be a group homomorphism. We proved in this question that its kernel is infinite. We are proving now that it is also uncountable.

**Solution**

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

splits.