HW #1 p21 #3, 8, 10, 11, 14, 16, 20, 22, 24, 52
3. (in book)
8.
Proof. Let a, b ∈ Z such that a|c, b|c. Assume gcd(a, b) = 1.
Then ∃s, t ∈ Z such that as + bt = 1 and ∃j, k ∈ Z such that
aj = c = bk.
So, c(as + bt) = c ⇒ bkas + ajbt = c ⇒ ab(ks + jt) = c ⇒ ab|c.
Example 0.1. a = 2, b = 4. Then for c = 4, a|4, b|4, but ab = 8 - 4.
10.
Proof. Let a, b ∈ Z and d = gcd(a, b). Assume a = da0 and b = db0 .
Then ∃s, t ∈ Z such that as + bt = d = da0 s + db0 t = d
⇒ a0 s + b0 t = 1 ⇒ gcd(a0 , b0 ) = 1
11.
Proof. By definition of modulo arithmetic, ∃q1 , q2 ∈ Z such that
a = nq1 + a0 and b = nq2 + b0
Thus, a + b = nq1 + nq2 + a0 + b0 = n(q1 + q2 ) + a0 + b0
⇒ (a + b) mod n = (a0 + b0 ) mod n
ab = (nq1 + a0 )(nq2 + b0 ) = n2 q1 q2 + nq1 b0 + a0 nq2 + a0 b0
⇒ ab mod n = a0 b0 mod n
14.
Proof. By the Euclidean algorithm, gcd(a, b) = gcd(b, r) where a =
bq + r, 0 ≤ 4 < b.
7n + 4 = (5n + 3)(1) + (2n + 1) ⇒
gcd(7n + 4, 5n + 3) = gcd(5n + 3, 2n + 1)
5n + 3 = (2n + 1)(2) + (n + 1)
= gcd(2n + 1, n + 1)
2n + 1 = (n + 1)(1) + (n)
= gcd(n + 1, n)
n + 1 = n(1) + 1
= gcd(n, 1) = 1
Since gcd(7n + 4, 5n + 3) = 1, 7n + 4 and 5n + 3 are relatively prime.
1
2
16. 71000 mod 6 = (72 )500 mod 6 = (72 mod 6)500 = 1500 = 1
61001 mod 7 = (62 )500 · 6 mod 7 = 1500 · 6 mod 7 = 6
20.
Proof. By the division algorithm, p1 p2 . . . pn +1 = pj (p1 p2 . . . pj−1 pj+1 . . . pn )+
1 the remainder is 1 so we know pj - p1 p2 . . . pn + 1.
22.
Proof. Let n = 1. Then 21 n(n + 1) = 21 (1)(2) = 1. So the statement
holds.
Assume the statement is true for some n ≥ 1 (WTS: Statement is true
for n + 1)
(1 + 2 + . . . n) + (n + 1) = 12 n(n + 1) + (n + 1) = (n + 1)( 21 n + 1) =
1
(n + 1)(n + 2)
2
By the principal of mathematical induction (PMI), the statement is
true for all n ≥ 1.
24.
Proof. Let n = 1. Then 2n 32n − 1 = 17 and 17 | 17 so the statement is
true.
Assume the statement is true for some n ≥ 1 (WTS: Statement is true
for n + 1)
2n+1 32(n+1) − 1 = 2n · 2 · 32n · 9 − 1 − 17 + 17
= 18 · 2n · 32n − 18 + 17
= 18(2n 32n − 1) + 17
Since 17 | 17 and 17 | (2n 32n − 1),
17 | 18(2n 32n − 1) + 17 = 2n+1 32(n+1) − 1. Thus the statement holds for
all n ≥ 1.
52.
Proof. 1. a ∈ R, a − a = 0 ∈ Z. Then a a (reflexive)
2. Suppose a b. Then a − b ∈ Z ⇒ −(a − b) = b − a ∈ Z. Then b a
(symmetric)
3. a, b, c ∈ R such that a b, b c. Then a − c = a − b + b − c ∈ Z ⇒ a c
(transitive)
For a ∈ R, [a] = {b ∈ R|a b} = {b ∈ R|a − b = n ∈ Z} =
{b ∈ R|a − n = b, n ∈ Z} = {a + n|n ∈ Z}
3
Extra Problem: Let S be a set and G = {f : S → S|f is bijective}.
