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12. a. Yes. Bijection since c. No. σ(1) = 12 = 1 and α(4) = 42 = 1, so α is not one-to-one. e. Yes. α(g + h) = 2(g + h) = 2g + 2h = α(g)α(h); α is a bijection: g. Yes. α(gh) = (gh)2 = α(g) · α(h); bijection because . i. Yes. α(2k + 2m) = 3(k + m) = α(2k) + α(2m); α is one-to-one because α(2k) = α(2m) ⇒ 3k = 3m ⇒ k = m. 50 13. If , then , . Thus G = {I, A, A2, A3} and and A4 = I. Similarly {1, i, − 1, − i} = {1, i, i2, i3} and i4 = 1. They are both cyclic of order 4. 15. 4. Hence σ is well defined (by ⇒), and as a bonus σ is one-to-one (by ⇐) .

It is a homomorphism because It is clearly not onto, but it is one-to-one because α(r) = α(s) implies so r = s. 3. If α is an automorphism, then a−1b−1 = α(a) · α(b) = α(ab) = (ab)−1 = b−1a−1 for all a, b. Thus G is abelian. Conversely, if G is abelian, 49 so α is a homomorphism; α is a bijection because α−1 = α. 5. σa = 1G if and only if aga−1 = g for all g R, if and only if ag = ga for all g R. 7. Let be given by . Then o(1) =∞ in , but in . 8. If is a homomorphism, let α(1) = m. Then Thus α is multiplication by m, and each such map is a homomorphism .

1 n−2) and (n−2, n−1). Continue. The result is (c). 28. a. σ=(1 2 3 4 . . 2k−1 2k) so σ2=(1 3 5 . . 2k−1)(2 4 6 . . 2k). c. The action of σ is depicted in the diagram, and carries k→k+1→k+2. .. If k+m>n, the correct location on the circle is given by the remainder r when k+m is divided by n, That is k+m≡4(modn. Now the action of σm is σmk=k+m, so σmk≡k+mmodn. 32 29. Each of σ and τ may be either even or odd, so four cases arise. They are the rows of the following table. The parity of στ in each case is clear, and so the result follows by verifying, sgn σ·sgnτ= sgn(τ) in every case.