Download CM121A, Introduction to Abstract Algebra, Sample Clas Test 2

CM121A, Introduction to Abstract Algebra, Sample Clas Test 2
Questions
Decide if the following statements are TRUE or FALSE. If False, explain
why by reasoning or providing a counterexample. (+5 for every correct
answer, -1 for every wrong answer)
1. If p, q, r are distinct prime numbers, then gcd(pqr, pq + qr + pr) = 1.
Solution: True. Let g be the greatest common divisor of pqr and
pq + qr + pr. Assume g 6= 1. Then g has at least one prime factor
`. Since g|pqr, the prime factor ` has to be either p, q or r. If ` = p,
then p|g and hence p|pq + qr + pr. Since pq + pr is divisible by p,
this implies that p|qr. Since p is prime, we must have p|q or p|r. This
is impossible since q, r are primes different from p. The possibilities
` = q or ` = r are ruled out int he same way (In fact wince the
problem is symmetric with respect to p, q, r, it is conventional to say,
by symmetry, we can assume ` = p without loss of generality, instead
of writing three identical proofs, one for the case ` = p, one for ` = q,
and one for ` = r). This contradiction shows that g = 1.
2. If p, q, r are primes and p|qr, then either p = q or p = r.
Solution: True. Since p is prime, if p|qr, it follows that p|q or p|r. If
p|q then p is a divisor of a prime number q, and hence p ∈ {±1, ±q}.
The only possibility is p = q. Similarly, if p|r, then p = r.
3. If p, q are primes and p + q is odd, then either p = 2 or q = 2.
Solution: True. If p + q is odd, both p, q can’t be odd. So at least
one is even. The only even prime number is 2 (because an even prime
number is divisible by 2 which is impossible unless 2 it itself.) This
implies that p = 2 or q = 2.
4. If p, q, r are distinct primes, then gcd(p2 qr2 , pq 2 r2 ) = p2 q 2 r2 .
Solution: False. gcd(p2 qr2 , pq 2 r2 ) = pqr2 .
5. If p is a prime, then it has exactly 4 divisors.
Solution: True. The divisors of p are ±1, ±p.
6. The least common multiple of integers a, b equals ab/gcd(a, b).
Solution: True. Let g = gcd(a, b). We have a = a0 g and b = b0 g,
where gcd(a0 , b0 ) = 1 (proven in the lecture). If m is a common
multiple of a, b, that is a|m and b|m, we can write
m = ax = a0 gx
and
m = by = b0 gy.
Therefore a0 x = b0 y. Now a0 |b0 y and gcd(a0 , b0 ) = 1 imply that a0 |y.
Therefore
b0 ga0 |b0 gy = m.
In other words, any positive common multiple of a, b is divisible by
a0 b0 g = ab/g, and hence at most ab/g. On the other hand, ab/g is a
common multiple of a, b (because ab/g = ab0 = ba0 ). It follows that
ab/g is the least common multiple of a, b.
7. The number of positive divisors of pn is n.
Solution: False. The positive divisors of pn are {1, p, p2 , .., pn }. There
are n + 1 of them.
8. a ∈ [b]n if and only if b ∈ [a]n .
Solution: True, as we have proved in the lecture, both of these statements are equivalent to a ≡ b (mod n).
9. 7 belongs to [11]9 [5]9 + [3]9 .
Solution: False. We write
[11]9 [5]9 + [3]9 = [55]9 + [3]9 = [58]9 = [4]9 .
But 7 6∈ [4]9 since 7 6≡ 4 (mod 9).
10. 104 belongs to [11]29 [5]39 .
Solution: True. We write
[11]29 = [2]29 = [2]9 [2]9 = [4]9 .
Also
[5]39 = [5]9 [5]9 [5]9 = [25]9 [5]9 = [−2]9 [5]9 = [−10]9 = [8]9 .
Therefore,
[11]29 [5]39 = [4]9 [8]9 = [32]9 = [5]9 .
Finally, 104 ∈ [5]9 since 104 ≡ 5 (mod 9).
11. If a ≡ 2 (mod 6), then a is even.
Solution: True. If a ≡ 2 (mod 6), then a = 6k + 2 for some integer
k. In particular, a = 2(3k + 1) is even.
12. If a2 ≡ 1 (mod 12), then a ≡ ±1 (mod 12).
Solution: False. For example, 52 ≡ 1 (mod 12), but 5 6≡ ±1 (mod 12).
