## § 3.2 Subgroups: Solved problems

23. Find all cyclic subgroups of Z24×.

Solution: You can check that x2 = 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 Z24× = { 1,5,7,11,13,17,19,23 }, use ±1, ±5, ±7, and ±11 to represent the congruence classes.

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24. In Z20×, find two subgroups of order 4, one that is cyclic and one that is not cyclic.

Solution: To find a cyclic subgroup of order 4, we need to check the orders of elements in
Z20× = { ± 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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25. (a) Find the cyclic subgroup of S7 generated by the element (1,2,3)(5,7).

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 S7 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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26. In G = Z21×, show that

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

are subgroups of G.

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 Z21×.

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27. Let G be an abelian group, and let n be a fixed positive integer. Show that

N = { g in G | g = an   for some   a in G }

is a subgroup of G.

Solution: First, the subset N is nonempty since the identity element e can always be written in the form
e = en. Next, suppose that g1 and g2 belong to N. Then there must exist elements a1 and a2 in G with
g1 = a1n and g2 = a2n, and so
g1 g2 = a1n a2n = (a1 a2)n. The last equality holds since G is abelian. Finally, if g belongs to N, with
g = an, then g-1 = (an)-1 = (a-1)n, and so g-1 has the right form to belong to N.

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28. Suppose that p is a prime number of the form p = 2n + 1.

(a) Show that in Zp× the order of [2]p is 2n.

Solution: Since 2n + 1 = p, we have 2n -1 (mod p), and squaring this yields
22n 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
2k 1 (mod p) since 2k - 1 < 2n + 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
| Zp× | = p-1 = 2n, so by part (a) this implies that n is a divisor of 2n-1, and therefore n is a power of 2.

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29. In the multiplicative group C× of complex numbers, find the order of the elements

=   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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30. Let K be the following subset of GL2 (R).

K =

Show that K is a subgroup of GL2 (R).

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 a2 + 2 b2, which is always positive.

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31. Compute the centralizer in GL2 ( R) of the matrix .

Solution: Let A = , and suppose that X = belongs to the centralizer of A in GL2 (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 GL2 ( R) of the matrix is the subgroup

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32. Let G be the subgroup of GL2 (R) defined by G = .
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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