text: # 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.
  
  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.
  
  Proof:
  
  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).
  
  ## 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}.
  
  Proof:
  
  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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explanation: The document is an educational text that explains the concept of the kernel of a group homomorphism in algebra. It includes a definition, a theorem proving that the kernel is a normal subgroup, and another theorem establishing the injectivity criteria for homomorphisms. The text does not contain a specific question or a question-and-answer pair.
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distilabel_metadata.content: # 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.
  
  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.
  
  Proof:
  
  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).
  
  ## 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}.
  
  Proof:
  
  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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distilabel_metadata.raw_output_q_a_detector_0: The document is an educational text that explains the concept of the kernel of a group homomorphism in algebra. It includes a definition, a theorem proving that the kernel is a normal subgroup, and another theorem establishing the injectivity criteria for homomorphisms. The text does not contain a specific question or a question-and-answer pair.
  
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