Prove that f ◦ g is bijective for all f, g ∈ G.
Proof. (onto) Let s ∈ S (WTS: ∃t ∈ S such that f ◦ g(t) = s)
Since f is onto, ∃a ∈ S such that f (a) = s
Since g is onto ∃t ∈ S such that g(t) = a
Therefore, f (g(t) = s.
(1-1) Let a, b ∈ S such that f ◦g(a) = f ◦g(b). Then f (g(a)) = f (g(b)).
Since f is 1-1, g(a) = g(b). Since g is 1-1, a = b.
Thus, f ◦ g is 1-1.
4
HW #2 p52 #5, 14, 17, 20, 24, 25, 26, 35
5. Done in class and in book.
14.
Proof. Let x, y ∈ G and define a = x, b = yx, c = xy. Then ab =
ca ⇒ xyx = xyx ⇒ yx = xy ⇒ G is abelian.
17.
Proof. (⇒) Assume G is abelian. Let a, b ∈ G. Then (ab)−1 =
b−1 a−1 = a−1 b−1 since G is abelian.
(⇐) Assume (ab)−1 = a−1 b−1 ∀a, b ∈ G. Let x, y ∈ G. Then xy =
(y −1 x−1 )−1 = yx. So, G is abelian.
−1
−1 −1
20. (a1 a2 a3 . . . an )−1 = a−1
n an−1 . .. a2 a1 since
−1
−1
(a1 a2 . . . an ) a−1
n an−1 . . . a2 a1 −1 = e.
·12 1 5 7 11
1 1 5 7 11
24. 5 5 1 11 7
7 7 11 1 5
11 11 7 5 1
28.
e
a
b
c
d
e
e
a
b
c
d
a
a
b
c
d
e
b
b
c
d
e
a
c
c
d
e
a
b
d
d
e
a
b
c
26.
Proof. Let a, b ∈ G. Then (ab)(ab) = aabb ⇒ ba = ab by left and right
cancellation.
35.
Proof. Let x, y ∈ G. Then x2 = e = y 2 . Then xy ∈ G and (xy)2 = e.
So (xy)2 = x2 y 2 ⇒ yx = xy from problem 26.
5
HW #3 p65 #1,2,4,6,18,19,22,23,25,28,46,56
1. in book
2. h 12 i = {. . . , −2, −3
, −1, −1
, 0, 12 , 1, 23 , 2, . . .} in Q
2
2
1
h 12 i = {. . . , 16, 8, 4, 2, 1, 21 , 41 , 18 , 16
, . . .} in Q∗
4.
Proof. Suppose o(a) = n. Then n is the smallest positive integer such
that an = e.
So, (a−1 )n = a−n = (an )−1 = e−1 = e. So, o(a−1 ) ≤ n.
WTS: n is the smallest positive integer such that (a−1 )n = e.
If ∃m ∈ Z+ s.t. m < n and (a−1 )m = e then
(am )−1 = e ⇒ am = e, which is a contradiction since m < n.
Therefore, o(a−1 ) = n.
6. Suppose a6 = e. Then o(a) = 1, 2, 3, or 6.
18.
Proof. Let H, K ≤ G (WTS: H ∩ K ≤ G).
By subgroup test, WTS ab−1 ∈ H ∩K ∀a, b ∈ H ∩K. Let a, b ∈ H ∩K.
Then a, b ∈ H ⇒ ab−1 ∈ H since H ≤ G
Furthermore, a, b ∈⇒ ab−1 ∈ K since K ≤ G.
Therefore, ab−1 ∈ H ∩ K.
19.
Proof. Let x ∈ Z(G). Then xa = ax ∀a ∈ G ⇒ x ∈ C(a) ∀a ∈ G.
So, x ∈ ∩a∈G C(a). Therefore, Z(G) ⊆ ∩a∈G C(a).
Now, let y ∈ ∩a∈G C(a). Then ya = ay ∀a ∈ G ⇒ y ∈ Z(G).
