Download 3.3 Constructing Examples

3.3
3.3
J.A.Beachy
1
Constructing Examples
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
From this point on, if the modulus n is fixed throughout the problem, we will write a rather
than [a]n for elements of Zn .
19. Show that Z5 × Z3 is a cyclic group, and list all of the generators of the group.
Solution: By Proposition 3.3.4 (b), the order of an element ([a]5 , [b]3 ) in Z5 ×Z3 is the
least common multiple of the orders of the components. Since [1]5 , [2]5 , [3]5 , [4]5 have
order 5 in Z5 and [1]3 , [2]3 have order 3 in Z3 , the element ([a]5 , [b]3 ) is a generator if
and only if [a]5 6= [0]5 and [b]3 6= [0]3 . There are 8 such elements, which can easily be
listed.
Comment: The other 7 elements in the group will have at least one component equal
to zero. There are 4 elements of order 5 (with [0]3 as the second component) and 2
elements of order 3 (with [0]5 as the first component). Adding the identity element
to the list accounts for all 15 elements of Z5 × Z3 .
20. Find the order of the element ([9]12 , [15]18 ) in the group Z12 × Z18 .
Solution: Since gcd(9, 12) = 3, we have o([9]12 ) = o([3]12 ) = 4. Similarly, o([15]18 ) =
o([3]18 ) = 6. Thus the order of ([9]12 , [15]18 ) is lcm[4, 6] = 12.
21. Find two groups G1 and G2 whose direct product G1 × G2 has a subgroup that is not
of the form H1 × H2 , for subgroups H1 ⊆ G1 and H2 ⊆ G2 .
Solution: In Z2 × Z2 , the element (1, 1) has order 2, so it generates a cyclic subgroup
that does not have the required form.
22. In the group G = Z×
36 , let H = {[x] | x ≡ 1 (mod 4)} and K = {[y] | y ≡ 1 (mod 9)}.
Show that H and K are subgroups of G, and find the subgroup HK.
Solution: It can be shown (as in Problem 3.2.35) that the given subsets are subgroups.
A short computation shows that H = {1, 5, 13, 17, 25, 29} and K = {1, 19}. Since
x · 1 6= x · 19 for x ∈ G, the set HK must contain 12 elements, and so HK = G since
1 2
G = Z×
36 has ϕ(36) = 36 · 2 · 3 = 12 elements.
23. Let F be a field, and let H be the subset of GL2 (F ) consisting of all invertible upper
triangular matrices. Show that H is a subgroup of GL2 (F ).
a11 a12
b11 b12
a11 b11 a11 b12 + a12 b22
Solution: Since
=
, and
0 a22
0 b22
0
a22 b22
a11 a22 6= 0 and b11 b22 6= 0 together imply that (a11 b11 )(a22 b22 ) 6= 0, it follows that
H is closed under multiplication. It is also closed under formation of inverses, since
−1 −1
a11 −a−1
a12 a−1
a11 a12
11
22
. The identity matrix is certainly in H.
=
0 a22
0
a−1
22
3.3
J.A.Beachy
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Comment: This comment is directed to the readers who remember some linear algebra.
We will prove a more general result.
Proposition: For a field F , the set H of upper triangular matrices in GLn (F ) is a
subgroup of GLn (F ).
