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Modern Algebra Concepts, Essays (university) of Abstract Algebra

Asia Pacific Institute of Information Technology (APIIT)Abstract Algebra

It is about Homomorphism in Modern Algebra.

Typology: Essays (university)

2019/2020

Uploaded on 12/02/2020

sohil-khokhar
sohil-khokhar 🇮🇳

2 documents

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bg1
Lecture 13
1 Homomorphism
•Def:
Let hG, +iand hG0,×i be two groups. A map f:G→G0is said to be a homomorphism if
f(a+b) = f(a)×f(b),∀a, b ∈G.
Note: Here the mapping fneed not be one-one but the group multiplication is preserved by
the images in G0under the mapping f.
•Eg:1
∀x∈G, f(x) = e0. This is a homomorphism from Ginto G0.
•Eg:2
G=hZ,+i, G0=hZn,⊕i
f:G→G0is defined as f(nk +r) = r0≤r < n
Let x1=nk1+r1, x2=nk2+r2
f(x1) = r1, f(x2) = r2
f(x1+x2) = f(n(k1+k2)+(r1+r2)) = r1⊕r2=f(x1)⊕f(x2)
•Eg: 3
G=S3={e, σ, σ2, τ, σ τ, σ2τ}, G0={e, τ }
f(σiτj) = τj, i = 0,1,2. j = 0,1
f(e) = f(σ) = f(σ2) = e, f(τ) = f(στ ) = f(σ2τ) = τ
f(τ·στ ) = f(σ2) = e
f(τ)·f(στ ) = τ·τ=e
f(σ·στ ) = f(σ2τ) = τ
pf3
pf4
pf5

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Lecture 13

1 Homomorphism

  • DefLet :〈G, +〉 and 〈G′, ×〉 be two groups. A map f : G → G′ (^) is said to be a homomorphism if f (a + b) = f (a) × f (b), ∀ a, b ∈ G. Note:the images in Here the mapping G′ (^) under the mapping f need not be one-one but the group multiplication is preserved by f.
  • Eg: ∀x ∈ (^1) G, f (x) = e′. This is a homomorphism from G into G′.
  • Eg: G = (^2) 〈Z, +〉, G′ (^) = 〈Zn, ⊕〉 f : G → G′^ is defined as f (nk + r) = r 0 ≤ r < n Let x 1 = nk 1 + r 1 , x 2 = nk 2 + r 2 f (x 1 ) = r 1 , f (x 2 ) = r 2 f (x 1 + x 2 ) = f (n(k 1 + k 2 ) + (r 1 + r 2 )) = r 1 ⊕ r 2 = f (x 1 ) ⊕ f (x 2 )
  • Eg: G = (^3) S 3 = {e, σ, σ (^2) , τ, στ, σ (^2) τ }, G′ (^) = {e, τ } f (σiτ j^ ) = τ j^ , i = 0, 1 , 2. j = 0, 1 f (e) = f (σ) = f (σ^2 ) = e, f (τ ) = f (στ ) = f (σ^2 τ ) = τ

f (τ · στ ) = f (σ^2 ) = e f (τ ) · f (στ ) = τ · τ = e f (σ · στ ) = f (σ^2 τ ) = τ

f (σ) · f (στ ) = e · τ = τ yields a homomorphism.^ Verify that this mapping

  • Eg 4 G is the same as above,: G′ (^) = {e, σ, σ (^2) }.

f (σiτ j^ ) = σi. Then f (τ · στ ) = f (σ^2 τ 2 ) = f (σ^2 ) = σ^2 f (τ )f (στ ) = eσ = σ ∴ f (τ · στ ) 6 = f (τ )f (στ ). This is not a homomorphism.

  • In a homomorphism φ : G → G′ φ(e) = e′^ and φ(x−^1 ) = (φ(x))−^1
  • Def: Kernel Let is called the kernel f : G → G′^ be a homomorphism. The set of elements in Kf of the homomorphism f , i.e, G that are mapped to e′^ ∈ G′

Kf = {a ∈ G|f (a) = e′}

  • Eg: Consider f : G → G′ (^) where G = 〈Z, +〉 and G′ (^) = 〈Zn, ⊕〉:

f we have seen that this is a homomorphism. (nk + r) = r, 0 ≤ r < n The kernel of this homomorphism is Kf = nZ = {.... − 2 n, −n, 0 , n, 2 n, ......} We can see that Kf is a subgroup of G.

y = hx for some h ∈ Kf. =⇒ f (y) = f (hx) = f (h)f (x) = e′f (x) = f (x). So every element of a coset of Kf is mapped to the same element in G.

  • The homomorphism f : G → G′^ partitions the set G into the cosets of Kf in G.
  • Since Kf is a normal subgroup we can form the factor group G/Kf.
  • We construct a mapping from ψ : G/Kf → G′^ as follows

ψ(Kf a) = f (a)

  • Is this mapping well defined?
  • If Kf a = Kf b then a and b belong to the same coset. ⇒ f (a) = f (b). So this mapping is well defined.
  • We show that ψ : G/Kf → G′^ is an isomorphism.
  • ψ(Kf aKf b) = ψ(Kf (ab)) = f (ab) = f (a)f (b) = ψ(Kf a)ψ(Kf b)
  • Now if ψ(Kf a) = ψ(Kf b) then f (a) = f (b) ⇒ f (ab−^1 ) = f (a)(f (b))−^1 = e′ ⇒ ab−^1 ∈ Kf ⇒ a ∈ Kf b ⇒ a and b are in the same coset of Kf. i.e, Kf a = Kf b. So ψ : G/Kf → G′^ is an isomorphism.
  • The following proposition says this
  • Proposition 9: Let f : G → G′ (^) be a homomorphism with kernel Kf. Then G/Kf is isomorphic to G′.
  • This theorem tells us precisely how many different homomorphism we can draw on G. This depends upon the number of different normal subgroups we can have in G. For each of these normal subgroups, N , we have a homomorphism from G onto G/N. Any homomorphism that we draw fromof the homomorphism from G → G/N for some normal subgroupG to some other group G′^ has to be identical to oneN of G.