Show that a permutation can be an exponent of a cycle Show that a permutation can be a power of a cycle if and only if it is a multiplication of disjoint cycles(with the same length).
I still have no idea, how to prove it, I would appreciate any kind of help.
 A: For a general cycle $g=(a_1 \ a_2 \ \cdots \ a_n)$, you can compute $g^k$ explicitly for a few values of $k$ to see why $g^k$ is the product of disjoint cycles of equal size. (It will depend on $\gcd(k,n)$.)
Here is a formal but non-constructive proof.
Note that $g^k$ can be written as the product of disjoint cycles involving $a_1,\ldots,a_n$ (like any permutation of $a_1,\ldots,a_n$).
The fact that the cycles have the same size comes from a symmetry argument, since
$$g = (a_1 \ a_2 \cdots a_n) = (a_2 \ \cdots a_n \ a_1) = (a_3 \cdots a_n \ a_1 \ a_2).$$
(Specifically, suppose otherwise that $g^k$ is the product of disjoint cycles that do not have equal size. By the alternate forms of writing $g$ above, you can cycle the positions of the $a_i$ in the cycle representation of $g^k$ and get another representation of $g^k$. This is impossible if the cycles do not have the same lengths.)

The computations I suggested in my first sentence above may help you see why the converse holds as well.
Given a product of disjoint cycles of equal size, "interlacing" them can give you the original large cycle. For example,
$$(1\ 2\ 3)(4\ 5\ 6) (7\ 8\ 9)(10\ 11\ 12)$$
can be written as
$$(1\; 4\; 7\; 10\; 2\; 5\; 8\; 11\; 3\; 6\; 9\; 12) ^ 4.$$
