文档介绍：page 1 of Solutions Chapters 1-5
SOLUTIONS CHAPTERS 1-5
Section
1. Under multiplication, the positive integers form a monoid but not a group, and the
positive even integers form a semigroup but not a monoid.
2. With |a| denoting the order of a, we have |0| =1, |1| =6, |2| =3, |3| =2, |4| = 3, and
|5| =6.
3. There is a subgroup of order 6/d for each divisor d of 6. We have Z6 itself (d=1),
{0}(d =6), {0, 2, 4}(d = 2), and {0, 3}(d =3).
4. S forms a group under addition. The inverse operation is subtraction, and the zero
matrix is the additive identity.
5. S∗ does not form a group under multiplication, since a nonzero matrix whose determinant
is 0 does not have a multiplicative inverse.
6. If d is the smallest positive integer in H, then H consists of all multiples of d. For if
x ∈ H we have x = qd + r where 0 ≤ r<d. But then r = x − qd ∈ H,sor must be 0.
7. Consider the rationals with addition mod 1, in other words identify rational numbers
that differ by an integer. Thus, for example, 1/3=4/3=7/3, etc. The group is infinite,
but every element generates a finite subgroup. For example, the subgroup generated by 1/3
is {1/3, 2/3, 0}.
mn m n n m
8. (ab) =(a ) (b ) = 1, so the order of ab divides mn. Thus |ab| = m1n1 where
m1divides m and n1 divides n. Consequently,
am1n1 bm1n1 = 1 (1)
mn1
If m = m1m2, raise both sides of (1) to the power m2 to get b = 1. The order of b,
namely n, must divide mn1, and since m and n are relatively prime, n must divide n1. But
n1 divides n, hence n = n1. Similarly, if n = n1n2 we raise both sides of (1) to the power
n2 and conclude as above that m = m1. But then |ab| = m1n1 = mn, as asserted.
If c belongs to both <a>and <b>then since c isapowerofa and also a power of
b, we have cm = cn = 1. But then the order of c divides both m and n, and since m and n
are relatively prime, c has order 1, ., c =1.
9. Let |a| = m, |b| = [m, n] is the mon multiple, and (m, n) the greatest
common divisor, of m and