Proof: Of course, the identity matrix is upper triangular, so it belongs to H.
Suppose that A = [aij ] and B = [bij ] belong to H. This condition is expressed by
that fact that aij = 0 if i > j and bij = 0 if i > j.
entries of the product matrix
PThe
n
[cij ] = [aij ][bij ] are given by the formula cij = k=1 aik bkj . Now look at cij when
i > j. Since aik = 0 for all k < i and bkj = 0 for all k > j, we must have aik bkj = 0
for all 1 ≤ k ≤ n because i > j. This argument shows that the set of upper triangular
matrices in GLn (F ) is closed under multiplication.
Finally, we need to show that the inverse of an upper triangular matrix is again upper
1
adj(A), where adj(A) is the adjoint of A, and
triangular. Recall that A−1 =
det(A)
that adj(A) is the transpose of the matrix of cofactors of A. It can then be checked
that the adjoint of an upper triangular matrix is again upper triangular.
Second proof: We will use induction (which
doesn’t
require using the adjoint). If an
A B
invertible matrix has the block form
, then A and C must be invertible
0 C
−1 −1
A B
A
−A−1 BC −1
matrices, and a direct calculation shows that
=
.
0
C −1
0 C
The induction begins with the 2 × 2 case proved above. Now suppose
that [aij ] is
A X
an n × n upper triangular matrix. We can write [aij ] =
, where A is an
0 ann
(n − 1) × (n − 1) upper triangular matrix, and X is an (n − 1) × 1 column. Given the
induction hypothesis that the inverse of an (n − 1)× (n − 1) upper triangular
matrix is
−1 −A−1 Xa−1
A
nn
again upper triangular, it is clear that [aij ]−1 =
is also upper
0
a−1
nn
triangular.
24. Let p be a prime number.
(a) Show that the order of the general linear group GL2 (Zp ) is (p2 − 1)(p2 − p).
Hint: Count the number of ways to construct two linearly independent rows.
Solution: We need to count the number of ways in which an invertible 2 × 2 matrix
can be constructed with entries in Zp . This is done by noting that we need 2 linearly
independent rows. The first row can be any nonzero vector, so there are p2 −1 choices.
There are p2 possibilities for the second row, but to be linearly independent of the
first row, it cannot be a scalar multiple of that row. Since we have p possible scalars,
we need to omit the p multiples of the first row. Therefore the total number of ways
to construct a second row independent of the first is p2 − p.
(b) Show that the subgroup of GL2 (Zp ) consisting of all invertible upper triangular
matrices has order (p − 1)2 p.
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J.A.Beachy
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Solution: An upper triangular 2 × 2 matrix has nonzero determinant if and only if
the elements on the main diagonal are nonzero. There are (p − 1)2 choices for these
entries. Since the third entry can be any element of Zp , there are p choices for this
entry.


