Let \(G\) and \(H\) be groups. A mapping

\[\phi:G\rightarrow H\]

is called a group homomorphism if

\[\phi(xy)=\phi(x)\phi(y)\qquad\text{for all }x,y\in G.\]

Thus, a group homomorphism is a mapping that preserves the group operation.

Types of Group Homomorphisms

• If \(\phi\) is injective, then it is called a monomorphism.
• If \(\phi\) is surjective, then it is called an epimorphism.
• If \(\phi\) is bijective, then it is called an isomorphism.
• If \(\phi\) is a homomorphism from \(G\) to itself, then it is called an endomorphism of \(G\).
• If \(\phi\) is a bijective endomorphism of \(G\), then it is called an automorphism of \(G\).

Basic Properties of a Group Homomorphism

Let \(G\) and \(H\) be groups with identity elements \(e\) and \(e'\), respectively. If \(\phi:G\rightarrow H\) is a homomorphism, then:

• The identity element of \(G\) is mapped to the identity element of \(H\): \[\phi(e)=e'.\]
• The inverse of an element is mapped to the inverse of its image: \[\phi(x^{-1})=\bigl(\phi(x)\bigr)^{-1}\qquad\text{for every }x\in G.\]
• For every \(x\in G\) and \(n\in\mathbb{Z}\), \[\phi(x^n)=\bigl(\phi(x)\bigr)^n.\]

Kernel of a Homomorphism

Let \(G\) and \(H\) be groups, and let \(\phi:G\rightarrow H\) be a homomorphism. The kernel of \(\phi\) is the set of all elements of \(G\) that are mapped to the identity element of \(H\).

It is denoted and defined by

\[\ker\phi=\{x\in G:\phi(x)=e'\},\]

where \(e'\) is the identity element of \(H\).

Criterion for Injectivity

Proposition: A group homomorphism \(\phi:G\rightarrow H\) is injective if and only if

\[\ker\phi=\{e\},\]

where \(e\) is the identity element of \(G\).

Proof: Suppose that \(\phi\) is injective. If \(x\in\ker\phi\), then

\[\phi(x)=e'=\phi(e).\]

Since \(\phi\) is injective, \(x=e\). Therefore, \(\ker\phi=\{e\}\).

Conversely, suppose that \(\ker\phi=\{e\}\). Let \(x,y\in G\) and assume that \(\phi(x)=\phi(y)\). Then

\[\phi(x)\bigl(\phi(y)\bigr)^{-1}=e'.\]

Using the homomorphism property, we obtain

\[\phi(xy^{-1})=e'.\]

Therefore, \(xy^{-1}\in\ker\phi\). Since \(\ker\phi=\{e\}\), we have \(xy^{-1}=e\), which gives \(x=y\). Hence, \(\phi\) is injective.