Thus, ∩a∈G C(a) ⊆ Z(G). By containment argument, ∩a∈G C(a) =
Z(G).
1
2
20. The lower left corner is:
3
4
2
1
4
3
4
3
1
2
3
4
2
1
23. (a) C(1) = {1, 2, 3, 4, 5, 6, 7, 8},
C(2) = {1, 2, 5, 6},
C(3) = {1, 4, 7, 3},
C(4) = {1, 4, 5, 8}
C(5) = {1, 2, 3, 4, 5, 6, 7, 8},
C(6) = {1, 2, 5, 6},
C(7) = {1, 3, 5, 7},
C(8) = {1, 4, 5, 8}
6
(b) Z(G) = {1, 5}
(c) o(1) = 1, o(2) = 2, o(3) = 4, o(4) = 2, o(5) = 2, o(6) = 2, o(7) =
4, o(8) = 2
o(a) | |G|
25. Proved in class.
28. Yes, elements of the center are defined to be those which commute
with all other elements. So, two elements of the center, x, y must commute with each other (i.e. xy = yx). So, Z(G) is abelian.
46. (a) |U (4)| = 2, |U (5)| = 4, |U (12)| = 4
(b) |U (5)| = 4, |U (7)| = 6, |U (35)| = 24
(c) |U (4)| = 2, |U (5)| = 4, |U (20)| = 8
(d) |U (3)| = 2, |U (5)| = 4, |U (15)| = 8
(e) |U (r)| = m, |U (s)| = n, |U (rs)| = mn
56. 1 + 2i ∈ H and −2 − i ∈ H, but (1 + 2i) + (−2 − i) = −1 + i ∈
/H
since (−1)(1) = −1 < 0. Therefore H G since H is not closed.
7
HW #4 p81 #2,3,4,8ab,10,19,21,22,24,32,38,52
2. hai = ha5 i
hbi = hb3 i = hb5 i = hb7 i
hci = hc3 i = hc7 i = hc9 i = hc11 i = hc13 i = hc17 i = hc19 =i
3. h10i = {10, 20, 0} = h20i
ha10 i = {a10 , a20 , 9, 12, e} = ha20 i
4. h3i = {3, 6, 9, 12, 15, 0} = h15i
ha3 i = {a3 , a6 , a9 , a12 , a15 , e} = ha15 i
8. (a) o(a3 ) = 5 o(a6 ) = 5 o(a9 ) = 5 o(a12 ) = 5
(b)o(a5 ) o(a10 ) = 3
10. h3i = {3, 6, 9, 12, 15, 18, 21, 0} = h9i = h15i = h21i since U (8) =
{1, 3, 5, 7}.
hai = ha3 i = ha5 i = ha7 i
19. In book
21. In book
22. Suppose G = e, a, b. Then ab ∈ {a, b, e}. If ab = b ⇒ a = e, which
is not possible. If ab = a ⇒ b = e, which is also not possible. Then,
ab = e So, a−1 = b and hai = a, a−1 , e = a, b, e = G. Therefore, G is
cyclic.
24. WTS: hai ≤ C(a)
First, we need to show hai ⊆ C(a)
If x ∈ hai then x = ai for some i ∈ Z. Now,
ax = aai = a1+i = ai+1 = ai a = xa. So, x ∈ C(a). Then we have
hai ⊆ C(a).
Next, we need to show that hai passes the subgroup test. Begin by
letting x, y ∈ hai.
Then x = ai and y = aj for some i, j ∈ Z.
xy −1 = ai (aj )−1 = ai a−j = ai−j ∈ hai With xy −1 ∈ hai, hai is a subgroup.
