HW1 solution
February 19, 2016
Chapter 0
12
Suppose gcd(5n + 3, 7n + 4) = d. Then by definition d|5n + 3 and d|7n + 4.
From the observation that 7(5n + 3) 5(7n + 4) = 1, we have d|1. Therefore by
definition d=1, which means these two numbers are co-prime.
HW2 solution
February 19, 2016
Chapter 1
3(a)
R03 = R0
3
R90
= R270
3
R180
= R180
3
R270
= R90
H3 = H
V3 =V
D3 = D
D03 = D0
Therefore, the only element in D4 satisfies X 3 = V is V
5
We are going to consider two cases: n is even and n is odd:
.
There are
Math 419/519, Spring 2016, HW3 solutions
Ch3. # 17.
1. U4 (20) = cfw_1, 9, 13, 17.
2. U5 (20) = cfw_1, 11.
3. U5 (30) = U10 (30) = cfw_1, 11.
Next, we shall show that Uk (n) is a subgroup of U (n) provided that k > 1
is a divisor of n. Since U (n) is a fi