Solutions to Selected Homework Problems
1.26 Claim: ∃α : S → S which is 1-1 but not onto ⇔ ∃β : S → S which is onto
but not 1-1.
Proof. (⇒) Since α is 1-1, ∃β : S → S such that β ◦ α = idS . Since β ◦ α = idS is
onto, β is onto. β is not invertible since otherwise α = β −1 would be invertible. So
β is not 1-1.
(⇐) Since β is onto, ∃α : S → S such that α ◦ β = idS . Since β ◦ α = idS is 1-1,
α is 1-1. α is not invertible since otherwise β = α−1 would be invertible. So α is
not onto.
2.24
Proof. Since α : T → U is 1-1, ∃δ : U → T such that δ ◦ α = idT . So
β = (δ ◦ α) ◦ β = δ ◦ (α ◦ β) = δ ◦ (α ◦ γ) = (δ ◦ α) ◦ γ = γ.
3.31
Proof. We have
x ∗ (y ∗ z) = (x ∗ z) ∗ y
∀x, y, z ∈ S.
Letting x = e in the above, ⇒
y∗z =z∗y
∀y, z ∈ S.
Hence ∗ is commutative. Now
x ∗ (z ∗ y) = x ∗ (y ∗ z) = (x ∗ z) ∗ y
∀x, y, z ∈ S.
Hence ∗ is associative.
4.8
Proof. (a) ∀αa,0 , αb,0 ∈ B where a, b ∈ R, a 6= 0, b 6= 0, αa,0 ◦αb,0 = αab,0 ∈ B where
0 6= ab ∈ R. Hence ◦ is an operation on B. It is associative and commutative. α1,0
is the identity.
(b) ∀α1,b , α1,c ∈ C where b, c ∈ R, α1,b ◦ α1,c = α1,b+c ∈ C. Hence ◦ is an
operation on C. It is associative and commutative. α1,0 is the identity.
(c) In fact, αa,b = α1,b ◦ αa,0 .
5.12
Proof. Let G = {2m 3n : m, n ∈ Z}. ∀2m1 3n1 , 2m2 3n2 ∈ G,
2m1 3n1 · 2m2 3n2 = 2m1 +m2 3n1 +n2 ∈ G.
Hence G is closed under multiplication. 20 30 is the identity of G. ∀a = 2m 3n ∈
G, a−1 = 2−m 3−n ∈ G. Hence G is a group.
5.21
1
2
Solution. The Cayley table must be
∗ x y z
x x y z
y y z x
z z x y
6.4
Solution. (a)
1
1
2
6
3
4
4
5
5
3
6
= (2, 6)(3, 4, 5).
2
(b) (1, 4)(1, 3)(1, 2) = (1, 2, 3, 4).
(c) (1, 2, 3)−1 (2, 3)(1, 2, 3) = (3, 2, 1)(1, 2)(3, 2, 1)−1 = (2, 1).
(d) (2, 4, 5)(1, 3, 5, 4)(1, 2, 5) = (1, 4)(2, 5, 3).
6.9
Proof. Let α = (a1 , a2 , . . . , ak ), β = (a1 , ak ) · · · (a1 , a3 )(a1 , a2 ). ∀x ∈ {1, . . . , n}, we
will show that α(x) = β(x).
Case 1. x ∈
/ {a1 , . . . , ak }. Then α(x) = x and β(x) = x.
Case 2. x = ai for some 1 ≤ i ≤ k − 1. We have α(x) = ai+1 and
β(x) = (a1 , ak ) · · · (a1 , ai ) (ai )
= (a1 , ak ) · · · (a1 , ai+1 ) (a1 )
= (a1 , ak ) · · · (a1 , ai+2 ) (ai+1 )
= ai+1 .
Case 3. x = ak . We have α(x) = a1 and β(x) = a1 .
7.2
Proof. (a) It is a subgroup.
(b) It is a subgroup.
(c) It is not a subgroup. (1, 2, 3, 4, 5)2 is not in the set.
(d) It is a subgroup.
