Some solutions to problems from Chapter 5

6. Show that A₈ contains an element of order 15.

The product of a 5-cycle and a 3-cycle will belong to A₈ and will have order 15.

(1 2 3 4 5) (6 7 8) will do.

8. What is the maximum order of any element of A₁₀?

We can see that (1 2 3 4 5 6 7) (8 9 10) has order 21. The element (1 2 3 4 5)(6 7 8)(9 10) has order 30, but is odd, so not in A₁₀. It would be best to show all orders.

10. Prove that a function from a finite set S to itself is 1-1 if and only if it is onto.

Suppose |S| = n, and let be a function. If the function is 1-1, then is a subset of S with n elements, so must be all of S, and the function is onto. Conversely, if the function is onto, then, if it fails to be 1-1, there are a, b in S, with . Then can have no more than n-1 elements, but has the same image as S, so the function cannot be onto.

The result may fail for infinite sets. The function sending the natural numbers to itself by mapping n to n + 1, for each n, is 1-1 but not onto, for example.

12. Prove: If a permutation is even then its inverse is even; if a permutation is odd, its inverse is odd.

Let α be even. Then it is the product of r transpositions, for some even r . By the s-s lemma, its inverse may be written as a product of the same transpositions, in the opposite order, and must be even. The same statement, with even changed to odd, proves the rest of the assertion

A prettier argument says, let α and its inverse be products of r and s transpositions, respectively, then the product of the two gives a representation of the identity as a product of r + s transpositions, so r + s must be even, by the lemma in the book. Then r and s must both be even or both be odd.

28. Let β = (123) (145). Write β⁹⁹ in disjoint cycle form.

First, β = (14523). Then β⁹⁹ = (β⁵)¹⁹β⁴ = β⁴ = (13254)

36. In S₄, find a cyclic subgroup of order 4 and a non-cyclic subgroup of order 4.

{(1234), (1234)² = (13) (24), (1234)³ = (1432), (1234)⁴ = (1)} is a cyclic subgroup of order 4.

{(12), (34), (12)(34), (1)} is a non-cyclic subgroup of order 4.

46. Suppose that H is a subgroup of S_n of odd order. Prove that H is a subgroup of A_n.

This result follows quickly from problem 19. I shall prove it using the argument for problem 19 on the way. If a subgroup of S_n is not a subgroup of A_n, then it has both even and odd permutations—the identity is even, of course. But then, if we let α be an odd permutation, the function λ_α, as defined in the proof of the Cayley representation theorem—just using this to set notation; we already knew about “translation functions”—gives a 1-1 mapping from the odd permutations in H onto the even permutations in H. Thus, exactly half of the elements of H are even. Since H is of odd order, this can’t happen, so H is a subgroup of A_n.