Cycles in permutation groups 
Let $p$ be an odd prime and let $n$ be an integer such that $n \ge 3p$. Let $\alpha$  be a $2p$-cycle in $S_n$.
a) For which values of $j \in  \{ 1, . . . , 2p − 1 \}$ is $\alpha^j$ a $2p$-cycle?
b) If $1 \le j \le 2p−1$ and $\alpha^j$ is not a $2p$-cycle, what is the cycle decomposition of $\alpha^j$?
c) Let $\beta$  be a $p$-cycle in $S_n$. Assume that $\alpha$ and $\beta$  are disjoint. Let 
  $H = \{ \alpha^j\beta^i \mid 0 \le j \le 2p − 1, 0 \le i \le p − 1\}$. Prove that $H$ is an abelian subgroup of $S_n$.
d) Let $\beta$ be as in part c). Suppose that $K$ is a subgroup of $S_n$ that contains $\alpha$ and $\beta$. Prove that $K$ is not cyclic.

for a) If I take $p = 3$, then shouldnt all elements 1,2,3,4,5 result in $\alpha^j$ be a 6 cycle?
Im not sure of my first part so im stuck for the rest of the problems, Im not expecting answers but I would really appreciate guidance as to which direction I should go on.
Thanks
 A: For the first and second parts, try a couple of examples.  Take a 6-cycle and raise it to a few powers and see what happens.  Now try the same with a 10-cycle.  Do you see a pattern?
A: There are three cases to consider: (i) $j$ and $2p$ are relatively prime; (ii) $j=p$; (iii) $j$ is a multiple of $2$.
For Case (i), consider the powers $(\alpha^j)^k$, as $k$ ranges from $1$ to $2p-1$. Suppose that for some $k_1, k_2$, with $1\le k_1\lt k_2\lt 2p$, we have $(\alpha^j)^{k_1}=(\alpha^j)^{k_2}$. Then $\alpha^{j(k_2-k_1)}$ is the identity permutation. Thus $2p$ divides $j(k_2-k_1)$. Since $j$ and $2p$ are relatively prime, it follows that $2p$ divides $k_2-k_1$, which is impossible.
Thus as $k$ ranges from $1$ to $2p-1$, $(\alpha^j)^k$ ranges over all $\alpha^t$, with $1\le t\le 2p-1$. It follows that $\alpha^j$ is a $2p$-cycle.
For Case (ii), we have $(\alpha^p)^2=\alpha^{2p}$. This is the identity, and since $\alpha^p$ is not the identity, $\alpha^p$ is a $2$-cycle.
For Case (iii), the same sort of argument as in Case (i) shows that $p$ is the smallest positive $k$ such that $(\alpha^j)^k$ is the identity. Thus if $j$ is even then $\alpha^j$ is a $p$-cycle. 