7.24
Proof. Let a, b ∈ Z(G). ∀x ∈ G, we have
(a ∗ b) ∗ z = a ∗ (b ∗ z) = a ∗ (z ∗ b) = (a ∗ z) ∗ b = (z ∗ a) ∗ b = z ∗ (a ∗ b).
Thus a ∗ b ∈ Z(G). Hence Z(G) is closed under ∗.
Clearly, e ∈ Z(G).
Let a ∈ Z(G). Then ∀x ∈ G, a ∗ x = x ∗ a. Thus
a−1 ∗ (a ∗ x) ∗ a−1 = a−1 ∗ (x ∗ a) ∗ a−1 ,
i.e., x ∗ a−1 = a−1 ∗ x. Hence a−1 ∈ Z(G).
8.5
3
Solution. Let pk = ek
is determine M(T ) .
2π
√
5
−1
, k = 0, 1, 2, 3, 4 and let T = {p0 , . . . , p4 }. The question
y
6 p1
•
p2
•
p0 x
•p3
•
•
p4
Let
α = the counterclockwise rotation of 72◦ ,
β = the reflection with respect to the x-axis.
We claim that
M(T ) = {αi β j : 0 ≤ i ≤ 4, j = 0, 1}.
Proof of the claim. It suffices to show that every γ ∈ M(T ) is of the form αi β j for
some 0 ≤ i ≤ 4 and 0 ≤ j ≤ 1. There is a j (0 ≤ j ≤ 4) such that αj (p0 ) = γ(p0 ).
Thus (α−j γ)(p0 ) = p0 . Since (α−j γ) is an isometry such that leaves T invariant, it
follows that
(α−j γ)({p1 , p4 }) = {p1 , p4 }.
Case 1. (α−j γ)(p1 ) = p1 and (α−j γ)(p4 ) = p4 . Using the fact (α−j γ) is an
isometry such that leaves T invariant, it follows that (α−j γ)(pk ) = pk for all k =
0, . . . 4. Thus (α−j γ) = id. Therefore,
γ = αj = αj β 0 .
Case 2. (α−j γ)(p1 ) = p4 and (α−j γ)(p4 ) = p1 . Then (βα−j γ)(pk ) = pk for
k = 0, 1, 4. By the arguments in Case 1, we have (βα−j γ)(pk ) = pk = id. Hence,
γ = αj β 1 .
9.8
Proof. (a) Reflexivity. ∀a ∈ N, a = a · 100 , hence a ∼ a.
Symmetry. Assume a, b ∈ N such that a ∼ b. Then a = b · 10k for some k ∈ Z.
We have b = a · 10−k , hence b ∼ a.
Transitivity. Assume a, b, c ∈ N such that a ∼ b and b ∼ c. Then a = b · 10k and
b = c · 10l for some k, l ∈ Z. It follows that a = c · 10k+l , hence a ∼ c.
(b) {a ∈ N : 10 - a} is a complete set of equivalence class representatives.
9.17
4
Proof. (a) Routine.
(b) [(x, y)] consists of all points in R2 obtained by shifting (x, y) and integral
units horizontally and an integral units vertically.
(c) [0, 1) × [0, 1) is a complete set of equivalence class representatives.
10.16
Proof. Since a ≡ b (mod n) and c ≡ d (mod n), we have a = b + un and c = d + vn
for some u, v ∈ Z. Thus,
ac = (b + un) + (d + vn) = bd + bvn + und + uvn2 = bd + (bv + ud + uvn)n
where bv + ud + uvn ∈ Z. Hence ac ≡ bd (mod n).
10.18
Solution. Not true. Example: 4 ≡ 1 (mod 3), but 42 6≡ 12 (mod 32 ).
11.3
Solution. (f) [17] [76] = [2] [1] = [2].
11.14
Disproof. [2] [2] ∈
/ Z]4 .
14.7
Solution. (a) h(1, 2, 3, 4)i = {id, (1, 2, 3, 4), (1, 3)(2, 4), (4, 3, 2, 1)}.
(b) h(1, 2, 3, 4, 5)i = {id, (1, 2, 3, 4, 5), (1, 3, 5, 2, 4), (4, 2, 5, 3, 1), (5, 4, 3, 2, 1)}.
(c) |h(1, 2, · · · , n)i| = n.
