Algebra: Abstract and Modern

CHAPTER 5

EXERCISE 5(A)

2. If f{ab⁻¹) = f(a)f(b)⁻¹ for all a, b ∈ G, then f(e) = f(aa⁻¹) = f(a)f(a)⁻¹ = e', f(b⁻¹) = f(eb⁻¹) = f(e)f(b)⁻¹ = e'f(b)⁻¹ = f(b)⁻¹ and f(ab) = f(a(b⁻¹)⁻¹) = f(a)(f(b)⁻¹)⁻¹ = f(a)f(b) for all a and b ∈ G. The converse is clear.

4. G is abelian ⇔ ab = ba for all a and b ∈ G

 

                          ⇔(ab)⁻¹ = b^−la⁻¹ = a⁻¹b⁻¹ for all a, b ∈ G

 

 

                          ⇔f(ab)= f(a)f(b)

 

6. For any homomorphism f : ℤ → ℤ₂, f{n) = nf{1) for all n ∈ ℤ and hence

 

f(1) = 0 ⇒ f(n) = 0 for all n ∈ ℤ

and f(1) = 1 ⇒ f(n)= 0 for all even n and

f(n) = 1 for all odd n

(since 1 1 1 = 0 in ℤ₂)

8. Let f : ℤ → ℤ be a nontrivial endomorphism. Then, f{n) = nf{1) for all n ∈ ℤ and hence f ≠ 0 so that

 

f{n) = f(m) ⇒ nf(1) = mf(1) ...