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Math 546
Homework 1
Due Wednesday, January 25.
This homework has two types of problems.
• 546 Problems. All students (students enrolled in 546 and 701I) are required to turn these in.
• 701I Problems. Only students enrolled in 701I are required to turn these in. Students not
enrolled in 701I are welcome to turn these in as well. I especially welcome students looking
for a challenge to attempt these.
Note: Part 3 below contains solutions to some of the unassigned problems from Sections 1 and 2
of Saracino’s text. I am providing these as examples of how to write solutions. Please have a look
at these, and make sure to read Sections 1 and 2 of the text.
1
546 Problems.
1.3 and 1.6. For each of the following sets S and functions ∗ on S×S, determine whether ∗ is
a binary operation on S. If ∗ is a binary operation on S, determine whether it is commutative
and whether it is associative.
a) S = Z, a ∗ b = a + b2 .
e) S = Z, a ∗ b = a + b − ab.
f) S = R, a ∗ b = b.
g) S = {1, −2, 3, 2, −4}, a ∗ b = |b|.
1.4. Is division a commutative operation on R+ ? Is it associative?
Note: If S ⊆ R, then we define S + = {s ∈ S : s > 0}.
1.9. Let S = {a, b, c, d}. The following table defines a binary operation ∗ on S.
Is ∗ commutative? Is it associative?
*
a
b
c
d
a
a
c
b
d
b
c
a
d
b
c
b
d
a
c
d
d
b
c
a
Note: For s1 , s2 ∈ S, we define s1 ∗ s2 to the be the element in the row that contains s1 and
the column that contains s2 . For example, we have c ∗ b = d.
2.1. Which of the following are groups? Why or why not?
d) The set {1, −1} under multiplication.
f) R × R = {(x, y) : x, y ∈ R} under the operation (x, y) ∗ (z, w) = (x + z, y − w).
g) R × R× = {(x, y) : x, y ∈ R and y 6= 0} under the operation
(x, y) ∗ (z, w) = (x + z, yw).
Recall that R× = R \ {0}.
h) R \ {1} under the operation a ∗ b = a + b − ab.
2.5. Let S = {a, b, c}. The following table defines a binary operation ∗ on S.
*
a
b
c
a
a
b
c
b c
b c
b c
c c
Is (S, ∗) a group? Why or why not?
2.8. Let G = {f : R → R : for all x ∈ R, f (x) 6= 0}. For all f , g ∈ G, define × by
(f × g)(x) = f (x)g(x) for all x ∈ R.
Is (G, ×) a group? Prove or disprove.
2.11. Let G =
a 0
: a, b 6= 0 in R . Show that G forms a group under matrix multi0 b
plication.
2
701I Problems.
1.10. How many binary operations are there on a set S with n elements? How many of these
are commutative?
2.1 e) Let S = {q ∈ Q+ :
√
q ∈ Q}, ∗ is ×. Is (S, ∗) a group? Why or why not?
2.6. Let S = {a, b, c}. The following table defines a binary operation ∗ on S.
*
a
b
c
a
a
b
c
b c
b c
a c
b a
Is (S, ∗) a group? Why or why not?
2.10. Let
a b
2
2
G= M =
: a, b ∈ R, det M = a + b 6= 0 ⊂ GL(2, R).
−b a
Let × denote ordinary matrix multiplication. Show that (G, ×) is a group.
3
Examples.
1.3 and 1.6. For each of the following sets S and functions ∗ on S×S, determine whether ∗ is
a binary operation on S. If ∗ is a binary operation on S, determine whether it is commutative
and whether it is associative.
b) S = Z, a ∗ b = a ∗ b = a2 b3 .
Claim: ∗ is a binary operation on S.
Proof: For all a, b ∈ Z, we have a ∗ b = a2 b3 ∈ Z, so ∗ is a binary operation.
Claim: ∗ is not commutative on S.
Proof: We note that b ∗ a = b2 a3 . Therefore, we have a ∗ b = b ∗ a if and only if
a2 b3 = a3 b2 , which holds if and only if a = b or either of a, b = 0. We observe that
1 ∗ 2 = 12 23 = 8 6= 4 = 22 13 = 2 ∗ 1.
