## § 3.5 Cyclic Groups: Solved problems

20. Show that the three groups Z6, Z9×, and Z18× are isomorphic to each other.

Solution: First, we have | Z9× | = 6, and | Z18× | = 6.
In Z9×, we have
22=4,
23=8 1,
and so [2] must have order 6, showing that
Z9× is cyclic of order 6. Our theorems tell us that
Z9× Z6. In Z18×,
52 7,
53 17 1,
and so [5] must have order 6, showing that
Z18× is cyclic of order 6. Our theorems tell us that
Z18× Z6. Thus all three groups are isomorphic.




21. Is Z4 × Z10 isomorphic to Z2 × Z20?

Solution: It follows from Theorem 3.5.4 that
Z10 Z2 × Z5, and that Z20 Z4 × Z5.
It then follows from Problem 3.4.23 that
Z4 × Z10 Z4 × Z2 × Z5, and Z2 × Z20 Z2 × Z4 × Z5. Finally,
it is possible to show that the obvious mapping from
Z4 × Z2 × Z5 onto Z2 × Z4 × Z5 is an isomorphism. Therefore
Z4 × Z10 Z2 × Z20.




22. Is Z4 × Z15 isomorphic to Z6 × Z10?

Solution: As in Problem 21, Z4 × Z15 Z4 × Z3 × Z5, and
Z6 × Z10 Z2 × Z3 × Z2 × Z5. The two groups are not isomorphic
since the first has an element of order 4, while the second does not.




23. Give the lattice diagram of subgroups of Z100.

Solution: The subgroups correspond to the divisors of 100, and are given below:

Note that n Z100 is used to mean all multiples of n in Z100.




24. Find all generators of the cyclic group Z28.

Solution: By Proposition 3.5.3(b), the generators correspond to the numbers less than 28 and relatively prime to 28. The Euler -function allows us to compute how many there are:
(28) = (1/2) · (6/7) · 28 = 12.
The list of generators is { ± 1, ± 3, ± 5, ± 9, ± 11, ± 13 }.




25. In Z30, find the order of the subgroup < [18]30 >; find the order of < [24]30 >.

Solution: Using Proposition 3.5.3(a), we first find that
gcd (18,30) = 6. Then < [18]30 > = < [6]30 >,
and so the subgroup has 30 / 6 = 5 elements.

Similarly, < [24]30 > = < [6]30 >, and so we actually have
< [24]30 > = < [18]30 >.




26. Prove that if G1 and G2 are groups of order 7 and 11, respectively, then the direct product G1 × G2 is a cyclic group.

Solution: Since 7 and 11 are primes, the groups are cyclic. If a has order 7 in G1 and b has order 11 in G2, then (a,b) has order lcm [7,11] = 77 in G1 × G2. Thus G1 × G2 is cyclic since it has an element whose order is equal to the order of the group.




27. Show that any cyclic group of even order has exactly one element of order 2.

Solution: If G is cyclic of order 2n, for some positive integer n, then it follows from Theorem 3.5.2 that G is isomorphic to Z2n. Since isomorphisms preserve orders of elements, we only need to answer the question in Z2n. In that group, the elements of order 2 are the nonzero solutions to the congruence 2x 0 (mod 2n), and since the congruence can be rewritten as x 0 (mod n), we see that [n]2n is the only element of order 2 in Z2n.




28. Use the the result in Problem 27 to show that the multiplicative groups Z15× and Z21× are not cyclic groups.

Solution: In Z15×, both [-1]15 and [4]15 are easily checked to have order 2. In Z21×, we have [8]212 = [64]21 = [1]21, and so [8]21 and [-1]21 have order 2.




29. Find all cyclic subgroups of the quaternion group. Use this information to show that the quaternion group cannot be isomorphic to the subgroup of S4 generated by (1,2,3,4) and (1,3).

Solution: The quaternion group Q = { ± 1, ± i, ± j, ± k } is defined in Example 3.3.7 of the text (see page 108 or the online table). The elements satisfy the following identities:

i2 = j2 = k2 = -1

and

i j = k,   j k = i,   k i = j,   j i =-k,   k j =-i,   i k =-j.

The cyclic subgroups

< -1 > = { ± 1 },
< ± i > = { ± 1, ± i },
< ± j > = { ± 1, ± j }, and
< ± k > = { ± 1, ± k }

can be found by using the given identities. For example,
i2 = -1, i3 = i2 i = - i, and i4 = i2 i2 = (-1)2 = 1.

In S4, let (1,2,3,4) = a and (1,3) = b.
Since a is a cycle of length 4, it has order 4, with
a2 = (1,3)(2,4) and a3 = a-1 = (1,4,3,2).
To find the subgroup generated by a and b, we have

ab = (1,2,3,4)(1,3) = (1,4)(2,3),
a2b = (1,3)(2,4)(1,3) = (2,4), and
a3b = (1,4,3,2)(1,3) = (1,2)(3,4).

On the other side, we have

ba = (1,3)(1,2,3,4) = (1,2)(3,4) = a3 b,
ba2 = (1,3)(1,3)(2,4) = (2,4) = a2 b, and
ba3 = (1,3)(1,4,3,2) = (1,4)(2,3) = ab.

This shows that the subgroup generated by a and b consists of the 8 elements
{ e, a, a2, a3, b, ab, a2 b, a3 b }.
Furthermore, from the cycle structures of the elements we can see that the only cyclic subgroup of order 4 is the one generated by a (and a3). In any isomorphism, cyclic subgroups would correspond to cyclic subgroups, and so it is impossible for this group to be isomorphic to the quaternion group, which has 3 cyclic subgroups of order 4.




30. Prove that if p and q are different odd primes, then Zpq× is not a cyclic group.

Solution: We know that [-1]pq has order 2, so by Problem 27 it is enough to find one other element of order 2. The Chinese remainder theorem (Theorem 1.3.6) states that the system of congruences
x 1 (mod p) and x -1 (mod q)
has a solution [a]pq, since p and q are relatively prime. Because q is an odd prime,
[-1]pq is not a solution, so [a]pq [-1]pq. But
a2 1 (mod p) and a2 1 (mod q), so a2 1 (mod pq) since p and q are relatively prime, and thus [a]pq has order 2.







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