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 .

Irreducible function on Q[z]

Let be irreducible with degree . If has a root on the unit circle then is even and

Solution

Let be a root of with . Since has real coefficients is also a root of . The product is a polynomial in of degree ( its leading coefficient is ) with root . By the irreducibility of we have

(1)

for some non zero rational number . Setting we have that . Since   , by our hypotheses , hence . Setting we get that and because we deduce that is even.

Note: The above tells us that can be expressed in terms of .

Inequality with roots

Let be positive real numbers. Prove that

Solution

We apply the AM – GM inequality, thus:

Hence it suffices to prove that which holds because it is equivalent to .

On permutation

Let be vectors of – dimensional Euclidean space such that . Prove that there exists a permutation of the integers such that

for each .

Solution

We define inductively. Set .Assume is defined for and also

(1)

Note that is true for . We choose in a way that is fulfilled for instead of . Set and . Assume that for all .  Then and in view of ones gets which is impossible. Hence , there is such that

(2)

Put . Then using and we have

which verifies for . Thus we define for every . Finally from we get