32. Z12 = h1i ⇒ a = 1, n = 12
8
n
k
ak
ha k i
12
1 1 = 0 h0i = {0}
2 16 = 6 h6i = {6, 0}
3 14 = 4 h4i = {4, 8, 0}
4 13 = 3 h3i = {3, 6, 9, 0}
6 12 = 2 h2i = {2, 4, 6, 8, 10, 0}
12
1
h1i = Z12
(subgroup lattice not included)
·20 4 8 12 16
4 16 12 8 4
38. 8 12 4 16 8
12 8 16 4 12
16 4 8 12 16
e = 16, h8i = h12i
52. |U (49)| = 42. Then |U (42)| = |{1, 5, 11, 13, 15, 17, 19, 23, 25, 29, 31, 37, 41}| =
13
9
HW #5 p113 #2,3,5,8,9,11,18,21,23,24,28,30,36,56,58
2. (a) (15)(234)
(b)(124)(35)
(c) (1423)
3. in book
5. in book
8. One permutation on length 3 and another of length 7 results in a
permutation of order 21.
9. in book
11. in book
18. (a) α = (12345)(678)
β = (23847)(56)
αβ = (1245736)
(b) α = (15)(14)(13)(12)(68)(67)
β = (27)(24)(28)(23)(56)
αβ = (16)(13)(17)(15)(14)(12)
21. in book
23. in book
24.
7
5
= 21
28. β = (14523)
β 99 = (β 95 )4 = β 4 = beta−1 = (13254)
30. (a1 an an−1 . . . a2 )
36. h(1234)i = {(1234), (13)(24), (1432), (1)}
56. βγ = (1432)
γβ(1243)
β(1423), γ = (423)
β(1) = 4
a = (13)(24), b = (234). Then ha, bi = {(13)(24), (234), (243), (132), (143),
(134), (123), (142), (124), (234), (1)} = {a, b, b2 , ab, ba, ab2 , b2 a, ab2 a, aba, (1)}
10
HW #6 p36 #11,13,15
p52 #10,34
p64 #1(D4), 41
11. RV = V R3 , but R 6= R3
13. Let V be vertical reflection and R be rotation by 180 degrees clockI
V
R VR
I
I
V
R VR
V
I VR R
wise. V
R
R VR I
V
VR VR R
V
I
15. in book
10. H = {I, R2 }, K = {I, R2 , V, V R, V R2 , V R3 }
34. In D4 , F R−2 F R5 = F R2 F R = F 2 R3 = R3
In D5 , R−3 F R4 F R−2 = R2 F R4 F R3 = F R7 R2 F = F R4 F = R
In D6 , F R5 F R−2 = F 2 R−1 F = R5 F
1. in book
41. in book
11
HW #7 p133 #2,4,6,7,10,14,18,24,26,31,38,40
2. Aut(Z) = {id, φ−1 } where φ−1 (1) = −1
4. U (8) = {1, 3, 5, 7} is not cyclic since o(3) = o(5) = o(7) = 2.
U (10) = {1, 3, 7, 9} is cyclic since o(3) = 4. Therefore, U (10) U (8).
6. Suppose G ∼
= H and H ∼
= K. So, there exists an isomorphism
φ : G → H and isomorphism ψ : H → K. Then ψφ is 1-1 and onto.
We have showed many times before, that the composition of two bijective functions is bijective.
Let a, b ∈ G. Then, ψφ(ab) = ψ(φ(a)φ(b)) = ψφ(a)ψφ(b). Thus, ψφ is
an isomorphism.
7. In book.
10. G, group. Then α(g) = g −1 is an automorphism ⇔ G is abelian.
(⇒) α(gh) = (gh)−1 = h−1 g −1 and α(g)α(h) = g −1 h−1 .
Since α is an isomorphism, h−1 g −1 = g −1 h−1 ⇒ gh = hg ⇒ G is
abelian.
(⇐) Suppose G is abelian. Then we need to show the mapping α is an
automorphism where α(g) = g −1 for all g ∈ G.
1. Suppose α(g) = α(h) ⇒ g −1 = h−1 ⇒ g = h ⇒ α is 1-1.
2. Let h ∈ G ⇒ h−1 ∈ G ⇒ α(h−1 ) = (h−1 )−1 = h ⇒ α is onto.
3. α(gh) = (gh)−1 = h−1 g −1 = g −1 h−1 = α(g)α(h)
1, 2, 3 ⇒ α is an automorphism.
14. Aut(Z6 ) = {id, φ5 } where φ5 (1) = 5.
18. Z, R
24. U (20) = {1, 3, 7, 9, 11, 13, 17, 19} where U (20) has 3 elements of
order 4. They are 3, 13, and 17.
