Homework Due February 22 - answers
Ch. 5, Problem 12: If α is even, prove that α−1 is even. If α is odd, prove that
α−1 is odd.
Solution: Write α as a product of 2-cycles, α = τ1 τ2 · · · τm . The permutation α is even
−1
if and only if m is even. We have that α−1 = (τ1 τ2 · · · τm )−1 = τm
· · · τ2−1 τ1−1 . Note that
for any 2-cycle τ , we have τ −1 = τ so α−1 = τm · · · τ2 τ1 . If m is even this latter expression
shows that α−1 is even. If m is odd, this latter expression shows that α−1 is odd.
Ch. 5, Problem 18: Let
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
α=
and β =
2 1 3 5 4 7 6 8
1 3 8 7 6 5 2 4
Write α and β as
a. products of disjoint cycles
b. products of 2-cycles.
Solution: α = (12)(45)(67); β = (23847)(56) = (27)(24)(28)(23)(56).
Ch. 5, Problem 21: Do the odd permutations in Sn form a group?
Solution: No. The identity is an even permutation and so cannot belong to the set of odd
permutations. Also the product of two odd permutations must be even.
Ch. 5, Problem 26: Prove that (1234) is not the product of 3-cycles.
Solution: A cycle of length m is even if m is odd and odd if m is even. Therefore (1234)
is an odd permutation and any 3-cycle is an even permutation. Since the product of even
permutations is even, we have the product of 3-cycles is even and so cannot equal (1234).