[0]5
and
[b]3 [0]3.
There are 8 such elements, which can easily be listed.
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 [0]5
and
[b]3 [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.
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17. 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.
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18. Find two groups G1 and G2 whose direct product G1×G2 has a subgroup that is not of the form H1×H2, for nontrivial 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.
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19. In the group G = Z36×, let
H = { [x] | x 
1 (mod 4) }
and
K = { [y] | y 1 (mod 9) }.
Solution:
It can be shown (as in Problem 3.2.26)
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] x · [19]
for x in G,
the set HK must contain 12 elements.
The group G = Z36×
has 12 elements since
(36)
= 36 (1-1/2)(1-1/3) = 12,
and so HK = G.
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20. Show that if p is a prime number, then the order of the general linear group GLn(Zp) is
(pn -1)(pn - p) · · · (pn - pn-1).
Solution: We need to count the number of ways in which an invertible matrix can be constructed. This is done by noting that we need n linearly independent rows. The first row can be any nonzero vector, so there are pn - 1 choices.
There are pn 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 pn - p.
For the third row, we need to subtract p2, which is the number of vectors in the subspace spanned by the first two rows that we have chosen. Thus there are pn - p2 possibilities for the third row. This argument can be continued, giving the stated result. (A more formal proof could be given by induction.)
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21. Find the order of the element A =
in the group GL3 (C).
Solution:
For any diagonal 3 × 3 matrix we have
.
The order of A must be the least common multiple of the orders
of the diagonal entries i, -1, and -i.
Therefore o(A) = 4.
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22. Compute the centralizer in GL2 ( Z3 ) of the matrix
.
Solution:
Let A = ,
and suppose that
X = 
belongs to the centralizer of A in GL2 (Z3) .
Then XA = AX, and so
XA =
=
and
AX =
=
.
Equating corresponding entries shows that we must have
2a = 2a+c,
a+2b = 2b+d,
2c = 2c, and
c+2d = 2d .
The first equation implies that c = 0,
while the second equation implies that a = d.
It follows that the centralizer in
GL2 ( Z3) of the matrix
is the subgroup
.
Comment:
The centralizer contains 6 elements,
while it follows from Problem 20
in this section
that GL2 ( Z3) has
(32-1)(32-3) = 48 elements.
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23. Compute the centralizer in GL2 ( Z3 ) of the matrix
.
Solution:
Let A = ,
and suppose that
X =
belongs to the centralizer of A in GL2 (Z3) .
Then XA = AX, and so
XA =
=
and
AX =
=
.
Equating corresponding entries shows that we must have
2a+b = 2a+c,
a+b = 2b+d,
2c+d = a+c, and
c+d = b+d .
The first equation implies that c = b,
while the second equation implies that d = a-b.
By listing all possibilities,
you can check that a matrix of this form
is invertible if either a or b is nonzero.
It follows that the centralizer in
GL2 ( Z3) of the matrix
is the subgroup
.
Comment: In this case the centralizer contains 8 of the 48 elements in GL2 ( Z3).
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24. Let H be the following subset of the group G = GL2 (Z5).
.
Solution:
Since in the matrix
there are two choices for m and 5 choices for b,
we will have a total of 10 elements.
The set is closed under multiplication since
and it is certainly nonempty, and so it is a subgroup since the group is finite.
(b) Show that if we let
A = 
and
B = 
,
then BA = A-1B.
Solution:
We have
BA =
=
and
A-1B =
=
.
(Click here
for more details about finding the inverses.)
(c) Show that every element of H can be written uniquely in the form Ai Bj, where i = 0,1,2,3 or 4 and j = 0 or 1.
Solution:
Since
=
,
the cyclic subgroup generated by A
consists of all matrices of the form
.
Multiplying on the right by B will create 5 additional elements,
giving all of the elements in H.
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