*Solution:*
You can check that x^{2} = 1 for all elements of the group.
Thus each nonzero element generates a subgroup of order 2,
including just the element itself and the identity [1]_{24}.

*Hint:*
Instead of writing **Z**_{24}^{×}
= { 1,5,7,11,13,17,19,23 },
use ±1, ±5, ±7, and ±11
to represent the congruence classes.

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*Solution:*
To find a cyclic subgroup of order 4,
we need to check the orders of elements in

**Z**_{20}^{×} =
{ ± 1, ± 3, ± 7, ± 9 }.

It is natural to begin with [3],
which turns out to have order 4,
and so < [3] > is a cyclic subgroup of order 4.

The element [9] = [3]^{2} has order 2.
It is easy to check that the subset

H = { ± [1], ± [9] } is closed under multiplication.
Since H is a finite, nonempty subset of a known group,
Corollary 3.2.4
implies that it is a subgroup.
Finally, H is not cyclic since no element of H has order 4.

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*Solution:*
The elements (1,2,3) and (5,7) commute, since they are disjoint cycles,
so

((1,2,3)(5,7))^{n}
= (1,2,3)^{n}(5,7)^{n}.
Thus we have

((1,2,3)(5,7))^{2} = (1,3,2),

((1,2,3)(5,7))^{3} = (5,7),

((1,2,3)(5,7))^{4} = (1,2,3),

((1,2,3)(5,7))^{5} = (1,3,2)(5,7),

((1,2,3)(5,7))^{6} = (1).

These elements, together with (1,2,3)(5,7), form the cyclic subgroup generated by (1,2,3)(5,7).

(b) Find a subgroup of S_{7} that contains 12 elements.
You do not have to list all of the elements
if you can explain why there must be 12,
and why they must form a subgroup.

*Solution:*
We only need to find an element of order 12,
since it will generate a cyclic subgroup with 12 elements.
Since the order of a product of disjoint cycles
is the least common multiple of their lengths,
the element

(1,2,3,4)(5,6,7) has order 12.

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H = { [x]_{21} | x
1 (mod 3) }
and
K = { [x]_{21} | x
1 (mod 7) }

*Solution:*
The subset H is finite and nonempty (it certainly contains [1]_{21}),
so by Corollary 3.2.4
it is enough
to show that H is closed under multiplication.
If [x]_{21} and [y]_{21} belong to H, then

x 1 (mod 3) and
y 1 (mod 3),
so it follows that

xy 1 (mod 3),
and therefore

[x]_{21} · [y]_{21} = [xy]_{21} belongs to H.

A similar argument shows that K is a subgroup of
**Z**_{21}^{×}.

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N = { g in G | g = a^{n} for some a in G }

*Solution:*
First, the subset N is nonempty since the identity element e
can always be written in the form

e = e^{n}.
Next, suppose that g_{1} and g_{2} belong to N.
Then there must exist elements a_{1} and a_{2} in G with

g_{1} = a_{1}^{n} and
g_{2} = a_{2}^{n}, and so

g_{1} g_{2}
= a_{1}^{n} a_{2}^{n}
= (a_{1} a_{2})^{n}.
The last equality holds since G is abelian.
Finally, if g belongs to N, with

g = a^{n}, then
g^{-1}
= (a^{n})^{-1}
= (a^{-1})^{n},
and so g^{-1} has the right form to belong to N.

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(a) Show that in **Z**_{p}^{×}
the order of [2]_{p} is 2n.

*Solution:*
Since 2^{n} + 1 = p, we have
2^{n} -1 (mod p),
and squaring this yields

2^{2n} 1 (mod p).
Thus the order of [2] is a divisor of 2n,

and for any proper divisor k of 2n we have
k n, so

2^{k} 1 (mod p)
since 2^{k} - 1 < 2^{n} + 1 = p.

This shows that [2] has order 2n.

(b) Use part (a) to prove that n must be a power of 2.

*Solution:*
The order of [2] is a divisor of

| **Z**_{p}^{×} | = p-1 = 2^{n},
so by part (a) this implies that n is a divisor of 2^{n-1},
and therefore n is a power of 2.

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= and = .

*Solution:*
It is probably easiest to change these complex numbers
from rectangular coordinates into polar coordinates.
(See Appendix A.5 for a discussion of the properties of complex numbers.)
Each of the numbers has magnitude 1, and you can check that

= cos ( 3 / 4 ) + i sin ( 3 / 4 ) and = cos ( 5 / 4 ) + i sin ( 5 / 4 ).

We can use Demoivre's Theorem to compute powers of complex numbers. It follows from this theorem that

^{8}
=
( cos ( 3 / 4 ) +
i sin ( 3 / 4 ))^{8}
=
cos ( 6 ) +
i sin ( 6 )
= 1,

and so
has order 8 in **C**^{×}.
A similar argument shows that
also has order 8.

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K =

Show that K is a subgroup of GL*Solution:*
The closure axiom holds since

The identity matrix belongs K, and

so K is closed under taking inverses. Click here for more details about inverses of 2 × 2 matrices.

*Comment:*
We don't need to worry about the condition
ad - bc 0,
since for any element in H the determinant
is a^{2} + 2 b^{2}, which is always positive.

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*Solution:*
Let A = ,
and suppose that
X =
belongs to the centralizer of A in GL_{2} (**R**) .
Then we must have XA = AX,
so doing this calculation shows that

XA = =

and

AX = = .

Equating corresponding entries shows that we must have

a+b = a,

0 = b,

c+d = a, and

0 = b.

The first, second, and last equations just say that b = 0,
while the third equation says that c = a-d.
On the other hand,
any matrix of this form commutes with A,
so the centralizer in GL_{2} ( **R**) of the matrix
is the subgroup

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Let A = and B = .

Find the centralizers C(A) and C(B), and show that C(A) C(B) = Z(G), where Z(G) is the center of G.

*Solution:*
Suppose that
X =
belongs to C(A) in G.
Then we must have XA = AX,
and doing this calculation shows that

XA = =

and

AX = = .

Equating corresponding entries shows that we must have m+b = b+1, and so m = 1.
On the other hand, any matrix of the form
commutes with A, and so

C(A) = .

Now suppose that X = belongs to C(B) in G. Then we must have XB = BX, and doing this calculation shows that

XB = =

and

BX = = .

Equating corresponding entries, we get b = 0.
On the other hand, any matrix of the form
commutes with A, and so

C(B) =
.

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.

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