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Homework #2 Solutions
Here are solutions to some selected problems.
p. 20: 43,44,45: For a relation which is reflexive, symmetric, but not transitive, take the set {a,b,c}, and put aRa, bRb,
cRc, aRb, bRc, bRa, cRb. It does not include aRc. For a relation which is reflexive, transitive, but not symmetric, take
the set {a,b}, and put aRa, bRb, aRb. For a relation which is symmetric, transitive, but not reflexive, take the set {a,b},
and put aRa.
p. 49 18: To prove G is Abelian we must show that ab=ba for all a and b in G. Suppose, by contradiction, there is a
pair a,b of elements of G with aba^(-1) not equal to b. Then aba^(-1)=c for some c not equal to b. Multiplying this
equation on the right by a gives ab=ca. By hypothesis, we have c=b, which is a contradiction.
p. 50 24: Several people were confused on this problem. The book asks for the inverse of a product of n elements,
a1*a2*.
..*a_n, which is the product of the inverses in the opposite order, a_n^(-1)*.
..*a2^(-1)*a1^(-1).
p. 50 25: If you have a group containing 5 and 15 mod 56 under multiplication, you must have all powers and
products of these numbers too. This is how the book gets the answer.
p. 50 30: By hypothesis, abab=(ab)^2=ab, so multiplying on the left by a^(-1) and the right by b^(-1) gives ba=e, so
ab=e=ba also.
p. 51 38: Let a and b be in G, we must show that ab=ba. By hypothesis, (ab)^2=abab=e, and we know that
a^2=e=b^2. So multiplying abab=e on the left by a and the right by b gives ba=ab.
Homework #3 Solutions
p. 63 4: Suppose that g has order n, so that g^n=e. Taking inverses, we see that g^(-n)=e also, so the order of g^(-1),