## Group homomorphism of uncountable kernel

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.

## Group homomorphism with an infinite kernel

Let be a homomorphism. Prove that kernel of is infinite.

Solution

If is a finite subgroup then all must have norm , i.e. . ( Otherwise otherwise is an infinite sequence of distinct elements in . )

Suppose , that is finite and is the kernel of . Then let and

So,  . But this is a contradiction since .

Hence,  no finite subgroup of can be a kernel.

Note: The homomorphisms that can easily be described are for the form .

## Zero determinant of a matrix

Suppose that for the complex square matrices it holds

(1)

Prove that .

Solution

If , then is invertible. So we get

But this equality is impossible: taking trace on both sides, we get

We get a contradiction, which shows that is singular, i.e. .

## On a power of matrix

Let be a natural number such that . Evaluate the power

Solution

This is a very standard exercise in diagonalisation of matrices and there would be no reason to post it here , if it did not include the Fibonacci result. We are proving that

where denotes the – th Fibonacci number. The proof now follows with an induction on .

## Multiple integral on a zero measured set

Let be a Jordan measurable set of zero measure. Prove that .

Solution

Since there exists a sequence of closed rectangles of such that and it is where is the volume of the rectangle . Then foreach