1. Since alpha is an r-cycle, the alpha=(i_1,i_2,...,i_r), where i_k means
i sub k. The cycle (i_1,i_2,...,i_r) means that alpha(i_1)=i_2, alpha(i_2)=i_3,
..., alpha(i_k)=i_(k+1), ..., alpha(i_r)=i_1.
For example, alpha=(123), the alpha(1)=2, alpha(2)=3, alpha(3)=1.
We consider alpha^k (alpha to the kth power) and focus on i_1.
alpha(i_1)=i_2, alpha^2(i_1)=alpha(alpha(i_1))=alpha(i_2)=i_3.
And so on. Then alpha^k(i_1)=i_(k+1) if k<r.
If k=r, then alpha^r(i_1)=alpha(alpha^(r-1)(i_1))=alpha(i_r)=i_1.
Thus alpha^k(i_1)=i_1. similarly, alpha^k(i_t)=i_t for all 1<=t<=r.
Therefore, alpha^k=1.
2. There are many ways to define an even permutation. An easy way is to count
the number of swaps in which a permutation can be express. If the number is
even, then the permutation is even.
For example, (123)=(13)(12) has 2 swaps, then (123) is even.
(1234)=(14)(13)(120 has 3 swaps, then (1234) is odd.
For an r-cycle alpha=(i_1,i_2,...,i_r), it is easy to verify that
(i_1,i_2,...,i_r)=(i_1,i_r)(i_1,i_(r-1))...(i_1,i_2) and thus alpha can be
presented in r-1 swaps. So an r-cycle is even if and only if r-1 is even, if
and only if r is odd.
3. It seems that you did not finish the statement of this problem.