Question

state and peove fundamental theorem of homomorphism

Answer 1

In abstract algebra, the fundamental theorem on homomorphisms, also known as the fundamental homomorphism theorem, relates the structure of two objects between which a homomorphism is given, and of the kernel and image of the homomorphism. The homomorphism theorem is used to prove the isomorphism theorems
*PROVING*
We start by recalling the statement of FTH introduced last time.
Theorem (FTH). Let G, H be groups and ϕ : G → H a homomorphism.
Then
G/Ker ϕ ∼= ϕ(G). (∗ ∗ ∗)
Proof. Let K = Ker ϕ and define the map Φ : G/K → ϕ(G) by
Φ(gK) = ϕ(g) for g ∈ G.
We claim that Φ is a well defined mapping and that Φ is an isomorphism.
Thus we need to check the following four conditions:
(i) Φ is well defined
(ii) Φ is injective
(iii) Φ is surjective
(iv) Φ is a homomorphism
For (i) we need to prove the implication “g1K = g2K ⇒ Φ(g1K) = Φ(g2K).”
So, assume that g1K = g2K for some g1, g2 ∈ G. Then g
−1
1
g2 ∈ K by
Theorem 19.2, so ϕ(g
−1
1
g2) = eH (recall that K = Ker ϕ). Since ϕ(g
−1
1
g2) =
ϕ(g1)
−1ϕ(g2), we get ϕ(g1)
−1ϕ(g2) = eH. Thus, ϕ(g1) = ϕ(g2), and so
Φ(g1K) = Φ(g2K), as desired.
For (ii) we need to prove that “Φ(g1K) = Φ(g2K) ⇒ g1K = g2K.” This
is done by taking the argument in the proof of (i) and reversing all the
implication arrows.
(iii) First note that by construction Codomain(Φ) = ϕ(G). Thus, for
surjectivity of Φ we need to show that Range(Φ) = Φ(G/K) is equal to
ϕ(G). This is clear since
Φ(G/K) = {Φ(gK) : g ∈ G} = {ϕ(g) : g ∈ G} = ϕ(G).
(iv) Finally, for any g1, g2 ∈ G we have
Φ(g1K · g2K) = Φ(g1g2K) = ϕ(g1g2) = ϕ(g1)ϕ(g2) = Φ(g1K)Φ(g2K)
where the first equality holds by the definition of product in quotient groups.
Thus, Φ is a homomorphism.
So, we constructed an isomorphism Φ : G/Ker ϕ → ϕ(G), and thus G/Ker ϕ
is isomorphic to ϕ(G).

Answer 2

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