What do all the $k$-cycles in $S_n$ generate?

Question

Why don't $3$-cycles generate the symmetric group? was asked earlier today. The proof is essentially that $3$-cycles are even permutations, and products of even permutations are even.

So: do the $3$-cycles generate the alternating group? Similarly, do the $k$-cycles generate the alternating group when $k$ is odd?

And do the $k$-cycles generate the symmetric group when $k$ is even? I know that transpositions ($2$-cycles) generate the symmetric group.

Answer 1

Yes, $k$ -cycles generate the symmetric group when $k$ is even and alternating group, when $k$ is odd. As you've said, for $k=2$ you know the answer. Suppose $k>2$.
$(1\space 2\space\ldots,k)(k\space\ldots 3\space 1\space 2)=(1\space 3\space 2)$
Similarly you can get any $3$-cycle.

Suppose $a$ is an even element of $S_n$. Then, as you know, $a$ is a product
$(k_1\space k_2)(k_3\space k_4)\ldots(k_{4l-1}\space k_{4l})$
of $2l$ transpositions. But product of $2$ transpositions can be written using $3$-cycles:
$$(k_1\space k_2)(k_2\space k_3)=(k_1\space k_2)(k_2\space k_3)\circ(k_2\space k_3)(k_3\space k_4).$$ Any of two products, separated by $\circ$ in right hand side, is either a $3$-cycle or unity.

If $a$ is odd and $k$ is even, than $a$ multiplied by any $k$-cycle is even, so we can apply the previous algorithm to it.

Answer 2

If $n\geq5$, then the only normal subgroups of the symmetric group $S_n$ are the trivial group, the alternating group and the symmetric group itself. Since the $k$-cycles form a full conjugacy class, it follows that the subgroup they generate is normal. This determines everything if $n \geq 5$.

More specifically: the $k$-cycles in $S_n$ generate the alternating group if $k$ is odd and $k \ne 1$; they generate the full symmetric group if $k$ is even.