i
0
0
25. Find the order of the element A =  0 −1
0  in the group GL3 (C).
0
0 −i
Solution: For any diagonal 3 × 3

a 0
 0 b
0 0
matrix we
n 
0
0  =
c
have

an 0 0
0 bn 0  ,
0 0 cn
It follows immediately that the order of A is the least common multiple of the orders
of the diagonal entries i, −1, and −i. Thus o(A) = 4.
26. Let G be the subgroup of GL2 (R) defined by
m b m 6= 0 .
G=
0 1 1 1
−1 0
Let A =
and B =
. Find the centralizers C(A) and C(B), and
0 1
0 1
show that C(A) ∩ C(B) = Z(G), where Z(G) is the center of G.
m b
Solution: Suppose that X =
belongs to C(A) in G. Then we must have
0 1
XA = AX, and doing this calculation shows that
m m+b
m b
1 1
1 1
m b
m b+1
=
=
=
.
0
1
0 1
0 1
0 1
0 1
0
1
Equating corresponding entries shows that we must have m + b = b + 1, and so m = 1.
On the other hand, any matrix of this form commutes with A, and so
1 b b∈R .
C(A) =
0 1 m b
Now suppose that X =
belongs to C(B). Then XB = BX, and so
0 1
−m b
m b
−1 0
−1 0
m b
−m −b
=
=
=
.
0 1
0 1
0 1
0 1
0 1
0
1
Equating corresponding entries shows that
m 0
C(B) =
0 1
we must have b = 0, and so
0 6= m ∈ R .
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J.A.Beachy
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This shows that C(A) ∩ C(B) is the identity matrix, and since any element in the
center of G must belong to C(A) ∩ C(B), our calculations show that the center of G
is the trivial subgroup, containing only the identity element.
1 −1
27. Compute the centralizer in GL2 (Z3 ) of the matrix
.
0
1
1 −1
a b
Solution: Let A =
, and suppose that X =
belongs to the cen0
1
c d
a −a + b
a b
1 −1
tralizer of A in GL2 (Z3 ). Then XA = AX, and so
=
=
c −c + d
c d
0 1
1 −1
a b
a−c b−d
=
. Equating corresponding entries shows that
0
1
c d
c
d
we must have a = a − c, −a + b = b − d, and −c + d = d. The first equation implies that c = 0, while the second equation
implies that a = d. It fol1 −1
lows that the centralizer in GL2 (Z3 ) of the matrix
is the subgroup H =
0
1
a b a, b ∈ Z3 and a 6= 0 . Note that the modulus 3 has played no role here.
0 a Comment: The centralizer contains 6 elements, while it follows from Problem 24 that
GL2 (Z3 ) has (32 − 1)(32 − 3) = 48 elements. H =
1 0
0 1
,a =
1 1
0 1
−1
1
−1 −1
−1
0
1 −1
, a2 b =
, ab =
,b =
, a2 =
0 −1
0 −1
0 −1
0
1
Note that the multiplication table for H will look like that of S3 , since it has order 6
and is not abelian (ab 6= ba). (See Exercises 15-16 in the text.)
1 −1
.
28. Compute the centralizer in GL2 (Z3 ) of the matrix
−1
0
1 −1
a b
Solution: Let A =
, and suppose that X =
belongs to the
−1
0
c d
a − b −a
centralizer of A in GL2 (Z3 ). Then XA = AX, and so
=
c − d −c
a b
1 −1
1 −1
a b
a−c b−d
=
=
. Equating correc d
−1
0
−1
0
c d
−a
−b
sponding entries shows that we must have a − b = a − c, −a = b − d, c − d = −a, and
−c = −b. The first equation implies that c = b, while the second equation
implies
1 −1
that d = a + b. It follows that the centralizer in GL2 (Z3 ) of the matrix
−1
0
a
b a, b ∈ Z3 and a 6= 0 or b 6= 0 .
is the subgroup
b a+b Comment: In this case the centralizer contains 8 of the 48 elements in GL2 (Z3 ). H =
1 0
0 1
−1 1
−1 −1
0 −1
−1
0
1 −1
1
1
0 1
,
,
,
,
,
,
,
1 0
−1
1
−1 −1
0 −1
−1
0
1 −1
1 1
Note that H is cyclic, as given above.
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29. Let H be the following subset of the group G = GL2 (Z5 ).
m b
H=
∈ GL2 (Z5 ) m, b ∈ Z5 , m = ±1
0 1
(a) Show that H is a subgroup of G with 10 elements.
m b
Solution: Since in the matrix
there are two choices for m and 5 choices
0 1
for
totalof 10 elements.
b, we will
have a The set is closed under multiplication since
±1 b
±1 c
±1 b ± c
=
, and it is certainly nonempty, and so it is a
0 1
0 1
