Kernal of a Homomorphism

The kernal of a homomorphism plays a crucial role in understanding the properties of the homomorphism and the relationship between the two groups involved. Let's break down the options and explain why option 'C' is the correct answer.

a) K is a Complex of G

A complex is a set of elements in a group that satisfy a certain property. In this case, the kernal K is defined as the set of elements in G that map to the identity element in G under the homomorphism F. However, the kernal K does not necessarily form a complex of G. It is possible for K to be a complex, but it is not a guaranteed property.

b) K is a Subgroup of G but not a Normal Subgroup of G

A subgroup is a subset of a group that itself forms a group under the group operation. In this case, the kernal K is indeed a subgroup of G. To show this, we need to verify that K satisfies the three conditions to be a subgroup:

1. Closure: For any two elements a, b in K, their product ab must also be in K. This can be shown by considering the homomorphism F(a) and F(b), both of which are in G. Since G is a group, the product F(a)F(b) = F(ab) is also in G. Therefore, ab is in K.

2. Identity: The identity element of G, denoted as e, is always in K since F(e) = e', where e' is the identity element of G. Therefore, K contains the identity element.

3. Inverse: For any element a in K, its inverse a^(-1) must also be in K. This can be shown by considering F(a^(-1)), which is the inverse of F(a) in G. Since G is a group, the inverse element F(a^(-1)) is also in G. Therefore, a^(-1) is in K.

However, K may or may not be a normal subgroup of G. A normal subgroup is one that is invariant under conjugation by elements of G. In other words, for any element g in G and any element k in K, the conjugate gkg^(-1) is also in K. Whether K is normal or not depends on the specific homomorphism and the groups involved.

c) K is a Normal Subgroup of G

The correct answer is option 'C'. The kernal K is always a normal subgroup of G when considering a homomorphism F. To prove this, we need to show that for any element g in G and any element k in K, the conjugate gkg^(-1) is also in K.

Let's consider an arbitrary element g in G and an arbitrary element k in K. We know that F(g) and F(k) are in G since F is a homomorphism. Now, let's consider the homomorphism of the conjugate element:

F(gkg^(-1)) = F(g)F(k)F(g^(-1))

Since F(g) and F(k) are in G, their product F(g)F(k) is also in G. Similarly, the inverse F(g^(-1)) is also in G. Therefore, the product F(g)F(k)F(g^(-1)) is in