Download Introduction to Abstract Algebra, Spring 2013 Solutions to Midterm I

Introduction to Abstract Algebra, Spring 2013
Solutions to Midterm I
1. a. What is an equivalence relation on a set?
b. How many distinct equivalence relations are there on a set with three elements?
c. Let S be a non-empty set and write P(S) for the set of all subsets of S. For X and
Y in P(S), define X ∗ Y = X ∩ Y . Is P(S) a group under this operation?
a. An equivalence relation ∼ on a set S is a reflexive, symmetric, transitive relation on
S. That is, for all x, y, z ∈ S,
i. x ∼ x (∼ is reflexive),
ii. if x ∼ y then y ∼ x (∼ is symmetric),
iii. if x ∼ y and y ∼ z then x ∼ z (∼ is transitive).
b. The number of equivalence relations on a set is the number of partitions of the set
(that is, the number of ways in which the set can be written as a disjoint union of
non-empty subsets). If S = {a, b, c} then the distinct partitions of S are given by
{a, b, c},
{a, b} ∪ {c}, {a, c} ∪ {b}, {b, c} ∪ {a},
{a} ∪ {b} ∪ {c}.
Thus there are 5 distinct equivalence relations on S.
c. For any X ⊂ S, we have X ∩ S = X = S ∩ X. This means that S is an identity
element. Suppose X $ S (for example, take X = ∅). Then, for any Y ⊂ S,
X ∩ Y ⊂ X 6= S.
Thus X does not have an inverse and hence P(S) cannot be a group.
2. Let G be a group and let H be a subset of G.
a. What does it mean to say that H is a subgroup of G?
b. Suppose that H is a finite non-empty subset of G that is closed under the binary
operation in G (i.e., if x, y ∈ H then xy ∈ H). Show that H is a subgroup of G.
c. Consider the group Z16 of residues modulo 16 (under addition modulo 16). How many
subgroups does this group have? Explain your answer.
a. It means that H is a nonempty subset of G that is closed with respect to the binary
operation on G (i.e., if x, y ∈ H then xy ∈ H) and such that the induced binary
structure on H makes H into a group.
Comment. We showed in class that this is the same as saying that H satisfies the
conditions:
i. if x, y ∈ H then xy ∈ H;
ii. 1 ∈ H (i.e., the identity element of G belongs to H)
iii. if h ∈ H then also h−1 ∈ H.
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b. By hypothesis, H is closed under the binary operation in G. We’ll show that 1 ∈ H
and that if h ∈ H then h−1 ∈ H. By the preceding comment, this suffices to show
that H is a subgroup of G. For h ∈ H, the infinite sequence
h, h2 , h3 , . . . , hn , . . .
consists of elements of H. Since H is finite, we have hk = hl , for distinct positive
integers k and l with say k < l. Thus hl−k = 1, and so hm = 1, for some positive
integer m. Hence 1 ∈ H and h−1 = hm−1 ∈ H.
c. The group Z16 is cyclic. From class, a finite cyclic group of order n has a unique
subgroup of order d for each (positive) divisor d of n and no other subgroups. Thus
# subgps. of Z16 = # divisors of 16 = 5.
3. a. What does it mean to say a group is (i) cyclic, (ii) abelian?
b. Show that every cyclic group is abelian.
c. Consider the multiplicative group C× of non-zero complex numbers and let
µ∞ = {z ∈ C× : z n = 1 for some n ∈ Z}.
i. Show that µ∞ is a subgroup of C× and that µ∞ is not cyclic.
ii. Show that every finite subgroup of µ∞ is cyclic. (You may use without proof the
fact that the group of N -th roots of unity is cyclic (for any positive integer N ).)
a. A group G is cyclic if there is an a ∈ G such that G = {an : n ∈ Z}. In other words,
for any g ∈ G, then there is an integer n (depending on g) such that g = an . In this
case, a is said to generate G or to be a generator of G.
A group G is abelian if gh = hg, for any g, h ∈ G.
b. Suppose G is cyclic and that a generates G. Let g, h ∈ G. Thus there are integers k
and l such that g = ak and h = al . Hence
gh = ak al = ak+l = al+k = al ak = hg,
and so G is abelian.
c.
i. Clearly, 1 ∈ µ∞ . Let z and w be elements of µ∞ so that z m = 1 and wn = 1 for
integers m and n. We have
(zw)mn = z mn wmn = (z m )n (wn )m = 1n 1m = 1,
and so µ∞ is closed under multiplication. Further,
(z −1 )m = (z m )−1 = 1−1 = 1,
so µ∞ is also closed under taking inverses. This proves that µ∞ is a subgroup of
C× .