U (24) = {1, 5, 7, 11, 13, 17, 19, 23} where U (24) has no elements of order
4. Therefore, U (20) U (24).
√
a 2b
26. G = {a + b 2 | a, b ∈ Q} and H = {
| a, b, ∈ Q}.
b a
√
a 2b
Define φ : G → H by φ(a + b 2) =
b a
12
√
√
a 2b
c 2d
Suppose φ(a + b 2) = φ(c + d 2) ⇒
=
⇒ a = c, b =
b a
d c
d.
√
√
So, a + b 2 = c + d 2 and φ is 1-1.
√
√
a 2b
a 2b
For
∈ H, a + b 2 ∈ G and φ(a + b 2) =
. Thus φ is
b a
b a
onto. Then φ is onto and 1-1.
√
√
√
a + c 2(b + d)
Now, φ((a+b 2)+(c+d 2)) = φ(a+c+(b+d) 2) =
b+d a+c
√
√
a 2b
c 2d
=
+
= φ(a + b 2) + φ(c + d 2)
b a
d c
So, φ is an isomorphism. Also,
17. In book
31. In book
38. Define φ : G → H by φ(2k) = 3k, k ∈ Z.
Then, φ(2k) = φ(2j) ⇒ 3k = 3j ⇒ k = j ⇒ φ is 1-1.
For 3k ∈ 3Z, 2k ∈ 2Z and φ(2k) = 3k ⇒ φ is onto.
φ(2k + 2j) = φ(2(k + j)) = 3(k + j) = 3k + 3j = φ(2k) + φ(2j)
. So, φ is an isomorphism.
If G = hmi, H = hni then φ : G → H by φ(km) = kn, for all k ∈ Z.
13
HW #8 p149 #3,6,7,13,14,15,17,20,27,40
3. In book
6. There are n cosets. They are:
H = {0, ±n, ±2n, ±3n . . .}
1 + H = {1, ±n + 1, ±2n + 1, ±3n + 1 . . .}
2 + H = {2, ±n + 2, ±2n + 2, ±3n + 2 . . .}
..
.
(n − 1) + H = {n − 1, ±n + (n1), ±2n + (n − 1), ±3n + (n − 1) . . .}
7. In book
13. In book
14. | K |= 42 must divide | H | and | H | must divide | G |= 420.
Since, K is proper in H and H is proper in G, | H |6= 42 or 420. So,
| H |= 84 or 210.
15. Let H be a proper subgroup of G. Then H 6= {e} or G. By Lagrange’s Theorem, | H | divides | G |= pq. So, | H |= p or q. Since p
and q are prime, H is cyclic.
17. In book
20. Since H ∩K ≤ H and H ∩K ≤ K, we will use Lagrange’s Theorem
to say | H ∩ K | divides 12 and | H ∩ K | divides 35. Since 12 and 35
are relatively prime, | H ∩ K |= 1. Thus, H ∩ K = {e}. In general, if
the gcd(| H |, | K |) = 1 then | H ∩ K |= 1.
27. In book
40. Let G be finite and abelian and gcd(n, | G |) = 1. Then there exists
s, t ∈ Z such that ns+ | G | t = 1 ⇒ ns = 1− | G | t.
Define φ : G → G by φ(a) = an .
Suppose φ(a) = φ(b). Then
an = b n ⇒
ans = bns ⇒
14
a1−|G|t = b1−|G|t ⇒
a(a|G|(−t) ) = b(b|G|(−t) ) ⇒
a=b
Thus, φ is 1-1.
Let x ∈ G. Then x = x1 = xns+|G|t = xns x|G|t = xns .
Then φ(xs ) = (xs )n = xsn = x by above. Thus, φ is onto.
φ(xy) = (xy)n = (xy)(xy) . . . (xy) = xn y n = φ(x)φ(y).
15
HW #9 p # 1,2,4,6,10,11,12,14,17,18,21,27,39,43,53
1. In book
2. | An |= 21 n! and | Sn |= n! so (An : Sn ) = 2. Thus, An / Sn .
1 2
1 2
4. Let
∈ GL(2, R) and
∈ H.
