1a) Since x is of order k, consider $ x, x^2, x^3, ... x^{n-1}, e $. These k elements are distinct from each other by x being order k. Further, r(x) takes $ x $ to $ x^2 $, $ x^2 $ to $ x^3 $ and so on, to $ x^{k-1} $ to $ e $ and $ e $ to $ x $. Thus, this is a cycle of length k.
Next, choose an element a not previously listed. Then if r(x) takes $ a $ to $ xa $, it also takes $ xa $ to $ x^{2}a $ and so on, similar to the previous cycle. Since x is still of order k, this is also a k cycle, and since a is not the identity, the elements in this cycle are distinct from the ones in the previous cycle. Thus the cycles are disjoint.
One can continue this process until all the elements are accounted for. Since the elements were chosen to be not previously listed, the cycles are disjoint, and because the order of an element divides the group, each element can be used exactly once. As a result, since there are #G elements and k elements in each group, there must be n distinct k-cycles.
1b) Note that a k-cycle is made up of k-1 transpositions. Since the order of an element divides the order of the group, the order k of any element $ x $ must be odd, and thus k-1 is even. Thus, each of the disjoint cycles making up r(G) is made up of an even number of transpositions, so r(G) itself must have an even number of transpositions. Thus r(G)$ \in{A_{G}} $.
2) Suppose $ x $ is conjugate to <math<x^{-1}</math>. Then $ \exists {g} $ where $ gxg^{-1}=x^{-1} $. Multiplying on the left by $ x $, and on the right by $ g $ gives $ xgxg^{-1}g=xx^{-1}g $ which shows $ xgx=g $. The element $ g $ must have odd order, as the group has odd order. Thus for some $ n $, $ g^{n}=e $ so $ g^{n+1}=g=gxg $, where $ n+1 $ is even. This leads to a contradiction, since the right side can only be expressed as an odd power of g. Thus, $ x $ can't be conjugate to $ <x^{-1} $.
