Math 0430
Solutions to Homework 6
August 2, 2012
Section 19 Number 27
Let D be a subdomain of the integral domain D. This means that D is
a subring of D which is itself an integral domain. We rst show that the
identity element 1 of D is also the identity
Math 0430
Solutions to Homework 5
July 29, 2012
Section 13 Number 25
Let be any homomorphism from Z into Z. Then for any n Z we have
(n) = (n 1) = n(1). Thus, the range of consists of all integer multiples
of (1). This gives all of Z if and only if (1) =
Math 0430
Solutions to Homework 4
July 21, 2012
Section 9 Number 27
(a) First, note that any cycle of length m can be written as a product of
m 1 transpositions:
(a1 , . . . , am ) = (a1 , am ) (a1 , a3 )(a1 , a2 )
Now let Sn with not the identity. (The c
Math 0430
Solutions to Homework 3
July 12, 2012
Section 7 Number 1
2, 3 contains 3 2 = 1, so it contains 1 . Therefore, 2, 3 = Z12 .
Section 7 Number 2
Since 2 = 6 4 is in 4, 6 , the subgroup 2 is contained in 4, 6 . On the
other hand, every member of 4,
Math 0430
Solutions to Homework 2
July 10, 2012
Section 4 Number 28
The notation used in this problem is ambiguous, since the symbol is used
for two dierent things. (Does e denote the inverse of the identity element e
of the group G, or does it denote the
Math 0430
Solutions to Homework 1
July 1, 2012
Section 2 Number 2
(a b) c = b c = a
and
a (b c) = a a = a.
We cannot conclude that the operation is associative, since we have not
veried that (x y ) z = x (y z ) for every choice of x, y , and z .
Section 2