MODEL ANSWERS TO THE THIRD HOMEWORK
1. (b) Circles centre the origin.
(c) The real line union
∞
, where the number
m
∈
R
∪
{
∞
}
represents
the slope.
2. Chapter 2, Section 4: 9.
[0] = 0 +
H
=
{
[0]
,
[4]
,
[8]
,
[12]
}
[1] = 1 +
H
=
{
[1]
,
[5]
,
[9]
,
[13]
}
[2] = 2 +
H
=
{
[2]
,
[6]
,
[10]
,
[14]
}
[3] = 3 +
H
=
{
[3]
,
[7]
,
[11]
,
[15]
}
2. Chapter 2, Section 4: 10. Four.
2. Chapter 2, Section 4: 13. First we write down the elements of
U
18
.
These will be the left cosets, generated by integers coprime to 18. Of
the integers between 1 and 17, those that are coprime are 1, 3, 5, 7,
11, 13 and 17.
Thus the elements of
U
18
are [1], [5], [7], [11], [13] and [17]. We calculate
the order of these elements.
[1] is the identity, it has order one.
Consider [5].
[5]
2
= [5
2
] = [25] = [7]
,
as 25 = 7
mod 18. In this case
[5
3
] = [5][5
2
] = [5][7] = [35] = [17]
,
as 35 = 17
mod 18.
We could keep computing. But at this point, we can be a little more
sly. By Lagrange the order of
g
= [5] divides the order of
G
. As
G
has
order 6, the order of [5] is one of 1, 2, 3, or 6. As we have already seen
that the order is not 1, 2 or 3, by a process of elimination, we know
that [5] has order 6.
As [17] = [5]
3
, [17]
2
= [5]
6
= [1].
So [17] has order 2.
Similarly, as
[7] = [5]
2
, [7]
3
= [5]
6
= [1]. So the order of [7] divides 3. But then the
order of [7] is three.
It remains to compute the order of [11] and [13]. Now one of these is
the inverse of [5]. It must then have order six. The other would then be
[5]
4