# Elements of $S_n$ can be written as a product of $k$-cycles.

Let $$k\leq n$$ be even. Prove that every element in $$S_n$$ can be written as a product of $$k$$-cycles.

I really have no idea how to go about this. My initial intuition was to proceed by induction first on $$n$$ for the base case of $$k=2$$ (i.e. first showing $$S_n$$ is generated by its transpositions) and then inducting on $$k$$. But I have no idea how to show that, assuming the statement is true for $$k=2i$$ for some i$$\in\mathbb{Z}^+$$, that it also holds for $$k=2(i+1)$$.

The conjugate of a $$k$$-cycle is a $$k$$-cycle. So $$G$$, the group generated by $$k$$-cycles is a normal subgroup of $$S_n$$. For $$n\ne 4$$, the normal subgroups of $$S_n$$ are $$S_n$$, $$A_n$$ and $$\{\text{id}\}$$. The only one of these that contains $$k$$-cycles for even $$k$$ is $$S_n$$.
For $$n=4$$, $$S_4$$ has an additional normal subgroup to consider.
Since we know that every permutation is a product of transpositions, it will suffice to show that, for even $$k$$, a transposition can be written as a product of $$k$$-cycles. In fact it is the product of three $$k$$-cycles; e.g., the transposition $$(1\ 2)$$ is equal to each of the following: $$(1\ 2\ 3\ 4)^2(4\ 2\ 3\ 1),$$ $$(1\ 2\ 3\ 4\ 5\ 6)^2(6\ 4\ 2\ 5\ 3\ 1),$$ $$(1\ 2\ 3\ 4\ 5\ 6\ 7\ 8)^2(8\ 6\ 4\ 2\ 7\ 5\ 3\ 1),$$ etc. In general, if $$k=2h\le n$$, then $$(a_1\ b_1)=(a_1\ b_1\ a_2\ b_2\ \cdots\ a_{h-1}\ b_{h-1}\ a_h\ b_h)^2(b_h\ b_{h-1}\ \cdots\ b_2\ b_1\ a_h\ a_{h-1}\ \cdots\ a_2\ a_1).$$