Download MATH1022 ANSWERS TO TUTORIAL EXERCISES III 1. G is closed

MATH1022
ANSWERS TO TUTORIAL EXERCISES III
ac
a c
× =
and if a, b, c, and d
b d
bd
1
are all odd, so are ac and bd. The identity 1 is of the correct form since 1 = ,
1
a
b
a
and the inverse of is . Hence G is a group. It is not cyclic, since if is a
b
a
b
proposed generator, and p is an odd prime which doesn’t divide either a or b,
a
then p ∈ G, but p cannot be expressed as a power of .
b
In {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} under × mod 11 we find that 22 = 4, 23 = 8,
24 = 5, 25 = 10, 26 = 9, 27 = 7, 28 = 3, 29 = 6, 210 = 1. Since every element
is equal to a power of 2, this group is cyclic.
1. G is closed under multiplication, since
2. From Exercises II Qus 4, 5 we know that T ET and HEX are groups of
order 12. They are non-abelian since in T ET , for example, the products of the
rotation through 2π/3 fixing A, and which takes B to C to D and back to B,
and that through 2π/3 fixing B, and which takes A to C to D and back to A,
are different half-turns depending on which order they are performed, and in
HEX, the product of two distinct rotations that interchange top and bottom
are different rotations about the vertical axis of symmetry again depending on
the order. The two groups are non-isomorphic since HEX has an element of
order 6 but T ET does not.
3. {2n : n ∈ Z} is a cyclic group, generated by 2 (or by 1/2), since, by
definition, any element of the group is of the form 2n for some integer n.
4. Z2 × Z7 is cyclic, since the element (1,1) has order 14 (we can check
(1, 1), (1, 1) + (1, 1) = (0, 2), (1, 1) + (1, 1) + (1, 1) = (1, 3), . . . and find that
all group elements arise. The reason is that 2 and 7 are coprime). Since Z14
is also cyclic of order 14, and any two cyclic groups of the same order are
isomorphic, Z2 × Z7 ∼
= Z14 . Z24 is not isomorphic to Z4 × Z6 since the latter
has no element of order 24 (all its elements have order a factor of 12).
1 0
0 1
2
ω 0
ω 0
= R,
= S,
0 ω 2
0 ω
ω
= B. The group table for the
0
Let us write
= I,
0 1
0 ω2
0
= A,
= C,
1 0
ω 0
ω2
group in (ii) is then as follows:
I R S A B C
I I R S A B C
R R S I B C A
S S I R C A B
A A C B I S R
B B A C R I S
C C B A S R I
We check by inspection that this matches up perfectly with the group table
given for D3 when we replace I by I, R by R, S by S, A by A, B by B, and
C by C throughout. (Note that we had to label B and C in the reverse order
to that in which they were given in the question, to achieve the isomorphism.)
Hence the groups in (i) and (ii) are isomorphic. Neither is however isomorphic
to the group in (iii), since that is abelian (cyclic actually), and they are not.
5.
6. In Z8 the possible generators are 1, 3, 5, and 7, making 4 in all, and in
Z9 the possible generators are 1, 2, 4, 5, 7 and 8, making 6. (In general, one
just has to count how many numbers between 1 and n − 1 are coprime with
n.)
7. Let G be this set of matrices. We first check
group. It is
that itis a a 0
b 0
ab 0
closed under matrix multiplication, since
=
,
0 a
0 b
0 ab
a 0
the identity matrix lies in G (put a = 1), and the inverse of
is
0 a
1/a 0
. An isomorphism from the group of non-zero reals to G is
0 1/a
a 0
given by θ where θ(a) =
.
0 a
8. We map R to (0, ∞) by θ, where θ(x) = ex . This is 1–1 and onto, and
since θ(x + y) = ex+y = ex ey = θ(x)θ(x), θ is an isomorphism.
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