Download solutions - UCI Math

MATH 120A MIDTERM EXAM SOLUTIONS
(WHITE PAPER)
WINTER 2015
Problem 1 (10 points). Mark each statement ‘T’ for true (meaning always true) or ‘F’ for
false (meaning sometimes false). You do NOT need to justify your answers to this problem.
T F “If every element of the group G1 is its own inverse, and G2 is
isomorphic to G1 , then every element of G2 is its own inverse.”
True. It is routine to show that this is a structural property. (Also,
it is equivalent to the property that square of every element is
the identity element, which we have already seen is a structural
property.)
T F “The set of all positive rational numbers forms a group under mul
tiplication.” True, by basic facts about the rational numbers.
T F “The set of all 2 × 2 real matrices of positive determinant is
closed under matrix multiplication.” True, because det(AB) =
det(A) det(B) and the product of two positive real numbers is positive.
F “The binary structure (S, max) has an identity element, where S =
T {x ∈ R : 0 < x < 1}.” False: for every x ∈ S we have x/2 ∈ S and
max(x, x/2) 6= x/2, so x is not an identity element for the structure.
F “The group (R∗ , ·) of nonzero real numbers under multiplication
T is isomorphic to the group (R+ , ·) of positive real numbers under
multiplication.” False, because the first group has an element of
order 2, namely −1, but every element of the second group has
order 1 or infinite order, and having an element of order 2 is a
structural property.
F “A binary operation on a set S is a function from S × S to S × S.”
T False: it is a function from S × S to S.
T F “There is a bijection from N to Q.” True: see Figure 0.15 on p. 5
of the textbook for an idea of how to prove this.
F “If H is a proper subgroup of G, then |H| < |G|.” False. For
T example, 2Z is a proper subgroup of Z, but there is a bijection
φ : Z → 2Z defined by φ(n) = 2n, so |2Z| = |Z|.
1
T F “Some element of the group (Z12 , +12 ) has order 4.” True: the
elements 3 and 9 have order 4 in Z12 .
F “Every abelian group is cyclic.” False. Some counterexamples are
T (Q, +), (R, +), and the Klein 4-group.
Problem 2 (4 points). Define the italicized terms:
(a) What does it mean for a binary operation ∗ on a set S to be commutative?
(b) What does it mean for a binary operation ∗ on a set S to be associative?
Solution.
(a) a ∗ b = b ∗ a for all a, b ∈ S.
(b) (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ S.
Problem 3 (4 points). Let (G, +) be an abelian group and define a function φ : G → G by
φ(a) = −a. Prove that φ is an isomorphism from (G, +) to (G, +).
Solution.
φ is injective: Let a, b ∈ G. If φ(a) = φ(b), then −a = −b, so we have
a = −(−a) = −(−b) = b.
φ is surjective: Let b ∈ G. Then φ(−b) = −(−b) = b.
φ is a homomorphism: Let a, b ∈ G. Then φ(a + b) = −(a + b) = (−a) + (−b) =
φ(a) + φ(b).
Remark. The step −(a + b) = (−a) + (−b) can also be written as −(a + b) = −a − b, where
in general c − d is defined as c + (−d). In any case, this step uses the assumption that G
is abelian; otherwise all we could say (in multiplicative notation) would be (ab)−1 = b−1 a−1 ;
the order would be reversed. One reason that additive notation is not used for non-abelian
groups is that the equation −(a + b) = −a − b looks so natural and tempting.
Problem 4 (4 points). Consider the group
G = {a ∈ Q : a = 2n or a = −(2n ) for some n ∈ Z},
with the operation of multiplication. (You may assume that this is a group.)
Prove that G is not cyclic.
Solution. Let a ∈ G. We want to show that hai =
6 G.
n
Case 1: Assume a = 2 where n ∈ Z. Then
hai = {(2n )k : k ∈ Z}.1
Then −1 is in G but not in hai.
Case 2: Assume a = −(2n ) where n ∈ Z. Then
hai = {(−(2n ))k : k ∈ Z}.
If n = 0 then hai = h−1i = {−1, 1} 6= G. If n 6= 0 then −1 6= (−(2n ))k for all k ∈ Z
(consider k = 0 and k 6= 0 separately) so again hai =
6 G. (This case is a bit tricky to
write down.)
Problem 5 (4 points). For each part, make sure to justify your answer:
(a) Give an example of a subset of R that is closed under addition, but not closed under
multiplication.
(b) Give an example of a subset of R that is closed under multiplication, but not closed
under addition.
Solution.
(a) Define the set S = {x ∈ R : x < 0}. This set is closed under addition because the sum of
two negative numbers is negative. It is not closed under multiplication because −1 ∈ S
but (−1)(−1) = 1 ∈
/ S (for example.)
(b) Define the set S = {1}. This set is closed under multiplication because 1 · 1 = 1. It is
not closed under addition because 1 + 1 = 2 ∈
/ S.
Solution (alternative).
(a) Define S = {n/2 : n ∈ Z}. This set is closed under addition because (m/2) + (n/2) =
(m + n)/2 (and if m and n are integers then so is m + n.) It is not closed under
multiplication because 1/2 ∈ S but (1/2)(1/2) = 1/4 ∈
/ S.
2
(b) Define S = {n : n ∈ N}. This set is closed under multiplcation because (m2 )(n2 ) =
(mn)2 (and if m and n are natural numbers then so is mn.) It is not closed under
addition because 1, 4 ∈ S but 1 + 4 = 5 ∈
/ S.
1It
is not correct to say hai = {(2n )n : n ∈ Z}. The exponent that determines a is already fixed, but the
exponent in the definition of “cyclic subgroup generated by ...” is allowed to vary, so we must introduce a
new symbol for it.
Problem 6 (4 points). Let G be the group {1, 3, 7, 9} with the operation ·10 of multiplication
modulo 10. (You may assume this is a group.)
(a) List all of the subgroups of G.
How do you know that your list is complete?
(b) Draw the subgroup diagram of G.
Solution.
(a) Start by computing the cyclic subgroups:
h1i = {1}
h3i = {1, 3, 9, 7}
h7i = {1, 7, 9, 3} = h3i
h9i = {1, 9}.
This must be all of the subgroups because G is cyclic (we have seen that 3 and 7 are
generators) and every subgroup of a cyclic group is cyclic. (Alternatively, one can show
that the subsets {1, 3}, {1, 7}, {1, 3, 7}, {1, 3, 9}, and {1, 7, 9} are not closed under the
operation.)
(b) Here is one way to draw the subgroup diagram:
G
{1, 9}
{1}