14.14
Proof. (a) (ab)mn = am bn = e.
(b) It follows from Theorem 14.3 (b).
(c) [2] ∈ Z4 with o([2]) = 2. But o([2][2]) = o([0]) = 1 6= 2 · 2.
14.20
∗
e
a
a2
a3
a4
14.31
e
e
a
a2
a3
a4
a
a
a2
a3
a4
e
a2
a2
a3
a4
e
a
a3
a3
a4
e
a
a2
a4
a4
e
a
a2
a3
Proof. Since (a−1 ba)n = a−1 bn a, we see that (a−1 ba)n = e ⇔ a−1 bn a = e ⇔ bn =
e. Thus o(a−1 ba) = o(b).
15.2
Proof. Since 24, −36, 54 ∈ h6i ⇒ h24, −36, 54i ⊂ h6i. Since 6 = 54 − 2 · 24 ∈
h24, −36, 54i ⇒ h6i ⊂ h24, −36, 54i.
15.19
Proof. Z2 × Z2 is not cyclic since it does not contain any element of order 4.
5
15.25
Proof. (⇒) Since [1] ∈ Zn = h[a]i ⇒ [1] = u[a] for some a ∈ Z. Thus ua ≡ 1
(mod n), i.e., ua + vn = 1 for some v ∈ Z. It follows that gcd(a, n) = 1.
(⇐) We have 1 = gcd(a, n) = ua + vn for some u, v ∈ Z. Thus [1] = u[a] ∈ h[a]i.
Hence Zp = h[1]i ⊂ h[a]i.
16.8
Solution. The distinct right coset of Z in R are
Z + x,
Z + x,
x ∈ [0, 1).
x ∈ [0, 1) is obtained by shifting Z x units to the right.
16.17
Solution.
h((1, 2), [1])i · (id, [0]) ={(id, [0]), (1, 2), [1])},
h((1, 2), [1])i · (id, [1]) ={(id, [1]), (1, 2), [0])},
h((1, 2), [1])i · ((1, 3), [0]) ={((1, 3), [0]), ((1, 3, 2), [1])},
h((1, 2), [1])i · ((1, 3), [1]) ={((1, 3), [1]), ((1, 3, 2), [0])},
h((1, 2), [1])i · ((2, 3), [0]) ={((2, 3), [0]), ((1, 2, 3), [1])},
h((1, 2), [1])i · ((2, 3), [1]) ={((2, 3), [1]), ((1, 2, 3), [0])},
16.20
Proof. Assume that HaK ∩ HbK 6= ∅. Then ∃h1 , h2 ∈ H and k1 , k2 ∈ K such
−1
that h1 ak1 = h2 bk2 . So a = h−1
1 h2 bk2 k1 . Now for any h ∈ H and k ∈ K, we
−1
−1
−1
have hak = (hh1 h2 )b(k2 k1 ) ∈ HbK since hh−1
∈ K. So
1 h2 ∈ H and k2 k1
HaK ⊂ HbK. By symmetry, HbK ⊂ HaK.
17.15
Solution. Trivial subgroups: {([0], [0])}, Zp ×Zp . Subgroups of order p: h([1], y)i, y ∈
Zp and h([0], [1])i.
18.14
Proof. Define f : Zmn −→ Zm × Zn by
f ([x]mn ) = ([x]m , [x]n ),
x ∈ Z.
Note that if [x]mn = [y]mn , then x ≡ y (mod mn). Hence x ≡ y (mod m) and
x ≡ y (mod n), i.e., [x]m = [y]m and [x]n = [y]n . Thus f is well defined.
For any [x]mn , [y]mn ∈ Zmn , we have
f ([x]mn [y]mn )
= f ([xy]mn ) = ([xy]m , [xy]n )
= ([x]m [y]m , [x]n [y]n )
= ([x]m , [x]n ) · ([y]m , [y]n )
= f ([x]mn )f ([y]mn ).
Hence f is a homomorphism.
6
Suppose f ([x]mn ) = f ([y]mn ). Then [x]m = [y]m and [x]n = [y]n , i.e., x ≡ y
(mod m) and x ≡ y (mod n). Since gcd(m, n) = 1, we have x ≡ y (mod mn), i.e.,
[x]mn = [y]mn . Therefore, f is 1-1. Since Zmn and Zm × Zn are finite of the same
cardinality, f is a bijection.
