Introduction to Abstract Algebra MAT02b3
Solutions to selected tutorial problems: Week 5 (Lectures 9 and 10)
Chapter 4
23 Not a solution but a clear hint: After reading the statement of the Fundamental Theorem
on Cyclic Groups you should be able to answer the first question without on Z without any
difficulty.
For the second question on Z consider a subgroup H of Z. Let m be a generator of H. Then
H = hmi = {km | k ∈ Z} = {0, ±m, ±2m, ±3m, . . .}.
29 List all the elements of order 8 in Z8000000 : Recall from the lecture that Nn has φ(8) elements
of order 8 if 8 divides n. Since 8 divides 8000000, we have φ(8) = 4 elements of order 8.
By the Fundamental Theorem on Cyclic Groups, Z8000000 has only one subgroup of order 8,
the subgroup h1000000i = {0, 1000000, 2000000, 3000000, 4000000, 5000000, 6000000, 7000000}.
All elements of order 8 are in this subgroup, and also generators of this subgroup. Now
1000000 has order 8 and is a generator of this subgroup. Since k × 1000000 is a generator
of this subgroup iff gcd(k, 8) = 1, and since 1, 3, 5, 7 are the integers in Z8 that are relatively
prime to 8, the elements of order 8 are 1000000, 3000000, 5000000, 7000000.
32 To draw the subgroup lattice of Z12 first determine the subgroups of Z12 using the full
statement of the Fundamental Theorem for Cyclic Groups: For each divisor of 12 (these are
1, 2, 3, 4, 6, 12) there exists exactly one subgroup, and there are no other subgroups. The
subgroup of order 1 is {0}, the subgroup of order 2 is generated by 12
2 = 6, the subgroup of
order 3 is generated by 12
=
4
and
so
on.
To
decide
which
subgroups
are contained in each
3
other, for example if h6i is contained in h2i, or if h6i is contained in h4i, one needs to observe
that 6 ∈ h2i and so h6i ≤ h2i, and that 6 ∈
/ h4i, so h6i is not contained in h4i. Checking all
possibilities, we obtain the following subgroup lattice (the biggest group, Z12 , is on top, and
the smallest group, {0}, at the bottom):
Z12
h2i
h3i
h4i
h6i
{0}
46 We know from the lecture that the number of elements of order k in a (not necessarily cyclic)
group is a multiple of Φ(k). Hence the number of elements of order 21 is a multiple of
Φ(21) = 12. Since 21600 is the only multiple of 12 among the three numbers, it is the only
possible number of elements of order 21.
54 If G is cyclic and 6 divides |G|, then by a theorem in the lecture G has Φ(6) = 2 elements
of order 6. If G is cyclic and 8 divides |G|, then by a theorem in the lecture G has Φ(8) = 4
elements of order 8. If a is an element of G of order 8, then the other elements of order 8 are
a3 , a5 and a7 .
Chapter 5
1
1 (a) The inverse of α is α. (This is a coincidence.)
(b, c) To determine βα, note that α maps 1 to 2, and β maps 2 to 1, so βα(1) = 1. Also, α
maps 2 to 1, and β maps 1 to 6, so βα(2) = 6. Continuing this, and repeating the same for
αβ, we obtain
1 2 3 4 5 6
1 2 3 4 5 6
βα =
, αβ =
.
1 6 2 3 4 5
6 2 1 5 3 4
2 (a) Write α as a product of disjoint cycles: α = (12345)(678).
(b) Write α as a product of 2-cycles as we did in the lecture: α = (12345)(678) = (15)(14)(13)(12)(68)(67).
3 (a) (1235)(413) = (15)(234).
4 (b) |(147)| = 3.
2