Homomorphism

Question 1

What is a homomorphism?

Answer 1

Let (G, *) and (H, °) be groups. If a mapping f: G → H exists such that,
f(x * y) = f(x) ° f(y), where x, y ∈ G, then f is said to be a group homomorphism.

Question 2

What is the kernel of a homomorphism?

Answer 2

The kernel of a homomorphism, Ker(f), is defined as:
Ker(f) = { x ∈ G: f(x) = e_H}
Where G and H are groups, and e_H is the identity element of the group H.

Question 3

What is an isomorphism?

Answer 3

Simply, it is a homomorphism that is bijective (one-to-one and onto mapping).

Question 4

What is a subgroup?

Answer 4

Let H be a non-empty subset of a group (G, *). If H is also a group under the same operation *, then H is said to be a subset of G. That is denoted by, H ≤ G

Question 5

What is the centre of a group?

Answer 5

The centre of a group (G, *), denoted by Z(G), is a subset of elements of G that are commutative with all other elements of G under the operation *.
That is,
Z(G) = { a ∈ G: x * a = a * x for all x ∈ G}

Question 6

What is the two-step proof for subgroups (definition)?

Answer 6

Suppose V is a nonempty subset of a group (G, *), then V is a subgroup if its elements are closed under the group operation, and there exists an inverse for every element in V.

Question 7

What is the two-step proof for subgroups (steps)?

Answer 7

1. Identify the defining properties (P) of the subset.
2. Show the subset is a nonempty set by proving the identity element of G has the property (checking to see if the identity element of G is defined by P/exists under P and thus exists in the subset.)
3. Check for closure under the binary operation.
4. Check for closure under inverses.

Question 8

Prove that the kernel of a homomorphism is a subgroup.

Answer 8

1. Recall Ker(f) = {x ∈ G: f(x) =e_H}
   f(e_G) = e_H by the property of homomorphism. Thus e_G ∈ Ker(f) => Ker(f) is nonempty.
2. Assume a and b ∈ Ker(f)
   => f(a) = e_H, f(b) = e_H
   We must show f(a * b) = e_H.
   Recall f(a * b) = f(a) * f(b)
   => f(a * b) = e_H * e_H = e_H. Thus it is closed under the group operation.
3. We must prove that for all a ∈ Ker(f), a^-1 ∈ Ker(f). Note that since a∈G, a^-1∈G, thus it exists so we can use it in the proof.
   f(a^-1) = (f(a))^-1 by properties of homomorphism.
   Since f(a) = e_H,
   f(a^-1) = (e_H)^-1
   => f(a^-1) = e_H. Thus a^-1∈Ker(f) for all a in ker(f).