19.9
Reason. S3 × Z2 has 2 elements of order 3; A4 has 8 elements of order 3.
19.18
Proof. Let G = hai be an infinite cyclic group. Then o(a) = ∞. Define
f : Z −→ G
n 7−→ an
Since f (m + n) = am+n = am an = f (m)f (n), f is a homomorphism. Clearly f
is onto. Assume f (m) = f (n) for some m, n ∈ Z. Then am = an , i.e., am−n = e.
Since o(a) = ∞, it follows that m − n = 0, i.e., m = n. Thus f is 1-1.
19.26
Proof. Clearly, θ is a bijection. ∀a, b ∈ G, we have
θ(ab) = (ab)−1 = b−1 a−1 = a−1 b−1 = θ(a)θ(b).
So θ is a homomorphism.
20.3
Solution.
an element α ∈ h(1, 2, 3)i
id
(1, 2, 3)
the permutation θ(α)
id
id
(1, 2, 3) (1, 3, 2)
(1, 2, 3) (1, 3, 2)
id
id
(1, 2, 3) (1, 3, 2)
(1, 3, 2)
id
(1, 2, 3)
(1, 3, 2)
21.5
Proof. (a) Assume that [a]6 = [b]6 . Then a ≡ b (mod 6). Hence a ≡ b (mod 3),
i.e., [a]3 = [b]3 .
(b) ∀[a]6 , [b]6 ∈ Z6 , we have
θ([a]6 + [b]6 ) = θ([a + b]6 ) = [a + b]3 = [a]3 + [b]3 = θ([a]6 ) + θ([b]6 ).
(c) ker θ = h[3]6 i.
21.11
Proof. ∀x, y ∈ G, we have
(β ◦α)(xy) = β(α(xy)) = β(α(x)α(y)) = β(α(x))β(α(y)) = (β ◦α)(x) (β ◦α)(y) .
Hence β ◦ α is a homomorphism.
21.30
7
Proof. (a) E is the identity. I 2 = J 2 = K 2 = −E. Additional rules of multiplication in Q8 are illustrated by the diagram below. The product of any two elements
in the counterclockwise direction equals the third element; the product of any two
elements in the clockwise direction equals the third element with a negative sign.
E.g, IJ = K, JK = I, IK = −J, etc.
K
I
@
I
@
@
- J
Thus Q8 is closed under multiplication and (±E)−1 = ±E, (±I)−1 = ∓I, (±J)−1 =
∓J, (±K)−1 = ∓K. Hence Q8 is a subgroup of M (2, C).
(b) Subgroup of order 1: {E}.
Subgroup of order 2: {±E}.
Subgroups of order 4:
hIi = {±E ± I},
hJi = {±E ± J},
hKi = {±E ± K}.
Subgroup of order 8: Q8 .
22.6
Proof. Let G = hai and N / G. Every element in G/N is of the form N x for some
x ∈ G. Since x = an for some n ∈ Z, we have N x = N (an ) = (N a)n . Hence
G/N = hN ai.
22.9
Proof. (a) Since |G/N | = p is a prime, G/N ≈ Zp is cyclic.
(b) Disproof. Z/6Z ≈ Z6 is cyclic. Yet [Z : 6Z] = 6 is not a prime.
22.12
Proof. Every element of Q/Z is of the form Z + r for some r ∈ Q. Write r =
where m, n ∈ Z and n 6= 0. Then
m
n,
n(Z + r) = Z + nr = Z + 0.
Hence o(Z + r) ≤ n.
∀n ∈ Z with n > 0, o(Z + n1 ) = n.
23.4
Solution. The homomorphic images of S3 are S3 , Z2 , and {e}.
S3 ≈ S3 /{id},
Z2 ≈ S3 /A3 ,
{e} ≈ S3 /S3 .
