Math 415.502
Exam 1
Solutions
October 12, 2010
As can easily be expected, the solutions provided below are not the only ways to solve these
problems, and other solutions may be completely valid. If you have questions about what went
wrong (or right!) with your own solutions, please talk to me about it so that you will improve on
the next exam and the final.
1. [14 points]
(a) Let H = { 2am | a, m ∈ Z, m ≥ 0}. Show that H is a subgroup of Q under addition.
(b) Show that Q+ is not a subgroup of Q under addition.
Solution: (a) To show that H is a subgroup of Q, we need to show that (i) H is closed
under addition; (ii) 0 ∈ H; and (iii) H is closed under additive inverses. To see (i), we let
a
b
+
2m , 2n ∈ H, with a, b ∈ Z and m, n ∈ Z . We then see
a
b
a · 2n + b · 2m
+
=
.
2m 2n
2m+n
Since a·2n +b·2m ∈ Z and m+n ≥ 0, we see that 2am + 2bn ∈ H. For (ii), certainly 0 =
For (iii), if 2an ∈ H, then its additive inverse is − 2an = −a
2n , which is also in H.
0
21
∈ H.
(b) There are essentially two ways to solve this problem. For the first way, we know that if
Q+ is a subgroup of Q under addition then they must share the same identity element. The
identity of Q is 0, but 0 ∈
/ Q+ . For the second way, we show that Q+ is not closed under
additive inverses in Q. For example, 2 ∈ Q+ , but −2 ∈
/ Q+ .
a 0
∈ M2 (R) ad 6= 0 .
2. [(a) 8 points; (b) 5 points] Let L =
c d
(a) Show that L is a group under matrix multiplication. (You may assume that matrix
multiplication is associative.)
(b) Is L abelian? Why or why not?
Solution: (a) Essentially we show that L is a subgroup of GL(2, R). So we need to show
that (i) L is closed under matrix multiplication;
L contains an identity element; and (iii)
(ii)
e 0
a
0
L is closed under inverses. For (i), we let c d , g h ∈ L. We multiply and find
a 0
c d
e 0
ae
0
=
.
g h
ce + dg dh
Since (ae)(dh) = (ad)(eh) and ad 6= 0, eh 6= 0, we see that (ae)(dh) 6= 0, and so the product
is indeed in L. Thus L is closed under matrix multiplication. For (ii), we see that the identity
matrix I = ( 10 01 ) is clearly in L. For (iii), if we are given A = ac d0 ∈ L, we want to show that
A−1 ∈ L. (We know that A is invertible in GL(2, R) because it has non-zero determinant,
but we need to show that the inverse is actually in L.) To this end, we calculate that
A
−1
−1 1
a 0
a
=
=
c
− ad
c d
0
,
1
d
which is lower triangular and has non-zero determinant. Thus A−1 ∈ L.
(b) L is not abelian. To see this we provide a counterexample:
1 0
2 0
2 0
=
1 1
−1 1
1 1
whereas
2 0
−1 1
1 0
2 0
=
.
1 1
0 1
[It is not sufficient to simply say that “matrix multiplication is not commutative,” because
L does not contain all 2 × 2 matrices. For example, if L had happened to be the set of all
diagonal matrices (so putting a 0 in the lower left entry as well as the upper right), then those
matrices form an abelian subgroup of GL(2, R).]
3. [14 points] In this problem let α = e2πi/5 ∈ U5 .
(a) Find all solutions of z 5 = i for z ∈ C. Write your answer(s) in the form eiθ , θ ∈ R.
(b) Define an isomorphism φ : Z5 → U5 such that φ(3) = α (where Z5 is a group under +5 ).
Explain how you arrive at your answer, but you need not supply a formal proof.
1/5
Solution: (a) We know that i = eπi/2 , and so eπi/2
= eπi/10 is a 5-th root of i. [In fact,
i already a 5-th root of itself, but it was not necessary to observe this.] Therefore, the roots
of z 5 = i are the five numbers
πi/10
e
, αeπi/10 , α2 eπi/10 , α3 eπi/10 , α4 eπi/10 = eπi/10 , e5πi/10 , e9πi/10 , e13πi/10 , e17πi/10 .
(b) We make use of the homomorphism property of φ, namely that for all x, y ∈ Z5 , we have
φ(x +5 y) = φ(x)φ(y). Since we require that φ(3) = α, it then follows that we must have
φ(1) = φ(3 +5 3) = φ(3)φ(3) = α2 ,
φ(4) = φ(3 +5 3 +5 3) = φ(3)3 = α3 ,
φ(2) = φ(3 +5 3 +5 3 +5 3) = φ(3)4 = α4 ,
φ(0) = φ(3 +5 3 +5 3 +5 3 +5 3) = φ(3)5 = α5 = 1.
4. [14 points] Fun with binary operations.
√
(a) Suppose we define a binary operation on R+ by a ∗ b = a + b. Is this operation
associative? Why or why not?
