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 Sn of odd order.  Prove that H is a subgroup of An.

This result follows quickly from problem 19.  I shall prove it using the argument for problem 19 on the way.  If a subgroup of Sn is not a subgroup of An, 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 λα, 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 An.