23.16
8
Proof. 1◦ A < G. ∀x, y ∈ A, θ(a) ∈ B and θ(a) ∈ B. Hence θ(xy) = θ(x)θ(y) ∈ B,
i.e., xy ∈ A. Thus A is closed under multiplication. We also have θ(x−1 ) =
(θ(x))−1 ∈ B. Hence x−1 ∈ A. Thus A is also closed under inversion. It follows
that A < G.
2◦ A / G. ∀a ∈ A and xg ∈ G, we have
θ(g −1 ag) = (θ(g))−1 θ(a)θ(g) ∈ B
since θ(a) ∈ B and B / H. Therefore, g −1 ag ∈ A, which proves that A / G.
23.20
Proof. (a) ∀h1 k1 , h2 k2 ∈ HK, where h1 , h2 ∈ H and k1 , k2 ∈ K, we have
(h1 k1 )(h2 k2 ) = (h1 h2 )(h−1
2 k1 h2 )k2 ∈ HK
since h1 h2 ∈ H, (h−1
2 k1 h2 ) ∈ K and k2 ∈ K. Thus HK is closed under multiplication. We also have
−1
−1 −1
(h1 k1 )−1 = k1−1 h−1
1 = h1 (h1 k1 h1 ) ∈ HK.
Hence HK is also closed under inversion. It follows that HK < G.
Clearly, K ⊂ HK, hence K < HK. ∀g ∈ HK, since K /G, we have g −1 Kg = K.
Hence K / HK.
(b) ∀h1 , h2 ∈ H, we have
θ(h1 h2 ) = Kh1 h2 = (Kh1 )(Kh2 ) = θ(h1 )θ(h2 ).
So θ is a homomorphism. ∀h ∈ H and k ∈ K, we have Khk = hkK = hK = Kh =
θ(h). So θ is onto.
(c) ∀a ∈ H ∩ K, we have θ(a) = Ka = K. So a ∈ ker θ. Thus H ∩ K ⊂ ker θ.
∀a ∈ ker θ, we have K = θ(a) = Ka. So a ∈ K. Since the domain of θ is H, we
also have a ∈ H. So a ∈ H ∩ K. Thus ker θ ⊂ H ∩ K.
By the fundamental homomorphism theorem,
HK/H ∩ K = HK/ ker θ ∼
= HK/K.
53.6
Proof. (a)
Orb(1) = Orb(2) = Orb(3) = Orb(4) = {1, 2, 3, 4}.
Orb(5) = Orb(6) = {5, 6}.
(b)
G1 = G2 = G3 = G4 = {id}.
G5 = G6 = {id, (1, 3)(2, 4)}.
(c) |Orb(1)| = 4, |G|/|G1 | = 4/1 = 4. |Orb(5)| = 2, |G|/|G5 | = 4/2 = 2.
53.8
Proof. ∀a, b ∈ Gs , we have πa (s) = s and πb (s) = s. Hence
πab (s) = (πa πb )(s) = πa (πb (s)) = πa (s) = s.
Thus ab ∈ Gs . So Gs is closed under multiplication. We also have πa−1 (s) =
πa−1 (s) = s since πa (s) = s. Thus a−1 ∈ Gs . Therefore Gs is closed under inversion.
It follows that Gs < G.
9
53.17
Proof. (a) ∀a, b ∈ G, we want to show that πab = πa πb . ∀gH ∈ S, where g ∈ G, we
have
(πa πb )(gH) = πa πb (gH) = πa (bg)H = (abg)H = πab (gH).
Hence πab = πa πb .
(b) Let a ∈ ker π. Then πa = id. Thus for any g ∈ G, πa (gH) = gH, i.e.,
−1
(ag)H = gH. It followsTthat g −1 (ag) ∈ H, i.e., a ∈ gHg
this is true for
T . Since
−1
−1
all g ∈ G, we
have
a
∈
{gHg
:
g
∈
G}.
So,
ker
π
⊂
{gHg
:
g
∈ G}.
T
Let a ∈ {gHg −1 : g ∈ G}. Then ∀g ∈ G, a ∈ gHg −1 . Thus g −1 (ag) ∈ H,
hence
(ag)H = gH, i.e., πa (gH) = gH. Therefore πa = id. Hence a ∈ ker π. So,
T
{gHg −1 : g ∈ G} ⊂ ker π.