1 1
0 1
1 2 1 2 −1 2
1 2 1 0
3 −2
Then,
=
=
∈
/ H.
1 1 0 1
1 −1
1 1 1 −1
2 −1
Thus, H is not normal in G.
6. h3i = {0, ±3, ±6, ±9, . . .} and h12i = {0, ±12, ±24, ±36, ldots}. So,
h3i/h12i = {h12i, 3 + h12i, 6 + h12i, 9 + h12i} where
3 + h12i = {. . . , −33, −21, −9, 3, 15, 27, 39, . . .}
6 + h12i = {. . . , −30, −18, −6, 6, 18, 30, 42, . . .}
9 + h12i = {. . . , −27, −25, −3, 9, 21, 33, 45, . . .}
Now, h3i/h12i is generated by 3 + h12i since
(3 + h12i)2 = (3 + h12i) + (3 + h12i) = 6 + h12i and
(3 + h12i)3 = (6 + h12i) + (3 + h12i) = 9 + h12i and
(3 + h12i)4 = (9 + h12i) + (3 + h12i) = 12 + h12i = h12i.
So, o(3 + h12i) = 4. So, h3i/h12i ∼
= Z4 .
Similarly, h8i/h48i = {h48i, 8 + h48i, 16 + h48i, 24 + h48i, 32 + h48i, 40 +
h48i} is generated by 8 + h48i and has order 6. So, h8i/h48i ∼
= Z6 .
In general, if n | k then nj = k for some j ∈ Z. Now, hni/hki =
{hki, n + hki, 2n + hki, 3n + hki, . . . , (j − 1)n + hki} is generated by
n + hki. Furthermore, hni/hki ∼
= Zk/n .
10. Let G =< a > be a cyclic group and H / G. Then we want to
show G/H = haHi. Let bH ∈ G/H. Then b = ak for some k ∈ Z. So,
bH = ak H = (aH)k ∈ haHi. So, G/H = haHi.
11. No, let G = D5 and H = {R, R2 , R3 , R4 , I} then G/H = {H, (V )H}.
So, H and G/H are abelian, but G is not.
12. Let G be an abelian group and H / G. Let aH, bH ∈ G/H. Then
(aH)(bH) = (ab)H = (ba)H = (bH)(aH). Thus, G/H is abelian.
14. o(14 + h8i) = 4 since
(14 + h8i) + (14 + h8i) = (14 + 14)h8i = 4 + h8i and
(14 + h8i)3 = (4 + h8i) + (14 + h8i) = 18 + h8i and
(14 + h8i)4 = (18 + h8i) + (14 + h8i) = 8 + h8i = h8i.
16
17. In book
18. Z60 /h15i = {h15i, 1 + h15i, 2 + h15i, 3 + h15i, . . . , 14 + h15i}. So,
the order of the factor group is 15.
The rest of the answers are in the back of the book.
17
HW #10 p167 #4,5,6,8,10,15,25,32,39,42,47
p211 #1,5,8,14,15,25,35
p226 #1,2,4,10,13,15
4.
Proof. (⇒) Suppose G ⊕ H is abelian. Let g1 , g2 ∈ G and h1 , h2 ∈ H.
So, (g1 , h1 )(g2 , h2 ) = (g2 , h2 )(g1 , h1 ). Simplifying the left and right
hand sides yields (g1 g2 , h1 h2 ) = (g2 g1 , h2 h1 ). Thus, g1 g2 = g2 g1 and
h1 h2 = h2 h1 . Therefore, G and H are abelian.
(⇐) Suppose G, H are abelian. Let (g1 , h1 ), (g2 , h2 ) ∈ G ⊕ H. Then
(g1 , h1 )(g2 , h2 ) = (g1 g2 , h1 h2 ) = (g2 g1 , h2 h1 ) = (g2 , h2 )(g1 , h1 )
So, G ⊕ H is abelian.
5. In book.
6. Z8 ⊕Z2 has an element of order 8. Namely, (1, 0). Now, Z4 ⊕Z4 cannot have an element of order 8 since lcm(o(a), o(b)) ≤ 4 when a, b ∈ Z4 .
