*Solution:*
By Proposition 3.3.4(b),
the order of an element
([a]_{5},[b]_{3}) in
**Z**_{5}×**Z**_{3}
is the least common multiple of the orders of the components.
Since [1]_{5},[2]_{5},[3]_{5},[4]_{5}
have order 5 in **Z**_{5}
and [1]_{3},[2]_{3} have order 3 in **Z**_{3},
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
**Z**_{5}×**Z**_{3}.

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*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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*Solution:*
In **Z**_{2}×**Z**_{2}, the element (1,1) has order 2,
so it generates a cyclic subgroup that does not have the required form.

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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 3.2.26)
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 = **Z**_{36}^{×}
has 12 elements since

(36)
= 36 (1-1/2)(1-1/3) = 12,
and so HK = G.

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(p^{n} -1)(p^{n} - p) · · ·
(p^{n} - p^{n-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 p^{n} - 1 choices.

There are p^{n} 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 p^{n} - p.

For the third row,
we need to subtract p^{2},
which is the number of vectors in the subspace
spanned by the first two rows that we have chosen.
Thus there are p^{n} - p^{2} 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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*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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*Solution:*
Let A = ,
and suppose that
X =
belongs to the centralizer of A in GL_{2} (**Z**_{3}) .
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
GL_{2} ( **Z**_{3}) of the matrix
is the subgroup

.

*Comment:*
The centralizer contains 6 elements,
while it follows from Problem 20
in this section
that GL_{2} ( **Z**_{3}) has

(3^{2}-1)(3^{2}-3) = 48 elements.

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*Solution:*
Let A = ,
and suppose that
X =
belongs to the centralizer of A in GL_{2} (**Z**_{3}) .
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
GL_{2} ( **Z**_{3}) of the matrix
is the subgroup

.

*Comment:*
In this case the centralizer contains 8
of the 48 elements in GL_{2} ( **Z**_{3}).

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H = .

(a) Show that H is a subgroup of G with 10 elements.

*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^{-1}B.

*Solution:*
We have
BA =
=
and

A^{-1}B =
=
.

(Click here
for more details about finding the inverses.)

(c) Show that every element of H can be written uniquely in the form
A^{i} B^{j},
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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