Hence, ∗ is not commutative.
Claim: ∗ is not associative on S.
Proof: We compute
(a ∗ b) ∗ c = (a2 b3 ) ∗ c = (a2 b3 )2 c3 = a4 b6 c3 ,
a ∗ (b ∗ c) = a ∗ (b2 c3 ) = a2 (b2 c3 )3 = a2 b6 c9 .
Therefore, we have (a ∗ b) ∗ c = a ∗ (b ∗ c) if and only if a4 b6 c3 = a2 b6 c9 , which holds
for a, b, c 6= 0 if only if a2 = c6 . Taking square roots, we see that associativity follows
if and only if a = c3 . We note that
(2 ∗ 1) ∗ 1 = 24 16 13 = 16 6= 4 = 22 16 19 = 2 ∗ (1 ∗ 1).
Hence, ∗ is not associative.
h) S = {1, 6, 3, 2, 18}, a ∗ b = ab.
Claim: ∗ is not a binary operation on S.
Proof: We note that 6, 2 ∈ S, but that 6 ∗ 2 = 12 6∈ S. Therefore, ∗ is not a binary
operation.
Since ∗ is not a binary operation, we do not consider whether or not ∗ is commutative
or associative.
2.1. Which of the following are groups? Why or why not?
b) S = 3Z = {3n : n ∈ Z}, the set of integers that are multiples of 3, ∗ is +.
Claim: (S, ∗) is a group.
Proof: It suffices to verify the axioms:
(i) ∗ is a binary operation on S. To see this, let a = 3x, b = 3y ∈ 3Z. Then we have
a + b = 3x + 3y = 3(x + y) ∈ 3Z,
so ∗ is a binary operation on S.
(ii) ∗ is associative on S. Since S = 3Z ⊂ Z, we see that 3Z inherits associativity
under + from Z.
(iii) (S, ∗) has an identity. We claim that 0 = 3 · 0 ∈ 3Z is an identity for (S, ∗). To see
this, let a = 3x ∈ 3Z. Then we have
a + 0 = 3x + 0 = 3x = 0 + 3x = 0 + a,
so 0 is an identity for (S, ∗).
(iv) (S, ∗) has inverses. To see this, let a = 3x ∈ 3Z. Then −a = −3x = 3(−x) ∈ 3Z
has −a + a = 0 = a + (−a), so −a is an inverse for a.
It follows that (S, ∗) is a group. Furthermore, (S, ∗) inherits commutativity from Z
since S ⊆ Z, so (S, ∗) is an abelian group.
i) S = Z, a ∗ b = a + b − 1.
Claim: (S, ∗) is a group.
Proof: It suffices to verify the axioms.
(i) ∗ is a binary operation on S. To see this, let a, b ∈ Z. Then we have
a ∗ b = a + b − 1 ∈ Z,
so ∗ is a binary operation on S.
(ii) ∗ is associative on S. To see this, let a, b, c ∈ Z. Then we have
(a ∗ b) ∗ c = (a + b − 1) ∗ c = (a + b − 1) + c − 1 = a + (b + c − 1) − 1
= a ∗ (b + c − 1) = a ∗ (b ∗ c),
so ∗ is associative on S.
(iii) (S, ∗) has an identity. We claim that 1 ∈ Z is an identity for (S, ∗). To see this, let
a ∈ Z. Then we have
a ∗ 1 = a + 1 − 1 = a = 1 + a − 1 = 1 ∗ a,
so 1 is an identity for (S, ∗).
(iv) (S, ∗) has inverses. To see this, let a ∈ Z. Then we have 2 − a ∈ Z and
a ∗ (2 − a) = a + (2 − a) − 1 = 1 = (2 − a) + a − 1 = (2 − a) ∗ a,
so 2 − a is an inverse for a.
It follows that (S, ∗) is a group. Furthermore, for all a, b ∈ Z, we have
a∗b=a+b−1=b+a−1=b∗a
implies that (S, ∗) is an abelian group.