0
1
subgroup since the group is finite.
1 1
−1 0
(b) Show that if we let A =
and B =
, then BA = A−1 B.
0 1
0 1
−1 0
1 1
−1 −1
Solution: We have BA =
=
and
0 1
0 1
0
1
1 −1
−1 0
−1 −1
−1
A B=
=
.
0
1
0 1
0
1
(c) Show that every element of H can be written uniquely in the form Ai B j , where
0 ≤ i < 5 and 0 ≤ j < 2.
1 b
1 c
1 b+c
Solution: Since
=
, the cyclic subgroup generated by
0 1
0 1
0
1
1 b
A consists of all matrices of the form
. Multiplying on the right by B will
0 1
create 5 additional elements, giving all of the elements in H.
30. Let H and K be subgroups of the group G. Prove that HK is a subgroup of G if and
only if KH ⊆ HK.
Note: This result strengthens Proposition 3.3.2.
Solution: First assume that HK is a subgroup of G. If k ∈ K and h ∈ H, then we
have k = ek ∈ HK and h = he ∈ HK, so the product kh must belong to HK. This
shows that KH ⊆ HK.
Conversely, suppose that KH ⊆ HK. It is clear that e = ee ∈ HK since H and K are
subgroups. To show closure, let g1 , g2 ∈ HK. Then there exist elements h1 , h2 ∈ H
and k1 , k2 ∈ K with g1 = h1 k1 and g2 = h2 k2 . Since k1 h2 ∈ KH, by assumption there
exist k1′ ∈ K and h′2 ∈ H with k1 h2 = h′2 k1′ . Thus g1 g2 = h1 k1 h2 k2 = h1 h′2 k1′ k2 ∈ HK
since both H and K are closed under multiplication. Finally, (g1 )−1 = (h1 k1 )−1 =
−1
−1
k1−1 h−1
1 ∈ HK since k1 ∈ K and h1 ∈ H and KH ⊆ HK.
ANSWERS AND HINTS
31. What is the order of ([15]24 , [25]30 ) in the group Z24 × Z30 ? What is the largest
possible order of an element in Z24 × Z30 ?
3.3
J.A.Beachy
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Answer: The element ([15]24 , [25]30 ) has order lcm[8, 6] = 24. The largest possible
order of an element in Z24 × Z30 is lcm[24, 30] = 120,
32. Find the order of each element of the group Z4 × Z×
4.
Answer: (0, 1) has order one; (2, 1), (0, 2), (2, 2) have order two; the remaining 4
elements have order four.
33. Check the order of each element of the quaternion group in Example 3.3.7 by
using the matrix form
of the element.
−1
0
i
0
0 1
0 i
Answer: Order 2:
Order 4: ±
,±
,±
0 −1
0 −i
−1 0
i 0
×
35. Let G = Z×
10 × Z10 .
(a) If H = h(3, 3)i and K = h(3, 7)i, list the elements of HK.
Answer: HK = {(1, 1), (3, 3), (9, 9), (7, 7), (3, 7), (9, 1), (7, 3), (1, 9)}
(b) If H = h(3, 3)i and K = h(1, 3)i, list the elements of HK.
Answer: HK = G.
37. In G = Z×
15 , find subgroups H and K with |H| = 4, |K| = 2, HK = G, and
H ∩ K = {1}.
Answer: Answer: Let H = h2i and K = h−1i.
1 2
1 2
41. Find the orders of
and
in GL2 (Z5 ).
3 4
4 3
Answer: The first matrix has order 8; the second matrix is not in GL2 (Z5 ).
m b
42. Let K =
∈ GL2 (Z5 ) m, b ∈ Z5 , m 6= 0 .
0 1
(b) Show, by finding the order of each element in K, that K has elements of order 2
and 5, but no element of order 10. 1 b
Answer: The 4 elements of the form
, with b 6= 0, each have order 5; the 5
0 1
4 b
2 b
3 b
elements
each have order 2; the 10 elements
or
each have
0 1
0 1
0 1
order 4.


0 0 1 0
 0 0 0 1 

45. Find the cyclic subgroup of GL4 (Z2 ) generated by 
 0 1 1 0 .
1 0 1 0



 
 

1 0 0 0
0 0 1 0
0 1 1 0
* 0 0 1 0 + 


 0 0 0 1 
 
 

 =  0 1 0 0 ,  0 0 0 1 ,  1 0 1 0 ,
Answer: 
 0 1 1 0 
 0 0 1 0   0 1 1 0   0 1 1 1 



1 0 1 0
0 0 0 1
1 0 1 0
0 1 0 0

 
 

0 1 1 1
1 1 0 1
1 0 0 1 

 0 1 0 0   0 0 0 1   1 0 1 0 

, 
, 

 1 1 0 1   1 0 0 1   1 0 0 0 


0 0 0 1
1 0 1 0
0 1 0 0
47. List the orthogonal matrices in GL2 (Z3 ) and find the order of each one.
3.3
J.A.Beachy
1 0
0 1
Answer: Case 1: ±
,±
have determinant 1.
0 1
−1 0
1
0
0 1
Case 2: ±
,±
have determinant −1.
0 −1
1 0
The symmetric orthogonal matrices have order 2 (except for I2 ), and the remaining
two matrices have order 4.
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