state and prove basic homomorphism theorem
Answers
Answer:
If φ: G → H is a homomorphism, then Im(φ) ∼= G/ Ker(φ). The FHT says that every homomorphism can be decomposed into two steps: (i) quotient out by the kernel, and then (ii) relabel the nodes via φ. If φ: G → H is a homomorphism, then Im(φ) ∼= G/ Ker(φ).
Answer:
If φ: G → H is a homomorphism, then Im(φ) ∼= G/ Ker(φ). The FHT says that every homomorphism can be decomposed into two steps: (i) quotient out by the kernel, and then (ii) relabel the nodes via φ. If φ: G → H is a homomorphism, then Im(φ) ∼= G/ Ker(φ).
Step-by-step explanation:
Let f:G→H denote a surjective homomorphisms of groups. Then f¯:G/ker(f)→H is an isomorphism
Proof: We have to first show that the map is well-defined. In particular, suppose gK=g′K where K=ker(f). Then we have to show that f¯(g)=f¯(g′). In particular, we have to show that f(g)=f(g′). But, gK=g′K is equivalent to g′−1g∈K. It follows that e=f(g′−1g)=f(g′)−1f(g) and therefore f(g′)=f(g). Next, one can easily check that f¯ is a homomorphism. Finally, we have to check that it is injective and surjective. Let h∈H. Then, since f is surjective, there is a g∈G such that f(g)=h. Then, by definition, f¯(gK)=h, and therefore f¯ is surjective. For injectivity, it is enough to show that ker(f¯)=eK. Let gK∈ker(f¯). Then e=f¯(gK)=f(g). It follows that g∈K, but this is equivalent to gK=eK. ■