Kernel of a Homomorphism

The kernel of a group homomorphism is an interesting subgroup of the domain group. This subgroup (kernel) determines whether it is an injective homomorphism or not. In this article, we will learn about the kernel of group homomorphisms.

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What is a kernel (Algebra)?

Definition of a Kernel of a Homomorphism

Let Φ: (G, 0) → (G′, *) be a group homomorphism, where G and G′ are two groups. Then the kernel of Φ, denoted by ker(Φ), is defined by

ker(Φ) = {a ∈ G : Φ(a)=eG′}.

Thus, the kernel of a group homomorphism Φ is the set of all elements of G that are mapped to the identity element of G′ under the map Φ.

Kernel is a Normal Subgroup

Theorem: The kernel of a homomorphism is a normal subgroup.


Let Φ: (G, 0) → (G′, *) be a group homomorphism. We will show that ker(Φ) is a normal subgroup of G. The following steps have to follow.

Step 1: As Φ(eG)=eG′, we have eG ∈ ker(Φ). Thus, ker(Φ) is a non-empty subset of G.

Step 2: To show ker(Φ) is a subgroup of G.

Let a, b ∈ ker(Φ).

Thus Φ(a) = eG′, Φ(b) = eG′

Now since Φ is a homomorphism, we have

Φ(a $\circ$ b-1) = Φ(a) * Φ(b-1) = eG′*eG′ = eG′

⇒ a $\circ$ b-1 ∈ ker(Φ) by the definition of kernel.

This shows that ker(Φ) is a subgroup of G.

Step 3: To show ker(Φ) is normal in G.

Let g ∈ G and h∈ ker(Φ) ⇒ Φ(h) = eG′.

Now Φ(g $\circ$ h $\circ$ g-1) = Φ(g) * Φ(h) * Φ(g-1) as Φ is a group homomorphism.

= Φ(g) * eG′ * Φ(g-1) as Φ(h) = eG′

= Φ(g) * Φ(g-1)

= Φ(g $\circ$ g-1) by the fact that Φ is a homomorphism.

= Φ(eG)=eG′

⇒ Φ(g $\circ$ h $\circ$ g-1) = eG′

⇒ g $\circ$ h $\circ$ g-1 ∈ ker(Φ) ∀ g ∈ G and h∈ ker(Φ).

Also Read:

Order of a Group: The order of a group and of its elements are discussed here with formulas.

Abelian Group: Definition, Properties, Examples

Cyclic Group: The definition, properties, and related theorems on cyclic groups are discussed.

First Isomorphism Theorem: Proof and Application

Injectivity criteria for homomorphism

Theorem: Let Φ: (G, 0) → (G′, *) be a group homomorphism. Then Φ is one-to-one if and only if ker(Φ) = {eG}.


First, we assume that Φ is one-to-one. We know that Φ(eG)=eG′. So eG is a preimage of eG′ under the map Φ. Since Φ is one-to-one, it is the only preimage of eG′. Therefore, the set {a ∈ G : Φ(a)=eG′} = {eG}. In other words, ker(Φ) ={eG}, that is, ker(Φ) is trivial.

Next, we assume that ker(Φ) = {eG}. We need to show that Φ is one-to-one.

Let a, b ∈ ker(Φ) be such that Φ(a)=Φ(b).

⇒ Φ(ab-1)=eG′

⇒ ab-1 ∈ ker(Φ)= {eG}

⇒ ab-1 = eG

⇒ a=b.

Hence Φ is one-to-one. This completes the proof of the theorem.

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