8. No. Z3 ⊕ Z9 is not cyclic since gcd(3, 9) 6= 1.
o(a) o(b) (a,b)
3
9 (1, 1) (1, 2) (1, 4) (1, 5) (1, 7) (1, 8) (2, 1) (2, 2) (2, 4) (2, 5) (2, 7) (2, 8)
1
9 (0, 1) (0, 2) (0, 4) (0, 5) (0, 7) (0, 8)
So, there are 18 elements of order 9.
10.
15. G ⊕ H is cyclic implies that there exists some generator (a, b).
Furthermore, for all (g, h) ∈ G ⊕ H there exists an integer, k such that
(a, b)k = (g, h). Now, (a, b)k = (ak , bk ) = (g, h). Thus, ak = g and
bk = h making both G, H cyclic. In general if G1 ⊕ G2 ⊕ . . . ⊕ Gn is
cyclic then so is Gi for i = 1, 2, . . . n.
25. In book.
32. Taking all elements of order 3 and the identity yields
H = {(4, 0, 0) (8, 0, 0) (0, 0, 5) (0, 0, 10) (4, 0, 5) (4, 0, 10) (8, 0, 5) (8, 0, 10) (0, 0, 0)}
which is a nine element subgroup.
39.
24
|U (6)|
= 24/2 = 12 subgroups of order 6.
18
42. Skip
47. Recall, that we just have to map a generator from Z12 to the generators of Z4 ⊕Z3 . So, we can map 1 ∈ Z12 to (1, 1), (3, 1), (1, 2), or (3, 2).
Thus, there are 4 isomorphisms.
1. In book.
5. In book
8. Let sgn : G → {1, −1} by sgn(σ) = +1 if σ is even. If σ is odd,
then sgn(σ) = −1.
Case 1: σ, ρ are even. Then sgn(σρ) = +1 = (+1)(+1) = sgn(σ)sgn(ρ).
Case 2: σ, ρ are odd. Then sgn(σρ) = +1 = (−1)(−1) = sgn(σ)sgn(ρ).
Case 3: σ even, ρ odd. Then sgn(σρ) = −1 = (+1)(−1) = sgn(σ)sgn(ρ).
Case 4: σ odd, ρ even. Then sgn(σρ) = −1 = (−1)(+1) = sgn(σ)sgn(ρ).
Thus, in all cases sgn is a homomorphism. Ker(sgn) = {σ ∈ G | σ
is even } = An . Since the kernel is a normal subgroup, An must be
normal as well.
14. Note that σ(9 +12 4) = σ(1) = 3, but σ(9) +10 σ(4) = 27 +10 12 = 9.
Since 3 6= 9, σ is not a homomorphism.
15. φ : Z30 → Z30 and ker(φ) = {0, 10, 20} Since φ(23) = 9, φ−1 (9) =
23 + ker(φ) = {3, 13, 23}.
25. Suppose φ(1) = a. Then o(a) | 10 and o(a) | 20. Thus, o(a) =
1, 2, 5, 10 and a = 0, 10, 4, 8, 12, 16, 2, 6, 14, 18. So 10 of the mappings
are to. The mappings that are onto will have o(a) = 10. These are
a = 2, 6, 14, 18. So 4 are onto.
25. In book
1. In book.
2. n = 8 will result in Z8 , Z2 ⊕ Z4 , Z2 ⊕ Z2 ⊕ Z2
19
4.
Order
2
Z4 ⊕ Z2 ⊕ Z2
(0,0,0) (0,1,0) (0,0,1) (2,0,1)
(2,1,0) (2,1,1) (0,1,1)
4
4, 12 (2,0) (6,0) (2,1) (6,1) (1,0) (3,0) (0,1) (0,3) (1,0,0) (1,1,0) (1,0,1) (1,1,1)
(1,3) (3,1)(1,2) (3,2) (3,0,0) (3,1,0) (3,0,1) (3,1,1)
(2,3) (2,1) (1,1) (3,3)
