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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 $k=2(i+1)$.

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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.

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Here is a direct proof, requiring no knowledge of the normal subgroups of the symmetric group.

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 $$(a_1\ b_1)=(a_1\ b_1\ a_2\ b_2\ \cdots\ a_{h-1}\ b_{h-1}\ a_h\ b_h)^2(b_h\ b_{h-1}\ \cdots\ b_2\ b_1\ a_h\ a_{h-1}\ \cdots\ a_2\ a_1).$$

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