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##### EXERCISE 5(A)

2. If f{ab−1) = f(a)f(b)−1 for all a, bG, 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 bG. The converse is clear.

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

⇔(ab)−1 = b−la−1 = a−1b−1 for all a, bG

⇔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) ...

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