Group Homomorphism

In abstract algebra, a group homomorphism from G to H is a function f between two groups (G,·) and (H,*), written as $f:G \rightarrow H$ , that satisfies the following property for all elements a, b in G: $\forall \ a,b \in G \Rightarrow f(a \ · \ b)=f(a)*f(b)$

In a group homomorphism, each element g∈G maps to a unique element h=f(g)∈H.

$\forall \ g \in G \Rightarrow h=f(g) \in H$

This relationship is called a homomorphism from G to H.

Note: A group homomorphism is a specific type of homomorphism in abstract algebra that creates a correspondence between two sets with the same algebraic structure—here, two groups.

If, in addition, every element h∈H is the image of some element under f:G→H,

$\forall \ h \in H \ \exists \ g \in G \ \ | \ \ h=f(g)$

then the homomorphism is said to be from G onto H, and H is called the homomorphic image of G.

A Practical Example
Theorems on Group Homomorphisms

A Practical Example

Consider the additive group (Z,+) and the multiplicative group (Z,·), along with the imaginary unit (i) of complex numbers.

The function f:n→iⁿ is a homomorphism from (Z,+) onto (Z,·)

$f:n \rightarrow i^n$

because the following condition holds for any a, b∈Z:

$\forall \ a,b \in Z \Rightarrow f(a \ + \ b)=f(a)·f(b)$

Proof

Given the relation f=iⁿ, we have f(a+b)=i^a+b and f(a)·f(b)=i^a·i^b

$\forall \ a,b \in Z \Rightarrow f(a \ + \ b)=f(a)·f(b)$

$\forall \ a,b \in Z \Rightarrow i^{a \ + \ b} =i^a·i^b$

The identity i^a+b=i^a·i^b holds for any a, b due to the properties of exponents with the same base, where i^x·i^y equals i^x+y.

Verification

For instance, consider the values a=3 and b=2 in Z:

$f(3 \ + \ 2)=f(3)·f(2)$

$f(5)=f(3)·f(2)$

Since f(n)=iⁿ, we have f(5)=i⁵, f(3)=i³, and f(2)=i².

$i^5=i^3·i^2$

Now, let's verify if this equation holds true.

We know that the square of the imaginary unit is i²=-1, so i⁵=i²·i²·i=(-1)·(-1)·i=i.

Substituting i⁵=i into the equation, we get:

$i^5=i^3·i^2$

$i=i^3·i^2$

Since i²=-1, then i³=i²·i=(-1)·i=-i.

Substituting i³=-i and i²=-1 into the equation, we have:

$i=i^3·i^2$

$i=(-i)·(-1)$

$i=i$

The identity is thus satisfied for the values a=3 and b=2 in Z.

Theorems on Group Homomorphisms

Here are some theorems and corollaries related to group homomorphisms:

In a homomorphism between two groups G and H, both groups share the same identity element.
If in a homomorphism between two groups G and H, two elements g∈G and h∈H are equal (g=h), then their inverses are also equal.
The homomorphic image of a cyclic group is also cyclic.

And so on.