10. 360 = 23 · 32 · 5
Isomorphism classes: Z2 ⊕Z2 ⊕Z2 ⊕Z3 ⊕Z3 ⊕Z5
Z2 ⊕Z2 ⊕Z2 Z32 ⊕Z5
Z22 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z5
Z22 ⊕ Z2 ⊕ Z32 ⊕ Z5
Z23 ⊕ Z3 ⊕ Z3 ⊕ Z5
Z23 ⊕ Z32 ⊕ Z5
Z16
8
13. In book
15. In book.
Z8 ⊕ Z2
(0,1) (4,0) (4,1)
Z4 ⊕ Z4
(2,0) (0,2) (2,2)
20
HW #11 p243 #1,2,10,18,20,22,23,40
p255 #2,4,8,10,20,26,39
p269 #5,11,13
1. In book
2. 6
10, In example 8, 0 − 0 ∈ {0} and 0 · 0 ∈ {0}. Also a − b ∈ R and
ab ∈ R
In example 9, any two elements have a difference that is either 0, 2, or
4. Also, the product is either 0, 2, or 4.
In example 10, nx−ny = n(x−y) ∈ nZ. Also, (nx)(ny) = n(xny) ∈ nZ
In example 11, (a + bi) − (c + di) = (a − c) + (b − d)i ∈ Z[i]. Also
(a + bi)(c + di) = (ac − bd) + (ad + bc)i ∈ Z[i]
In example 12, (f − g)(0) = f (0) − g(0) = 0 − 0 = 0 ⇒ (f − g)(x) ∈ S.
Also, (f g)(0) = 0 ⇒(f g)(x)
∈ S a 0 c 0
a−c
0
c 0
a 0
=
∈ S Also,
=
−
In example 13,
0 b 0 d
0
b−d
0 d
0 b
ac 0
∈ S.
0 bd
18. x, y ∈ S ⇒ a(x − y) = ax − ay = 0 ⇒ x − y ∈ S and
a(xy) = (ax)y = (0)y = 0 ⇒ xy ∈ S.
20. GL(2, Z).
22. x, y ∈ U (R) ⇒ (xy)−1 = y −1 x−1 So, xy is a unit and U (R) is
closed. Furthermore, x(yz) = (xy)z since rings are associative with
multiplication. Also, R has a unity, 1. So U (R) has an identity, 1. By
definition of a unit, each element has an inverse. So, U (R) is a group.
23. In book
1 2 0 1
2 3
40.
=
∈
/ R. This counterexample shows R is not
2 1 1 1
1 4
a subring of M2 (Z).
2. Only example 5 is a field.
21
4. The zero divisors are: 2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18. The zero divisors of Z20 are the nonzero, non-unit elements.
8. The zero divisors are {(0, x, y), (0, x, 0), (0, 0, y), (z, 0, 0), (z, x, 0), (z, 0, y) |
z, y ∈ Z∗ , x ∈ Q∗ }
The units are {(1, x, 1), (−1, x, 1), (1, x, −1), (−1, x, −1) | x ∈ Q∗ }
= 2−1 = 4
−2
= (−2)(3−1 ) = 5(5) = 4
3
√
−3 = 4 So 3 = 2, 5 since 22 = 4 = −3 and 52 = 4 = −3
−1
= (−1)(6−1 ) = 6(6) = 1
6
10.
1
2
20. (1 + 2i)(1 + 3i) = 1 − 6 + (3 + 2)i = 0
√
√
d]
is
a
ring.
So,
(a
+
db) + (x +
26.
First
we
need
to
show
R
=
Q[
√
√
dy) = (a + x) + (b + y) d ∈ R
This is a subset of R, so Ris associative.
0∈R √
√
−(a + db) = −a + d(−b) ∈ R Again, R ⊆ R so R is abelian. All
the above show that R is an abelian group under +.
As a subset of R, R is associative with multiplication and distribution
will hold.
√
√
√
−b
a
Now, for 0 6= a + b d ∈ R, (a + b d)−1 = a2 −b
d and
2 d + a2 −b2 d
2
2
a − b d 6= 0 since a and b are not both 0. So, each nonzero element is
a unit (i.e. it has a multiplicative inverse).
39. Check orders of elements and note that o(1 + 2i) = 8 = o(2i + 2).
So, the group is isomorphic to Z8 .
5. In book.
11. In book.
13. In book.