13. If a2 ≡ 1 (mod 19), then a ≡ ±1 (mod 19).
Solution: True. Assume a2 ≡ 1 (mod 19), then 19|a2 − 1, that is,
19|(a − 1)(a + 1). Since 19 is a prime number 19|a − 1 or 19|a + 1. If
19|a − 1, then a ≡ 1 (mod 19). If 19|a + 1, then a ≡ −1 (mod 19).
(Note: in the previous problem 12 is not a prime, so we can not make
the above argument.)
14. If a ≡ 3 (mod 13) and b ≡ −100 (mod 13), then ab has remainder 12
on division by 13.
Solution: True. We have ab ≡ 3 × (−100) ≡ 300 ≡ 12 (mod 13).
Hence ab has remainder 12 on division by 13.
15. a ≡ α (mod n) and b ≡ β (mod n) implies that ab ≡ αβ (mod n).
Solution: False. 2 ≡ 2 (mod 3), and 1 ≡ 4 (mod 3), but 21 6≡ 24
(mod 3).
16. a ≡ α (mod n) and b ≡ β (mod n) implies that ab ≡ αβ (mod n).
Solution: Ture. Proven in class.
17. a ≡ α (mod n) implies that aM ≡ αM (mod n).
Solution: True. Follows from last part by induction.
18. There is an element x ∈ Z6 such that x2 = [10]6 .
Solution: True. First note that [10]6 = [4]6 , and hence we are looking
for solutions to the equation
x2 = [4]6 .
There are only 6 elements in Z6 , so we can easily check all of them
and see if any of them satisfies the desired equation:
x = [0]6 =⇒ x2 = [0]6
x = [1]6 =⇒ x2 = [1]6
x = [2]6 =⇒ x2 = [4]6
x = [3]6 =⇒ x2 = [9]6 = [3]6
x = [4]6 =⇒ x2 = [16]6 = [4]6
x = [5]6 =⇒ x2 = [25]6 = [1]6
It follows that x = [2]6 and x = [4]6 are solutions.
19. There is an element x ∈ Z7 such that x2 = [10]7 .
Solution: False. First note that [10]7 = [3]7 , and hence we are looking
for solutions to the equation
x2 = [3]7 .
There are only 7 elements in Z7 , so we check all of them to see if any
of them satisfies the desired equation:
x = [0]7 =⇒ x2 = [0]7
x = [1]7 =⇒ x2 = [1]7
x = [2]7 =⇒ x2 = [4]7
x = [3]7 =⇒ x2 = [9]7 = [2]7
x = [4]7 =⇒ x2 = [16]7 = [2]7
x = [5]7 =⇒ x2 = [25]7 = [4]7
x = [6]7 =⇒ x2 = [36]7 = [1]7
Hence, there are no solutions.
20. There is a nonzero element x ∈ Z5 such that x2 + x = [0]5 .
Solution: True. Again, we check all the 5 elements in Z5 to see if any
of them satisfies the desired equation:
x = [0]5 =⇒ x2 + x = [0]5
x = [1]5 =⇒ x2 + x = [2]5
x = [2]5 =⇒ x2 + x = [6]5
x = [3]5 =⇒ x2 + x = [12]5 = [2]5
x = [4]5 =⇒ x2 + x = [20]5 = [0]5
We see that there are two solutions x = [0]5 and x = [4]5 . Hence there
is a nonzero solution (i.e., x = [4]5 ).
21. Every nonzero element in x ∈ Z7 has a multiplicative inverse. (For an
element x ∈ Z7 , a multipicative inverse is an element y ∈ Z7 such that
xy = yx = [1]7 .)
Solution: True. We know that [a]n has a multiplicative inverse in Zn
if and only if gcd(a, n) = 1. If x = [a]7 is nonzero, then 76 |a, which
means gcd(a, 7) = 1. Hence [a]7 has a multiplicative inverse in Z7 .
22. Every nonzero element in x ∈ Z8 has a multiplicative inverse.
Solution: False. [2]8 has no multiplicative inverse as gcd(2, 8) 6= 1.
23. Every nonzero element in x ∈ Z9 has a multiplicative inverse.
Solution: False. [3]9 has no multiplicative inverse as gcd(3, 9) 6= 1.
24. Let n be a natural number. For any x ∈ Zn , there is y ∈ Zn such that
x + y = [0]n .
Solution: True. If x = [a]n , take y = [−a]n .