To see that µ∞ cannot be cyclic, note first that µ∞ is infinite. For example, for
any positive integer N , µ∞ contains the cyclic group µN of order N consisting
of the complex N -th roots of unity. By definition, each element of µ∞ has finite
order and thus the cyclic group generated by any element if µ∞ is finite. In
particular, no element of µ∞ is a generator of µ∞ , that is, µ∞ is not cyclic.
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ii. Let ν be a finite subgroup of µ∞ . Note there is a positive integer N such that
z N = 1 for all z ∈ ν. (For example, we could take N to be the product of the
orders of the elements of ν, or more efficiently we could take N = |ν|, the order
of ν (recall from class the order of each element of ν divides |ν|).) Thus ν is a
subgroup of µN , the cyclic group of N -th roots of unity in C× . It follows that ν
must be cyclic (from class, any subgroup of a cyclic group is cyclic).
a b
4. Let G = {
: a, b ∈ Z with a = ±1}.
0 1
a. Show that G is a group under matrix multiplication.
b. Is G abelian? Explain your answer.
c. Describe all elements of order two in G.
a b
c d
0
a. Let g =
and g =
be elements of G. Then
0 1
0 1
a b c d
ac ad + b
gg 0 =
=
0 1 0 1
0
1
Note ac = ±1 (since a = ±1 and c = ±1) and clearly ad + b ∈ Z. Thus gg 0 ∈ G.
In other words, matrix multiplication defines a binary operation on G. This is an
associative operation (since matrix multiplicationfor matrices
with integer entries is
1 0
associative). The usual 2 × 2 identity matrix I2 =
is an identity element (since
0 1
a b
AI2 = A = I2 A, for any 2 × 2 matrix A with say integer entries). For any g =
0 1
1/a −b/a
in G, the matrix g is invertible with inverse g −1 =
. (Check this!) Since
0
1
a = ±1, we see that 1/a = ±1 and −b/a ∈ Z and thus the matrix g −1 belongs to G.
This proves that G is a group under matrix multiplication.
b. No, G is not abelian. For example,
−1 1 1
0 1 0
1
−1
=
1
0
1
0
1
−1
=
1
0
2
,
1
while
1
0
1
1
−1
0
so that
−1
0
c. An element
1 1
1 0
1
1
6=
1
0
1 −1
1
0
1
.
1
a b
in G has order two if and only if it not the identity and
0 1
2
a b a b
a ab + b
1 0
=
=
.
0 1 0 1
0
1
0 1
So we must have a = ±1 and b(a + 1) = 0. Note b 6= 0 implies a = −1. On the
other hand, b = 0 and a = 1 gives I2 which has order one, not two. Hence a = −1
4
for
of order two. Thus the elements of order two are exactly the matrices
all elements
−1 b
as b varies through Z.
0 1
5. a. State Lagrange’s theorem.
b. Show that every group of prime order is cyclic.
c. The set of ordinary integers Z is a subgroup of the additive group of rational numbers
Q. Show that Z has infinite index in Q (that is, there are infinitely many (left or
right) cosets of Z in Q).
a. Let G be a finite group and let H be a subgroup of G. Then the order of H divides
the order of G.
b. Suppose |G| is prime. Let g ∈ G with g 6= 1. Then hgi = {g n : n ∈ Z} is subgroup of
G. By Lagrange’s Theorem, |hgi| divides |G|. Note |hgi| > 1 as g ∈ hgi (and 1 ∈ hgi).
Since |G| is prime, it follows that |hgi| = |G|. Thus hgi = G, and G is cyclic.
c. Let x, y ∈ Q with 0 ≤ x < 1 and 0 ≤ y < 1. We have x + Z = y + Z if and only if
x − y = n, for some n ∈ Z. Since x and y are each in the interval [0, 1), this is only
possible if x = y. Thus the cosets x + Z, for 0 ≤ x < 1, form an infinite family of
distinct cosets of Z in Q. (In fact, these are all the cosets of Z in Q.)