Z 1
(b) Suppose we define a binary operation on R+ by a∗b =
bxeax dx. Is this a well-defined
binary operation? Why or why not?
0
Solution: (a) No this operation is not associative. We provide a counterexample, taking
a = 1, b = 3, c = 6. Then
q
√
√
a ∗ (b ∗ c) = 1 + 3 + 6 = 1 + 3 = 2,
whereas
(a ∗ b) ∗ c =
q
√
1+3+6=
√
2+6=
√
8.
(b) Yes, it is well-defined. First, given a, b ∈ R+ , the function f (x) = bxeax is a continuous
function and f (x) ≥ 0 for all x in [0, 1]. Therefore,
Z
1
a∗b=
bxeax dx
0
is always a non-negative real number. Thus we need only show that a ∗ b > 0. There are
various ways to do this (integrate by parts, estimation, etc.), but here is one way. Since
f (x) = 0 only for x = 0 in [0, 1], the integral above represents the area of between the graph
of y = f (x) in the first quadrant above the interval [0, 1]. This is necessarily positive.
5. [21 points] Consider the group Z42 under +42 .
(a) Determine the order of the subgroup h35i.
(b) Which of 8, 19, and 21 generate Z42 . Why?
(c) Determine all elements of Z42 of order 7.
Solution: (a) By a theorem from class (or Thm. 6.14 in Fraleigh), we know that for a cyclic
group hai of order n, we have
n
|har i| =
.
gcd(r, n)
Therefore in Z42 = h1i, which is written additively, we have
|h35i| =
42
42
=
= 6.
gcd(35, 42)
7
(b) One of the corollaries to the theorem above is that the generators of the cyclic group hai
are all elements of the form ar with gcd(r, n) = 1. Therefore, the elements of Z42 that are
relatively prime to 42 will generate Z42 . Among the numbers we are given, 19 is relatively
prime to 42 (and so is a generator), but 8 and 21 are not (and so are not generators).
(c) Using the same theorem from part (a), we see that a ∈ Z42 has order 7 if and only if
42
gcd(a,42) = 7, which is true if and only if gcd(a, 42) = 6. The numbers in Z42 = {0, 1, . . . , 41}
that have gcd 6 with 42 are 6, 12, 18, 24, 30, and 36.
6. [(a) 4 points; (b) 10 points; (c) 4 points] Suppose G is a group of order 8 containing two
elements, A and B, such that
• A has order 4, and B has order 2,
• BAB = A3 ,
• G = {e, A, A2 , A3 , B, AB, A2 B, A3 B}.
(a) Show that BA2 B = A2 and BA3 B = A.
(b) Write down the multiplication table for G. You do not need to justify all of your work,
but do write elements of G in the first row and column in the order e, A, A2 , A3 , B,
AB, A2 B, A3 B.
(c) (Extra Credit) Is G isomorphic to the quaternion group H8 ? Why or why not?
Solution: (a) We know that A4 = e and B 2 = e, and in particular A−1 = A3 and B −1 = B.
Now using BAB = A3 , we see that
(BAB)(BAB) = (A3 )(A3 ),
BAB 2 AB = A6 ,
BA2 B = A2 .
Likewise, if we invert both sides of BAB = A3 , we see that
(BAB)−1 = (A3 )−1 ,
B −1 A−1 B −1 = A−3 ,
BA3 B = A.
(b) To write down the multiplication table for G, we use the following relations obtained
from part (a) and the given information: A4 = e, B 2 = e, BA = A3 B, BA2 = A2 B, and
BA3 = AB. The multiplication table is then the following.
∗
e
A
A2
A3
B
AB
A2 B A3 B
e
e
A
A2
A3
B
AB
A2 B A3 B
A
A
A2
A3
e
AB
A2
A2
A3
e
A
A2 B A3 B
A3
A3
e
A
A2
A3 B
B
B
AB
AB
AB
A2 B A2 B
AB
A3 B A3 B A2 B
B
B
AB
B
AB
A2 B
e
A3
A2
A
A3 B A2 B
A
e
A3
A2
B
A3 B
A2
A
e
A3
AB
B
A3
A2
A
e
A3 B A2 B
B
A2 B A3 B
7. [10 points] Suppose G is an abelian group, and let H = {a ∈ G | a = a−1 }. Show that H ≤ G.
Solution: We need to show that (i) H is closed under multiplication in G; (ii) e ∈ H; and
(iii) H is closed under inverses. To show (i), we let a, b ∈ H. Then
(ab)−1 = b−1 a−1 ,
= ba
(because a, b ∈ H),
= ab
(because G abelian).
Therefore ab ∈ H, and so H is closed under multiplication in G. To see (ii), we simply
observer that e−1 = e, and so yes, e ∈ H. For (iii), we let a ∈ H. Since a = a−1 by the
definition of H and a ∈ H, indeed a−1 ∈ H.