25. The binary operation on Z given by a ∗ b = ab + a + b is associative.
Solution: True.
(a ∗ b) ∗ c = (ab + a + b) + c + (ab + a + b)c = a + b + c + ab + ac + bc + abc
a ∗ (b ∗ c) = a + (b + c + bc) + a(b + c + bc) = a + b + c + bc + ab + ac + abc
26. The binary operation on Z given by a ∗ b = ab + a + b is commutative.
Solution: True. a ∗ b = ab + a + b = ba + b + a = b ∗ a.
27. There is a binary operation on Z such that a ∗ b = 0 for all a, b.
Solution: True. Just define a∗b = 0 for all possible choices of a, b ∈ Z.
28. Z7 − {[0]7 } is a group under multiplication.
Solution: True. We know that for any n > 0, Z×
n is a group under
multiplication. If n is a prime number, then gcd(a, n) = 1 if and only
if a is not divisible by n, in other words, every integer that is not
relatively prime to n lies in [0]n . In particular, for a prime number n
we have Z×
n = Zn − {[0]n }. It follows that Z7 − {[0]7 } is a group under
multiplication.
29. Z12 − {[0]12 } is a group under multiplication.
Solution: False. For example [2]12 has no multiplicative inverse
(why?)
Note: (see previous solution as well) We know that for any n > 0,
×
Z×
n is a group under multiplication. But in general Zn does not equal
Zn − {[0]n }. For example, in this case n = 12, we have
Z×
12 = {[1]12 , [5]12 , [7]12 , [11]12 },
but
Z12 −{[0]n } = {[1]12 , [2]12 , [3]12 , [4]12 , [5]12 , [6]12 , [7]12 , [8]12 , [9]12 , [10]12 , [11]12 }.
In fact Z×
n = Zn − {[0]n } if and only if n is a prime (prove this).
30. Let G be a group and g, h, h0 ∈ G. Let e be the identity element of G.
(a) ghg −1 h−1 = e.
Solution: False. Assume that for all g, h ∈ G we have
ghg −1 h−1 = e.
(∗)
Multiplying the left hand side of (∗) by hg from the right, we get
ghg −1 h−1 hg = ghg −1 eg = ghg −1 g = ghe = gh
Multiplying the right hand side of (∗) by hg from the right, we
get
ehg = hg.
Setting these sides equal, we find that for all h, g ∈ G, we must
have gh = hg. This means that the group is abelian. So this
statement is not true for a general group.
(b) g 2 = g implies g = e.
Solution: True. Multiply both sides by g −1 from the right to
get g = g 2 g −1 = gg −1 = e.
(c) g 2 = e implies g = g −1
Solution: True. Multiply both sides by g −1 from the right to
get g = g 2 g −1 = eg −1 = g −1 .
(d) ghg = g implies h = g −1 .
Solution: True. Multiply both sides by g −1 from the left to get
hg = g −1 ghg = g −1 g = e. Then, multiply both sides by g −1 from
the right to get h = hgg −1 = eg −1 = g −1 .
(e) If G is commutative, then ghg −1 = h.
Solution: True. Since G is abelian, for all g, h ∈ G, we have
gh = hg. Multiplying both sides by g −1 from the right, we get
ghg −1 = hgg −1 = he = h.
(f) g 2 h2 = (gh)2
Solution: False. Assume that for all g, h ∈ G we have
g 2 h2 = (gh)2
(∗)
Multiplying the left hand side of (∗) by g −1 from the left, and
h−1 from the right, we get
g −1 g 2 h2 h−1 = g −1 gghhh−1 = eghe = gh.
Multiplying the right hand side of (∗) by g −1 from the left, and
h−1 from the right, we get
g −1 (gh)2 h−1 = g −1 ghghh−1 = ehge = hg.
Setting these sides equal, we find that for all h, g ∈ G, we must
have gh = hg. This means that the group is abelian. So this
statement is not true for a general group.
(g) If G is commutative/abelian g 2 h2 = (gh)2 .
Solution: True. As shown in the last part.
(h) (gh)−1 = g −1 h−1 .
Solution: False. Multiplying both sides with gh from the left,
we get e = ghg −1 h−1 for all g, h ∈ G. We showed that this is
false for a general group earlier above. (Part (a)).
(i) ghg −1 = gh0 g −1 implies h = h0 .
Solution: Ture. By cancellation of g and g −1 from the left and
the right, respectively.