Homework SourseThe coordinates for each point is according to the axes Calc
Solution
Ans-
Let
U
(20) =
f
1
;
3
;
7
;
9
;
11
;
13
;
17
;
19
g
b e the group of p ositive integers coprime
to 20 whose op eration is multiplication mo d 20. Then
U
5
(20) =
f
x
2
U
(20) :
x
= 1
mo d
5
g
=
f
1
;
11
g
:
This is a subgroup of
U
(20)
. It is normal b ecause
U
(20)
is Ab elian. Since
j
U
(20)
j
= 8
and
j
U
5
(20)
j
= 2
, by Corollary 1 on page 142, there are
8
2
= 4
distinct cosets of
U
5
(20)
. They are:
1
U
5
(20) =
U
5
(20) =
f
1
;
11
g
=
f
11
;
1
g
=
f
11
;
11
11
g
= 11
U
5
(20)
3
U
5
(20) =
f
3
;
33
g
=
f
3
;
13
g
=
f
13
;
3
g
=
f
13
;
13
11
g
= 13
U
5
(20)
7
U
5
(20) =
f
7
;
77
g
=
f
7
;
17
g
=
f
17
;
7
g
=
f
17
;
17
11
g
= 17
U
5
(20)
9
U
5
(20) =
f
9
;
99
g
=
f
9
;
19
g
=
f
19
;
9
g
=
f
19
;
19
11
g
= 19
U
5
(20)
Rememb er, we are working mo d 20. Hence
U
(20)
=U
5
(20) =
ff
1
;
11
g
;
f
3
;
13
g
;
f
7
;
17
g
;
f
9
;
19
gg
:
We want to write our cosets as
aU
5
(20)
. Let us use values of
a
b etween 1 and
9:
U
(20)
=U
5
(20) =
f
U
5
(20)
;
3
U
5
(20)
;
7
U
5
(20)
;
9
U
5
(20)
g
:
Multiplication in
U
(20)
=U
5
(20)
works like the multiplication on
U
(20)
:
aU
5
(20)
bU
5
(20) = (
ab
)
U
5
(20)
2
(make sure to reduce
ab
mo d 20 as well).
When making the Cayley table, we only want to use
U
5
(20)
;
3
U
5
(20)
;
7
U
5
(20)
;
9
U
5
(20)
.
So, if we were multiplying and somehow got
13
U
5
(20)
, we should replace it by
the same coset
3
U
5
(20)
.
U
5
(20)
3
U
5
(20)
7
U
5
(20)
9
U
5
(20)
U
5
(20)
U
5
(20)
3
U
5
(20)
7
U
5
(20)
9
U
5
(20)
3
U
5
(20)
3
U
5
(20)
9
U
5
(20)
U
5
(20)
7
U
5
(20)
7
U
5
(20)
7
U
5
(20)
U
5
(20)
9
U
5
(20)
3
U
5
(20)
9
U
5
(20)
9
U
5
(20)
7
U
5
(20)
3
U
5
(20)
U
5
(20
