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13.02 (Determine whether or not φ is a homomorphism.) Let φ : R → Z under addition be given by φ(x) = the greatest integer ≤ x. Solution: It is not a homomorphism, since
φ(1.5) + φ(1.5) = 1 + 1 = 2 6 = 3 = φ(1.5 + 1.5).
13.04 (Determine whether or not φ is a homomorphism.) Let φ : Z 6 → Z 2 be given by φ(x) = the remainder of x when divided by 2, as in the division algorithm. Solution: It is a homomorphism: φ(a + b) = φ(a) + φ(b) for all a, b ∈ Z 6.
13.06 (Determine whether or not φ is a homomorphism.) Let φ : R → R∗, where R is additive and R∗^ is multiplcative, be given by φ(x) = 2x. Solution: It is a homomorphism: φ(a + b) = 2a+b^ = 2a 2 b^ = φ(a)φ(b) for all a, b ∈ R.
13.08 (Determine whether or not φ is a homomorphism.) Let G be any group and let φ : G → G be given by φ(g) = g−^1 for g ∈ G. Solution: If G is not abelian then φ is not a homomorphism. Because when G is not abelian, there exist a, b ∈ G such that ab 6 = ba. Then φ(ab) = (ab)−^1 6 = (ba)−^1 = a−^1 b−^1 = φ(a)φ(b).
13.10 (Determine whether or not φ is a homomorphism.) Let F be the additive group of all continuous functions mapping R into R. Let R be the additive group of real numbers, and let φ : F → R be given by
φ(f ) =
0
f (x) dx.
Solution: It is a group homomorphism, since for any f, g ∈ F ,
φ(f + g) =
0
(f + g)(x) dx =
0
f (x) dx +
0
g(x) dx = φ(f ) + φ(g).
13.18 Compute ker(φ) and φ(18) for φ : Z → Z 10 such that φ(1) = 6.
Solution: For any integer k,
φ(k) = φ(1 + 1 +︸ ︷︷ · · · + 1︸ k copies
) = φ(1) + φ(1) + · · · + φ(1) ︸ ︷︷ ︸ k copies
= 6k mod 10.
So
ker φ = {k ∈ Z | 6 k = 0 mod 10} = 5Z, φ(18) = 6 ∗ 18 = 108 = 8 ∈ Z 10.
13.21 Compute ker(φ) and φ(14) for the homomorphism φ : Z 24 → S 8 where φ(1) = (2, 5)(1, 4 , 6 , 7). Solution: By homomorphism property,
φ(k) = φ(1 + 1 +︸ ︷︷ · · · + 1︸ k copies
) = φ(1)φ(1) · · · φ(1) ︸ ︷︷ ︸ k copies
= φ(1)k^ = (2, 5)k(1, 4 , 6 , 7)k.
Therefore, φ(14) = (2, 5)^14 (1, 4 , 6 , 7)^14 = (1, 4 , 6 , 7)^2 = (1, 6)(4, 7), and
ker(φ) = 4Z 24.
13.29 Let G be a group, and let g ∈ G. Let φg : G → G be defined by φg(x) = gxg−^1 for x ∈ G. For which g ∈ G is φg a homomorphism? Solution: For every x, y ∈ G we have
φg(xy) = gxyg−^1 = (gxg−^1 )(gyg−^1 ) = φg(x)φg(y).
So for every g ∈ G the map φg is a homomorphism.
13.32 a. T.
b. T. Let φ : G → G′^ such that φ(g) := e′^ for g ∈ G. c. F. d. T. e. F. When |G| is finite, |φ[G]| is a divisor of |G|. f. F. g. T. For example the trivial homomorphism. h. T. For example the trivial homomorphism. i. F. Kernel contains at least the identity. j. F. Let φ : Z 2 → Z 2 × Z by φ(a) := (a, 0).
13.51 Let G be any group and let a be any element of G. Let φ : Z → G be defined by φ(n) = an. Show that φ is a homomorphism. Describe the image and the possibilities for the kernel of φ. Solution: First we show that φ is a homomorphism. Given m, n ∈ Z,
φ(m + n) = am+n^ = aman^ = φ(m)φ(n).
So φ is a homomorphism. Second we describe the image of φ:
φ[Z] = {φ(n) | n ∈ Z} = {an^ | n ∈ Z} = 〈a〉 (≤ G).
Finally we describe the possibilities for the kernel of φ. Let e be the identity of G. If a has a finite order m (so that |〈a〉| = m), then am^ = e and so ker φ = φ−^1 ({e}) = {n ∈ Z | an^ = e} = mZ. If a has infinte order, then there exists no positive integer m such that am^ = e. So
ker φ = φ−^1 ({e}) = {n ∈ Z | an^ = e} = { 0 }.
13.52 Let φ : G → G′^ be a homomorphism with kernel H and let a ∈ G. Prove the set equality {x ∈ G | φ(x) = φ(a)} = Ha. Solution: Let e′^ be the identity of G′. Then
{x ∈ G | φ(x) = φ(a)} = {x ∈ G | φ(x)φ(a)−^1 = e′} = {x ∈ G | φ(xa−^1 ) = e′} = {x ∈ G | xa−^1 ∈ H} = {x ∈ G | x ∈ Ha} = Ha.
